R a DeCarlo and P Lin Linear Circuit Analysis s

PROBLEM SOLUTIONS CHAPTER 1.

Solution 1.1. (a) Charge on one electron: -1.6019×10-19 C. This means that charge on 1013 electrons

is: -1.6019×10-6 C. Net charge on sphere is: 1.6019×10-6 C (POSITIVE).

Solution 1.2. (a) 1 atom ≡ -4.646×10-18 C. By proportionality, 64g ≡ NA atoms.

3.1g ≡ ? atoms ⇒ 3.1g ≡3.1NA

64 atoms.

Total Charge = −4.646 ×10−18 C

atom×

3.1× 6.023 ×1023

64atoms =−1.3554 ×105 C

(b) Total charge per atom is -4.646×10-18 C. Total charge per electron is –1.6019×10-19 C. Therefore,there are 29 electrons per atom of copper.

(c) 0.91 A ≡ 0.91 C/s. i =∆Q

∆t ⇒ ∆t =

∆Q

i=

1.36 ×105

0.91=1.49 ×105 sec .

(d) We know there are 3.1NA

64= 2.9174 ×1022 atoms in the penny. Removing 1 electron from

0.05 ×3.1NA

64 atoms means removing 0.05 ×

3.1NA

64 electrons. Therefore,

Net charge = 0.05 ×3.1NA

64×1.6019 ×10−19 = 234C

Solution 1.3 (a) 7.573 ×1017 × −1.6019 ×10−19( ) = −0.1213C

(b) Current =0.1213

10−3 = 121.3A flowing from right to left.

(c) Again, use proportionality:

10 A =x ×1.6019 ×10−19

60sec ⇒ x =

10 × 60

1.6019 ×10−19 = 3.75 ×1021

(d) i t( ) =dq

dt= 1− e−5t A. This is an exponential evolution with an initial value of 0, a final value of 1,

and a time-constant of 1/5 (signal reaches ~63% of it’s final value in one time-constant).

0.2 time in sec

i(t) 1

(e) Current is the slope of the charge waveform. Therefore, by inspection:

Solution 1.4 (a) 6.023×1023×(-1.6019×10-19) = –9.65×104 C.

(b) Current flows from right to left (opposite electrons), and:

I =9.65 ×104

10−3 = 9.65 ×107 A

(c) Using proportionality:

5A =x × 1.6019 ×10−19

60sec ⇒ x =

5 × 60

1.6019 × 10−19 =1.87 ×1021

(d) i t( ) =dq

dt= 1+ 0.5πcos πt( ) ⇒ i 1sec( ) = 1− 1.57 = −0.57A. Current flows from left to right.

Solution 1.5 (a) i t( ) = 1− 4e−2 t + 3e−3t t ≥ 0. Then

q t( ) = i t( )dt0

t

∫ = 1 − 4e−2τ + 3e−3τ( )d0

t

∫ τ = τ]−∞t − 4 e−2τdτ

0

t

∫ + 3 e−3τdτ0

t

∫

= t − 4 −0.5e−2τ[ ]0

t+ 3 −0.333e−3τ[ ]0

t= t + 2e−2t − e−3t −1

(b) By inspection:

(c) q t( ) = 120cos 120πt( ) . Hence

i t( ) =dq

dt= −120π× 120sin 120πt( ) =−14400πsin 120πt( ) A

Solution 1.6. (a) i(t) = 1 − cos(πt) A. Hence

q( t) = i(τ)dτ0

t

∫ = 1 − cos(πτ)( )dτ0

t

∫ = t −1

πsin(πτ)

0

t= t −

1

πsin(πt) C

(b) Charge is integral of current. Graphically, the charge at time t is the area under the current curve up to

time t: (note the quadratic nature between 2 and 4 seconds)

Solution 1.7

Again, Q is the running area under the current curve. Between 0 and 3 seconds, current decreases linearly

until zero. So, Qtot = 7.5 C. From 0 to 6: Qtot = 7.5 + Q3_6 = 7.5 -1/1×0.5 + -1/1×0.5 + -1×1 = 5.5 C,

where the curve from 3 to 6 was divided into two triangular sections and one rectangular one.

Solution 1.8 Charge is the area under the current curve. Thus, Q = 0.1*4 – 0.1*2 = 0.2 C.

Solution 1.9 Calculate the change in energy for the electron: ∆E = Q ∆V = 3.218×10-15.

Equate this to kinetic energy:

3.218 ×10−15 =1

2mv2 ⇒ v = 8.4 ×107m / s

where the mass of an electron, 9.1066×10-31 has been substituted.

Solution 1.10 P = VI. Hence I = P/V = 2×103/120 = 16.6667 A

PROBLEM Solution 1.11 (a) It is necessary to integrate the i(t) curve to obtain q(t). We do this

interval by interval:

(i) 0 ≤ t ≤ 1 ms, q( t) = 0 + dτ0

t

∫ = t µC

(ii) 1 ms ≤ t ≤ 2 ms, q( t) = 1 − 2 dτ1

t

∫ = 3 − 2t µC

(iii) 2 ms ≤ t ≤ 3 ms, q( t) = −1+ dτ2

t

∫ = −3 + t µC

(iv) 3 ms ≤ t ≤ 5 ms, q( t) = 0 + 8 − 2τ( )dτ3

t

∫ = 8t − t2 −15 µC

(v) 5 ms ≤ t, q( t) = 0 µC

0 1 2 3 4 5 6-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Cha

rge

in m

icro

C

TextEnd

Time in ms

(b) Voltage is the ratio of the power and current curve. In this case, the division can be done graphically

by inspection. Note that the ratio of a quadratic function and a linear function is a linear function:

0 1 2 3 4 5 6-1

-0.5

0

0.5

1

1.5

2

Vol

tage

in V

TextEnd

Time in ms

Solution 1.12 (a) VA = P/I = 20/4 = 5 V

(b) PB = VI = 2×7 = 14 W

(c) VC = P/I = -3W/3A = -1V

(d) ID = P/V = -27W/3V = -9A

(e) IE = P/V = 2/1 = 2A

(f) PF = VI = -4×5 = -20W

In all of the above, note that the direction of the current flow relative to the polarity of the voltage across a

device determines whether power is delivered or absorbed. Power is absorbed when current flows from the

positive terminal of the device to the negative one.

Solution 1.13 (a) By inspection: Circuit Element (CE) 1 absorbs –5W, and CE 2 absorbs 6W.

(b) Compute power absorbed by all elements including independent sources:

I3A: -15

CE1: -5

V3V: -12

CE2: +6

V5V: 10

I2A: 16

----------

Sum: 0 (Verifies conservation of power.)

Solution 1.14 (a) Compute power absorbed:

I5A: -85

CE1: 98

V3V: 33

CE2: 16

V7V: -42

I2A: -20

-------------

Sum: 0

(b) Add all terms:

I-source: Pabsorbed = −3 1− e−t( ) = −3 + 3e−t watts

V-source: Pabsorbed = −2 3e−t −1( ) = −6e−t + 2 watts

CE1: Pabsorbed = 3e−t × 3 1− e−t( ) = 9e−t − 9e−2t watts

CE2: Pabsorbed = 3e−t −1( ) 3e−t −1( ) = 9e−2t − 6e−t +1 watts

Simple algebraic manipulation of the the sum of all the above terms reveals that the result is zero.

Solution 1.15 (a) When IL = 1, P = VLIL = (16-4)×1 = 12 W. When IL = 2, P = VLIL = (16-16)×1 = 0.

(b) P = (16-4IL2)IL. Differentiate this w.r.t. IL and set to zero: 16 – 12IL

2 = 0. Therefore, IL = 1.155A.

Solution 1.16 (a) When IL = 2, P = (16-4)×2 = 24W. When IL = 3, P = (16 - 9)×3 = 21 W.

(b) Maximum occurs in the interval from 0 to 4: P = (16 - IL2) IL

Differentiate w.r.t. IL and set to zero: 16 – 3IL2 = 0.

Therefore, IL = 2.31 A.

Solution 1.17 (a) Power is the product of the current and voltage. We can compute the product

graphically:

0 0.5 1 1.5 2 2.5 30

2

4

6

8

10P

ower

in W

atts

TextEnd

Time in s(b)

W t( ) = p t( )dt0

t

∫ = 10 −10e−7τ( )dτ0

t

∫ = 10τ]0t − − 10

7 e−7τ[ ]0

t=10t +

10

7e−7t −

10

7

This can be used as an aid to plot the work function:

0 0.5 1 1.5 20

1

2

3

4

5

6

7

8

9

Time in s

Ene

rgy

in J

TextEnd

Solution 1.18 (a) Since, i t( ) = 115 − 23t mA ,

q 7( ) = i τ( )dτ0

7

∫ = 115t −23t2

2

0

7

×10−3 = 0.2415 C

(b) Energy is the integral of power:

E = p t( )dt0

7

∫ = v t( ) × i t( )dt0

7

∫ = 25 i t( )dt0

7

∫

= 25 × 0.2415 = 6.0375 C

Solution 1.19 (a) ∆t = 100oF, Rate of temp. increase is 2.5 Wh/oF per gallon:

Energy = 2.5Wh/oF/gallon×100 oF×30gallons = 7500 Wh = 2.7×107J.

(b) Heater generates P = 120×10 = 1200 W. We want 7500 Wh. Therefore, the total number of hours

needed is 7500Wh/1200W = 6.25 h.

Solution 1.20 First compute the change in temperature required, in oF:

∆t = 80-25 = 55oC = 55×9/5 oF= 99oF

Next, compute the energy spent every hour, which means on 40 gallons of water:

E = 2.5 Wh/oF/gallon×99oF×40gallons = 9900 Wh

Since the heater is not 100 % efficient, we spend more energy than is actually needed to heat the water:

E_spent = 9900 Wh/0.9 = 11000 Wh

So, far, this was the energy spent every hour. Over six hours, the total energy spent is:

E6h = 11000×6 = 66,000 Wh

Finally, the total energy spent per month is Em = 66,000×30 = 1980 kWh

and the bill is 1980 kWh×0.14$/kWh = $277.2

Solution 1.21

Energy = 120 W × 6 h = 720 Wh = 0.72 kWh

Therefore, cost per day = 0.72 kWh × 8 = 5.76 cents, and cost per month is 5.76×31 = $1.785.

Solution 1.22

We need to compute the difference between the inner diameter of the tube and the outer one in order to get

the cross-sectional area:

area = πRout2 −πRin

2 = π0.0032 − π0.00182 =1.81 ×10−3 m2

Then, R = 1.7×10-5×(12/1.81) = 11.3 mΩ.

Solution 1.23 L = 20 m, W = 0.015m, H = 0.001 m. Thus, A = W×H, and R = 5.1× copper×L/A

= 0.116 Ω.

Solution 1.24. (a) 500 ft, 20 gauge wire: 10.35 Ω/1000 ft from table 1.3. This implies that

R = 5.175 Ω.

(b) 55 ft, 20 gauge, nickel wire:

R = 5.1×10.35

1000× 55 = 2.9 Ω

(c) Rtot = 2.9 + 5.175 = 8.08 Ω.

Solution 1.25. R T( ) = R 20( ) 1 +α T − 20( )[ ] . Substituting at T = –10 yields:

21 = R 20( ) 1+ 0.0039 −30( )[ ] or R 20( ) = 23.78 Ω

Evaluating at T = +10 yields,

R 10( ) = 23.78 + 23.78 × 0.0039 × −10( ) or R 10( ) = 22.85Ω

Solution 1.26. For tungsten, we know that α = 0.0045. Therefore:

R 150( ) = R 20( ) 1 +α T − 20( )[ ] = 200 1 + 0.0045 150 − 20( )[ ]= 317Ω

Rate of change of resistance is (317-200)/(150-20) = 0.9 Ω/oC.

Solution 1.27. Plug numbers directly into the same formula as problem 1.26:

0.0022 = 0.002 + 0.002×0.0039(T-20)

Rearrange to obtain: T = 45.64o C.

Solution 1.28. (a) Power in a wire: P = I2R. Rearranging, we can express the current as

I = PR .

Substitute given P and R to obtain I = 0.707 mA.

(b) Use the same formula for current obtained above to get 50 A.

Solution 1.29. Use formula for power: P = V2/R. Rearranging, R = V2/P = 96 Ω.

Solution 1.30 (a) I = V/R = 12 A, out of the positive terminal of the battery.

(b) Up through the resistor.

(c) Absorbed power by resistor: P = V2/R = 14.4 W. Same power is delivered by source.

(d) From table 1.2 and 1.3, 1000 feet of 18 AWG aluminum wire has resistance:

»R1000ft = 1.6*6.51R1000ft = 1.0416e+01

By proportionality, 1000 × 0.1 = L ×10.416 . Hence,

»L = 100/10.416L = 9.6006e+00 meters.

Solution 1.31 (a) V = 10 V.

(b) P = V2/R, which means that R = V2/P = 100/25 = 4 Ω.

(c) I = V/R = 10/4 = 2.5 A. Current flow is downwards through resistor.

(d) Up through resistor.

(e) P = V2/R10 = 100/10 = 10 W. Hence, I10 = V/R10 = 1 A. Without applying material from a future

chapter, a legitimate way to obtain Isource is to apply conservation of power first and then compute Isource

from the power formula. Hence, Psource =10 + 25 = 35 watts. Using material from a later chapter, in

particular KCL, we may conclude that, Isrc = 2.5 + 1 = 3.5 A. Thus, Psource = VIsource = 10×3.5 = 35 W.

This approach indicates that power is conserved.

Solution 1.32 (a) From 0 to 1 s, i(t) = 10-3t. Thus, i2R = 10−6t2R is the power absorbed during this

interval. Integrating this expression for the power from 0 to 1 s gives us the total energy used:

10−6t3R

30

1

= 500010−6

3= 0.001667 J.

Finally, we need to multiply this by 2 to account for the interval from 1 to 2 seconds. Thus, the total energy

spent is 3.33 mJ.

(b) The same charge that got transported in one direction during the interval from 0 to 1 is being

transported back in the interval from 1 to 2 (by symmetry). Therefore, total charge transfer is zero.

Solution 1.33. (a) 60 W + 120 W = 180 W.

(b) P = IV þ I = P/V = 180/12 = 15 A.

(c) P = Energy/Time þ Time = 1.2 MJ/180 W = 6.67×104 sec = 1.85 h.

Solution 1.34. P = I2R. Therefore, 325 = 25×(5+4+2R). Solving for R, yields R = 2 Ω.

Solution 1.35. (a) Use definition of power and substitute given power:

V2 = P × R = 98 × 2 = 14 V

Similarly, I3 =P

R=

12

3= 2 A, V4 = P × R = 16 × 4 = 8 V, I5 =

768.8

5= 12.4 A, and

V6 = 486 × 6 = 54 V.

(b) Ptot = Pdissipated = 98 + 12 + 16 + 768.8 + 486 = 1380.8 W.

(c) Vin = V2 + V6 = 68 V. Iin + I3 = I5 + I4 and I4 =V4

4. Thus, Iin = I5 + I4 – I3 = 12.4 A.

Solution 1.36. (a) Sources are the top, right-most, and bottom left. The reason is that current flows out

of the positive terminal of the device.

(b) The 32/16 element is a 2 Ω resistor. The 54.5/18.167 element is a 3Ω resistor. The 13/2.167 element

is a 6 Ω resistor. The 93/2.833 element is a 32.827 Ω resistor. The 24/5 element is a 4.8 Ω resistor.

Solution 1.37. Power: 12 = Ix2R, which means that R = 12/Ix

2. Now, analyze the loop: 16 = Ix(R+4).

Substitute the value of R into this expression: 16 = Ix12

Ix2 + 4Ix . Hence: Ix

2 − 4Ix + 3 = 0 .

This equation has two solutions: one is at Ix = 1 A or R = 12 Ω. The other is at Ix = 3 A or R = 4/3 Ω.

Solution 1.38. (a) Conservation of power:

16Ix = 4Ix2 +12 + 10 − 6

Hence

0 = Ix2 − 4Ix + 4 = Ix − 2( )2

Thus, Ix = 2 A.

(b) 32Ix = 4Ix2 + 28 ⇒ Ix

2 −8Ix + 7 = Ix − 7( ) Ix −1( ) = 0 . Hence, Ix = 7A or Ix = 1A .

Solution 1.39. (a)

(i) AA: I = 36/12 = 3A

BB: I = 24/12 = 2A

CC: I = 14.4/12 = 1.2A

(ii) Sum = 6.2A

(iii) P = VI = 6.2×12 = 74.4W. This is equal to the sum of the powers absorbed by the bulbs.

(iv) R = V/I

AA: R = 12/3 = 4Ω

BB: R = 12/2 = 6Ω

CC: R = 12/1.2 = 10Ω

(b) Each AA bulb draws 3 A. Thus, up to five bulbs can be connected without blowing the fuse (5×3=15).

So, 6 or more would blow the fuse.

(c) Similar analysis suggests that 13 or more bulbs would blow the fuse. Intuitively, the bulbs draw less

current, so more of them can be used.

Solution 1.40. (a)

p(t) = i2 t( )R = 20cos 2πt( )[ ]2 ×10 = 40001 + cos 4πt( )

2= 2000 1 + cos 4πt( )( ) W

(b)

W t( ) = p t( )dt0

t

∫ = 2000t + 2000 cos 4πt( )dt0

t

∫ = 2000t +2000sin 4πt( )

4π J

0 0.5 1 1.5 2 2.5 30

1000

2000

3000

4000

5000

6000

Ene

rgy

in J

TextEnd

Time in s

Solution. 1.41. When the switch is closed, a constant current of 5/10000 = 0.5 mA flows through the

circuit. When the switch is open, no current flows. So, 50% of the time, a 0.5 mA current flows, and the

other 50% no current flows. The average current is therefore 0.25 mA.

Solution. 1.42 When the switch is at A, the current is 5/5000 = 1 mA. When the switch is at B, the

current is 5/10000 = 0.5 mA. Now, the switch is at position A 20% of the time (1ms out of a 5ms period,

after which the events repeat). So, the average current is 0.2*1 mA + 0.8*0.5 mA = 0.6 mA.

Solution. 1.43 The current in the load resistor is 2 A. So, the power is 22×RL = 8 W.

Solution 1.44. Vin = IinR1 è Iout = µVin/R2 = µIinR1/R2.

Solution 1.45 (a) I1 = Vin/R1. Hence, Vout = αVinR2/R1.

(b)

Vout

Vin=

αR2

R1=

100 ×10

R1= 5 ⇒ R1 = 200 Ω

(c)

Power − gain =

α2Vin2

R12 R2

Vin2

R1

=α2 R2

R1= 500

Solution 1.46 (a) V1 = 200 mA × 5 Ω = 1 V implies V2 = 0.8×8 = 6.4 V. Hence

Vout = 5×6.4 = 32 V and Iout = 32/64 = 0.5 A.

(b) Current Gain = 0.5/0.2 = 2.5.

(c) Power values for the 5, 8, and 64 Ω resistors are, respectively, P5 = 0.2 W, P8 = 5.12 W, P64 = 16 W.

Solution 1.47 (a) I1 = 5A è I2 = 3×5/3 = 5 A è Iout = 25 A, Vout = 50 V.

(b) Voltage Gain = 5.

(c) Pin = 5×5×2 = 50 W, P1 = 5×5×3 = 75W, P2 = 25×50 = 1250W.

Solution 1.48 I1 = Vin/10 = 0.1 A, VR = 10×(Vin/10)×R = R; Vout = 5R×10 = 50R =50RVin

Vout/Vin = 50R. If we want Vout/Vin to be 150, R has to be 3 Ω.

SOLUTIONS CHAPTER 2

SOLUTION 2.1. Using KCL at the center node of each circuit:

(a) I3 = I2 − I1 = −1 − 2 = −3A

(b) I3 = I1 + I2 − I4 = 2 −1 − 0.5 = 0.5A

SOLUTION 2.2. KCL at the bottom node gives I1 = −7 − 8 = −15A , and at the right node

I4 =−6 − 8 = −14 . From these, KCL at the top node gives I3 = I4 − 5 = −19A, and finally at the central

node gives I2 = 6 + I3 − 7 = −20 A,

SOLUTION 2.3. Use a gaussian surface on the top triangle. Performing KCL around this surface yields

1A − 2A + 3A + 4 A − 5A = I = 1A.

SOLUTION 2.4. Use a gaussian surface around the bottom rectangle. KCL yields

I1 = 2A +10A + 3A = 15A .

SOLUTION 2.5. Using KVL, V1 = 55V −15V + 105V −100V − 30V = 15V .

SOLUTION 2.6. Using KVL, Vx = 5V −1V −1V −1V +1V −1V = 2V .

SOLUTION 2.7. Using KVL once again.

v1 = 7 + 6 + 5 = 18V

v2 = 6 + 7 −8 = 5V

v3 = −5 − 6 = −11V

v4 = 8 − 7 =1V

SOLUTION 2.8. KVL is used to find the voltage across each current source, and KCL to find the current

through each voltage source.

I3V = 6A − 7A = −1A

I4V = I3V + 8A = 7A

I5V =−8A − 6A = −14A

V7 A = 4V + 3V = 7V

V8A = −4V + 5V = 1V

V6 A = V8 A − 3V = −2V

Chap 2 Probs P2 - 2 © R. A. DeCarlo, P. M. Lin

SOLUTION 2.9. Using the same method as before, the current and voltages are found through and across

each sources.

I5V = 9 +8 − 7 = 10 ⇒ P = 50W

I4V = −6 − I5V = −16 ⇒ P =−64W

I2V = 6 − 7 = −1A ⇒ P = −2W

I3V = − I2V − 9 = −8A ⇒ P =−24W

V8A = 4 − 5 = −1V ⇒ P =−8W

V9 A = 3 + V8 A = 2V ⇒ P = 18W

V7 A = 2 − V9 A = 0 ⇒ P = 0W

V6 A = 5 − V7A = 5 ⇒ P = 30W

Summing all the power give 0W, hence conservation of power.

SOLUTION 2.10. Doing KVL around the right loop does not balance out. Changing 8V to 5V would fix

this.

SOLUTION 2.11. Using KVL to determine the voltages, and KCL to determine the currents:

Vy = 8V

Vx = Vy − 4 = 4V

Ia = 4 A

Iy = 4 −14 + 2Ia = −2A

Ix = Ia − Iy = 6A

SOLUTION 2.12. First Vin = I2 ⋅ 8Ω = 24V . Then I1 = Vin / 3Ω = 8A and I3 = 12A − I1 − I2 =1A.

Therefore RL = Vin / I3 = 24Ω ⇒ P = I3 ⋅ Vin = 24W

SOLUTION 2.13. (a) First, from current division, get

I1 =1/ 3

1/ 3 +1/ 6 + 1/ RL

⋅ 12 − aI1( ) ⇒ I1 =

12 / 3

1 + a( ) / 3 +1/ 6 +1/ RL

.

(b) Using the previous equation and solving for 1/ RL = 12/ 3I1( ) −1/ 6 − 1+ a( ) / 3 = 0.5S or RL = 2Ω .

The power P = I32 RL =

1/ RL

1/ 3 +1/ 6 +1/ RL

⋅ 12 − aI1( )

⋅ RL = 18W

SOLUTION 2.14. For the power delivered by the source to be 60W, the voltage across it should be

V = P / 2A = 30V . Therefore the current through the 20Ω must be I20Ω = 30 / 20 = 1.5A , and by KCL the

current through IRL= 2 − I20Ω = 0.5A . From this, RL = V / IRL

= 60Ω .


SOLUTION 2.15. Writing KVL around the loop 25V − 4I −15V − 5I − I = 0 ⇒ I =1A, and

P5Ω = I 2R5Ω = 5W

SOLUTION 2.16. The total power supplied by the source is P = 50V ⋅ 0.5A = 25W . The power absorbed

by the resistor is P60Ω = 0.5A( )2 ⋅ 60Ω = 15W . Therefore by conservation of power, the power absorbed by

X is 10W.

SOLUTION 2.17. (a) As this loop is open, no current flows through it, so IR is 0A. The output voltage is

VOUT = −2V + 3V − 2V = −1V by KVL.

(b) Writing out the KVL equation around the loop 3 − 2 − IRR − 2 − IR2R − IRR = 0 ⇒ −1 = IR 4R.

Therefore IR =−1/ 4R and VOUT = IR R = −1/ 4V .

SOLUTION 2.18. Writing out KVL around the loop 60 − 30I − 30 − 20 + 60 − 40I = 0 ⇒ I = 1A . From

ohm's law R = V / I = 30Ω.

SOLUTION 2.19. (a) Using Ohms law Iin = V2 / 20 +12( ) = 0.75A, and V1 =12 ⋅ Iin = 9V . To find R, write

KCL and get VR = 30 − V2 = 6V . Therefore using Ohms law again, R = VR / IIN = 8Ω .

(b) Writing KVL around the loop, 30 = aV1 + IinR + Iin 20 + V1, and substituting Iin = V1 / 12,

V1 = 30 / R + 32( ) /12 + a[ ] is obtained. Next substitute back V1 =12Iin and solve for

R =30

12I IN

− a

⋅12 − 32 = 40Ω

SOLUTION 2.20. (a) i. Using R = Vxy / Ibat the value of each resistors starting with the top one are 2.7Ω,

0.6Ω, and 0.25Ω. Using the same relationship, the resistance for the motor is 1.25Ω.

ii. Using P = Vxy2 / R the power dissipated by each resistor is 16.875W, 3.75W, 1.5625W, and for the

motor 7.8125W.

iii. The relative efficiency is = 7.8125/ (12 ⋅2.5) ⋅100 = 26 %

(b) i. Performing voltage division across each resistor

VAB = 0

VBC =12 ⋅ RBC / (RBC + RCD + Rmotor ) = 3.43V

VCD = 12 ⋅ RCD / (RBC + RCD + Rmotor ) =1.43V

Vmotor = 12 ⋅ Rmotor / (RBC + RCD + Rmotor ) = 7.14V


ii. Ibat = 12 / (RBC + RCD + Rmotor ) = 5.71A

iii. The relative efficiency is = (Vmotor2 / Rmotor)/ (12 ⋅5.71) ⋅100 = 59.5 %

(c) i. Repeating the steps from (b), the voltages across the first two resistance are 0, then across the other

and the motor 2V, and 10V

ii. Ibat = 12 / (RCD + Rmotor) = 8A

iii. And the relative efficiency is = (Vmotor2 / Rmotor)/ (12 ⋅8) ⋅100 = 83.3 %

(d) What is the largest equivalent resistance of the motor that will draw 30A? R = 12/ 30A = 0.4Ω.

SOLUTION 2.21. (a) Observe that i =− IO , thus v = ki3 = −kI03 .

(b) Using KVL and previous equation, vx = (R1 + R2)IO + VO + kIO3 .

(c) The power is = IOvx = (R1 + R2)IO2 + VOIO + kIO

4

SOLUTION 2.22. I100Ω =0.04

100= 0.02 A. Therefore V300Ω = 0.02 × (100 + 200) = 6 V. By KCL,

I150Ω = 0.02 +6

300= 0.04 A. Req, seen by the source, is 300 Ω. Therefore Vs = 0.04 × 300 = 12 V.

SOLUTION 2.23. Using KCL IR = 5 − 3 = 2A , and KVL VR = 10 + 6 = 16V . Thus

R = VR / IR =16 / 2 = 8Ω.

SOLUTION 2.24. Using KCL, KVL, along with Ohm’s law,

I5Ω = 6 − 7 = −1A

I4V = 8 − I5Ω = 9A

I2Ω = 8 + 7 =15A

V6 A = 4 + 5I5Ω = −1V

V8A = −4 + 2I2Ω = 26V

V7 A = V8 A − 5I5Ω = 31V

Now, the power delivered or absorbed by each element is calculated:

P6A = I6A ⋅V6 A = −6W

P7A = I7A ⋅V7 A = 217W

P8A = I8A ⋅ V8A = 208W

P4V = I4V ⋅ 4 = 36W

P5Ω = I5Ω2 ⋅5 = 5W

P2Ω = I2Ω2 ⋅ 2 = 450W


Note that for passive elements, when the power is positive it is absorbed, while for independent sources it is

generated when the power is positive.

SOLUTION 2.25. Note that I1 = 6A . Thus by KCL

I3 = 6 − 0.5I1 = 3A

I2 = 2 + 0.2I1 = 3.2A

I4 = 8 − 0.3I1 = 6.2A

And finally using KVL

V2 = 8A ⋅1+ 4I4 + 3I3 = 41.8V

V1 = 2I2 −3I3 = −2.6V

SOLUTION 2.26. (a) Using KCL,

I4 = 5 − 4 =1A

I3 = I4 − 2 = −1A

I2 = 3 − 2 = 1A

I1 = − I2 − 5 = −6A

(b) Using KVL and Ohm’s law,

VIV

VIV

VIV

VIV

44

1010

1212

183

44

33

22

11

==−==

==−==

(c)

WVVVP

WVVP

WVVVP

WVVAP

A

A

A

A

60)(5

24)(4

40)(2

90)(3

1345

434

3212

123

=−+==−−=

−=−−==−=


SOLUTION 2.27. Write KVL around the outside loop, 40 = 500Ix + (400 + 200)i . And write KCL

equation i = Ix − 2I x . Solving yields Ix = −0.4A . The dependent source delivers 2I x ⋅(−600i) = 192W , and

the independent 40I x = −16W . Finally the resistors absorb 500Ix2 + 400i2 + 200i2 =176W verifying the

conservation of energy since the source generate 192W-16W=176W.

SOLUTION 2.28. By voltage division V2 =[((90 ||180) + 60)||40]

[((90||180) + 60)||40] +160⋅

60

60 + (90||180)⋅ Vs = 1/14 ⋅Vs .

Therefore Vs = 14V2 = 280V .

SOLUTION 2.29. By voltage division

vx = 9V ⋅18 + 3

(18 + 3) + 6= 7V

SOLUTION 2.30. By voltage division we get the following two equations in order to solve for the two

unknowns.

V2 = V1 ⋅R1

R1 + R2

V1 =100V ⋅R1 + R2

R1 + R2 + 60

Solving yields R1 = 40Ω , and R2 = 100Ω .

SOLUTION 2.31. Dividing 1400 in four gives 350. If we only need 1/4 and 2/4, the resistor string can be

made of three resistances: 350Ω, 350Ω, and 700Ω.

SOLUTION 2.32. Using voltage division, at t=0 vR = 15 ⋅2R

3R= 10V , and t = 5 s vR = 10V , and at t = 10

the voltage goes back to 0V.


0 5 10 150

1

2

3

4

5

6

7

8

9

10

time (sec)

Volts


Vb =Rb

Ra + RbVin and Vd =

Rd

Rc + RdVin

By KVL, if Vout = 0, then

0 = Vout = Vb − Vd =Rb

Ra + Rb−

Rd

Rc + Rd

Vin

For arbitrary Vin , this requires that Rb

Ra + Rb=

Rd

Rc + Rd or equivalently that Rb Rc = Ra Rd .

SOLUTION 2.34. First Geq = 1m +1.5m + 2m + 3m = 7.5mS . By current division

I2 =100mA ⋅1.5m

Geq

= 20mA , P = 100mA ⋅ I2 /1.5mS =1.33W .

SOLUTION 2.35. By current division, for I1 to be 2A then 160 + R = 300 ||600 for an even split. Thus

R = 40Ω .

SOLUTION 2.36. By current division, i1 = 0.4A ⋅1/10

1/10 + 1/ 40= 0.32 A . Therefore using KVL

vd =10i1 − 0.25i1 = 3.12V .

SOLUTION 2.37. (a) Req = (8k | |2k) + (9k ||1k) = 2.5kΩ


(b) Req = 2k ||[(2k | |2k) + (2k | |2 k)] = 1kΩ

SOLUTION 2.38. (a) Req = 2 + 15 +10 + 10 + 40 + 30 + 20 + 8 =135Ω .

(b) Four of the resistors are shorted, thus Req = 2 + 15 +10 + 8 = 35Ω .

(c) Lumping the series resistance together Req = 8 + [50 ||(50||25)] + 2 = 22.5Ω

SOLUTION 2.39. (a) Req = [2R + (4R | |4 R)]||[2 R + (4 R | |4 R)] = 2R

(b) Req = 2R | |2 R + (4R | |4 R | |4 R | |4 R) = 2R

SOLUTION 2.40. (a) First Req = 150 + [375||(250 + 500)] = 400Ω . Next Iin = 14V / Req = 35mA . The

power delivered by the source is then = 14Iin = 0.49W .

(b) Req = 150 + [375||(250 + 500)||1k] = 350Ω, and Iin = 14 / Req = 40mA . The power delivered by the

source is = 14Iin = 0.56W .

As the equivalent resistance decreases, more of it gets dissipated by it.

SOLUTION 2.41. Req1 > Req2. Without going into a detailed analysis using methods of Chapter 3, we

present the following intuitive argument. First note that the points a and b represent points on an

unbalanced bridge circuit meaning that the voltage between a and b would not be zero. Also note that when

two resistors are placed in parallel, the equivalent resistance becomes smaller than either resistance. The

addition of the resistor R in circuit 2 essentially creates an internal parallel resistance resulting in an Req2

lower than Req1.

SOLUTION 2.42. Req1 = Req2. As was the case in the previous problem, this is a balanced bridge circuit.

Hence no voltage appears between a and b making the additional resistor irrelevent.

SOLUTION 2.43. (a) Rin = [(20||20) +10]||(1/ 0.12)||(1/ 0.08) = 4Ω.

(b) Rin = 6R ||[(R || R ||0.5R) + 0.75R + (2R | |2 R)] = 1.5R

SOLUTION 2.44. (a) The infinite resistance are essentially open circuits, thus

Req = 1 + 2 + 3 + 4 + 2 + 4 + 3 + 2 +1 = 22Ω

(b) 0 resistances are short circuits. Labeling one branch x and the other y, it can be seen that the circuit is a

set of 3 resistor strings in parallel to each other between x and y, then added in series to the two 1 Ohm

resistor. Thus Req = [(2 + 3)||(4 + 2 + 4)||(2 + 3)]+1 +1 = 4Ω .


(c) Writing out Req = 1 + [Rx ||(2 + 3 + 4 + 2 + 4)] + 3 + 2 +1 = 7 + [Rx ||15] , and solving for Rx = 3.75Ω.

(d) No, it requires methods to be covered in the next chapter.

SOLUTION 2.45. Using the formulas for parallel resistances, the circuit of figure 2.45 reduces to

(a) RAC = (2 +1) / /6( ) + 8[ ] / /10 = 5 Ω

(b) RAB cannot be calculated by series parallel formulas, but RBC can be done.

RBC = (8 +10) / /6( ) + 2[ ] / /1 = 0.86667 Ω

SOLUTION 2.46. (a) Req = 300 + (R ||5.6k), thus R = 800Ω.

(b) Req = R + (R ||1.2K) , the following quadratic equation must be solved R2 +1.4k ⋅ R −1.2M = 0 . This

yields R = 600Ω .

(c) Req = 500 + 300 + (800||400 || R). Solving for R yields 800Ω .

SOLUTION 2.47.

(a) Using the fact that the resistance seen into terminal a-b is the same as that seen in terminal c-d, we can

obtain the following relationship. eqeq RRRR ||+= . This produces a quadratic equation whose solution is

Req = 1.618R .

(b) Using the previous argument )5(||105 eqeq RR ++= . Solving for Ω= 18.11eqR .

SOLUTION 2.48. By current division Ix =1/ 18k

1/18k +1/ 9k

⋅

1

6k + (9k ||18k)1

6k + (9k ||18k)+ 1

4k

⋅36m = 3mA


SOLUTION 2.49. The 500 Ω resistor has no effect on the current entering the circuit to its right.

0.15 =30

R+

30

600=

30

R+ 0.05

Hence R = 30/0.1 = 300 Ω.

SOLUTION 2.50. (a) First, express the total current as I =120

0.5 + (20 ||30||40 || RL2 ). Next, find RL2 that

will cause I to be 15A. Thus RL2 = 40Ω or less will cause the fuse to blow as this will cause the current to

be 15A or more.

(b) Repeating the previous procedure, RL2 = 20Ω .

(c) RL2 = 120Ω .

SOLUTION 2.51. At time 0, all switches are open and Vout =260

260 + 40⋅ 220 =190.7V .

Then at t = 5s, switch one closes and

Vout =260||260

(260 ||260) + 40⋅ 220 = 168.2V .

At t = 10s,

Vout =130 ||260||260

(130 ||260||260) + 40⋅ 220 =136.2V .

Finally at t =15 s,

Vout =65||130 ||260||260

(65||130 ||260||260) + 40⋅ 220 = 98.5V .

0 2 4 6 8 10 12 14 16 18 2080

100

120

140

160

180

200

Time (sec)

Volts


SOLUTION 2.52. (a) Lumping the two sources together and the resistors into an equivalent resistor gives

i1( t) =9cos(2 t) − 3cos(2t)

7k + 9k + 8k + (2k | |3k | |6 k)= 0.24cos(2t)mA .

(b) By current division i2 (t) =1/ 6k

1/ 2k +1/ 3k +1/ 6k

⋅ i1(t) = 40cos(2 t) A .

SOLUTION 2.53. (a) Starting with,

Req1 = 5||(10 + 10) = 4Ω

Req 2 = 10||(6 + Req1) = 5ΩReq 3 = 5 + Req2 = 10Ω

(b) Using the values R just obtained,

Va = 100 ⋅Req 2

Req3

= 50V

Vb = Va ⋅Req1

6 + Req1

= 20V

Vc = Vb ⋅ 1010 +10

= 10V

(c) Finally,

Iin =100

Req3

=10A

Id =Va

6 + Req1

= 5A

Ie = Vb

10 +10= 1A

.

SOLUTION 2.54. (a) Circuit a: Using voltage division,

vout (t) = vin (t) ⋅300||(20 + 30 + 50)

[300 ||(20 +30 + 50)] + 5

⋅

30

30 + 20 + 50

= 33.75sin(377t)V , and Ohm’s law

iout(t) = vout( t)/ 30 = 1.125sin(377t )A . The instantaneous power is then

P(t) = iout (t) ⋅ vout (t) = 37.969sin 2(377t)W .

Circuit b: By current division

iout(t) = iin (t) ⋅1/ (20 + 30 + 50)

1/ (20 + 30 + 50) +1/ 300 +1/ (50 +100)

= 60sin(377t)A , and from Ohm’s law

vout (t) = 50 ⋅ iout(t) = 3000sin(377t)V . The instantaneous power is P(t) = 180sin2 (377t)kW .

(b) No, since the current source forces the amount of current in the circuit.


SOLUTION 2.55. (a) Noting that i2 = v1 /10 = 6A, then we can write KCL at the top left node,

isource = i2 + v1 / 6 + (v1 − 5i2 )/ 5 = 22A. Thus P = 60 ⋅ 22 = 1.32kW .

(b) First, determine the current through each resistor:

i2 = 60 /10 = 6A

i2.5Ω =60

2.5 + (5||5)

= 12A

i5Ω = 1/ 2 ⋅i2.5Ω = 6A

Then calculate the power absorbed by each resistor:

P10Ω = 10i22 = 360W

P2.5Ω = 2.5i2.5Ω2 = 360W

P5Ω = 5i5Ω2 = 180W

SOLUTION 2.56. From Ohm’s law I1 = 100m / 200 = 0.5mA . By current division

IRL =20k

20k + 200

⋅150I1 = 75.257mA , and PRL = 200IRL

2 = 1.103W .

SOLUTION 2.57. First, using voltage division, Vx = Vs

2

2 +1

= (2 / 3)Vs . Then using KCL and the

previous equation, Is = (Vs / 3) − Vx = −(1/ 3)Vs . Finally using Ohm’s law Req = Vs / Is =−3Ω.

SOLUTION 2.58. Observing the following relationship, V1 = Vin , the following nodal equation can written:

Iin = Vin / 3 + Vin / 6 − 2Vin = −1.5Vin .

SOLUTION 2.59. Step 1. From voltage division

V1 =18

18 + 4 + 2Vs = 0.75Vs and Vin =

22

24Vs =

11

12Vs

Hence

Pin =Vin

2

22=

11×11

22 ×144Vs

2 =11

288Vs

2

Step 2. For the load, by current division


I2Ω =6

6 + 2AV1 =

3

4A

3

4Vs

=

9A

16Vs

Therefore

P2Ω = 2 × I2Ω2 = 2

81A2

256Vs

2 =81A2

128Vs

2

Step 3. P2Ω = 10 × Pin implies that

81A2

128Vs

2 =1011

288Vs

2

Hence

A =128 ×110

81 × 288= 0.7769

SOLUTION 2.60. By voltage division V1 =6

6 + 2Vin = (3/ 4)Vin . By current division, and substituting the

previous equation I2 =3

3 + 64V1 = Vin . Using voltage division and Ohm’s law, and substituting the previous

equation,

Vout = 4.5I2

1010 + 5

= 3Vin = 30V

Iout = 4.5I2 / (10 + 5) = 0.3Vin = 3A

Finally, from the previous equations | Vout / Vin |= 3 .

SOLUTION 2.61. Writing out KCL when the switch is closed, ibat = 150A +Vbat − 0.04ibat

240

. Solving

gives ibat = 150.02A and Vout ≈ 6V . When the switch is open Vout = Vbat

240

240 + 0.04

≈12V . Therefore,

the reason for the radio stopping is insufficient supply voltage.

SOLUTION 2.62. (a) Using the following relationship P = V 2 / R, the resistance of each headlight on low

beam is R = V 2 / P = 4.11Ω .

(b) Using the same relationship R = 2.22Ω .

(c) By voltage division, Vout =14.7240

240 + 0.04= 14.698V .

(d) Using voltage division, Vout = 14.7240 ||4.11||4.11

(240||4.11||4.11) + 0.04= 14.417V


(e) Using voltage division, Vout =14.7240||2.22||2.22

(240 ||2.22||2.22) + 0.04= 14.186V


11.96 =15

15 + R012 =

180

15 + R0

Therefore

R0 =180 −15 ×11.96

11.96= 0.050167 Ω

SOLUTION 2.64. (a) Using KVL Vt = 102 − 0.05 ⋅80 = 98 V.

(b) Using KVL Vt = 102 + 0.05 ⋅ 50 =104.5 V.

(c) P = Vt ⋅50 = 5.225. kW

SOLUTION 2.65. Minimum load means the minimum load resistance that the system can handle.

»MaxPwr = 0.8*50e6

MaxPwr = 40000000

»Vs = 750e3;

»Iline = MaxPwr/Vs

Iline = 5.3333e+01

»Rmin = Vs/Iline

Rmin = 1.4062e+04, i.e., Rmin = 14.062 kΩ.

SOLUTION 2.66. (a) Using the following general form for a non-ideal voltage source: vout = −Rsiout + Vs ,

one sees that for zero current vout = Vs = 40V . The slope of the line is −40

1000= −Rs = −0.04Ω , thus

Rs = 0.04Ω .

(b) This curve represents a resistor’s I-v characteristic, thus the slope 60

11

3

= R = 45Ω .

(c) The general form for a non-ideal current source is iout = −1

Rs

Vout + I s . When the voltage is zero,

iout = Is = 5A . From the slope of the line, −4000

5, Rs = 4000 / 5 = 800 Ω.


SOLUTION 2.67. Using the following formula: T

n

nI

Cn

= 1, solve for T, and get 0.625 hrs, or 37.5 min.

SOLUTION 2.68. Using the same equation as before and solving for Cn = nIT

n

1/

, with n=10, and

T=55/60 hrs, the capacity obtained is 20 Ah.

SOLUTION 2.69. C20 = 50 Ah

(a) In eq. 9, solving for I with n=20, and T=10, I=4.2A

(b) Calculate the capacity for n=10 and T=10, this yields 42 Ah.

SOLUTION 2.70. (a) Using a sequence of voltage division,

V1 = 50mV4850

= 48mV

V2 = 50V1

195

200= 2.34V

Vload = 2.5V2 = 5.85V

And the power is RRL= Vload

2 / RL = 2.278W .

(b) Following is the graph, and the script used to generate it.

0 10 20 30 40 50 60 700

1

2

3

4

5

Pow

er in

Wat

ts

0 10 20 30 40 50 60 700

200

400

600

800

Resistance in Ohms

Cur

rent

in m

A

%Script for Question 70 in chapter 2

RL=8:1:64;

V2=2.34;

IL=2.5*V2 ./ RL;

%Note the use of the ".*" which means that the division

%is performed for each value of RL.


PL=RL .* (IL .^ 2);

%Plot the Power versus RL

subplot(2,1,1);

plot(RL,PL);

ylabel('Power in Watts');

%Plot the Current versus IL

subplot(2,1,2);

plot(RL,1000.*IL);

xlabel('Resistance in Ohms');

ylabel('Current in mA');

%The use of subplot lets you subdivide the graphing

%window in two halfs.

SOLUTION 2.71. (a) Using the following script:

%Script for problem 2.71

R1=15; R2=4; R3=9; R4=2; R5=8;

R6=18;

Ra= R4+R5;

Ga= 1/Ra;

Gb= Ga+1/R1;

Rb= 1/Gb;

Rc= 1/(1/R6+1/R3)+Rb;

Gc= 1/Rc;

Geq= Gc+1/4;

Req= 1/Geq;

Irc= 20*Gc/Geq;

Vrb= Irc*Rb;

Vout= Irc*(Ga/Gb)*8;

Req

Vout

So (a) Req = 3Ω , and (b) Vout = 24V

SOLUTION 2.72. Using the following script:


R1=1e3; R2=2.2e3; R3=2e3; R4=5e3; R5=3e3; R6=R5;

R7=3.2e3; R8=1.2e3; R9=1.6e3;


Ga=1/R7+1/(R8+R9);

Ra=1/Ga;

Gb=Ga+1/R6;

Rb=1/Gb;

Gc=1/R4+1/(R5+Rb);

Rc=1/Gc;

Gd=1/R2+1/(R3+Rc);

Rd=1/Gd;

Geq=1/R1+Gd;

Req=1/Geq

%Going through the same step to find Vout

Id=200e-3*(Gd/(Geq));

Ic=Id*((1/(R3+Rc))/Gd);

Ib=Ic*((1/(R5+Rb))/Gc);

Ia=Ib*Ga/Gb;

Iout=Ia*((1/(R8+R9))/Ga)

Vout=Iout*R9

The following values are obtained:

Req =

5.9121e+02

Iout =

5.5431e-03

Vout =

8.8689e+00

R1

a b c d

R9 R7 R6 R4 R2

SOLUTION 2.73. Using the following script:


R1=20; R2=40; R3=60; R4=30; R5=10; R6=135;

R7=150; R8=300; R9=130; R10=200; R11=50;

Ga=1/R10+1/R11;

Ra=1/Ga;

Rb=Ra+R9+(1/(1/R7+1/R8));


Gb=1/Rb;

Gc=Gb+1/R6;

Rc=1/Gc;

Rd=Rc+R5+(1/(1/R3+1/R4));

Gd=1/Rd;

Ge=Gd+1/R2;

Re=1/Ge;

Rin=R1+Re

Ie=10/Rin;

Id=Ie*Gd/Ge;

I1=Id*(1/R6)/Gc

Ib=Id*Gb/Gc;

Vout=Ib*Ra

The following values are obtained:

Rin = 50ΩVout = 0.667V

I1 = 33.3mA

R1 e d

R2

R11

R10

R9 R8 R7

R6

R5

R3

R4 a c b

SOLUTION 2.74. An identical procedure to the one followed in the previous problem will yield the

following values:

Rin = 50.53ΩIout = 133.8mA


Solution 3.1. Select the bottom node as the reference node, and write a node equation at the positive

terminal of the V1 resistor:

V1 − V0

3R+

V1

6R+

V1 − 4V0

6R= 0

⇒ 2V1 − 2V0 + V1 +V1 − 4V0 = 0

⇒ 4V1 = 6V0

⇒ V1 = 1.5V0

Solution 3.2 Write a node equation at the top node:

0.6 −Vx

100−

2Vx

100−

Vx

50= 0

⇒ −Vx − 2Vx − 2Vx = −60

⇒ −5Vx = −60

⇒ Vx = 12V

Solution 3.3

0.6 −Vx

100+

25Vx

100−

Vx

50−

Vx − 0.2Vx

40= 0

⇒ −3Vx

100+

25Vx

100−

8Vx

400= −0.6

⇒ Vx = −0.6 × 400

80⇒ Vx = −3V

Solution 3.4 (a)

It is evident from the figure that Vc = 20. We need to write two equations in Va and Vb and put them in

matrix form. In this case, we can write the matrix equation by inspection. Note that the resistors are

identified by conductance values.

15m −5m

−5m 35m

Va

Vb

=

0.5

0.5

(b) Solve the matrix equation by inverting the left-most matrix:

Va

Vb

=

1525 − 25

35m 5m

5m 15m

0.5

0.5

=1

0.5

35 5

5 15

0.5

0.5

=40

20

(c) Vx = Vab = Va −Vb = 20V , Vda = -Va = -40, Vdb = -Vb = -20.

(d) Pi = 0.5×40 = 20W, Pv = 20×(20-20) = 0. Pdiss = 40×40×10m + 20×20×5m + 20×20×5m = 20 W.

Power delivered equals dissipated power.

Solution 3.6 Write two nodal equations:

Vs1 − V1

3000= Is3 +

V1 −V2

6000V2

30000=

Vs2 − V2

12000+

V1 − V2

6000

Rewrite equations as:

2Vs1 − 2V1 = 6000Is3 +V1 −V2

2V2 = 5Vs2 − 5V2 +10V1 −10V2

Cast into a matrix equation

−3 1

−10 17

V1

V2

=

6000Is3 − 2Vs1

5Vs2

Solving the matrix equation yields:

V1

V2

=

181.46

124.39

Power absorbed by the 6k resistor is (V1-V2)2/R = 0.5429W .

Similarly, Ps1 = (Vs1-V1)/3000×Vs1 = 4.7W, Ps2 = (Vs2 −V2) /12000 ×Vs2 = −0.32W

Ps3 = Is3 × (Vs2 − V1) = −1.21W

Solution 3.7 (a) Again, the matrix equation can be written by inspection:

G1 + G2 + G4 −G4

−G4 G3 + G4 + Gs

VB

VC

=

50G1

50G3

(b) Substituting the values of conductances and inverting the above matrix equation yields:

VB = 34.0132V

VC = 33.6842V

(c) Power delivered is 80.7566W . Using the Principle of Conservation of Power:

Pdel = P1 + P2 + P3 + P4 + P5

or,

Pdel = 50 ×VA −VB

20+

VA −VC20

= 80.7566W

(d) In this part, we take the above matrix equation and solve it for each value of Gs. If we do this, we can

get a feel for the behavior of VB and VC w.r.t. changes in Gs. The following plot is the voltage difference

between the two nodes as a function of Gs, and hence as a function of temperature.

As can be seen, in this figure, the voltage difference between B and C does not change linearly with Rs.

Since this resistance itself changes linearly with temperature, this means that VB-VC does not change

proportionally to temperature.

Solution 3.8 The answer is:

G1 + G2 −G1 −G2 0

−G1 G1 + G3 + G4 + G7 −G4 −G7

−G2 −G4 G4 + G5 + G6 + G2 −G6

0 −G7 −G6 G6 + G8 + G7

V1

V2

V3

V4

=

− Is1

0

0

Is2

Solution 3.9 Write the matrix equation by inspection:

4 /100 −1/100 −1/100

−1/100 4 /100 −1/100

−1/100 −1/100 4 /100

V1

V2

V3

=

Vs1100Vs1100Is2

Solving the equation in MATLAB, we get: V1 = 7, V2 = 7, V3 = 11, and P = 0.6.

P = Vs1 ×VS1 − V1

100+

VS1 −V2100

= 10 ×

310

+3

10

= 0.6W

Solution 3.10 (a)

Nodal equation for A:

VA −Vs110

+VA −VB

10+

VA −VC10

= 0

(b) At node B:

VB10

+VB −VA

10− Is2 = 0

(c) At node C:

Is2 − Is3 +VC − VA

10= 0

(d) Manipulate algebraically to cast as the following matrix equation:

3 −1 −1

−1 2 0

−1 0 1

VA

VB

VC

=Vs1

10Is2

10 Is3 − Is2( )

VA

VB

VC

=

−13 13

1313

−43 13

(e) P = (-10-13.333)/10×(-10) = 23.33W.

Solution 3.12

We are required to write the equations in matrix form. First, write a node equation at VA and Vout:

VA − 5 +VA

5+

VA −Vout

10= 0

Vout −VA

10− 7.5VA +

Vout

10= 0

Now group the coefficients for VA and Vout, and write the matrix equation:

1+1/5 +1/10 −1/10

−1/10 + 7.5 2/10

VA

Vout

=

5

0

VA

Vout

=

1

−37

where the matrix inversion was performed in MATLAB. The ratio of the output voltage to the input voltage

is -37/5.

Solution 3.13 (a) Nodes A and B are already labeled:

VA − 9( )0.1 + VA − 0.5VB( )0.2 + VA − VB( )0.3 = 0

VB −VA( )0.3 + VB − 9( )0.5 + 0.4VB = 0

This can be rearranged Into:

0.1+ 0.2 + 0.3( )VA − 0.3VB − 0.2 × 0.5( )VB = 0.9

−0.3VA + 0.3 + 0.5 + 0.4( )VB = 4.5

The matrix equation can now be easily obtained:

0.6 −(0.5)(0.2) − 0.3

−0.3 1.2

VA

VB

=

0.9

4.5

VA

VB

=

4.8

4.95

(b) Iin = -(VA-Vs)0.1 - (VB-Vs)0.5 = 2.445A

(c) Ps = VSIin = 22.005W, Pdep = 0.5 ×Vout × (0.5Vout −VA) × 0.2 = −1.151W .

(d) P = V2/R = (4.95)×(4.95)×0.4 = 9.801W.

Solution 3.14 (a) We write node equations at VA and VB:

−Is1 +VA

20k+ gm1VA +

VA −VB

10k= 0

VB − VA

10k− gm2 VA − VB( ) +

VB

2.5k+ Is2 − gm1VA = 0

Rearranging, we have:

1

20k+ gm1 +

1

10k

VA −

VB

10k= Is1

−1

10k− gm2 − gm1

VA +

1

10k+ gm2 +

1

2.5k

VB = −Is2

(b)

1/20 k + gm1 +1/10k −1/10k

−1/10k − gm2 − gm1 1/10k + gm2 +1/2.5k

VA

VB

=

Is1

− Is2

(c) The above matrix is inserted into MATLAB, with all the values substituted, to obtain:

VA

VB

=

9.722

5.972

(d) Vo = VA-VB = 3.75V

(e) P1 = VAIs1 = 0.0292W, Pgm1 = -Vogm1VA = -0.008W, Pgm2 = VBgm2Vo = 0.0112W, P2 = -VBIs2 = -

0.0119W.

Solution 3.15 I1 = 0.4. Write nodal equations at A and B

VA100

+VA − VB

20+ 0.03 VA − VB( ) = 0.4

VB − VA20

+VB40

+VB − 80VB

4040

= 0

Rearranging and casting into matrix form:

1/100 +1/20 + 0.03 −1/20 − 0.03

−1/20 1/20 +1/40 −1/40

VA

VB

=

0.4

0

VA

VB

=

40

40

It is obvious then, that Vx, the voltage between A and B, is zero.

Solution 3.16

1) VA = 3000ix = 3000 ×VA −VB

9000=

VA − VB3

Equation at node B:

VB − VA9000

+VB

6000+

VB −VD18000

= 0 is equivalent to:

2) −2VA + 6VB − VD = 0

Equation at node D:

VD − VB18000

+VD

9000+ IS = 0 which can be rewritten as:

3) 3VD −VB = −360 Solving the system formed by equations (1), (2) and (3) we obtain:

VA = 9V , VB = −18V , VD = −126V .

Solution 3.17 (a) Choose E as the reference node

VA = 2ix

At node B

6 = (VB-VC)/3 + (VB-VA)/2

Or 6 = 5/6VB –1/3VC –VA/2

At node C, VC = 2iy

At node D, VD = -12V

iy = (VD-VA)/2 = VD/2 –ix

From here on, the solution involves algebraic manipulations to solve the system of equations. MATLAB or

hand analysis can be performed to obtain:

VA = 48V, VB = 12V, VC = -60V.

(b) P6A = 6(12 = 72W

I12V = 30-8 = 22A ( P12V = 264W.

P2i x= 2ix × (−iz − iY ) = 2 × 24 × (+18 + 30) = 2304W

P2i y= 2iy × (−ix +

vc − vD6

) = 2 × (−30)× (−24 − 8) =1920W

(c) P3 Ω = ix × (VB −VC ) =1728W = 1728W

P6 Ω = 6×8×8 = 384W

P2 Ωy = 2×iy×iy = 1800W

P2 Ω = (VB-2ix) (VB-2ix)/2 = 648W.

(c) VD = -12V, VA = 2ix, VC = 2iz.

Substitute the above VA and VC into the node equation for node B:

iz = (VB-2ix)/2 = VB/2 –ix and

ix = (VB-2iz)/3 = VB/3 –2/3iz

Substitute iz into ix to obtain: ix = 0, VA = 0. Then, VB can be deduced to be 12.

Finally, VC = 12V.

Now, compute the powers.

P6A = 72W , P3 Ω = 0, P6 Ω = (VC-VD)2/6 = 96W, P2 Ωy = 72W, P2 Ωz = 72W.

P2ix = 0W , P2iy = 12 × 4 = 48W

Solution 3.18 The three node equations at A, C, and D are:

−0.8 − 0.3 = 0.015VA + 0.02VA − 0.02VC

Vc = 440

−0.8 + 2.5 = −0.005VD + 0.025VC − 0.025VD

As can be seen, these really reduce to only two equations in two unknowns. These can be solved rather

easily either by hand or by MATLAB to obtain: VA = 220V, VD = 310V . Note that all of these voltages

are already referenced to node B (i.e. VA = VAB, etc).

Solution 3.19 (a) Supernode is BC (50 V source).

(b) Only one node equation needs to be written:

P12V =12 × (VA − VD

2+

VC − VD6

) = 12 × (6 + 4) =120W,

VB90

−VA90

+VB10

+VC10

−VA10

+VC90

= 0

with the constraint that VC −VB = 50.

(c) The constraint equation can be substituted into the B node equation to obtain

VB = 125V. Thus, VC = 175V, and ix = (VC-VA)/10 = -12.5A.

(d) (VA-VB)/90 = 1.94A è P300 = 300×(1.94+12.5) = 4332W.

VC/90 = 1.94 è P50 = -528W.

P50 = 50 × (ix +VC90

) = −528W

Solution 3.20 (a) VB = VA – 440 and VC = VA – 460.

(b) Supernode is one including A, B, and C.

(c) VC − 40( )0.15 + 0.05VB + 0.25VA − 25 + 0.2 VA − 40( ) = 0

This can easily be rearranged to get VA = 200V.

(d) power Ps = VA × I = 200 × 25 = 5000W = 5KW

Solution 3.22 (a) VC = Vs2 = 6V

(b) Ix = 0.01VA

(c) Supernode at A,B, encompassing the controlled source. So, we have one equation:

Is1 = 0.01VA + 0.0125VB + 0.1VB − 0.1VC

(d) Substitute the constraint equations, VA –VB = 20Ix = 0.2VA , (equivalently: VA =VB0.8

) into the above

equation: VA =VB0.8

Is1 = 0.01VB

0.8+ 0.0125VB + 0.1VB − 0.1VC

⇒ VB = 6.4V

⇒ VA = 8V

(e) Ix = 0.08A.

(f) P0.0125Ω = VB2 / R = 0.512W

(g) P = Is1 ×VA = 0.2 × 8 =1.6W

Solution 3.23 (a) VC = Vs2 = 50V.

(b) ix = VA/100.

(c) Supernode A,B:

Is1 = 0.01VA + 0.05VB + 0.05VB − 0.05Vs2 + 0.09VA − 0.09Vs2

VA = VB + 300ix = VB + 3VA

(d) Solving the above two equations yields: VA = -90V, VB = 180V.

(e) ix =VA100

=−90100

= −0.9A

(f) (VB-VC) (VB-VC)/R = 845W.

(g) PS1= IS1

× VA = 2 × (−90) = −180W

Solution 3.24 (a) VB − VC = 3Vx = 3VB; VC = −2VB

(b) Supernode at B and C, encompassing controlled source.

(c)

Is2 = (VB10

−VA10

) +VB10

+VC10

+ (VC10

−VA10

); VC = −2VB; 10Is2= −2(VB +VA)

equivalently: 2VA + 2VB = −10

(d)

(0.1VA − 0.1Vs1) + (0.1VA − 0.1VB) + (0.1VA − 0.1VC ) = 0; 0.3VA + 0.1VB = 0.1VS1; equivalently:

3VA − VB = −10

(e) Again, any method can be used to simplify and solve the system of two equations. The solution is:

VA = -2.5V, VB = -2.5V.

Solution 3.25 (a) VA = Vs1 = 16V.

(b) Supernode at C and D, encompassing controlled voltage source.

(c) Is2 = (0.75mVD − 0.75mVB ) + (1mVC −1mVA)

(d) VC = 4VB + VD

(e)

0.75mVD + (0.75mVB − 0.75mVD) + (0.25mVB − 0.25mVA) = 0

or 1mVB − 0.25mVA = 0

(f) We now have three equations in VB, VC, and VD. These can be solved using any method. By inspection,

we can immediately deduce VB from VA using the last equation: VB = 4V.

The remaining two equations can be solved to obtain: VC = 20V and VD = 4V.

Solution 3.26 (a) The supernode is the combination of A, C, and the controlled voltage source.

(b) Write node equations starting at the supernode:

(G2VA − G2Vin ) + (G3VA ) + (G4VA − G4VB) + (G6VC ) + (G5VC − G5VB ) = 0

⇒ G2 + G3 + G4( )VA + −G4 − G5( )VB + G6 + G5( )VC = G2Vin

and

(2G6VC ) + (G4VB − G4VA ) + (G5VB − G5VC ) + (G1VB − G1Vin ) = 0

⇒ −G4( )VA + G1 + G4 + G5( )VB + 2G6 − G5( )VC = G1Vin

and

VA −VC = 3Vx, VA −VC = 3(Vin − VA ), 4VA − VC = 3Vin

In matrix form:

0.8

−0.4

4

−0.5

0.6

0

0.2

0.1

−1

VA

VB

VC

=6

6

180

(c) The above system of equations can be solved to obtain: VA = 38.75V, VB = 40V, VC = -25V.

(d) Iin = (G7Vin )+ (G2Vin – G2VA) + (G1Vin – G1VB) = 5A.

è Req = 12Ω and P = 300W.

P = Vin × Iin = 60 × 5 = 300W

(e) Iout = VCG6 = -2.5A è P = 62.5W.

Solution 3.27 (a) Supernode is A,B encompassing controlled voltage source.

(b)

(VA – Vs1) + 0.4VB + (0.2VB – 0.2VC) = 0

è VA + 0.6VB – 0.2VC = Vs1

(c) VA – VB = IB = 0.4 VB è VA = 1.4VB.

(d) Is2 = 0.2VB + (0.2VC – 0.2VB) = 0.2VC.

(e)

In matrix form:

1

1

0

0.6

−1.4

0

−0.2

0

0.2

VA

VB

VC

=8

0

2

The solution is: VC = 10V,VA = 7V ,VB = 5V .

(f) i = (VA-Vs1)1S = -1 è Pccvs = (VA −VB ) × (VA − Vs1 ) ×1S = −2W

Pvccs = (Vs1-VC)(0.2VB) = -2W.

Solution 3.28 (a) Supernode at A,C, CCVS.

(b) Node equation at supernode:

Is + 0.25mVA = G1VA + (G2VA –G2VB) + (G5VC) + (G4VC – G4VB)

è Is = (G1 + G2 – 0.25m)VA + (–G2 – G4)VB + (G4+G5)VC

Constraint:

VA – VC = 104ix = 104G3VB

è 0 = VA – 104G3VB – VC

At node B:

G3VB + (G2VB – G2VA) + (G4VB – G4VC) = 0

è –G2VA + (G3+G2+G4) – G4VC = 0

è (c) Matrix equation:

G1 + G2 − 0.25m −G2 − G4 G4 + G5

1 −104G3 −1

−G2 G3 + G2 + G4 −G4

VA

VB

VC

=Is

0

0

;

0

1000

−0.2

−1

−1000

1.1

1

−1000

−0.8

VA

VB

VC

=2

0

0

(d) Substitute the values of conductances and solve the above matrix equation in MATLAB to obtain:

VA = -38V, VB = -20V, VC = -18V.

(e) P = V × I = (104G3VB) × −Is + VAG1 + G2(VA − VB)[ ] = (−20) × −2m −1.9m − 3.6m[ ] = 0.15W

Solution 3.29

Loop equation: Vin = 2kI1 + 500(I1 + 20m)

è Vin = 2500I1 + 10 è I1 = 20mA.

Pvin = 20m×60 = 1.2W.

PI = 20m×(500I1 + 500×20m) = 0.4W.

P2k = I1×I1×R = 0.8W.

P500 = (I1 + 20m)2R = 0.8W.

total power absorbed by resistors: PR = 0.8 + 0.8 = 1.6

total power delivered by sources: Ps =1.2 + 0.4 W=1.6W

Conservation of power is verified.

Solution 3.30

Loop equation: 100(I1 − 0.5) + 200I1 + 500 × (I1 + 20m) = 0

è I1 = 0.05A.

P0.5A = I × V100Ω; V100Ω = 100 × (0.5 − 0.05); where V100Ω is the voltage on the 100Ω resistor.

P0.5A = 0.5(0.5×100 – 0.05×100) = 22.5W.

P20m = 20m(I1+20m)500 = 0.7W.

Solution 3.31

Loop equation: 3.3 = 50I1 + (50m + I1)100 + (I1 – 30m)40 + (I1 – 50m)60

è I1 = 0.01A.

The power delivered by the independent voltage source:

P = I1×3.3 = 0.033W.

Solution 3.32

Loop equation: 50 = 300I1 + (I1 – 0.4I1)500

50 = (300 + 500 – 200)I1 è I1 = 0.0833A.

Power absorbed by the 500Ω resistor.

P500 = (I1 – 0.4I1)2500 = 1.25W.

Solution 3.33

Loop equation: 1000(I1 − Is) + 4000I1 + 5000(I1 − gmVx ) = 0→10000I1 − 2Vx = 50

and

Vx = 1000(Is − I1) →1000I1 + VX = 50 .

Solve the above two equations in I1 and Vx to obtain: I1 = 12.5mA, Vx = 37.5V.

Thus, Req = Vx/Is = 750_,

P = Ivccs ×Vvccs = gm ×Vx × 5000 × (gmVx − I1)=0.1875W

Solution 3.34

Loop equation: Vin = 2Iin + 14Iin – 10V1

V1 = 2Iin

After replacing V1 in the loop equation we obtain:

è Vin = – 4Iin è

R1eg =VinIin

= −4Ω

Solution 3.35

Loop equation: Vs = 500I1 + 100(I1 + 0.5) + 400(I1 − 0.001Vx ) +100(I1 + 0.005Vy )

Vx = 500I1,Vy = 400I1 - 400×0.001Vx = 400I1 – 200I1 = 200I1

After replacing Vx and Vy in the loop equation we obtain:

Vs – 50 = 1000I1 è I1 = 0.1A

Vy = 200I1 = 20V è P400ohm = Vy2 /400Ω= 1W

Req = Vs/I1 = 150/0.1 = 1500Ω.

Solution 3.36

Select clockwise loop current I1 in the left loop. Select anti clockwise loop current I2 in the right loop.

The two mesh equations are:

12 = I1 +10(I1 + I2)

and 10(I2 + I1) + 2I2 + 12 = 0

The two simultaneous equations can be solved easily to obtain: I1 = 0.75A, I2 = +0.375A.

P10ohm = (I1 + I2)2/10 = 0.127W.

Battery 1 supplies more current. (I1 > I2)

Solution 3.37

(a) The equation for the left loop is:

660 = I1R + 1.296(I1 + I2)+ 590 + I1R

The equation for the right loop is:

660 = (0.3 – R) I2+ 1.296 (I1 + I2) + 590 + (0.3 – R) I2

Simplifying the two equations:

70 = 1.596I1 + 1.296I2

70 = 1.296I1 + 1.596I2

The solution of these two equations is: I1 = I2 = 24.2A.

(b) I1 + I2 = 48.4 , voltage across locomotive = 590 + 48.4×1.296 è power = 31592W.

(c) Because the locomotive is 1/3 distance from either station it follows that

R = 1/3×0.3 = 0.1Ω. The two equations become:

70 = I1(2R + 1.296) + 1.296I2

70 = 1.296I1 + (1.296 + 0.6 – 2R)I2

The solution of these two equations is: I1 = 32.64A, I2 = 16.32A.

Current in locomotive motor I1 + I2 = 48.96A.

Voltage across locomotive 590 + (I1 + I2) × 1.296

It follows that:

è P = (I1 + I2)(590 + 49×1.296) = 31993W.

Solution 3.38

(a)

(b) The three loop equations are:

660 – 590 = 0.1I1 + 1.296 (I1 – I2) + 0.1I1

0 = 1.296 ( I2 − I1) + 0.2I2 + 1.296 ( I2 − I3)

–70 = 1.296( I3 − I2) + 0.2I3

These three equations can be solved using any method to obtain:

I1 = 46.8A, I2 = 0, I3 = – 46.8A.

(c) Motor currents are 46.8A each.

(d) Ps = VI = 660×46.8 = 30.9kW. Each source supplies 30.9kW.

Solution 3.39 (a) Define three meshes with three mesh currents. The first, I1 , is a clockwise current

around the first mesh. The second, I2 , is a clockwise current around the middle loop of the circuit

(through the 10mS, 5ms, and 5ms conductances). The third, I3 , is a counterclockwise current through the

right-most loop containing the voltage source.

*current names shown above.

(b) I1 = 0.5A

(I2

10m−

I110m

) +I25m

+ (I25m

+I35m

) = 0.

(I3

25m+

I35m

) +I25m

= 20

These are two equations in two unknown currents. After grouping the terms, it can be verified that:

I2 = 0.1A, I3 = 0.

(c) Vx = 20V

Vad = (0.5 – I2)/10m = 40V

Vbd = 20V

(d) P0.5 = Va × 0.5 = 40×0.5 = 20W

P20V = 0W

Presistors = 2× I2 2/5m + (0.5 – I2)2/10m = 20W

The conservation of power is verified.

Solution 3.40 (a) We can either write down the equations or evaluate the matrix by inspection:

90 ×(I2 − 4.8m) + 10kI2 + 50 = 0

90kI3 + 10 × (I3 − 4.8m) = 50

OR

100k 0

0 100k

I2

I3

=

90k × 4.8m − 50

10k × 4.8m + 50

(b) The solution of the above equation is: I2 = 3.82mA, I3 = 0.98mA.

(c) Current source: P = 4.8m×[(4.8m – I2)90k + (4.8m – I3)10k] = 0.61W.

Voltage source: P = 50(I3 – I2) = – 142mW.

Solution 3.41 (a) By inspection:

112k −90k −10k

−90k 100k 0

−10k 0 100k

I1

I2

I3

=180

−60

60

(b) Using MATLAB:

I1 = 4.4mA, I2 = 3.36mA, I3 = 1.04mA

→ ix = − I2 = −3.36mA

(c) P180 = 180×4.4m = 0.792W, P60 = 60( I3 − I2) = -0.139W.

Solution 3.42 The matrix equation is:

8 −6 3

−6 8 −2

3 −2 4

I1I2

I3

=14

0

6

whose solution is: I1 = 4A, I2 = 3A, I3 = 0A

v = ( I1 + I3 + I2) ×2 = 2V

Solution 3.43 (a) First, note that two mesh currents are needed. Two clockwise currents are defined: I1

in the middle loop, and I2 in the right-most loop:

Middle loop equation:

100I1 – 100Is1 + 20Ix + 80I1 – 80I2 = 0, where Ix = Is1 – I1

and

Right-most loop equation:

80I2 – 80I1 + 10I2 + Vs2 = 0

These can easily be cast into the following matrix equation:

160

−80

−80

90

I1I2

=

16

−6

(b) The solution of this equation is: I1 = 0.12A, I2 = 0.04A.

(c) VA = 100(Is1 – I1) = 8V and VB = 80x(I1 − I2) = 6.4V .

(d) Ps1 = Is1VA = 1.6W.

(e) P0.0125S = (I1 − I2)2/0.0125 = 0.512W.

Solution 3.44 (a) Create two clockwise mesh currents in the top loop (I2) and the bottom-left loop (I1) .

The bottom-right loop has an independent current source. Writing the loop equations:

Vs1 = 200 (I1 – I2) + 200 (I1 + Is2)

200 (I2 – I1) + 100I2 + 300Ix + 200 (I2 + Is2) = 0, where Ix = I1 – I2

(b) Solving, we get: I1 = -0.1A, I2 = -0.7A, Ix = 0.6A.

(c) VB = (I1 + Is2)200 = 130V.

(d) Pvs1 = I1Vs1 = –25W, Pis2 = (VB + (Is2 + I2)200)Is2 = 105W, P300ix = (−I2)(300Ix) = 126W.

Solution 3.45 (a) Create two clockwise mesh currents in the top loop (I1) and the middle loop (I2) (all

resistor loop):

Top loop equation:

0.5vx = 500 (I1 - I2,) +500I1 where vx = –500I1

and

Middle loop equation:

600 (I2 – Is1) + 500 (I2 – I1) + 900 (I2 + Is2) = 0

(b) Solving, we get: I1 = 0.015A, I2 = 0.0375A, vx = -7.5V.

(c) Pis1 = Is1[0.5vx + (Is1 – I2)600] = 109.7W

P0.5vx = 0.5vx(I1 – Is1) = 1.63W

Pis2 = (I2 + Is2)×900×Is2 = 53.2W

Solution 3.46 Write the mesh equations in terms of R’s and then substitute the values from the matrix:

Mesh 1 equation:

v1 = R1 ( i1 − i3) + R2 (i1 – i2)-25i2From this equation, and the first row of the matrix equation, we can deduce that

R1 = 5 Ωand

R2 + 25 = 40 ⇒ R2 = 15Ω.

Similarly:

Mesh 3 equation: R1(i3 − i1) + i3R4 + R3(i3 − i2) =0

From which we can deduce:

R3 = 25 Ω and R4 = 5 Ω.

Solution 3.47

Modified loop 1 equation:

Vs1 = 3MI1 + v + 2MI1

Constraint equation:

– I1 + I2 = Is3

Modified loop 2 equation:

v = 2MI2 + Vs2 + 8MI2

Or in matrix form:

5M 0 1

−1 1 0

0 −10M 1

I1

I2

v

=Vs1

Is3

Vs2

Solving: I1 = -1.1 A, I2 = -0.95 A.

The power Ps3 = Is3 xv= 7.58 W

Solution 3.50 I2 = 2A, I3 = –7A

Loop 1 equation:

Vs = 3I1 + 3(I1 − I2) + 6(I1 + I3 − I2) + 2vy + 2(I1 + I3) =14I1 + 2vy − 74 ⇒ 14I1 + 2VY = 88

vy = 3(I1 – I2)= 3I1 − 6

Solving the above system, we obtain: I1 = 5A è vy = 9, v can be found from the loop 3 equation

v + (I1 + I3 − I2).6 + 2v y + 2(I3 + I1) = 0 . Solving, we obtain v =10V

Finally, Pvs = Vs×I1 = 70W.

Solution 3.51

Mesh 1 equation:

9kI1 + 3k (I1 – I4) + 6k (I1 – I3) + 12k (I1 – I2) = 0

where we have used the fact that Ix = I1 – I4 (and I4 = – 4mA)

Mesh 3 equation:

6k (I3 – I1) – v2 +2kI3 + v = 0

Mesh 2 equation:

2.4kI2 + 12k (I2 – I1) – v = 0

Constraint equations:

I2 – I3 = 0.5mA

I3 – I4 = 0.5Ix = 0.5I1 – −0.51I4 = 0.51I1 + 2mA ; 0.51I1 − I3 = −2mA

The above five equations need to be put into matrix form:

30k −12k −6k 0 0

−6k 0 8k 1 −1

−12k 14.4k 0 −1 0

0 1 −1 0 0

0.5 0 −1 0 0

I1I2

I3

v

v2

=

−12

0

0

0.5m

2m

The solution is:

I1

I2

I3

I1I2

I3

v

v2

=

−0.002A

−0.0025A

−0.003A

−12V

−24V

Ix = I1 − I4 = 2mA

Power delivered by 0.5mA current source: P1 = 0.5mxv = −6mW

Power delivered by the dependent current source: P2 = 0.5Ix ⋅ v2 = 1m × (−24) = −24mW

Power delivered by 4mA current source: P3 = 4m × (3k ⋅ Ix − V2) = 120mW

Solution 3.52

Write the following equations:

Mesh 2 equation:

100 (I2 – I1) + 150 (I2 – I4 – I6) + v = 0

Mesh 4 equation:

3vx + 30 (I4 – I3) + 150 (I4 - I2 + I6) = 0

Mesh 5 equation:

800I5 – 3vx + 10( I5 − I1 + I6) =0

Mesh 6 equation:

250I6 +150(I6 + I4 − I2) +10(I6 + I5 − I1) = 0

Constraint equation:

I1 I2 I3

I4

I5

I6

Is2= −I2 + I3

Substituting the values of I1 = Is1 = 1.15 and I3 = Is3 = 0.95 , and noting that

vx = (I1 – I6 – I5)10

we can write:

250 −150 0 −150 1

−150 180 −30 120 0

0 0 840 40 0

−150 150 10 410 0

1 0 0 0 0

I2

I4

I5

I6

v

=

100I130I3 − 30I1

40I1

10I1I3 − IS2

I2

I4

I5

I6

v

=

0.65

0.45

0.05

0.1

35

Finally, vx = 10V and v is as given above.

Solution 3.53 (a)

Replace the voltage source by current sources:

At node 1

is1 = (V2 − V1) /12k

At node 2

(V2 − V1) /12k + (V2 − V3) /90k + (V2 − V4) /10k = 0

At node 3

is2 = (V3 – V2) /90k + V3/10k

At node 4

is2 + (V4 −V2) /10k + V4/90k = 0

V1 = 180

V4 – V3 = 60

(b) In matrix form:

−1/12k 1/12k 0 0 −1 0

−1/12k 1/12k +1/90 k +1/10k −1/90k −1/10k 0 0

0 −1/90 k 1/90k +1/10k 0 0 −1

0 −1/10k 0 1/90 k +1/10k 0 1

1 0 0 0 0 0

0 0 −1 1 0 0

V1

V2

V3

V4

is1is2

=

0

0

0

0

18060

The solution from MATLAB is

180.0000

127.2000

33.6000

93.6000

-0.0044

0.00232

(c) Power delivered by S1 is:

Ps1 = Vs1 × (−Is1) = 0.792W

Power delivered by S2 is:

Ps2 = Vs2 × (−Is2) = −0.139W

Solution 3.54 (a) Replace the 100 ohm resistor, the controlled voltage source, and vs2 by current source.

Then write the node equations:

Is1 = ix + I1 + 0.09 (V1 – V3)

I1 = V2/20 + 0.05 (V2 – V3)

– I2 = 0.05 (V3 – V2) + 0.09 (V3 – V1)

V3 = Vs2

V1 – V2 = 300ix = 300V1/100

In matrix form:

0.1

0

−0.09

0

2

0

0.1

−0.05

0

1

−0.09

−0.05

0.14

1

0

1

−1

0

0

0

0

0

1

0

0

V1

V2

V3

I1I2

=

2

0

0

50

0

(b) Using MATLAB to solve the above system the solution is:

V1 = -90.0000V

V2 = 180.0000V

V3 = 50.0000V

I1 = 15.5000A

I2 = -6.1000A

(c) Power delivered by the current source is

Ps1 = Is1 × V1 = −180W

Power delivered by the voltage source is:

Ps2 = Vs2 × (−I2) = 305W

Solution 3.55 Modify the circuit so that it looks like the following:

The modified node equations are:

Is1 = - Iy + Ix

Ib = Iy + 0.2 (VC – VB)

Is2 = 0.2 (VC – VB) + Ix

The equations describing the constitutive relationships of elements in the original network are:

VA – VB = Ib

Ix = 0.2Vb

Ib = 0.4Vb

Iy = 8 −VA

These can be cast into a matrix equation and solved easily to obtain the same result as previously arrived

at.

In matrix form:

0

0

0

1

0

0

1

0

−0.2

−0.2

−1

0.2

0.4

0

0

0.2

0.2

0

0

0

0

1

0

1

0

−1

0

0

−1

1

0

0

0

0

1

0

−1

0

−1

0

−1

0

−1

0

0

0

0

0

0

VA

VB

VC

Ix

Iy

Ib

Is1

=

0

0

2

0

0

0

8

VA

VB

VC

Ix

Iy

Ib

Is1

=

7V

5V

10V

1A

1A

2A

0A

Solution 3.56 Modify the circuit as follows:

The modified node equations are:

At node A: Is = G1VA + G2(VA −VB ) + Ia

At node B: G2(VA −VB ) = Ix + G4(VB −VC )

At node C: 0.25mVA + Ia = G4(VC − VB) + G5VC

The equations describing the constitutive relationships of elements in the original network are:

Ix = G3VB

VA – VC = 104Ix

These can be cast into a matrix equation that can be solved in MATLAB.

In Matrix form:

0.25m

−0.2m

−0.25m

0

1

−0.2m

1m

−0.8m

−0.1m

0

0

−0.8m

1m

0

−1

1

0

−1

0

0

0

1

0

1

−104

VA

VB

VC

Ia

Ix

=

2m

0

0

0

0

The solution is:

VA

VB

VC

Ia

Ix

=

−38V

−20V

−18V

0.0075A

−0.002A

We observe that we have obtained the same results as in problem 3.28.

Solution 3.57 Replace dependent source by i35 (from 3 to 5). Also, replace voltage source by i10 (from 1

to 0). Now, write the modified noted equations. The reference node is O:VO = 0V :

At node 1: i10 = (V6 – V1) + (V2 – V1)

At node 2: 2 = (V2 – V1) + (V2 – V3)

At node 3: i35 = (V4 – V3) + (V2 – V3)

At node 4: 2 = 2 + (V4 – V3) + V4

At node 5: i35 = (V5 – V6) + V5

At node 6: 2 = (V5 – V6) – V6

Constraints:

V3 – V5 = 15vx = 15V4

V1 = 5

The following matrix equation is obtained:

−2 1 0 0 0 1 −1 0

−1 2 −1 0 0 0 0 0

0 1 −2 1 0 0 0 −1

0 0 −1 2 0 0 0 0

0 0 0 0 2 −1 0 −1

0 0 0 0 1 −2 0 0

0 0 1 −15 −1 0 0 0

1 0 0 0 0 0 0 0

V1

V 2

V 3

V 4

V 5

V 6

i10

i35

=

0

2

0

0

0

2

0

5

The solution of this equation is obtained from MATLAB:

V1 =5.0000V

V2 =3.3571V

V3 =0.2857V

V4 = -0.1429V

V5 =1.8571V

V6 = -0.0714V

i10 = -6.7143A

i35 = 3.7857A

The power delivered by the dependent voltage source connected between nodes 3 and 5:

P35 =15vx (−i35) = 15 ×V4 × (−i35) = 8.115W

The power delivered by the current source connected between nodes 2 and 4:

P24 = (2A) × (V2 −V4 ) = 7W

The power delivered by the current source connected between nodes 4 and 6:

P46 = (2A) × (V4 −V6) = −0.143W

The power delivered by the voltage source connected between nodes 1 and 0:

P10 = 5V × (I12 + I16) = 5x (V1 −V2) + (V1 − V6)[ ] = 33.57W

Solution 3.59 Using the appropriate element stamps for each element of the circuit, we obtain the

following system:

0.15 + 0.2 −0.15 −0.2 0

−0.15 0.15 + 0.05 0 −1

−0.2 0 0.25 + 0.2 1

0 −1 1 0

VA

VB

VC

Ix

=

−8 − 3

3

25

440

Solution 3.60

1/20 k +1/10k + gm1 −1/10k

−1/10k − gm1 − gm2 gm2 +1/10k +1/2.5 k

V1

V2

=

Is1

−Is2

The solution is the same as that of problem 3.14.

Solution 3.62 (a) Because RT (T ) can be approximated by a straight line between (250Ω,0oC) and

(80Ω,50oC) it follows that:

RT (T )= – 3.4T + 250

(b) For T = 25oC , RT = 165Ω

(c) The voltage across the RT + RL series combination can be obtained from voltage division:

VT ,L =RT + RL

RT + 2RL + R⋅12 = 4.7857V

This is the same as the voltage across Rx because the meter is at zero deflection. Thus,

RxRx + R

⋅12 = 4.7857. It follows that Rx =165.84Ω.

(d) We first denote the nodes:

A - the node common to R,Rx and the voltmeter;

B - the node common to R,R and the voltage source;

C - the node common to RL,RT and RL;

D - the node common to Rx,RL and the voltage source.

The reference node is D:VD = 0. It follows that VB =12V .

We also have: vout = VA − VC

The node equations are:

12

0

At node A:VA −VC

Rm+

VA −12R

+VARx

= 0

Equivalently: VA(RRx + RmRx + RmR) - VC ⋅ RRx = 12RmRx

VA × 4199.86 − VC × 41.46 =19900.8 (1)

At node C:VC −12R + RL

+VC

RT + RL+

VC − VARm

= 0

Equivalently: (VC −12) × 0.004 +VC

RT + 2.5+ (VC −VA) ⋅10−4 = 0

VC (0.004 ⋅ RT + 0.01 +1+ 0.00025 + RT ×10−4) −

−VA(RT ×10−4 + 0.00025) = 0.48 × (RT + 2.5)

The last equation can be rewritten as:

VC (0.0041RT +1.01025) - VA(RT ×10−4 + 0.00025) = 0.48(RT + 2.5) (2)

From (1) and (2), we obtain:

(0.0041RT +1.01025) ×−19900.8 + VA × 4199.86

41.46−

− VA × (RT ×10−4 + 0.00025) = 0.48(RT + 2.5)

Equivalently: VA × (RT × 0.415 +102.337) = 2.448 × RT + 486.12

It follows that VA =2.448 × RT + 486.120.415 × RT +102.337

From the equation at node A:

vout = VA − VC = −Rm ×VA −12

R+

VARx

=

2.448 × RT + 486.120.415 × RT +102.337

×(−100.3) + 480

At T = 0oC: RT = 250Ω. It follows that vout = −54.4415V

At T = 50oc: RT = 80Ω. It follows that vout = 80V

(e): The formula has been derived at part d):

T RT vout

0oC 250Ω −54.4415V

5oC 233Ω −52.4136V

10oC 216Ω −50.2368V

15oC 199Ω −47.8938V

20oC 182Ω −45.3650V

25oC 165Ω −42.6273V

30oC 148Ω −39.6537V

Solution 3.63

Place a source Vin between C and D, and calculate the current drawn from the source as below:

Loop 1 equation:

I1R1 + (I1 − I2)R2 + (I1 − I3)R3 = 0

Equivalently:

I1(R1 + R2 + R3) − I2R2 − I3R3 = 0

Loop 2 equation:

(I2 − I1)R2 + I2R4 −1 = 0

Equivalently:

−I1R2 + I2(R2 + R4) =1

Loop 3 equation:

1 + I3R5 + (I3 − I1)R3 = 0

Equivalently:

−R3I1 + I3(R3 + R5) = −1

We obtain the following system of equations:

30I1 − 4I2 − 6I3 = 0

−4I1 + 6I2 =1

−6I1 +14I3 = −1

⇒ I1 = 0.0096A , I2 = 0.1731A , I3 = −0.0673A;

Iin = I2 − I3 = 0.2404 A

Reg,CD =VinIin

=1

0.2404= 4.16Ω

Solution 3.64

The node equation at node A is:

VAG1 + (VA − VB)G2 + (VA −VC )G3 = 0

Equivalently:

(G1 + G2 + G3)VA −VBG2 − VCG3 = 0

The supernode is identified by a Gaussian surface enclosing the controlled voltage source. The supernode

equation is:

G2(−VA + VB) − 6 + G4VC + G3(VC − VA ) = 0

Equivalently, we have:

−VA(G2 + G3) + G2VB +VC (G3 + G4 ) = 6

One way of obtaining the solution to the problem is:

We multiply the above two equations by 30.

−30(G2 + G3)VA + 30G2VB + 30(G3 + G4 )VC =180

and

30(G1 + G2 + G3)VA − 30G2VB − 30G3VC = 0

By equating the coefficient of the above two equations with the coefficients of the first and second given

equations, we obtain:

30G2 = 30 ⇒ G2 = 0.1S

30G3 = 2 ⇒ G3 = 0.067S

30(G1 + G2 + G3) = 11⇒ G1 = 0.2S

30(G3 + G4 ) = 32 ⇒ G4 = 0.87S

can be obtained as follows:

VC −VB = VX = (VC − VA )

Equivalently:

VA −VB + (1− )VC = 0

By comparing with the third given equation ⇒ = 3.

Solution of 3.66

(a)

At node A: (VA-VC)/2 + (VA-VB)/2 + (VA-VD)/2 = 14

At node B: (VB-VA)/2 + (VB-VC)/2 + (VB-VD)/2 = 7

At node C: (VC-VA)/2 + (VC-VB)/2 + (VC-VD)/2 + 2VC = 0

At node D: (VD-VA)/2 + (VD-VB)/2 + (VD-VC)/2 + 0.5VD = 0

These can be solved in MATLAB to obtain:

22.0000

18.5000

7.5000

12.0000

(b) Mesh analysis would result in the same voltages

The loops and their current loops are:

A,C,14A:I1A,C,D:I2

A,B,D:I3

B,C,D:I4

C,D,reference node:I5

B,D,7 A:I6

I1 = 14A,I6 = 7A

Loop ACDequation: 2(I2 − I1) + 2(I2 − I3) + 2(I2 + I4 + I5) = 0

Loop BCD equation: 2(I3 + I6 + I4 ) + 2(I2 + I4 + I5) + 2I4 = 0

Loop ABD equation: 2(I3 + I4 + I6) + 2(I3 − I2) + 2I3 = 0

Loop CDref node equation: 2(I5 + I2 + I4 ) + 2(I5 − I6) + 0.5(I5 + I1) = 0

In matrix form:

6

2

−2

2

−2

2

6

0

2

6

2

2

2

2

0

4.5

I2

I3

I4

I5

=

28

−14

−14

7

;

I2

I3

I4

I5

=

6.75

1.75

−5.5

1

VA = 2(I1 − I2) + 0.5(I1 + I5) = 22V

VB = 2(I4 + I3 + I6) + 2(I6 − I5) = 18.5V

VC = 0.5(I1 + I5) = 7.5V

VD = 2(I6 − I5) =12V

(c) Mesh analysis requires more work.

(d) The removal of the top resistor will result in more node equations than loop equations. The addition of

a resistor between node A and reference node will result in more loop equations than node equations.

Thevenin Probs, 7/11/01 - P4.1 - @R.A. Decarlo & P. M. Lin

PROBLEM SOLUTIONS CHAPTER 4

SOLUTION 4.1. First, find Vout / Vs for each circuit. Then solve for R knowing

Vout = P ⋅10 =±14.142V .

(a) Writing KCL at the inverting terminal, 1 /1k(v− − vs) =1/ R(Vout − v− ) ⇒ Vout / Vs = −R / 1k , since

the inverting terminal is a virtual short. Solving for R = −Vout ⋅1k / Vs = 2.828kΩ .

(b) Writing KCL at the inverting terminal, Vs /1.5k = (Vout − Vs)/ R ⇒ Vout / Vs = R / 1.5k +1, solving

for R = 1.5k(Vout / Vs −1) = 2.743kΩ.

(c) From (a) Vout / Vs = −12k / R , thus R = −12k ⋅Vs / Vout = 4.243kΩ.

(d) This is the same circuit as (b) except the output voltage is taken across two resistors. Thus

Vout =P

1010 + 6( ) = 22.627V . Using the general form from (b), R = 400 Vout / Vs −1( ) = 1.410kΩ

SOLUTION 4.2. (a) First, find the voltage at the non-inverting terminal as v+ = 1/ 2 ⋅Vs . Then write KCL

at the inverting terminal, and make use of the virtual short property,

(Vs / 2)/10 k = (Vout − Vs / 2) / 30k ⇒ Vout / Vs = 30k(1/ 20k +1/ 60k) = 2 .

(b) Relating the output of the amplifier to the output of the circuit, Vout = Vamp(500 / 800) . Then writing

KCL at the inverting terminal, Vs / 400 = (Vamp − Vs) /1.2k ⇒ Vamp / Vs = 1.2k / 400 +1 = 4. Therefore

Vout / Vs = (Vamp / Vs ) ⋅ (Vout / Vamp ) = 2.5.

(c) Note that since no current goes into the non-inverting terminal of the op-amp, the voltage at that node is

–Vs. KCL at the inverting terminal, −Vs / 4k = (Vout + Vs )/ 20k ⇒ Vout / Vs = −6.

SOLUTION 4.3. Write KCL for both terminals,

(V− −Vi ) /1k = (Vo − V− ) / 2k

V− /1k = (Vo − V− )/ 3k

Solving and doing the appropriate substitutions, Vo / Vi = −8 .

SOLUTION 4.4. This is essentially the basic inverting configuration, which is defined as

Vo / Vi = −2k / 1k = −2 .


SOLUTION 4.5. (a) By voltage division VL = 1V ⋅100

200= 0.5V . Using Ohm’s law

Is = IL =1

100 +100= 5mA .

(b) No current flows in the input terminal of an ideal op-amp, thus Is = 0A and VL = 1V . From Ohm’s

law Ia = IL = VL /100 = 10mA .

SOLUTION 4.6. (a) Using voltage division,

V1 = Vs32||(8 + 24)

[32||(8 + 24)] + 8

=

23

Vs

Vout = V124

24 + 8

= 0.5Vs

(b) By voltage division,

V1 = Vs32

32 + 40

= 0.8Vs

Vout = V124

24 + 8

= 0.6Vs

(c) Using voltage division, V1 = Vs32

32 +8

= 0.8Vs , as no current enters the non-inverting terminal of the

op-amp. Due to the virtual short property, Vout = V124

24 + 8

= 0.6V1 . This is indeed the same results as

(b), which should be expected because of the isolation provided by the ideal buffers.

SOLUTION 4.7. Write KCL at the inverting terminal,

−Vs1 /1k − Vs2 / 2k = Vout / 4k ⇒ Vout = −4Vs1 − 2Vs2 = 40mV .

SOLUTION 4.8. (a) The voltage at the non-inverting terminal is V+ = 3 / 2V , KCL at the inverting

terminal gives (1.5− 2.5) /10k = (Vout −1.5)/ 30k ⇒ Vout = −1.5V . The power is

P = Vout2 / 500 = 4.5mW .

(b) The voltage at the non-inverting terminal is 3V this time, thus KCL

(3 − 2.5)/ 10k = (Vout − 3) / 30k ⇒ Vout = 4.5V . The power is P = Vout2 / 500 = 40.5mW .

SOLUTION 4.9. (a) Define the point between the two op-amp as Vint. Observe that the first op-amp is in

the basic inverting configuration, and the second the non-inverting configuration. By inspection,


Vint / Vs = −R1 / 1k

Vout / Vint = (1+ R2 /1k )

Cascading the two stages, (Vint / Vs)(Vout / Vint ) = Vout / Vs =− R1 /1k(1+ R2 /1k) . Solving for

R1 = 20 ⋅1k / (1+ R2 /1k) = 5kΩ. The power absorbed is P = (20 ⋅0.5)2 / 8 =12.5W .

(b) Using the same equations as (a), solve for R2 = (20 ⋅1k / 2k −1)1k = 9kΩ .

(c) Rewriting the equation obtained in (a), R12 +1kR1 − 20M = 0, and solving the quadratic equation yields

R1 = R2 = 4kΩ .

SOLUTION 4.10. This is a cascade of two non-inverting configuration op-amp of the form

Vo / Vs = (1+10k /10k ) for each. Therefore 2 ⋅ 2 = 4.

SOLUTION 4.11. This system is made up of a non-inverting stage with a gain of 1+10k/10k, a voltage

divider of gain 8k/(2k+8k), and a second non-inverting stage of gain 1+10k/10k. The product of all three

yields Vout / Vin = (2)(0.8)(2) = 3.2.

SOLUTION 4.12. (a) By inspection, the gain of the first stage is –1. Then write KCL for the second stage

Vs1 / 2R − Vs2 / R = Vout / 2R ⇒ Vout = Vs1 − 2Vs2 = 10V .

(b) The first stage gain is –0.5, thus Vout = 2R(0.5Vs1) / 2R − 2R(Vs2)/ 0.5R =−7.5V , using the same

procedure as in (a).

SOLUTION 4.13. (a) This is a cascade of a summing amplifier with the following transfer characteristic,

Vo = −4Vs1 − 2Vs2 , and an inverting stage of gain –1.5. Thus Vout = 1.5(4Vs1 + 2Vs2 ) = 2.25V .

(b) Notice that the only difference is the gain of the inverting stage, which is now –2. Therefore

Vout = 2(4Vs1 + 2Vs2) = 3V .

SOLUTION 4.14. This circuit is a cascade of two summing amplifier where the output of the first is an

input of the second stage. The transfer function of the first stage is Vo = −2RVs1 / 2R − 2RVs2 / R , which

is substituted in the transfer function of the second stage to obtain

Vout = −R[−2RVs1 / 2R − 2RVs2 / R]/ R − RVs3 / R = Vs1 + 2Vs2 − Vs3 =−2V .

SOLUTION 4.15. Writing KCL at the inverting node, −V1 / R1 − V2 / R2 − V3 / R3 = Vout / Rf , and

solving for Vout = −Rf

R1V1 +

Rf

R2V2 +

Rf

R3V3

.


SOLUTION 4.16. Referring figure P4.15, the value of the resistance must satisfy the following

constraints: R1 = R2 = R3 = 3R

Rf = R

These will yield the inverted average. If polarity is a concern, a second inverting stage should be added

with a unity gain, i.e. both R’s equal.

SOLUTION 4.17. Using the topology of 4.12 the following parameters are chosen,

Ga1 = 3, Ga2 = 5, Gb1 = 2, Gb2 = 4

For the time being assume G f = 1. Now we calculate δ = (1+ 3 + 5) − (2 + 4) = 3, this sets Gg = 3.

(a) The requirement for G f = 10µS sets the scaling factor K = 10µ / 1 =10µ . This then yields the

following set of parameters,

Ga1 = 30µS, Ga2 = 50 µS, Gb1 = 20µS, Gb2 = 40µS, G f = 10µS, Gg = 30µS

(b) The requirement for G f = 2µS , sets the scaling constant to 2uS. So the following parameters are

obtained:

Ga1 = 6µS, Ga2 =10 µS, Gb1 = 4 µS, Gb2 = 8µS

Furthermore for Gg = 12µS , ∆G = 6µS in order to make the incident conductance equal at both terminal.

(c) Using the starting values from (a), one could choose a scaling constant of 5 µS. This will yield the

following resistances:

Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 100kΩ, Rb2 = 50kΩ, R f = 200kΩ, Rg = 66.67kΩ

These are all reasonable values for circuit implementation.

SOLUTION 4.18. (a) Choosing the following initial values:

Ga1 = 3S, Ga2 = 5S, Gb1 = 11S, Gb2 = 4S, G f =1S


then calculate δ = (1+ 3 + 5) − (11+ 4) =−6 . Thus Gg = 1S , and ∆G =1 + 6 = 7S . Scaling everything by

1µS, yield this final set of parameters, which meet the requirements.

Ga1 = 3µS, Ga2 = 5µS, Gb1 =11µS, Gb2 = 4µS, G f = 1µS, Gg =1µS, ∆G = 7uS

(b) The set of parameters remains unchanged, except for ∆G which now becomes 6uS in order to

maintain the equal termination conductance requirement due to Gg = 0S .

(c) Scale the initial parameters of (a) by 5uS, and get the following set of resistances:

Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 18.18kΩ, Rb2 = 50kΩ, Rf = 200kΩ, Rg = 200kΩ, ∆R = 28.57kΩ

SOLUTION 4.19. (a) Choosing the following initial set of parameters:

Ra1 = 1/ (4S) = 0.25Ω, Ra2 = 1/ (2S) = 0.5Ω, Rb1 = 1/ (5S) = 1/ 5Ω, Rb2 =1/ (4S) = 0.25Ω, R f = 1Ω

and δ = (1+ 4 + 2) − (5 + 4) = −2 , thus choose Rg =1/ (1S) = 1Ω and ∆R = 1/ (1+ 2) = 1/ 3Ω . To meet

the Rf = 50kΩ requirement, all the parameters must be scaled by 50k, which gives

Ra1 = 12.5kΩ, Ra2 = 25kΩ, Rb1 = 10kΩ, Rb2 =12.5kΩ, R f = 50kΩ, Rg = 50kΩ, ∆R = 16.67kΩ

(b) Same as (a) with a 100k scaling constant:

Ra1 = 25kΩ, Ra2 = 50kΩ, Rb1 = 20kΩ, Rb2 = 25kΩ, Rf = 100kΩ, Rg = 100kΩ, ∆R = 33.33kΩ

SOLUTION 4.20. (a) When the op-amp is in its active region vout / vs = −5 . Thus it will operate in its

active region when −3 ≤ vs ≤ 3, and will saturate at 15V when vs ≤ −3 , and at –15V when vs ≥ 3. SPICE

yield the following plot:


(b) Using SPICE the following plot is obtained:


SOLUTION 4.21. The first stage is in a summing configuration, thus its output is, assuming it's in the

active region of operation, -15 V which means it is just about to saturate. The second stage is in the

inverting configuration with a gain of -1.5, which means that the overall output will be saturated at 15V.

SOLUTION 4.22. When vin −80kvin +1.5

100k

> 0, or vin > 6 the output of the comparator saturates at

–15 V, when it is vin < 6 , it will saturate at 15 V. The following plot is obtained from SPICE.

SOLUTION 4.23. When vin −10kvin + 20

110k

> 0 , or vin > 2 the output of the comparator will be

saturated at -15V. Otherwise when it is < 2V the output saturates at 15V. In SPICE:


SOLUTION 4.24. Based on the same reasoning as the previous questions,

The output will be +Vsat, when vin < vref 1 −R1 + R2

R2

= −

R1

R2vref , and –Vsat for

vin > vref 1 −R1 + R2

R2

= −

R1

R2vref .

SOLUTION 4.25. Using the previously derived relationship, and the topology of figure P4.24, set

vref = −1.5V , and R1 = 2kΩ and R2 = 3kΩ . Set the power supplies to the Op-amp to +/– 10V to satisfy

the Vsat requirement. Also the input to the inverting and non-inverting terminal are reversed for fig. P4.24.

Verifying in SPICE we obtain the following,


SOLUTION 4.26. The design that fulfills the requirement is the same as for P4.25, with the input to the op

amp reversed. The following is obtained from SPICE,


SOLUTION 4.27. First, for the comparator to give +Vsat for the lower voltages, the inputs to the op amp

in the topology of P4.24 must be interchanged. Then the components are chosen to satisfy the following

relationship, vswitch = vref 1−R1 + R2

R2

=−

R1

R2vref . Choose vref = −1.5V , and R1 = 2k and R2 = 1k.

Verifying in SPICE,


SOLUTION 4.28. Write KCL at the inverting terminal, noting that the no current flows into it:

(V− − vin )/ R = (vout − V− )/ R . Use the following relationship vout = A(V+ − V− ) = −AV− . Solving

using the previous two equations yields vout / vin = −A

A + 2.

SOLUTION 4.29. (a) By inspection the voltage gain for the ideal case is –1. When A=1000, the gain

becomes –0.998, thus 0.2%.

(b) Repeating the method of P4.28, and setting vout / vin = −ARf

Rf + R1 + AR1to –1 and solving for

Rf =10.417kΩ.

(c) Solving the previous equation when the gain is –1, Rf / R1 = (A + 1)/ ( A −1).

SOLUTION 4.30. (a)The first part was obtained in P4.29. Rearranging the equation yields

vout / vin = −Rf

R1

1

1 + 1+ Rf / R1( ) / A

.


(b) The error is caused by (1+ Rf / R1) / A in the denominator, and may be defined, in percent, as

100 −1

1+ 1+ Rf / R1( ) / A

⋅100. Thus for the conditions listed in the problem, it will always be less than

2.05%. With A = 10000 it will be less than 0.21%.

SOLUTION 4.31. (a) Substituting the non-ideal model, and writing KCL at the inverting terminal,

(V− − vin )/ R1 + V− / Rin = (vout − V− ) / Rf is obtained. Now observe the following dependencies,

iout = vout / RL , and vout =− AV− − (iout + (vout − V− )/ R f )Rout . Using these three equations, substitute

the second into the third and then solve for vout / vinusing the last two. This yields

V− = vout + vinRf

R11/

Rf

R1+

Rf

Rin+1

= vout

−1 −Rout

RL−

Rout

R f

A − Rout

Rf

and

vout / vin = −Rf

R1

1

1+

1 + RoutRf

+ RoutRL

1+Rf

R1+

Rf

Rin

A −Rout

Rf

A gain of –9.988

(b) For an ideal op-amp the gain is −Rf / R1=-10.

(c) The error is about 0.1175%.

SOLUTION 4.32. The gain is –9.883, and the error 1.16%

SOLUTION 4.33. This derivation was performed in P4.31.


SOLUTION 4.34. Assume that the sliding contact is at the bottom of Rp. Then, writing KCL at the

inverting terminal yields vin / Ro = (vout − vin )/ Rp . This implies vout / vin = 1 + Rp / Ro . When the

slider is at the top, it is evident that vout = vin . Therefore 1 ≤ vout / vin ≤1 +Rp

Ro.

SOLUTION 4.35. Writing KCL at inverting input, and making use of voltage division,

−vin / R1 = βvout / Rf[ ] where β is the fraction of vout that appears across Rf. Hence, vout

vin=−

Rf

βR1.

When the slider is at the top β= 1 and vout

vin=−

Rf

R1. When the slider is at the bottom, the fraction of vout

appearing across Rf is β=Rf / / R0

Rf / / R0 + Rp=

Rf R0

Rf + R0×

1Rf R0

R f + R0+ Rp

=Rf R0

R f R0 + Rp(R f + R0). Hence

1

β=

Rf R0 + Rp(Rf + R0 )

R f R0=1 +

Rp

R0+

Rp

R f. It follows that

vout

vin=−

Rf

βR1=−

Rf

R11 +

Rp

R0+

Rp

Rf

.

Therefore the range of achievable voltage gain is

−Rf

R1≥

vout

vin≥ −

Rf

R11 +

Rp

R0+

Rp

Rf

SOLUTION 4.36. Using the basic non-inverting configuration of figure 4.10 characterized by

vout / vin = 1+Rf

R1

, i.e., µ = 1+

Rf

R1

.

SOLUTION 4.37. At first glance, one might use two inverting configurations, figure 4.5, in cascade.

However, such would not have infinite input resistance. To circumvent this problem we add a buffer

amplifier as per figure 4.7 at the front end of a cascade of two inverting configurations. The resulting

overall gain is µ =R f1

R11

Rf 2

R12

. Indeed, such a configuration can achieve theoretically any gain greater

than zero.


SOLUTION 4.38. Using a single inverting amplifier configuration, figure 4.5, preceded by a buffer stage

of figure 4.7. The gain is µ = −Rf

R1.

SOLUTION 4.39. By KVL for figure P4.39a, Vo = −i1Rf . Thus to achieve Vo = −i1rm in figure P4.39b,

we set Rf = rm .

SOLUTION 4.40. Writing KCL at the inverting node of the ideal op amp yields IL = Vi / Ra , which is

indeed independent of the load resistor which has no effect on the load current.

SOLUTION 4.41. The current through the LED is IL = 10R1

10k

/ 3.8k , so for (a) it is 1.32mA and for

(b) 2.11mA.

SOLUTION 4.42. Applying KCL at the inverting terminal, IL = vin / R1 . Again, ideally, RL does not

affect IL.

SOLUTION 4.43. (a) Defining a temporary voltage Vo at the output of the op-amp, we can write KCL at

the inverting and non-inverting terminal:

(V− − 2)/ 1k = (Vo − V− ) / 2k

V− /100 + (V− − Vo) / 200 = Iout

Substituting the first equation into the second and simplifying causes Vo to drop out and Iout = 20mA .

(b) The answer remains the same as the value of the load resistance was not used for finding the load

current.

SOLUTION 4.44. Using the same approach as for the previous question, but with resistor labels instead,

the following equations are obtained from KCL:

V− = R2Vs + R1VoR2 + R1

Iout = V−R2 + R1R1R2

−

VoR2

Substituting the first into the second yields Iout = Vs / R1 .


SOLUTION 4.45. (a) Vs = 5 V, (b) Iout = 10mA sets R1 = Vs / Iout = 500 Ω. (c) From KVL and Ohm's

law, Is = (Vs + RL Iout ) / αR1 . We require Is < 0.5 mA. This means that in the worst case, RL = 500 Ω,

(Vs + RLIout )

0.5 ×10−3R1=

5 + 500 × 0.01

0.25= 40 <α .

(d) From KVL and Ohm's,

Vo = −RLIout − (Iout + (RL Iout )/ R1)R2 ≥ −20 V

Hence

R2 ≤20 − RL Iout

(Iout + (RL Iout ) / R1)=

20 − 5

0.01 + 5/ 500= 750

Hence one design is to pick R2 = 750 Ω and α = 40 which impliesαR2 = 30 kΩ.

SOLUTION 4.46. The exact same design as P4.45 can be used with the isolation buffer of figure 4.7

placed at the input of it in order to provide the infinite input resistance needed by P4.46b.

SOLUTION 4.47. The general expression for this summing circuit is

Vout = −Rf

RoVo −

Rf

R1V1 −

Rf

R2V2 −

R f

R3.

(a) Using the expression above | Vout |=| −1 − 0 − 0 − 8 | E = 9E .

(b) | Vout |=| −0 − 2 − 4 − 0 | E = 6E .

(c) It has to be a linear combination of 8, 4, 2, 1, thus [1 1 0 1] would yield 13E.

(d) With the same approach, [0 1 1 1].

SOLUTION 4.48. For this implementation we add an extra R-2R branch along with an extra summing

input to the op amp. From the theory developed in Example 4.9 the total resistance seen by the source is

2R. For the total power supplied by the source to be less than 0.02 W, we require R ≥E2

2 0.02( ) = 2.5 kΩ.


SOLUTION 4.49. The same steps as in the previous questions are repeated. Because the resistance seen

by the source is unchanged no matter how many branches are added to the R-2R network,

R ≥E2

2 0.01( ) = 5 kΩ.

SOLUTION 4.50. (a) If the input is 3vmax / 8, then the first comparator will give –Vsat, keeping S2 down.

The next comparator will output +Vsat, causing S1 to go up. After subtraction, the input to the last

comparator is vmax / 8 yielding +Vsat at its output since its input is slightly above the reference input

level. Thus the logic output values are [0 1 1].

(b) Putting in 6vmax / 8, will cause +Vsat and S2 to go up. The input to the second comparator will be

2vmax / 8, which will cause +Vsat and S1 to go up. The input to the last comparator will be 0, thus it will

output –Vsat. The corresponding logic output is [1 1 0].

SOLUTION 4.51. Simply add a subtractor and switch to the last comparator, followed by an additional

comparator. The reference level to the new (additional) comparator will be vmax− /16 , and its output will be

the new least significant bit.

SOLUTION 4.52. (a) Writing the node equation for figure P4.52c,

Vout

RL' +

(Vout − V1)

10k=

A(0 − V1) − Vout

Rout

which implies that

Vout / V1 =

1

10k−

A

Rout

1

RL' + 1

10k+ 1

Rout

For figure P4.52a, the corresponding node equation is

Vout

RL=

A(0 − Vin ) − Vout

Rout

which leads to


Vout / Vin =−

A

Rout

1RL

+ 1Rout

Note that 1

RL=

1

RL1 +

1

10k, which when substituted into the later equation make both of them

approximately the same since the 1/10k term in the numerator of Vout / V1 has a negligible contribution.

(b) Writing the node equation for figure P4.52d, yields

Vout

RL' +

(Vout − V2)

10k=

A(0 − V1) − Vout

Rout

Hence

V1 = V2100|| Rin

(100 || Rin ) +10k

≈

V2

101

Solving produces Vout / V2 =

1

10k−

A

101Rout

1

RL1 + 1

10k+ 1

Rout

. Note that as in (a) the 1/10k term in the numerator is

negligible; after eliminating this negligible term, one sees that Vout

V2 is 101 time smaller than

Vout

Vin.

SOLUTION 4.53. (a) Using the equation just derived, after substituting in the values, the gain is –980.392

(b) From the previous equation, Vout / V1 = −980.382 ; write KCL at the non-inverting terminal to obtain,

Vin − V1

10k=

V1

Rin+

V1 − Vout

10k; substitute Vout = −980.382V1; solve for V1 / Vin , and then multiply both gains

to obtain (Vout / V1)(V1 / Vin ) = Vout / Vin = −0.9979 .

(c) They only differ by about 0.01%, thus they are very similar.

SOLUTION 4.54. Writing out the transfer equation, Vout =R2

R1Vs2 −

R2

R1Vs1 , thus R2 / R1 = 4 . Using

R2 = 100kΩ , R1 = 25kΩ . As expected SPICE shows to noticeable difference in outputs when the source

resistances are varied.

SOLUTION 4.55. Due to the ideal nature of the op-amp, the voltage VRb= Vs2 − Vs1 . By KVL


V2 = Vs2 + Ra(Vs2 − Vs1)/ Rb

V1 = Vs1 − Ra(Vs2 − Vs1) / Rb

Next, V1 − V2 = (Vs1 − Vs2)(1+ 2Ra / Rb) .

SOLUTION 4.56. (a) Noticing that the final stage is a summing op-amp in which Vout =R2

R1V1 −

R2

R1V2 .

From the previous question, Vout =R2

R1(V1 −V2) =

R2

R1(1+ 2Ra / Rb)(Vs1 − Vs2) . Thus

α =R2

R1(1+ 2Ra / Rb). The gain α can be varied by adjusting the single resistance Rb.

(b) Picking the set of values below will satisfy the requirement:

R2 = 100kΩ, R1 =100kΩ, Ra = 20kΩ, Rb = 10kΩ .

(c) Doing the SPICE simulation using the parameters from (b) yield 5 V at the output for

Vs1 −Vs2 = 2 −1 V. Setting Rb arbitrarily to 20 kΩ, the output now becomes 3 V, which agrees with the

relationships developed earlier.


Solution 5.1. (a) Vs = 10 V, P = 20 W and P = Vs×Is implies Is = 2 A.

(b) Rin = Vs/Is = 10/2 = 5 Ω

(c) By the linearity/proportionality property: Vs

new

Vsold =

Isnew

Isold which implies

210

=Isnew

2 implies

Isnew = 0.4 A.

(d) Pnew = Vsnew × Is

new = 2 × 0.4 = 0.8 watts. Observe that

Pnew

Pold =0.820

≠Vs

new

Vsold =

210

It follows that the proportionality property does not hold for power calculations.

Solution 5.2 First note that the ratio IR/VS is constant. With the given values of voltage and current, this

ratio is:

IR/VS = 0.25/25 = 0.01

Power dissipated in the resistor is

P = IR2R = 2.5 è IR

2 = 2.5/R = 0.25 è IR = 0.5

Since IR is always 0.01×VS, it follows that VS = 50V.

Solution 5.3 Label the resistances R1, R2, and so on in the manner shown in Example 5.11. In this

problem, we have R1 to R10 (the last being the 2 Ohm resistance at the voltage source). First, assume that

V1 (the voltage across R1) is 1V. Then evaluate the rest of the currents and voltages until you deduce the

resulting VS. It should be noted that the equivalent resistance looking into R3, R5, R7, and R9 is always

2Ω.

V1 =1 ⇒ I1 =V14

= 0.25 ⇒ I2 = 0.25 ⇒ V2 = I2 × 2 = 0.5 V

V3 = V1 + V2 = 1.5 è I3 =V33

= 0.5 è I4 = I3 + I2 = 0.75 è V4 = I4 × 4 = 3

V5 = V3 + V4 = 4.5 è I5 =V53

= 1.5 è I6 = I5 + I4 = 2.25 è V6 = I5 × 4 = 9

V7 = V6 + V5 =13.5 è I7 =V73

= 4.5 è I8 = I7 + I6 = 6.75 è V8 = I8 × 4 = 27

V9 = V8 + V7 = 40.5 è I9 =V93

= 13.5 è I10 = I9 + I8 = 20.25 è V10 = I10 × 2 = 40.5

VS = V9 + V10 = 40.5 + 40.5 = 81 V

Thus, an 81 V input produces a 1 V output è Vout = (1/81)×VS = 2 V.

Solution 5.4 Label the resistances R1 to R10 progressively from right to left just like in the previous

problem. Then, assume Iout = 1 and proceed as follows:

Iout = 1 è V1 = Iout × 4 = 4 è I2 = V14

= 1

I3 = I1 + I2 = 2 è V3 = I3 × 4 = 8 è V4 = V3 + V1 = 12 è I4 =V43

= 4

I5 = I4 + I3 = 6 è V5 = I5 × 4 = 24 è V6 = V5 + V4 = 36 è I6 =V63

= 12

I7 = I6 + I5 = 18 è V7 = I7 × 4 = 72 è V8 = V7 + V6 = 108 è I8 =V83

= 36

I9 = I8 + I7 =54 è V9 = I9 × 4 = 216 è V10 = V9 + V8 =324 è I10 =V103

= 108

IS = I10 + I9 = 162

è Iout/IS = 1/162 è Iout = (1/162)×40.5 = 0.25A.

Solution 5.5 (a) MATLAB code given in problem.

(b) Subsitute to obtain V1 = 10V.

(c) Req = VS/IS = 11.6667Ω.

(d) First, define r1 = 1:0.25:10;

then create an outermost loop around the code of part (a) as: for j=1:length(r1)

then, in the statement defining R, do R = [R1(j), R2, R3, R4, R5, R6, R7, R8]’;

Finally, replace the last statement with Vs(j) = V(n) + V(n-1); end;

The following is the resulting plot:

Solution 5.6 (a)

The following code can be used:

n = 9;

v = zeros(n,1);

i = zeros(n,1);

r = [r1 r2 r3 r4 r5 r6 r7 r8 r9]’;

i(1) = 1;

v(1) = i(1)*r(1);

i(2) = i(1);

for k=2:2:n-2

v(k) = r(k)*i(k);

v(k+1) = v(k)+v(k-1);

i(k+1) = v(k+1)/r(k+1);

i(k+2) = i(k+1) + i(k);

end;

v(8) = i(8)*r(8);

v(9) = v(8) + v(7);

i(9) = v(9)/r(9);

Is = i(9) + i(8);

It follows that Is =16.9877A.

(b) By the proportionality property: I1new

I1old =

Isnew

Isold → I1

new =1

16.9877× 200mA =11.77mA

(c) Req = v(9)/Is = 38.15Ω.

Solution 5.7 Va = 12V, iB = 60m

By inspection:

Vout_a = 300/900×12 = 4V

Vout_b = (300||600)×60m = 12

_ Vout = 4 + 12 = 16V.

*SOLUTION 5.8. Part 1: Set the 3 A current source to zero. This generates an open circuit in place of

the current source eliminating the effect of the series 0.1 S resistor. The equivalent circuit is:

By voltage division,

VL12V =

1

0.25 + 0.2 + 0.051

0.25 + 0.2 + 0.05+

1

0.1

×12 =2

2 +10×12 = 2 V

Part 2: Set the 12 V source to zero. This generates a short circuit in place of the voltage source which

shorts out the effect of the 0.5 S resistor. The equivalent circuit is:

Note that the 0.1 S resistor in series with the 3 A source is redundant to the calculation of VL. Hence, by

Ohm's law,

VL3A =

1

0.25 + 0.2 + 0.05 + 0.1× 3 =

3

0.6= 5 V

Therefore by superposition,

VL = VL12V + VL

3A = 2 + 5 = 7 V

Solution 5.9 Replace the dependent source by an independent voltage source VS:

In the following analysis, we have to always compute Va because It defines the constraint on Vs. So, when

only the 88V source Is active, VA Is the result of voltage division between the 60||30 Ω resistor and the

120||30 Ω resistor. So,

Va_1 = 40V

And, since deactivated VS, Vout_1 = 0.

Now, due to the 55V source, we have

VS

Now, the 120 and 60 Ω resistors are in parallel, and the same can be said about the 30 and 30 Ω resistors.

Thus, another voltage divider gives:

Va_2 = – 15V and Vout_2 = 0V.

Finally, when VS1 is active, the left part of the circuit consists only of resistances, so Va_3 = 0. Vout is given

by another divider formula:

Vout_3 = 90/100×VS

Now add all contributions:

Va = 40 –15 + 0 = 25V

Vout = 0 + 0 + 0.9VS, where VS = 2Va.

Vout = 0.9×2×25 = 45V.

Finally, P = V2/R = 22.5W.

Solution 5.10

Due to 3A source:

iout = 1A by current division between the two paths. So, vout_1 = 2 V.

Due to 1A source:

iout = 2/3A again by current division. So, vout_2 = 4/3V.

Due to 1vV source:

iout = 14/6 = 7/3A (by Ohm’s Law). So, vout_3 = 14/3V.

Finally, vout = 6/3 + 4/3 + 14/3 = 8V, and the power delivered by the source is 8×1 = 8 W.

Solution 5.11

Due to 22 V source:

Req = 900||225 = 180. Now, by voltage divider:

Vout_1 = 0.5×22 = 11 V.

Due to the 20 V source:

Req = 180||225 = 100 Ω. So, again, by voltage division:

Vout_2 = 900/(900+100)×20 = 18 V.

Finally, due to current source:

We have three resistances in parallel with a resistance equal to 90 Ω. So, Vout_3 = 0.1×90 = 9 V.

Vout = 11 + 18 + 9 = 38V and P = 38×38/900 = 1.6 W.

Solution 5.12

Find contribution to Vout :

First, note that no current flows through R1 - R3 because of the virtual ground property of the op-amp.

Thus, this circuit is identical to the inverting amplifier studied in Chapter 4. So,

0

0

0_ VR

RV f

out −=

Similarly, when each of the other sources is activated, the circuit will be an inverting amplifier. So,

Vout _1 = −R f

R1V1, Vout _2 = −

Rf

R2V2, Vout _3 = −

Rf

R3V3

Vout = −RfV0R0

+V1R1

+V2R2

+V3R3

Solution 5.13 Due to the 4 V source, the circuit looks like an inverting amplifier:

So, Vout_1 = -30/10×(-4) = 12 (from the results of Chapter 4).

Again, note here that no current flows through the two resistances connected to the + terminal of the op

amp. Since no current flows through them, then no voltage develops across them. So, the + terminal can be

assumed to be connected to ground, and this is why we say that the circuit looks like that of the inverting

amplifier.

Now, due to the 6 V source:

The voltage at the + terminal is established by a resistive divider between the two 10K resistors. So, this

voltage is 3V. Thus, the voltage at the negative terminal is also 3V. We can now use KVL on the inverting

side of the op-amp to get:

Vout_2 = 3 + 30k×0.3m = 12V

So, Vout = 12 + 12 = 24V, and P = (24)×(24)/500 = 1.15W.

Solution 5.14

(a) When VS2 is deactivated. The circuit looks like two inverting amplifiers in cascade.

Thus, by inspection, V1 = –2VS1 and Vout_1 = –3V1 = 6VS1 = 3V.

(b) Similarly, when VS1 is zero, V1 is zero because the first inverting amplifier has zero input. Thus, the

circuit consists just of the second inverting amplifier:

Vout_2 = – 3/2VS2 = – 4.5

(c) Vout = 3 – 4.5 = – 1.5V.

*SOLUTION 5.15. For Vs1 and Vs2 , the situation reduces to the analysis of two inverting amplifiers in

cascade. For Vs3 , the situation is simply a single stage inverting amplifier. Note that because of the

virtual ground at the inverting terminal of the op amp, when Vs1 and Vs2 are zero, they have no

contribution to the input of the second stage.

(a) With Vs2 and Vs3 set to zero,

Vouts1 =

−R

R

−2R

2R

Vs1 =Vs1 = 5 V

(b) With Vs1 and Vs3 set to zero,

Vouts2 =

−R

R

−2R

R

Vs2 = 2Vs2 = 2 × (−2.5) = −5 V

(c) With Vs1 and Vs2 set to zero,

Vouts3 =

−RR

Vs3 = −Vs3 = −2 V

(d) By superposition,

Vout = Vouts1 + Vout

s2 + Vouts2 = −2 V

Solution 5.16

If the op-amp were ideal, we would get:

Vout = – 4VS1 – 2VS2 = – 26

This is clearly beyond the linear range of operation of the op-amp. In other words, the amplifier responds

in a non-linear manner to this level of input. Hence, superposition, which relies on linearity, cannot be

used.

Solution 5.17

We know by linearity that

Vout = aIs1 + bVs2

Substitute the first measurement to obtain:

5 = 0 + b×10 è b = 0.5

Now, take the second measurement:

1 = a×10 + 0 è a = 0.1

So,

Vout = 0.1Is1 + 0.5Vs2

At 20A, 20V: Vout = 12V

Solution 5.18

Again Vout = aIs1 + bVs2

Substitute the two measurements to obtain:

5a + 10b = 15

2a + 5b = 10

These two simultaneous equations in a and b can easily be solved to obtain:

a = – 5 and b = 4

Therefore, at 1A and 5V,

Vout = – 5 + 20 = 15V

Solution 5.19 (a)

Again,

Vout = aIs1 + bVs2

Substituting the result of the first measurement into this equation yields:

a×4×cos(2t) + 0 = –2cos(2t) è a = – 0.5

Now, substitute the second measurement:

0 + 10b = 55 è b = 5.5

Therefore, for the given input current and voltage:

Vout = –cos(2t) –55cos(2t) = –56cos(2t) V

(b) Vout = 2cos(5t) + 110cos(5t) = 112cos(5t) V

Solution 5.20 (a)

First of all,

Iload = aVa + bIb

Now, substitute the two measurements into this equation:

7a + 3b = 1

9a + b = 3

Solving these two equations for the unknowns a and b, we get

a = 0.4 and b = – 0.6

(b) Iload = 0.4×15 –0.6×9 = 0.6A

Solution 5.21 (a)

We know that the output is going to be a linear combination of the three inputs:

Vout = aIs1 + bVs2 + cVs3

Now, substitute the three measurements into this relationship:

50ma – 2b + 5c = – 13

0 + 3b + 5c = 2

0 + 2b + 4c = 0

These equations can be written in matrix form and solved as follows:

0.05 −2 5

0 3 5

0 2 4

a

b

c

=−13

2

0

⇒

a

b

c

=100

4

−2

(b) Substitute the given values of the input sources to obtain:

Vout = 100×1 – 4×40 + 2×10 = – 40V

Solution 5.22

Vout = AIs1 + BVs2 + CVs3

Substituting the measurements into this equation results in a system of three equations and three

unknowns. This system can be written in matrix form by inspection:

30 ⋅10−3 2 −1

−20 ⋅10−3 4 2

−10 ⋅10−3 −3 1

A

B

C

=11.5

27

−14

⇒

A

B

C

=150

5.5

4

This can be solved to obtain:

Solution 5.23 (a)

The coefficient matrix is inverted, and both sides of the nodal equation are multiplied by it to obtain:

Va

Vb

=

43.0108 0.233

43.0108 −1.4337

0.02Vs1 − 0.00125Vs2

Vs2

Expanding the first row of the above equation gives:

Va = 0.8602Vs1 – 0.0538Vs2 + 0.233Vs2

This is exactly in the form required, where A = 0.8602 and B = 0.1792

(b) For this part, we expand the second row of the equation:

Vb = 0.8602Vs1 – 0.0538Vs2 – 1.4337Vs2

Again, this is in the desired form, where A = 0.8602 and B = – 1.4875

(c) Vab = Va – Vb = 0 + 1.667 Vs2

Thus, A = 0 and B = 1.667

Solution 5.24 (a)

Again, we invert the coefficient matrix to obtain

I1I2

I3

v1

v2

=

0.0022 0.0022 0.0022 0.6296 −0.0741

0.0022 0.0022 0.0022 −0.7037 0.2593

0.0022 0.0022 0.0022 −0.3704 −0.0741

−0.2222 0.7778 −0.2222 137.037 −92.5926

−0.3333 −0.3333 0.6667 −11.1111 77.7778

Vs1

0

0

Is2

0

Expanding the first row of this equation:

I1 = 0.0022Vs1 + 0.6296Is2

So, A = 0.0022, B = 0.6296

(b) Similarly, expanding the third row:

I3 = 0.0022Vs1 + -0.3704Is2

So, A = 0.0022, B = -0.3704

(c) By the same procedure, A = -0.3333 and B = -11.111

Solution 5.25

Invert the coefficient matrix and mujltiply both sides of the equation in the problem by this inverse matrix

to obtain

V1

V2

V3

Ia

Ib

Ic

Id

=

0 0 0 −3 2 −1 −4

0 0 0 −3 1 −1 −4

0 0 0 12 −3 3 16

0 0 0 −4 1 −1 −5

0 −1 0 −16.6 9.2 −5.2 −21.8

1 1 0 −42.1 −16.2 11.7 53.8

1 1 1 26.1 −12.2 7.7 32.8

Is1

0

0

0

Vs2

0

0

The second to last row can be expanded to get Ic = Is1 – 16.2Vs2 è A = 1, B = –16.2

Solution 3.26 (a)

First compute the response due to V1:

By voltage divider:

Vout_1 = 6/(2+6+1)×V1 = 2/3V1

Then, due to I2:

I2 flows through the 20 ohm resistor in series with 6||3. Thus,

Vout_2 = –2I2

Therefore, Vout = 2/3V1 – 2I2

(b) Vout = 8cos(10t) – 4

*SOLUTION 5.27. (a) By linearity Vout = A Vs1 + B Is2 .

To find A, let Is2 = 0. The circuit becomes a ladder network as follows.

Let VoutA = 1 V. Then

»I1 = 1/420 + 1/70

I1 = 1.6667e-02

»V400 = I1*40 + 1

V400 = 1.6667e+00

»I400 = V400/400

I400 = 4.1667e-03

»I2 = I1 + I400

I2 = 2.0833e-02

»Vs1 = I2*20 + V400

Vs1 = 2.0833e+00

»A = 1/Vs1

A = 4.8000e-01

To find B, let Vs1 = 0. The circuit becomes a ladder network as follows.

Again assume that VoutB = 1 V. Then

»I1 = 1/420 + 1/70

I1 = 1.6667e-02

»V400 = I1*40 + 1

V400 = 1.6667e+00

»I400 = V400/400

I400 = 4.1667e-03

»I20 = V400/20

I20 = 8.3333e-02

»Is2 = I20 + I400 + I1

Is2 = 1.0417e-01

»B = 1/Is2

B = 9.6000e+00Hence by linearity Vout = 0.48 Vs1 + 9.6 Is2 .

(b)

»Vout = A*20 + B* 0.5

Vout = 1.4400e+01

(c) Doubling resistances does not change voltage ratios hence A is the same. However, the doubling also

doubles the voltage to current ratio. Hence, B is doubled. It follows that if all resistances are

doubled, thenVout = 0.48 Vs1 + 19.2 Is2

Solution 5.28 (a) (b)

We solve parts (a) and (b) at the same time. First, we find the responses to Vs1:

Equivalent resistance across Vo2: 60||30 = 20

Now, by voltage divider: Vo2_1 = 20/140Vs1

This voltage now divides between Vo1 and the 10 ohm resistance:

Vo1_1 = 20/30Vo2_1 = 2/3×1/7Vs1 = 2/21Vs1

Now, compute the responses due to Is2:

Equivalent resistance across Vo2: 60||120 = 40Ω

By voltage division: Vo2_2 = 40/50Vo1_2

Where Vo1_2 = Is2×(50||20) = 14.286Is2

Now, add the contributions:

Vo1 = 0.0952Vs1 + 14.286Is2

Vo2 = 0.1429Vs1 + 11.4288Is2

Solution 5.29 (a)

Find contribution due to v1:

The parallel combination results in a 12 ohm resistance in series with the remaining two. Thus, by voltage

division:

vout1 = 24/48v1 = 0.5v1

Now, due to v2

The equivalent resistance across v in this figure is 36||(24+12) = 18, which means that by voltage divider:

v = 0.5v2

Similarly, by another voltage division application

vout2 = 24 / 36 × v= 0.66×0.5v2 = 0.333v2

Now, due to i3

Define Req1 = 24 + 36||18 = 36 Ω. This resistance is in parallel with the 12 ohm resistance to introduce

an equivalent of 9 Ω. The total voltage that develops across this 9 ohm resistance is –9is3. This voltage

divides between vout and the 36||18 resistance:

vout3 = 24/36v1 = -9×24/36is3 = -6is3

Finally, due to i4:

A similar analysis of this resistive network can reveal that vout4 = 6is4. Thus

vout = 0.5v1 + 0.333v2 – 6is3 + 6i s4

(b) For this part, note that scaling resistance values does not affect voltage ratios. This can be evident from

the application of any voltage divider formula. On the other hand, scaling resistances does affect current-

to-voltage or voltage-to-current ratios. This is by definition of a resistance! So, in the above equation,

doubling the resistances does not affect the first two terms, but doubles the second two terms.

Solution 5.30 (a)

Define two clockwise mesh currents: I1 in the bottom left loop and I2 in the top loop. The bottom right

loop has a current source, so it will not be considered:

Va – (I1-I2)3 – I2 – (I1 – ib) = 0

and

6I2 – I2 + (I2 – I1)3 = 0

Solving these two equations for the two currents gives:

I1 = 8 A and I2 = 3 A

The power delivered by the dependent source is: P = ix × I ix where I ix is the current leaving the ' +'

terminal of the dependent voltage: P = ix × (I2 − I1) = I2 × (I2 − I1) = −15W

è vout = I1 + ib = 8 + 26 = 34V

(b) Now, we express vout = Ava + Bib

Turning off ib, we still have two loops, in which we can define the same mesh currents as above to obtain:

va – 3(I1 – I2) – I2 – I1 = 0

I2 – 3(I2 – I1) – 6I2 = 0

Solving for the two currents, we get I1 = 4/13va which sets up vout across the 1 ohm resistor:

Vout1 = 4/13va

Now, turn off the voltage source:

Write a node equation at vout:

bout

b

voutoutout

iv

ivvv out

13

9

0316

2

6

=?

=−−

++

So, vout = 4/13va + 9/13ib

(c) Substitute into the above equation: vout = 35V

Solution 5.31 By inspection:

where the leftmost current is 0.25 with a resistance of 32Ω. Similarly, the downward current is 0.75 and its

resistance is 32Ω. This reduces to:

Solution 5.32

The circuit can be transformed as follows:

Write two nodal equations at 1 and 2:

05166.9

92

02

92

1

2212

211

=−+++−

=−+−+−

mk

vk

vm

kvv

kvv

mk

vm

These two equations can be solved using any method to obtain:

v1 = 11.2V and v2 = 2.4V

The power absorbed by the 9.6kΩ resistor is:

P =V2

2

9.6 ×103 = 0.6mW

Solution 5.33

Then

Output voltage is 5V

(b) P = 1.25W

(c) For a given resistance, doubling the voltage increases the current by two times. So, the current is

doubled. It follows that Voutnew = 2 × Vout

old = 10V

Solution 5.34

This circuit is easy enough to solve by inspection. Vs = 28V.

Solution 5.35

Now, write two node equations:

mv

vvv

mvvv

80125

032.0500500

12500500500

21

12

121

=++−

=+−

These two equations can be solved to get v1 = 2.8V, v2 = -0.4V.

Solution 5.36

Now, we can write the node equations:

02.01.011005050100

100100

095.020

4.02.025

100255050

2122

1121

=−−++−+−

=−+++−+−

vvvv

vvvv

Solving these two equations yields:

v1 = 25V and v2 = 20V

*SOLUTION 5.37. After a source transformation on the 30 V independent source and one on the 9Vx

dependent source we obtain the circuit below.

Transforming the two dependent voltage sources and combining yields the following circuit.

Writing a single node equation for Vx yields

7.5 =Vx

4+

Vx

2+

Vx

2= 1.25Vx

Hence, Vx = 6 V.

Solution 5.38

Replace the dependent source with a temporary independent source. When doing the analysis, always

compute Va in order to keep track of the constraint on the dependent source.

When Vt is not active, vout is obtained from a voltage divider between the 60||30 combination and the 15

ohm resistor:

vout1 = 4/7Vs

Similarly va1 = 3/7Vs

Now, compute the response due to the temporary source.

Straightforward voltage division also applies here to get:

va2 = -2/7Vt

and vout2 = -5/7Vt

So, vout = 4/7Vs – 5/7Vt

va = 3/7Vs – 2/7Vt, where Vt = µva è va = 3/7Vs – 2µ/7va

Rearranging,

sa Vv27

3

+=

Then,

vout =4

7Vs −

5µ7

va

vout =4 − µ7 + 2µ

Vs

Solution 5.39 Replace the dependent source by a temporary independent source:

When Vt is shorted, the result is a ladder network. The input resistance looking each of the vertical

branches is R. Label these vertical branches V1, Va, V2 from left to right. It follows that

V1 = R/3RVs

_ Va_1 = R/2R × R/3RVs = 1/6Vs

Also, Vout_1 = 0

Now, short VS and turn on Vt.

Again, the result is a ladder network (note the symmetry in the above figure). Thus, we can write by

inspection:

Va_2 = 1/6Vt

Vout_2 = Vt

Adding the contributions:

Vout = 0 + Vt = Vt = µVa where we have substituted the constraint on Vt

and Va = 1/6Vs + 1/6Vt = 1/6Vs + 1/6 µVa

è Va = [1/(6 – µ)]Vs

Substitute this Va into the expression for Vout:

sout VV−

=6

Solution 5.40

The first step is to replace the dependent source with a temporary independent source. Then, superposition

can proceed as usual.

Now, let’s short the temporary source, Vt. Again, this network is a ladder network, like the one in the

previous problem. However, now, the equivalent resistance looking into each of the vertical branches (from

left to right) is different. Now, it is

Req = 20R||(R+4R) = 20R×5R/25R = 4R

Now, again, define the voltages across these three vertical branches (from left to right) as V1, Va, V2. It

follows by voltage division that

V1 = 4/9Vs

è Va_1 = 4/5 × 4/9 × Vs

è Va_1 = 16/45 Vs

It should be noted that Vout = 0.

Now, short the input source and find the response due to Vt. Again, in this case, the circuit is identical to

the case when Vs was active (note the symmetry in the above figure). Therefore,

Va_2 = 16/45 Vt

Vout_2 = Vt

Adding the two contributions:

( ) sa

asa

VV

VVV

−=?

+=

1645

145

16

45

16

andsout

atout

VV

VVV

164516−

=

==

Solution 5.41

First, replace the controlled source by a temporary independent source:

Now, do a source transformation on the Vt source:

The equivalent resistance seen by the current source is 99Ω. Therefore:

Vout = -99/100Vt = -0.99Vt

Now, Vt = µVg and IL = -0.99Vt/990k where IL is the current through R2

è VR2 = ILR2 = (100k/990k)×(-0.99Vt) = -0.1Vt

By KVL

Vg = VR2 + Vs = -0.1Vt + Vs = -0.1µVg + Vs

Rearranging

sg VV11.0

1

+=

Finally, and substituting:

91.8

1.011

99.0

−=

+−=

s

out

s

out

VV

VV

Solution 5.42

Ra

Rbb

xaa

xbb

V

V

in

out

1

10

10

10

1++=

++=

At R = 0

20

201

0

0

=?

==

b

b

V

V

in

out

At infinite R

11

1

1

8

8

ab

a

b

V

V

in

out

=?

==

Finally, at R = 10

4

5.0

101018020

1

1

1

1

=?=?

=++

b

a

aa

Substituting these values for R = 2 yields

1425.01

2420 =↔+↔+=

in

out

V

V

SOLUTION 5.43. From the chapter

Vout

Vin=

b0 + b1µa0 + a1µ

=b0 + b1µ1+ a1µ

Equivalently,

−µVout

Vina1 + b0 + µb1 =

Vout

Vin

Plugging in the data yields three equations in three unknowns which in matrix form are:

0 1 0

−154 1 1

−168 1 2

a1

b0

b1

=264

154

84

»A=[0 1 0;-154 1 1;-168 1 2]

A =

0 1 0

-154 1 1

-168 1 2

»y = [264 154 84]'

y =

264

154

84

»Coefs = A\y

Coefs =

2.8571e-01

2.6400e+02

-6.6000e+01

Therefore, a1 = 0.28571, b0 = 264, and b1 = –66. Making a1 = 1, will yield a different set of answers.

When m = ∞, Vout/Vin = b1/a1 = –231.



SOLUTION 6.1. (a) Voc is found by removing RL and doing voltage division.

VVVOC 28100600

600

300)700||600(

)700||600(63 =

+×

+=

RTH is found by setting the source to zero and by calculating the equivalent resistance seen looking back

between the A and B terminal.

Ω=+= 200600||]100)600||300[(THR

(b) Using

LL IRP ⋅=

the power for each resistance may be found by substituting the appropriate RL in the following equation.

L

LTH

OC RRR

VP ⋅

+

=2

For 50 Ω, 200 Ω, and 800 Ω , the power obtained is 627.2 mW, 980.0 mW, and 627.2 mW respectively.

The use of Thevenin equivalent does reduce the effort in obtaining the answer.

SOLUTION 6.2. (a) To find RTH, open circuit the current source and short-circuit the voltage source. The

resulting resistance seen from terminal A-B is 1 kΩ. Using superposition, the contribution of the current

and voltage source at the open circuit output may be summed as 30 V (2 k/4 k) + 10 mA (2 k/4 k) (2 kΩ).

VOC is then 25 V and ISC = VOC/RTH is 25 mA.

(b) Following is a plot of

L

LTH

OC RRR

VP ⋅

+

=2

for RL from 100 Ω to 4 k Ω.


SOLUTION 6.3. (a) Turning off the two sources

RTH = (60 + 60)||40 = 30 Ω,

and using superposition

VOC = 6V40

40 + 60 + 60

+ 0.1A

60

40 + 60 + 60

40 = 3 V.

ISC is the obtained as 100 mA.

(b) Using

L

LTH

OC RRR

VP ⋅

+

=2

a load of 90 Ω will absorb 56.25 mW.

(c) It absorbs 75 mW; hence the 30 Ω resistor absorbs more power.

SOLUTION 6.4. As both resistor divider ration are the same (3/6), the voltage at A and B is the same

resulting in a VOC of 0 V.

RTH = (3K | |6 K) + (9K ||18K) = 8 kΩThe relation VOC/ISC cannot be used in this situation.

SOLUTION 6.5. Using superposition

VOC = 2020 ×103

20 ×103 + 5 ×103

−105 ×103

20 × 103 + 5 ×103

+ 20 ⋅ 20sin(50t) = 14 + 400sin(50t) V,

and


RTH = (20k | |5k) + 20k = 24 kΩ.

ISC can then be found using the VOC/RTH relationship as 0.58+20sin(50t) mA.

SOLUTION 6.6. Once again using superposition

VKmAKKKK

KmA

KKK

KVVOC 2042546

584

454

486

672 =⋅+⋅

+++

++= ,

and

Ω=+++= KKKKKKRTH 1024]6||)84[(

SOLUTION 6.7. Using source transformation, (a) is obtained from the original circuit. Then combining

in series the resistors and voltage sources, and retransforming them (b) is obtained. Finally adding the two

currents and transforming back the circuit to its Thevenin form (c) is obtained.

From (c),

WRIP

mAkkI

kR

VV

LL

L

TH

oc

15.0

5)66/(60

6

60

2 ==

=+=Ω=

=


SOLUTION 6.8. First, each source in series with 2R, can be replaced by an up going current source of

RVx 2/ in parallel with 2R. Then starting from the left, the two 2R in parallel are combined and then

retransformed to a voltage source of 2/oV in series with 2R once added to the series resistance. Repeating

the previous steps,

RRVVVVRRVRVRVRV

RVRRRVRVRVRVVV

RRVVVRRVRVRV

RVRRRVRVRVV

RRVVRRVRV

RVRRRVRV

oo

oo

oo

oo

oo

oo

+++++?+++++?+++

++++?+++?++

+++?+?+

2/4/8/16/||2/4/8/16/

2/||2||2||4/8/16/22/4/8/

2/4/8/||2/4/8/

2/||2||2||4/8/22/4/

2/4/||2/4/

2/||2||2||4/22/

321321

32121

2121

211

11

1

Thus 2/4/8/16/ 321 VVVVV ooc +++= and RRTH 2= .

SOLUTION 6.9. First we find RTH

RTH = 2K + [ 6K | |3K( ) + 15K ||10K( )]| |24 K = 8KΩ

Use nodal analysis to solve for VOC. At node a

1mA =1/ 6K(Va − Vc ) +1/ 3K(Va − Vd ) + 1/ 24K(Va) or

(1/ 6K +1/ 3K +1/ 24K)Va −1/ 6K(Vc) −1/ 3K(Vd ) = 1m

doing the same for the supernode, and the equation inside the supernode yields the following two

equations:

(1/ 6K +1/ 3K)Va − (1/ 6K +1/15 K)Vc − (1/ 3K +1/10 K)Vd = 0, and

Vc − Vd = 30

using these three equations, it is now possible to solve for the three unknowns VA, VC, and VD. Using

MATLAB they are respectively 4.5V, 22.8750V, and -7.1250V. VOC being VA, 4.5V.

a

c d

SOLUTION 6.10.

RTH = [(18K | |9K) + 66K]| |36K = 24K


By superposition, noting that all resistances are in k-ohms,

VOC =66+(9||18)

66+(9||18)+36⋅36 ⋅ 2.5 +

(66 + 36)||9

[(66 + 36)||9] +18⋅

36

36 + 66⋅18 −

36

36 + (9||18) + 6630 = 52 V

When a 2 kΩ is connected, the current IL becomes VOC/(RTH+2k) = 2mA; thus the power absorbed is 8

mW.

SOLUTION 6.11. (a) Introduce a test voltage source at the output, and write the nodal equations in matrix

form:

(1/100 +1/ 50) −1/ 50 −1/100

−1/ 100 −1/100 (1/ 100 +1/ 200 +1/100)

−1/ 50 (1/ 50 +1/100 +1/100) −1/100

Vtest

VC

VD

=itest

0

Vs / 100

Solving we obtain, Vtest = 100itest + 2Vs / 3. From eq. 6.10 RTH =100 Ω, voc = 2Vs / 3.

C D

(b) To obtained the power the following equation is used, P =voc

RTH + RL

2

⋅ RL .


0 20 40 60 80 100 120 140 160 180 2000.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

Load resistance in Ohms

Po

we

r/V

s in

mW

SOLUTION 6.12.

VOC = 0, as no independent sources are present. Writing the following nodal equation where vx is the

voltage across both ports,

ix = ((vx − iz ) − ix ) + (vx − iz )/ 2, RTH can be found as vx / ix =2.5 +

1.5,

SOLUTION 6.13.

Defining vi and ii as the voltage across and current into the input ports, writing the nodal equation at the

input: ii + gmvx = 1/ 200K(vi − vx ) . We can also get the following equation vx = 200K(gmvx + ii) . Using

the previous two equations we can solve for RTH = vi / ii =−200K 1+200K ⋅ gm

1 + 200K ⋅ gm

+1

1 + 200K ⋅ gm

.

This yields a gm of 10 µS.

SOLUTION 6.14.

ISC is null as no independent source are present. To find RTH vi and ii are defined as the voltage across and

current into the input ports. Writing the nodal equation we get:

ii = Vx / 1.8K + (1/ 200)(Vx − 3 / 4Vx) , and vi = Vx −300 ⋅ ii . Solving RTH = vi / ii =−600Ω .

SOLUTION 6.15.

First, write out the equation around loop 1:


Vx = i1 ⋅1200 − i2 ⋅100 − i3 ⋅300 . Then substituting the following relationships,

i1 = Ix , i2 = 2I x , and i3 = 0.01Vx + i2 = 0.01Vx + 2Ix , and solving for Vx / Ix = RTH = 100Ω . VOC = 0 as no

independent sources are present.

1

2

3

SOLUTION 6.16.

Introduce a test source at the output terminals, and write out the nodal equations in matrix for the top node,

and the supernode comprised of the current controlled voltage source (ccvs). vb is the node left of ccvs, and

vc the node to the right.

1/ 30 +1/ 20 −1/ 20 −1/ 30

−1/ 20 −1/ 30 + 0.4 1/ 20 − 0.4 2 / 30

1/ 6 −1 5/ 6

vtest

vb

vc

=itest

0

0

Solving, we obtain vtest =−90itest , thus RTH =−90 Ω.

b c

SOLUTION 6.17.

(a) Turn off independent source. Introduce a test source and write loop equation:

vx = 6ix − 4i1 +10ix . Note that i1 = ix . Now solve for vx / ix = RTH = 12Ω .


Short the input and write the loop and nodal equations:

3A = i1 + ISC

10i1 = 4i1 + 6ISC

Solving yields ISC = 1.5A, and VOC = ISC RTH = 18V .

(b) In MATLAB the following plot is generated: P =VOC

RTH + RL

2

⋅ RL for 1 ≤ RL ≤ 24Ω.

0 5 10 15 20 251

2

3

4

5

6

7

Resistance in Ohms

Pow

er

in W

atts

Maximum power is absorbed by 12 Ω load.

SOLUTION 6.18.

To find thevenin resistance, introduce a test source and write the following equations:

vs = 20i1 + 40i1 + 40i1 =100i1, and is = vs /100 + i1. Solving for vs / is = RTH = 50Ω .

Next, use the following nodal equation; 0.2A = i1 + ISC , and loop equation 20i1 + 40i1 = 40ISC . Solving

using these two equations yields ISC = 0.12A , and consequently VOC = RTHISC = 6V .

SOLUTION 6.19.

(a) Introduce a test source, vs, and get the following two equations: vs −15is =−Vx , and

kVx + Vx / 3 + Vx / 5 + is = 0. Solving yields vs / is = RTH =15 +1

k +8 /15

, or 95/6 Ω for k = 2/3.

Next, write the following nodal equation kVx + Vx / 3 + (Vx −1) / 5 = 0 and observe that VOC = 1V − Vx .

Thus solving yields VOC = 1−1/ 5

k + 8

15

= 5/ 6V for k = 2/3, and ISC = VOC / RTH =1/19 A .


(b) Solving the previously obtained equation for k when VOC = 0, yields k = -1/3, and consequently

RTH = 20 Ω.

SOLUTION 6.20.

Introduce a test source, vt, and get the following equations: it + ix = ix ⋅Vs − 300ix

300⇒ Vs = 300it + 300ix ,

and vt = Vs − 300ix +100it . Solving yields vt / it = RTH = 400Ω .

Observe how no current flows through the 300 resistor in parallel with the dependent source. Thus VOC

will always be 0 V and is independent of VS.

SOLUTION 6.21. (a) For this part, consider the modified circuit below.

Step 1: Applying KCL to node A, we have

1

400VA −100( ) +

VA

800+ 0.1 − Is = 0

Multiplying through by 800 yields

3VA = 800Is +120

Step 2. Computing Vs, we have

Vs = VA − 2000ix = VA −2000

400100 − VA( ) = 6VA − 500

Hence

Vs = 6800

3Is + 40

− 500 =1600Is − 260 = RthIs + Voc

Therefore Rth = 1.6 kΩ and Voc = −260 V.

(b) By linearity, Voc = −130 V.


SOLUTION 6.22.

Introduce a test source, vt,, and get the following two equations: vt = (30m − V1 / 100)400 − V1 = 12 − 5V1 ,

and it = 0.06V1 + 30m − V1 / 100 = 30m + 0.05V1 . Solving obtain vt = −100it +15. Thus RTH = −100Ω ,

and ISC = VOC / RTH =−150mA .

SOLUTION 6.23. Insert Itest as per text. Hence

1

40+

1

120

VC −

16

40= Itest

Solving, we obtain

VC = 30Itest +12

By KVL,

Vtest = VC + 30 × 2i1 + Itest( ) = VC +60

4016 − VC( ) + 30Itest =−0.5VC + 30Itest + 24

Substituting for VC yields

Vtest = 15Itest +18

Hence, Voc =18 V and Rth = 15 Ω. Thus i =18

75 + 15= 0.2 A. Further, P75 = 75(0.2)2 = 3 W.

C

SOLUTION 6.24.

(a) and (b)


vs

-

+

-

+

v 2

1

(c) Writing KVL around loop 1, loop 2, and finally relating v to vx ,

vs = 10(0.02 − I1) + v − 20I1v =−5I2 −15(I2 − 0.02)

3vx = I1 − I2 = 3[10(0.02 − I1)] = 0.6 − 30I1

(d) Rewriting these in matrix form,

−30 0 1

0 20 1

31 −1 0

I1

I2

v

=1

0

0

vs +

−0.2

0.3

0.6

(e), (f), and (g). Using MATLAB,

»A=[-30 0 1;0 20 1; 31 -1 0];

»b1 = [1 0 0]';

»b2 = [-.2 .3 .6]';

»I1 = inv(A)*b1

I1 =

-1.5385e-03

-4.7692e-02

9.5385e-01

»I2 = inv(A)*b2

I2 =

1.9231e-02

-3.8462e-03

3.7692e-01

»Rth=-1/I1(1)

Rth = 650

Note: Isc = I2(1) = 0.01923 A

»Voc= -Rth*I2(1)


Voc =

1.2500e+01

(h)

SOLUTION 6.25.

(a)

(b) Write four nodal equations,

is = (VA − VC ) / 2k

is + 1m = (VC − VD )/ 6k + (VC − VE )/ 3k

VE / 2k = (VC − VD) / 6k − VD /15k

VE / 2k = VE /10k + (VE − VC )/ 3k

(c)

1/ 2k −1/ 2k 0 0

0 1/ 2k −1/ 6k −1/ 3k

0 1/ 6k −(1/ 6k + 1/ 15k) −1/ 2k

0 −1/ 3k 0 −1/ 15k

VA

VC

VD

VE

=

1

1

0

0

is +

0

1m

0

0

(d) Solving in MATLAB VA = 5.818kis + 3.8182 , thus

RTH = 5.818kΩVOC = 3.8182V

(e) This only changes VOC = 3818 ⋅8m = 30.54V .

SOLUTION 6.26. For this circuit, no current flows though the 20 Ω resistor. Therefore VAB = VCB .

Further, from the examples in the chapter, VCB = 4Vs . Hence, Voc = VAB = VCB = 4Vs . Also, shorting

terminals A and B, yields Isc =VCB

20=

4Vs

20=

Vs

5. It follows that Rth =

Voc

Isc= 20 Ω. Note that the

Thevenin equivalent to the left of C-B is a voltage source of value 4Vs. Therefore the Thevenin equivalent

to the left of A-B is 20 Ω in series with 4Vs.

SOLUTION 6.27. (a) From the previous problem VCB = 4Vs . Thus by voltage division,

VOC = VAB = VCB

180

180 + 20

= 3.6Vs . Next find ISC = VCB / 20 = Vs / 5, and then RTH = VOC / ISC =18Ω .


(b) This changes the voltage division at the output, thus VAB / Vs = 4180||162

(180||162) + 20

= 3.24 .

SOLUTION 6.28. (a) Writing the following two KCL equations,

vtest / 4k = (Vo − vtest ) /12 k

Itest = (vtest − Vo ) /15k

where Vo is the voltage at the output of the op-amp. Doing the appropriate substitution get

vtest = −15k / 3 ⋅ Itest , thus

RTH = −5kΩVOC = 0V

Since no independent source exist right of A-B

(b) Applying Ohm’s law Is = 1/ (10k − 5k )Vs = 0.2mVs

SOLUTION 6.29. (a) Adding a test source at terminal A-B, and noting that the voltage at the output of the

op-amp is Vo = −5 / 2Vs . Write KCL at terminal A,

vtest / 900 = (Vo − v test )/ 100 + itest

vtest = 90itest − 2.25Vs

Where one sees by inspection that

RTH = 90ΩVOC = −2.25Vs

ISC = VOC / RTH = −0.025Vs

(b) Noticing the virtual short to ground provided by the ideal op-amp, RTH = 20kΩ , and VOC = 0V since no

independent sources are present right of the input terminal.

SOLUTION 6.30. The output voltage of the ideal op amp is −2.5Vs1 − 2Vs2 which drives a voltage divider

circuit. Hence

Voc = VAB = 0.9(−2.5Vs1 − 2Vs2 ) =−2.25Vs1 −1.8Vs2

Further,

Isc =−2.5Vs1 − 2Vs2

100= −0.0225Vs1 − 0.018Vs2

Finally,

Rth =Voc

Isc=

0.9(−2.5Vs1 − 2Vs2)(−2.5Vs1 − 2Vs2)

100

= 90 Ω


Equivalently if one sets Vs1 and Vs2 to zero, then the ouput terminal of the op amp goes to ground. Hence

Rth = 100 / /900 = 90 Ω

SOLUTION 6.31. Define the node at the output of the op-amp as Vo. Note how the circuit left of this

node is a general summing circuit as per text. Thus, Vo = 2Vs2 − 4Vs1. Hence we simply replace the op

amp circuit to the left of the 20 Ω resistor by an ideal voltage source of value V0. Hence

Voc =80

80 + 20V0 + 2I = 0.8V0 + 2

−V0

100

= 0.78V0 = 1.56Vs2 − 3.12Vs1

Alternately, one can introduce a test source at the output terminal and write out a set of equations using

KVL,

Vo = 2Vs2 − 4Vs1 = −20I − 80(I − itest )

vtest = 2I + 80(itest − I)

Solving yields vtest = 17.6itest + 0.78(2Vs2 − 4Vs1) , and by inspection,

RTH =17.6ΩVOC = 0.78(2Vs2 − 4Vs1) = 56Vs2 − 3.12Vs1

SOLUTION 6.32. Upon inspection when the op-amp is in active range, for inputs from –3V to 3V,

RTH =1kΩVOC = 0V

However when the input exceeds 3V, the output of the op-amp will saturate at –15V, and

Vs = −15V + 6kIs . Therefore from eq. 6.10,

RTH = 6kΩVOC = −15V

When the input is less than –3V, Vs = 15 + 6kIs , thus

RTH = 6kΩVOC = 15V

SOLUTION 6.33. (a) The op-amp configuration has a gain of –4. So when the input is between –3V and

3V it is operating in its active region, thus


RTH = 4kΩVOC = 0V

When the input is greater than 3V the output saturates at –12V and

Vs = 20kIs −12


When the input is less than –3V the output saturates at 12V,

Vs = 20kIs +12

RTH = 20kΩVOC = 12V

(b) When the input is in the active range,

RTH =16kΩVOC = −4Vs

When it is greater than 3V,


And when less than –3V,

RTH = 0kΩVOC = 12V

The last two obtained using figure 6.28.

The maximum power is when the output is in saturation, P = VOC( )2/ 28k = 6mW .

SOLUTION 6.34. From the table the following two equations can be written:

6 = 2RTH + voc

12 = 8RTH + voc

Putting in matrix form and solving,

1 1

8 1

RTH

voc

=

6

12

−1/ 6 1/ 6

4 / 3 −1/ 3

6

12

=

RTH

voc

Thus RTH =1 kΩ since current was in mA, and voc = 4 V.

SOLUTION 6.35. (a) From Ohm’s law, IAB = VAB / RL . Thus 0.2uA and 0.1uA.

(b) Note how using this topology VAB = voc − IABRTH , thus


1 −0.2u

1 −0.1u

voc

RTH

=

0.4

1

−1 2

−10M 10M

0.4

1

=

voc

RTH

Thus, voc = 1.6V , and RTH = 6MΩ .

SOLUTION 6.36. (a) Although the text describes finding Rth from a measurement or calculation of both

Voc and Isc, measurement of Isc is often impractical. Hence the procedure outlined in this problem

provides a more practical means of determining the Thevenin equivalent.

Since the internal meter resistance is 10 MΩ,

IAB (µA) =VAB

10+

VAB

RL

Hence, the completed table is:

RL(M Ω) vAB (V) IAB (µA)

2 0.4 0.24

10 1 0.2

(b) The terminal relationship assuming a Thevenin equivalent is given by

VAB = Voc − RthIAB

In matrix form with the data from the table

1 −0.24

1 −0.2

Voc

Rth

=

0.4

1

Hence Voc = 4 V and Rth = 15 MΩ. Note that since we used µA and V, Rth is in MΩ.

SOLUTION 6.37. (a) For this scenario, the circuit is essentially a voltage source with a resistance R in

series with the circuit under test, in parallel with a voltmeter measuring the voltage division between the


later two. Therefore if replacing R = 0 by R = R2 causes the voltage measured by the voltage meter to

drop by half, then by voltage division, Eo / 2 = Eo

RTH

R2 + RTH

, and RTH = R2 .

(b) Using the same reasoning and voltage division,

Eo = Vs

RTH

RTH + Rs

Eo / 2 = Vs

RTH

RTH + Rs + R2

Therefore RTH + Rs = R2 or equivalently RTH = R2 − Rs .

(c) Again by voltage division,

Eo = VsRTH || Rm

(RTH || Rm ) + Rs

Eo / 2 = VsRTH || Rm

(RTH || Rm ) + Rs + R2

Therefore (RTH || Rm ) + Rs = R2 or RTH Rm

RTH + Rm= R2 − Rs . Solving for RTH =

R2 Rm − RsRm

Rm + Rs − R2.

SOLUTION TO 6.38. (a) The voltmeter measures the voltage division between the two resistors involved

thus,

Eo = Voc

Eo / 2 = Voc

R2

RTH + R2

and RTH = R2 .

(b) Now

Eo = Voc

Rm

Rm + RTH

Eo / 2 = Voc

Rm || R2

(Rm || R2) + RTH

From the division of the former by the later R2 = RTH || Rm . And from the former

Voc = (Rm + RTH )Eo / Rm = (1+ RTH / Rm )Eo .

SOLUTION TO 6.39. Using the relation developed in problem 6.37 RTH =R2Rm − Rs Rm

Rm + Rs − R2

= 5kΩ

SOLUTION TO 6.40. Using the equation developed in problem 6.38,


RTH =R2 Rm

Rm − R2

= 4MΩ

Voc = (1+ RTH / Rm )Eo = 20V

SOLUTION TO 6.41. Writing the line equation in the following general form,

i = vab / RTH − isc

i = 2vab − 4

Thus RTH = 0.5Ω , and isc = 4A .

SOLUTION TO 6.42. (a) In this range the curve appears to be varying linearly between (0V,0mA) and

(0.2V, 0.1mA) pair, thus writing the line equation i = (0.1m / 0.2)v yields RTH = 2kΩ ⇒ Voc = 0V .

(b) Writing the line equation of the following form, i = v / RTH − isc , into a matrix equation,

0.2 −1

0.7 −1

1/ RTH

isc

=

0.1m

10.1m

and solving gives

RTH = 50Ωisc = 3.9mA

voc = 0.195V

(c) First make a guess as to which region of the curve, N will operate in. Assuming that it will operate in

the A-B region, then by KVL i(t) = (vs (t) + vb − voc )/ (R + RTH ) = (50sin(1000t) − 0.095) / 550. It can be

seen that this guess is wrong as the range of i(t) is not in the appropriate region. Assuming the 0-A region,

by KVL i(t) = (vs (t) + vb − voc )/ (R + RTH ) = (50sin(1000t)m + 0.1)/ 2500 = 0.02sin(1000t) + 0.04mA .

Note that the highest current is 0.06mA, thus still in the appropriate region of operation.

(d) Repeating the procedure above and guessing the region A-B, by KVL

i(t) = (vs (t) + vb − voc )/ (R + RTH ) = (200sin(1000t)m + 305m) /100 = 2sin(1000t) + 3.05mA . The

maximum and minimum current are 5.05 and 1.05 mA respectively, thus the assumption made was correct.

SOLUTION TO 6.43. (a) By the power transfer theorem RL = RTH . For circuit (a) RL = 80 ||240 = 60Ω,

and circuit (b) RL = (900||180) + 50 = 200Ω.

(b) Finding isc for each circuit: (a) isc = 36 / (80 + 240||60) ⋅ 240 / 300 = 225mA and (b)

isc = 60mA900 ||180

(900||180) + 50 + 200

= 22.5mA . The power is now obtained from P = isc / 2( )2

RL . Thus

759mW for (a) and 25.3mW for (b).


SOLUTION TO 6.44. Finding the Thevenin equivalent,

RTH = 30||15( ) +10[ ] ||80 = 16Ω

Voc = −32V

The value for the load resistance is 16Ω, and the power is P =Voc

2

2

/ RL =16W .

SOLUTION TO 6.45. (a) Find the Thevenin equivalent,

RTH =12k | |6k + 8k =12kΩ = RL

Voc = 8k(2mA) + 246

6 +12

= 24V

The power is P = Voc / 2( )2/ RL = 12mW .

(b) The maximum power will be transferred to the load, when the value of its load is closest to 12kΩ. Thus

the power is P =Voc

RTH +10k

2

⋅10k =11.9mW .

(c) Same reasoning as (b) the power is P =Voc

RTH +15k

2

⋅15k =11.9mW .

SOLUTION TO 6.46. (a) The Thevenin equivalent to the left of RL has Rth = 12 + 20 / /180 = 30 Ω and

Voc =1 ×18 +180

180 + 2040 = 54 V. Therefore, for maximum power transfer

GL + 2GL + 3GL = 6GL =6

RL=

1

30

Hence RL =180 Ω.

(b) For this part, let VL denote the voltage (top to bottom) across the load. With RL =180 Ω, then the

parallel combination equals Rth and hence VL = 27 V. It follows that

P180 =(27)2

180= 4.05 watt.

Since the terminal voltages are the same, the absorbed power is inversely proportional to the resistance.

Hence P90 = 8.1 watt, and P60 = 12.15 watt.


SOLUTION 6.47. To find the Thevenin equivalent introduce a test voltage source at the output, and write

KCL at the two node in the circuit. By inspection the following matrix expression is obtained:

(1/ 200 +1/ 400) −1/ 400

−1/ 400 0.0015 +1/1k +1/ 400

VA

Vtest

=

0.1

Itest

Solving Vtest = 240Itest + 8. And RTH = 240 Ω, Voc = 8 V. By voltage division, the voltage across the load

resistor is 4V, and the power delivered to it is 66.7mW.

VA Vtest

SOLUTION 6.48. Using KCL get ix =1mA , which independent of what is connected to the output. Thus

isc = 10ix =10mA

RTH = 3kΩ

The power is then P = 10mA / 2( )2 ⋅3k = 75mW .

SOLUTION 6.49. Performing a source transformation on (a), and combining the elements will simplify to

one current source going up of 2/3A in parallel with a 10Ω resistor. This is essentially the Norton

equivalent of the circuit,

isc = 2/ 3A

RTH =10Ω

For (b), combine the voltage sources and resistor in series. The circuit obtained is one voltage source of

5V in series with a 45Ω resistor. This is the Thevenin equivalent,

voc = 5V

RTH = 45Ω

(a) The value of the load resistor is simply the thevenin resistance obtained above.

(b) Using Ohm’s law for (a) VL = isc(10 ||10) = 10/ 3V , and voltage division for (b) VL = voc(1/ 2) = 2.5V

(c) Using the following formula, P = VL2 / RL , (a) absorbs 1.1W and (b) 139mW, thus (a) absorbs more

power.


SOLUTION 6.50. (a) Note that the circuit left of the terminal is already in its Thevenin form. The load

RL = R | |( R + 300)

R2 + R(300 − 2RL ) − 300RL = 0

Solving, R = 71.6Ω . By voltage division, the voltage across the load is 5V. The power absorbed isP = Voc / 2( )2

/ RL = 416.7mW .

(b) The following script can be used to plot the power absorbed by the load versus R:

%Script for problem 6.50b

R=0:2:400;

%Calculate Load resistance

RL= 1./((1./R)+1./(R+300));

%Calculate the power

P=(10.*(RL./(RL+60))).^2./RL;

%Plot the power versus R

plot(R,P);

ylabel('Power in Watts');

xlabel('Resistance in Ohms')

SOLUTION 6.51. First, find the Thevenin equivalent by writing out the transfer equation vab = 200i + 40.

Thus RTH = 200Ω⇒ Voc = 40V . The maximum power will then be P = (Voc / 2)2 / 200 = 2W .

SOLUTION 6.52. The assumption that all controlling voltages or currents for dependent sources within

Ni are assumed to be in Ni, implies that the nodal equation matrix of figure P6.52a has the partitioned

form:

G11 G12 0

G21 G22 G23

0 G32 G33

VN1

Vm

VN 2

=IN1

Im

IN 2

(*)

where VN1 is the vector of UNKNOWN and INDEPENDENT node voltages internal to N1 and VN2 is

the vector of UNKNOWN and INDEPENDENT node voltages internal to N2. The right side of the


equation consists of (effective) currents injected into the appropriate node. However, IN1 depends only on

sources in N1 and IN2 depends only on sources in N2.

At this point we must presume that the matrix equation (*) has a unique solution, i.e., the determinant

of the coefficient matrix is non-zero. Hence we can calculate VN1, Vm, and VN2 uniquely. As such, by

considering the first row of (*), we can assert that VN1 satisfies

G11VN1 = IN1 − G12Vm (**)

Note that we are not claiming that we can solve for VN1 from this equation.

Replacing N2 by a voltage source of value Vm results in the network of figure P6.52b. For this

network, the nodal equations are

G11VN1 = IN1 − G12Vm (***)

where Gij is the same as in (*). Clearly, this is the same as equation (**). Again we presume there is a

unique solution to this equation, i.e., the determinant of G11 is non-zero. If so, we can solve for VN1

uniquely and the result is the same as that obtained by solving (*).

This theorem can be extended to RLCM networks (to be studied in later chapters) or even

nonlinear networks under appropriate conditions.

To emphasize the subtlety of this result and the need for unique solvability in each network,

consider the following circuit.

The resulting nodal equations are:


2 −1 0 0

−1 0.5 −0.5 0

0 −0.5 2.5 −1

0 0 −1 2

V1

V2

V3

V4

=

−2

0

0

4

There exists a unique solution and from MATLAB, we find

Vm = V3 = 2 V

To apply the voltage source substitution, we replace N2 by a voltage source of 2 V and obtain thefollowing circuit.

The nodal equations here are

2 −1

−1 0.5

V1

V2

=

−2

1

Observe that the coefficient matrix has a zero determinant. Thus there is either no solution or many

solutions, i.e., no unique solution. This demonstrates that unique solvability of the larger network does not

imply the unique solvability of the smaller derived network.

SOLUTION 6.53. The assumption that all controlling voltages or currents for dependent sources within

Ni are assumed to be in Ni, implies that the loop equation matrix of figure P6.53a has the partitioned form:

R11 R12 0

R21 R22 R23

0 R32 R33

IN1

Im

IN 2

=EN1

Em

EN 2

(*)


where IN1 is the vector of unknown and independent loop currents internal to N1 and IN2 is the vector of

unknown and independent loop currents internal to N2 and Im a single independent loop current common

to N1 and N2. The right side of the equation represents the net contribution of voltage sources present in

the appropriate loop. However, EN1 depends only on N1 and EN2 depends only on N2.

At this point we must presume that the matrix equation (*) has a unique solution, i.e., the determinant

of the coefficient matrix is non-zero. Hence we can calculate IN1, Im, and IN2 uniquely. As such, by

considering the first row of (*), we can assert that IN1 satisfies

R11IN1 = EN1 − R12Im (**)

Note that we are not claiming that we can solve for IN1 from this equation.

Replacing N2 by a current source of value Im results in the network of figure P6.53b. For this

network, the loop equations are

R11IN1 = EN1 − R12Im (***)

where Rij is the same as in (*). Clearly, this is the same as equation (**). Again we presume there is a

unique solution to this equation, i.e., the determinant of R11 is non-zero. If so, we can solve for IN1

uniquely and the result is the same as that obtained by solving (*). For some subtlety in the proof refer to

the solution of 6.52.

This theorem can be extended to RLCM networks (to be studied in later chapters) or even

nonlinear networks under appropriate conditions.

SOLUTION 6.54. (a) The thevenin equivalent to the left of terminal A-B is

RTH = [(30||60) + 20]||10 = 8Ω , and using Ohm’s law along with voltage division

VOC = 15 −1015V

15 + 60

30

30 + 30

= 14V

(b) Doing the same for the circuit right of terminal A-B. RTH = [(30||15) + 10]||20 =10Ω and

VOC = 7.5 − 207.5V

15 +15

30

30 + 30

= 5V .

(c) Using superposition, VAB = 1410

18

+ 5

8

18

= 10V .

(d) (e) (vCB − 15) / 30 + vCB / 60 = (10 − vCB )/ 20 Hence, vCB = 10V .


SOLUTION TO P6.55. The proof is based on superposition. Let us consider the figure below where N1

and N2 are differently named but identical networks.

We first compute the contribution to ia from the independent sources in N1 with those of N2 deactivated.

Let this current be ia1 . The contribution to ia from the independent sources in N2 with those of N1

deactivated is ia2 . But because N1 and N2 are identical, ia

1 = −ia2. Hence by superposition

ia = ia1 + ia

2 = 0. By the current source substitution theorem we can replace the lines by current sources of

value 0-amp. This defines an open circuit and the connecting line can be replaced by an open circuit.

From the given network we also note by KVL that Vx + Vy = 0 which implies that Vx = –Vy . On the other

hand, since the networks are identical, Vx = Vy . Thus we conclude that Vx = Vy = 0. Thus we can replace

Vx and Vy by a voltage source of 0-volt (voltage source substitution theorem) which is the definition of a

short circuit.

SOLUTION 6.56. Label the potential between each line starting from the top as Vx1, Vx2 on the left, and Vy1

and Vy2 on the right. Now by superposition and linearity notice that

Vx1 = −Vy1

Vx2 = −Vy2

because the independent source is negative on the right side. Additionally, from KVL,

Vx1 = Vy1

Vx2 = Vy2

The only way all these condition can be met, is if all the voltages are 0 V, or short circuited.


SOLUTION 6.57. (a) Using the results from P6.55, no current flows between the two halves. So the right

hand side circuit may be analyzed as if it was stand alone. By voltage division then,

va =3 + 2

3 + 2 +1

18 = 15V .

(b) From the results of P6.56, all the lines crossing the symmetry line are shorted together. Consequently,

by voltage division, va =6 ||3

(6 ||3) +1

(−18) = −12V .

SOLUTION 6.58. Note how this circuit is the same as in P6.57: it is just redrawn with the neighboring

resistors added in parallel or in series. Using superposition, we can solve for va when the sources [vs1 vs2]

are [18 18], and then [18 –18]. By linearity adding the two contribution will be equivalent to solving for

[36 0] directly, since adding the source contributions [18 18]+[18 –18] =[36 0]. The contributions of

15V–12V were obtained in P5.57; thus va = 3V .

SOLUTION 6.59. Yes since

45 27[ ] = 36 36[ ] + 9 −9[ ] = 2 18 18[ ] +1/ 2 18 −18[ ] = 2(15) + (12) / 2 = 24V .

SOLUTION 6.60. For this proof we attach an arbitrary network N to each of the networks N1 and N2 in

figure P6.60 as shown below.

N may have internal independent sources, but we consider N1 and N2 external excitations to N and

we assume no violation of KVL in the attachment. Choose node 3 as a reference node. Then

V13 = Va − Vc and V23 = Vb − Vc for both figures. Hence N1 and N2 provide the identical external


excitations to N and hence all currents and voltages in N remain unchanged.

The extension of this result to 4 external nodes is shown in the figure below. The verification is

the same as above. The extension of course to n-terminals is clear.

SOLUTION 6.61. Using the E-shift theorem, remove the 9 V source from each branch and add it to the

4V source, and notice by inspection that VOC = −4 + 9 = 5 V.

SOLUTION 6.62. (a) Writing a KCL equation for each node in N2:

I1 = Ia

I2 = Ia − Ia = 0

I3 = Ia − Ia = 0

I4 =− Ia

Do the same for N1:

I1 = Ia

I2 = 0

I3 = 0

I4 =− Ia

This shows that the two have identical outcomes.

(b)


SOLUTION 6.63. Using the I-shift theorem, this circuit is essentially a –2A source in parallel with a

series combination of resistors, and a 5A source. Thus isc = 5 − 2 = 3A .

SOLUTION 6.64. (a) This can be done by inspection. An equal source is connected between A-C and C-

B; thus by the I-shift theorem, it is equivalent to the same source just connected between A-B.

(b) In figure 6.64c the VCCS is replaced by a resistor using the Ohm’s law relationship

100Ω = V1 / (0.01V1).

(c) (d) By voltage division V1 = Vs

100||900

(100 ||900) + 10

, and by Ohm’s law

Vout = Vs

100||900(100 ||900) +10

0.01 20k | |5k( ) = 36Vs

Vout / Vs = 36

L&C Probs, 11/15/01 P:7-1 © R. A. DeCarlo, P. M. Lin


SOLUTION 7.1. Given the coil has 48 turns and 12 turns/cm, we know that the length of the coil is 4 cm.Since the length of the coil is greater than 0.4 times its diameter, the formula given in the question can beused:

L =4 ×10−5 × 0.02( )2 × 48( )2

18 × 0.02( ) + 40 × 0.04( ) =18.81 H

SOLUTION 7.2. Part 1. Applying (7.1)The voltage vL(t) can be computed using the inductor v-i relationship:

vL (t) = LdiL (t)

dt

The calculations for vL(t) for t = 0s to 5s are summarized in the following table:

Time Interval d/dt (iin(t)) vL(t)0s< t ≤1s 2 As-1 1V1s< t ≤3s -2 As-1 -1V3s< t ≤4s 2 As-1 1V4s< t ≤5s 0 As-1 0V

Below is the plot of VL vs time.

Part 2. Applying (7.4)

wL (t) =12

(0.5 iin2( t))

In the time interval 0s < t ≤ 1s, iin (t) = 2t . Thus

wL (t) =12

0.5iin2 ( t)( ) = t2

In the time interval 1s < t ≤ 3s, iin (t) = 4 − 2t . Thus


wL (t) =12

0.5iin2 ( t)( ) =

14

(4 − 2t)2 = t2 − 4t + 4

In a similar way, wL (t) can be computed in the remaining intervals.The calculations for wL(t) for t = 0s to 5s are summarized in the following table:

Time Interval iin(t) wL(t)0s< t ≤1s 2t t2

1s< t ≤3s 4-2t t2-4t+43s< t ≤4s -8+2t t2-8t+164s< t ≤5s 0 0

Below is the plot of wL vs t.

SOLUTION 7.3. Applying (7.2)

iL ( t) = iL (0) +1L

vL ( )d0

t∫

It is assumed that iL(0) = 0 A. Using the preceding formula and the fact that vin(t) = t in the time interval0s < t ≤ 2s, the current is

iL ( t) =1

0.5d = t2

0

t∫ in the time interval 0s < t ≤ 2s

In the time interval 2s < t ≤ 3s vin (t) = 2 . Hence

iL ( t) =1

0.5vL ( )d

0

t∫ = iL (2) +

10.5

2dt2

t∫ = 4 + 4 2

t = 4t − 4

In a similar way iL ( t) can be computed in the remaining intervals.

The calculations for iL(t) for t = 0s to 6s are summarized in the following table:


Time Interval vin(t), V iL(t), A0s< t ≤2s t t2

2s< t ≤3s 2 4t-43s< t ≤5s -2 20-4t5s< t ≤6s 0 0

Below is the plot of iL vs t.

SOLUTION 7.4. Part 1. Using (7.1), we have

vin (t) = 0.2 ×10−3 ddt

is( t)[ ] = 0.2 ×10−3 ×1000cos(1000 t) = 0.2cos(1000t)mV

For the 2mH inductor

iout ( t) = iout (0) +1L

10vin0

t

∫ ( )d =1

2 ×10−3 2cos0

t

∫ (1000 )d = sin(1000t)mA

Below is a sketch of iout vs t.

Part 2. Instantaneous power delivered by the dependent source is given by

pL ( t) = vL (t) × iL (t) = 10vin ( t) × iout ( t) = 2cos(1000t) × sin(1000t) = sin(2000t) W


Part 3. The energy stored in the 2mH inductor is given by

WL (t) =12

LiL2(t) =

12

Liout2( t) = sin2(1000t)nJ

Below is a sketch of WL vs t

SOLUTION 7.5. Part 1 For the excitation in Figure P7.5b,

i1(t) = i1(0) +1

0.5vin

0

t

∫ ( )d , i2(t) = i2(0) +1

0.25vin

0

t

∫ ( )d = 2i1( t)

It is assumed that i1(0) = i2(0) = 0 A . In the interval 0s < t ≤ 1s, vin (t) = −10V . Hence, in this interval

i1(t) =1

0.5vin ( )d

0

t∫ =

10.5

(−10)d0

t∫ = −20t

i2( t) =1

0.25vin ( )d

0

t∫ =

10.25

(−10)d0

t∫ = −40t

In the interval 1s < t ≤ 2s, vin (t) = −5V . Hence, in this interval,

i1(t) = i1(1)+1

0.5vin( )d

1

t∫ = −20 +

10.5

(−5)d1

t∫ = −10t −10

i2( t) = i2(1)+1

0.25vin( )d

1

t∫ = −40 +

10.25

(−5)d1

t∫ = −20t − 20

In a similar fashion i1(t) and i2(t) can be computed in other intervals.

The calculations for i1(t) and i2(t) for t = 0s to 5s are summarized in the following table:


Time Interval vin(t), V i1(t), A i2(t), A0s< t ≤1s -10 -20t -40t1s< t ≤2s -5 -10-10t -20-20t2s< t ≤3s 0 -30 -603s< t ≤4s 5 -60+10t -120+20t4s< t ≤5s 10 100+20t 200+40t

Below are the plots of i1 vs t and i2 vs t.

Part 2 For the excitation in Figure P7.5c,

i1(t) = i1(0) +1

0.5vin

0

t

∫ ( )d , i2(t) = i2(0) +1

0.25vin

0

t

∫ ( )d = 2i1( t)

It is assumed that i1(0) = i2(0) = 0 A. In the interval 0s < t ≤ 1s, vin (t) =10t . Hence, in this interval

i1(t) = i1(0) +1

0.5vin( )d

0

t∫ =

10.5

(10 )d0

t∫ = 10t2

and

i2( t) = i2(0) +1

0.25vin( )d

0

t∫ =

10.25

(10 )d0

t∫ = 20t2

In the interval 1s < t ≤ 3s, vin (t) =10t − 20 . Hence, in this interval

i1(t) = i1(1)+1

0.5vin( )d

1

t∫ =10 +

10.5

(10 − 20)d1

t∫ =10t2 − 40t + 40

i2( t) = i2(1)+1

0.25vin( )d

1

t∫ = 20 +

10.25

(10 − 20)d1

t∫ = 20t2 − 80t + 80

The calculations for i1(t) and i2(t) for t = 0s to 3s are summarized in the following table:


Time Interval vin(t), V i1(t), A i2(t), A0s< t ≤1s 10t 10t2 20t2

1s< t ≤3s 10t-20 10t2-40t+40 20t2-80t+80

Below are the plots of i1 vs t and i2 vs t.

SOLUTION 7.6. For the circuit in Figure P7.6, the parallel combination of the 0.75mH and 1.5mHinductors can be replaced by an inductor with the inductance of (0.75||1.5)mH. The v-i relationship forthis inductor is

iin (t) = iin (0) +1

[(0.75) (1.5)] ×10−3vs( )d =

1

0.5 ×10−3 vs( )d0

t

∫ A0

t∫

The series combination of the 0.8mH and 0.6mH can be replaced by an (0.8mH + 0.6mH) inductor. Thev-i relationship for this inductor gives:

vout (t) = (0.8 + 0.6) ×10−3 ddt

6iin (t)[ ] =0.8 + 0.6

0.5× 6

ddt

vs0

t

∫ ( )d

= 16.8vs( t) V

Below is a plot of vout(t) vs t.


Assume iL(0) =0 for all inductors.

iout ( t) = 6iin (t) =6

0.5 ×10−3 vs0

t

∫ ( )d =12 ×103 vs0

t

∫ ( )d A

In the interval 0s < t ≤ 1s, vs(t) = 1mV . Hence,

iin (t) =1

0.5 ×10−3 vs( )d0

t∫ = 2t(A), iout ( t) = 6iin (t) = 12t(V )

The power p(t) on the same interval is computed as:

p( t) = 6iin ( t) × vout (t) = (12t) × (16.8×10−3) = 0.202t(W )

In the interval 1s < t ≤ 3s, vs(t) = 2mV . Hence,

iin (t) = iin (1) +1

0.5 ×10−3 vs( )d1

t∫ = 2 + 2 × 2 1

t = 4t − 2(A)

iout ( t) = 6iin (t) = 24t −12(A)and

p( t) = 6iin ( t) × vout (t) = (24 t −12) × (16.8× 2 ×10−3) = 0.806t− 0.403(W )

In a similar fashion, iout(t) and p(t) can be computed for the remaining intervals.

The calculations for iout(t) and the instantaneous power delivered by the dependent source, p(t), for t = 0sto 6s are summarized in the following table:

Time Interval vs(t), V vout(t), V iout(t), A p(t), W0s< t ≤1s 1×10-3 16.8×10-3 12t 0.202t

1s< t ≤3s 2×10-3 33.6×10-3 -12+24t 0.806t-0.403

3s< t ≤5s -2×10-3 -33.6×10-3 132-24t 0.806t-4.435

5s< t ≤6s 0 0 12 0


SOLUTION 7.7. For 0 ≤ t < 2 s, using the inductor v-i relationship, we have

iin (t) =1

0.5vin ( )d

0

t

∫ =2

0.25π−cos(0.25π )[ ]0

t =8π

1− cos(0.25πt)[ ] A

The associated energy stored as a function of t for this time interval is

W0.5(t) =16

π 2 1− cos(0.25πt)[ ]2 J

The energy for the second inductor remains zero over this interval.

For 2 s ≤ t, we have

iin (t) = i0.5(t) + i0.25( t) = i0.5( t) +1

0.25vin ( ) d

2

t

∫

=8π

1− cos(0.25πt)[ ] +4

0.25π−cos(0.25π )[ ]2

t =8π

1− cos(0.25πt)[ ] −16π

cos(0.25πt)

=8π

−24π

cos(0.25πt) A

Here the current

i0.5( t) =8π

1− cos(0.25πt)[ ] A

in which case the energy stored over the interval [2,t] is

W0.5(2,t) =12

0.5 i0.52 (t) − i0.5

2 (2)( ) =12

0.5 i0.52 (t) −

64

π 2

J

Further

i0.25( t) = −16π

cos(0.25πt) A

in which case the energy stored over the interval [2,t] is

W0.25(2,t) =12

0.25 i0.252 (t) − i0.25

2 (2)( ) =12

0.25i0.252 ( t) J

SOLUTION 7.8. Let the 5mH inductor be L1 and the 20mH inductor be L2.For 0 ≤ t < 3ms,

iL1(t) = iL1(0) +1L1

vs0

t

∫ ( )d =1

5 ×10−3 12cos0

t

∫ (500 )d = 4.8sin(500t) mA


iL 2(t) = iL2(0) +1L2

vs0

t

∫ ( )d =1

20 ×10−3 12cos0

t

∫ (500 )d = 1.2sin(500t) mA

For t ≥ 3ms,

iL1(t) = 4.8sin(500t) mA as L1 is still subjected to the same voltage.

on the other hand,

iL 2(t) = iL2(3− ms) +1

L2vL2

3+ ms

t

∫ ( )d = iL 2(3− ms) +1

L20

3+ ms

t

∫ d

= iL 2(3− ms)

=1.2sin(500 × 3 ×10−3) =1.197mA

For 0≤ t < 3ms, the energies stored in the inductors are given as follows:

WL1(t) =1

2L1 iL1(t)[ ]2 = 57.6sin 2(500t) nJ,

WL2( t) =1

2L2 iL2( t)[ ]2 = 14.4sin 2(500t) nJ

For t ≥ 3ms, the energies stored in the inductors are given as follows:

WL1(t) =1

2L1 iL1(t)[ ]2 = 57.6sin 2(500t) nJ,

WL2( t) =1

2L2 iL2( t)[ ]2 = 14.328 nJ

SOLUTION 7.9. Given the dielectric parameters and the dimensions of the capacitor, the capacitance of thepaper capacitor is given by

C = r 0Ad

= 3× 8.854 ×10−12 ×0.04 × 0.8

10−4 = 8.5nF

SOLUTION 7.10. . Part 1. Applying the capacitor v-i relationship:

iC ( t) = Cddt

vC ( t)( ) =1 ×100 ×1000 × (− sin(1000t)) = −0.1sin(1000t) A

Part 2. Applying the capacitor v-i relationship:


iC ( t) = Cd

dtvC ( t)( )

10 ×10−3 cos(1000t) = Cddt

sin(1000t)( )

10 ×10−3cos(1000t) =1000C cos(1000t)Therefore, C=10µF.

SOLUTION 7.11. Applying the capacitor v-i relationship for C1 and C2:

iC1(t) = (2mF)ddt

vin (t)( ), iC 2(t) = (6mF)ddt

vin ( t)( ) = 3iC1( t),

In the interval –1s < t ≤ 0s, vin(t) = 5t + 5. Hence, in this time interval

iC1(t) = 2 ×10−3 ×d(vin (t))

dt= 10−2 A

and

iC 2(t) = 6 ×10−3 ×d(vin ( t))

dt= 3×10−2 A

Using KCL, iin(t) can be computed as

iin (t) = iC1(t) + iC 2(t) = 4 ×10−2 A

In the interval 0s < t ≤ 1s, vin(t) = 5. Thus

iC1(t) = iC2( t) = 0 A

Using KCL, iin (t) = iC1(t) + iC 2(t) = 0 A.In a similar fashion iC1(t), iC2(t) and iin(t) can be computed for the remaining intervals.

The calculations for iC1(t), iC2(t) and iin(t) for t = -1s to 6s are summarized in the following table:

Time Interval d/dt (vin(t)), Vs-1 iC1(t), mA iC2(t), mA iin(t), mA-1s< t ≤0s 5 10 30 400s< t ≤1s 0 0 0 01s< t ≤3s -5 -10 -30 -403s< t ≤4s 20 40 120 1604s< t ≤5s 0 0 0 05s< t ≤6s -15 -30 -90 -120

Below are the plots of iC1(t), iC2(t) and iin(t).


SOLUTION 7.12. Applying (7.6)

vC (2) = vC (0) +1C

iC ( )d0

2

∫ = 4V +1C

d0

2

∫ = 5V

vC (3) = vC (2) +1C

iC ( )d2

3

∫ = vC (2) +1C

2d2

3

∫ = 5V +1V = 6V

Applying (7.11), the energies stored in the capacitor over the intervals [0,2] and [2,3] are given by

WC[0,2] =12

C vC2 (2) − vC

2 (0)[ ] = 9 J

WC[2,3] =12

C vC2 (3) − vC

2 (2)[ ] =11 J

SOLUTION 7.13. Part 1 Applying (7.6)


vC1(t) = vC1(0) +1

C1iin ( )d

0

t

∫ = 1

0.25iin ( )d

0

t

∫

vC 2( t) = vC2(0) +1

C2iin ( )d

0

t

∫ = 1

0.1iin ( )d

0

t

∫

In the interval 0s < t ≤ 1s, iin (t) = 2t . Hence, in this interval

vC1(t) =1

0.25iin ( )d =

10.25

( 2)o

t∫

0

t= 4 ×106 t2(V )

and

vC 2( t) =1

0.1iin ( )d =

10.1

( 2)o

t∫

0

t=10 ×106t2(V )

In the interval 1s < t ≤ 3s, iin (t) = 2 ×10−3 (A). Thus

vC1(t) = vC1(1) +1

0.25× iin ( )d =

1

t∫ 4 +

10.25

× (2 ×10−3 )1

t= 8 ×103t − 4(V )

and

vC 2( t) = vC2(1) +1

0.1× iin ( )d =

1

t∫ 10 +

10.1

× (2 ×10−3 )1

t= 20 ×103t −10(V )

In a similar fashion, vC1(t) and vC2(t) can be computed for the remaining intervals.

The calculations for vC1(t) and vC2(t) for t = 0s to 8ms are summarized in the following table:

Time Interval iin(t), A vC1(t), V vC2(t), V0s< t ≤1ms 2t 4×106 t2 10×106 t2

1ms< t ≤3ms 2×10-3 -4 + 8×103 t -10 + 20×103 t

3ms< t ≤5ms -2×10-3 44 - 8×103 t 110 - 20×103 t5ms< t ≤8ms 0 4 10

Below are the plots of vC1(t) and vC2(t).


Part 2 Applying (7.12),

WC1(t) =12

C1vC12(t) , WC 2(t) =

12

C2vC 22( t)

The expressions for WC1(t) and WC2(t) for t = 0ms to 8ms are listed in the following table:

Time Interval WC1(t), µJ WC2(t), µJ

0s< t ≤1ms 0.125 × (4 ×106 t2)2 0.05 × (10×106 t2)2

1ms< t ≤3ms 0.125 × (-4 + 8×103 t)2 0.05 × (-10 + 20×103 t)2

3ms< t ≤5ms 0.125 × (44 – 8×103 t)2 0.05 × (110 – 20×103 t)2

5ms< t ≤8ms 0.125 × (4)2 0.05 × (10)2

Below are the plots of WC1(t) and WC2(t).

Part 3 Since the current iC1(t) stays constant at 0A after t = 5ms,

vC1(∞) = vC1(5) = 4V

vC 2(∞) = vC 2(5) = 10V

SOLUTION 7.14. Part 1 Applying (7.6),


vin (t) = vin (0) +1C

iin ( )d0

t

∫ = 1

0.5 ×10−3 iin ( )d0

t

∫

In this part, we use the current excitation signal described in Figure P7.14b. In the interval 0s < t ≤ 1s,iin (t) = −10(mA). Thus, in this interval

vin (t) =1

0.5 ×10−3 (−10 ×10−3)do

t∫ = −20t(V )

In the interval 1s < t ≤ 2s, iin (t) = −5(mA). Hence, in this interval

vin (t) = vin (1)+1

0.5 ×10−3 (−5 ×10−3)d1

t∫ = −20 + (−10t +10) = −10t −10(V )

In a similar fashion, vin(t) can be computed for the remaining intervals. The calculations for vin(t) for t = 0sto 5s are summarized in the following table:

Time Interval iin(t), mA vin(t), V0s< t ≤1s -10 -20t1s< t ≤2s -5 -10 -10t2s< t ≤3s 0 -303s< t ≤4s 5 -60+10t4s< t ≤5s 10 -100+20t

Below is the plot of vin(t) vs. time.

Part 2. In this part, we use the current excitation signal described in Figure P7.14c. In the interval 0s < t≤ 1s, iin (t) = 10(mA) . Thus, in this interval

vin (t) =1

0.5 ×10−3 (10×10−3 × )do

t∫ = 10t2(V)

In the interval 1s < t ≤ 3s, iin (t) = 10t − 20(mA) . Hence, in this interval

vin (t) = vin (1)+1

0.5 ×10−3 10−3 × (10 − 20)do

t∫ = 10 +10(10 2 − 40 )1

t =10t2 − 40t + 40 (V)


The calculations for vin(t) for t = 0s to 3s are summarized in the following table:

Time Interval iin(t), mA vin(t), V0s< t ≤1s 10t 10t2

1s< t ≤3s 10t-20 10t2 - 40t + 40

Below is the plot of vin(t) vs. time.

SOLUTION 7.15. Part 1 Using the capacitor v-i relationship,

iin (t) = C1ddt

(vs(t)) = 20 × 6 ×1500 × cos(1500t) = 0.18cos(1500t) A

Then, we can find vout by applying (7.6),

vout (t) = vout (0) +1

0.5m2iin ( )d

0

t

∫

= 10 +2 × 0.18

0.5m ×1500sin(1500t)

= 10 + 0.48sin(1500t) V

Part 2 The instantaneous power delivered by the independent source is given by

p( t) = 2iin ( t) ×vout (t) = 0.36cos(1500t) × [10+ 0.48sin(1500t)] W

Part 3 Applying (7.11), the energy stored in the capacitor over the interval [0,t] is given by

WC1[0, t] =12

C1 vs2(t) − vs

2(0)[ ] = 360sin2(1500t) J


SOLUTION 7.16. Part 1 From the solution of problem 7.15,

vout (t) = vout (0) +1

0.5m2iin ( )d

0

t

∫

= vout (0) +1

0.5m2 × 20

dd

vs( )( )d0

t

∫

= vout (0) + 0.08dd

vs( )( )d0

t

∫

= vout (0) + 0.08vs( )

If we assume vout(0) = 0V, then

vout (t) = 0.08 vs( )

The following is a plot of vout(t) vs time.

Part 2 The instantaneous power delivered by the dependent source is given by

p( t) = vout ( t) × 2iin (t) = ( vout (0) + 0.08 vs( t)) × 2 × 20d

dtvs( t)( )

= 40( vout (0) + 0.08 vs( t)) d

dtvs(t)( ) W


p( t) = 3.2 vs( t) ddt

vs(t)( ) W

The calculations for d/dt (vS(t)) and p(t) for t = 0s to 6ms are summarized in the following table:


Time Interval d/dt (vS(t)), Vs-1 p(t), W0ms< t ≤1ms 2 × 10

312.8 × t

1ms< t ≤3ms -2 × 103

12.8 × (t-2)

3ms< t ≤4ms 2 × 103

12.8 × (t-4)4ms< t ≤6ms 0 0

The following is a plot of p(t) vs time.

Part 3 The energy stored in the 0.5-mF capacitor is given by

WC (t) =12

× 0.5m × vout2(t) = 0.25( vout (0) + 0.08 vs( t))2 mJ

If we assume vout(0)=0V, then

WC (t) = 1.6(vs(t))2 J

In the interval 0s < t ≤ 1ms, vs(t) = 2 ×103t(V ). Thus, in this interval

WC (t) = 6.4t2J

In the interval 1s < t ≤ 3ms, vs(t) = 2 ×103 × (2 − t)(V ). Hence, in this interval

WC (t) = 6.4 × (2 − t)2J

In a similar way WC(t) can be computed for the interval 3ms < t ≤ 4ms.

The following is a plot of WC(t) vs time.


SOLUTION 7.17. Part 1 From the solution of problem 7.16,

vout (t) = vout (0) + 0.08 vs( )


vout (t) = 0.08 vs( )

The following is a plot of vout(t) vs time.

Part 2 The instantaneous power delivered by the dependent source is given by.

p( t) = 40( vout (0) + 0.08 vs(t)) ddt

vs( t)( ) W


p( t) = 3.2 vs( t) ddt

vs(t)( ) W

The calculations for d/dt (vS(t)) and p(t) for t = 0s to 6ms are summarized in the following table:


Time Interval d/dt (vS(t)), Vs-1 p(t), W0ms< t ≤1ms 2 × 103 12.8 × t1ms< t ≤2ms 0 02ms< t ≤4ms -2 × 103 12.8 × (t-3)4ms< t ≤5ms 0 05ms< t ≤6ms 2 × 103 12.8 × (t-6)

The following is a plot of p(t) vs time.

Part 3 The energy stored in the 0.5-mF capacitor is given by

WC (t) = 0.25( vout (0) + 0.08 vs(t))2 mJ


WC (t) = 1.6(vs(t))2 J

In the interval 0ms < t ≤ 1ms, vs(t) = 2 ×103t(V ). In this interval

WC (t) = 6.4t2J

In the interval 1ms < t ≤ 2ms, vs(t) = 2(V ). In this interval

WC (t) = 6.4( J )

In a similar fashion WC(t) can be computed for the remaining intervals.

The following is a plot of WC(t) vs time.


SOLUTION 7.18. Part 1 Applying the capacitor v-i relationship for the equivalent capacitor of the seriescombination of 0.3 mF and 0.6 mF capacitors

vin (t) = vin (0) +1

0.3 ×10−3 0.6 ×10−3is( )d

0

t

∫

= (0) +1

0.3×10−3 0.6 ×10−360 ×10−3 sin(100 )d

0

t

∫

= 3 − 3cos(100t) V

Therefore,

iout ( t) = (0.2 ×10−3 + 0.8 ×10−3)ddt

(10vin (t)) = 3sin(100t) V

Below is a sketch of iout(t) vs time.

Part 2 The instantaneous power delivered by the dependent source is given by

p( t) = iout ( t) ×10vin (t) = 90sin(100t) × [1− cos(100t)] W


Part 3 The instantaneous energy stored in the 0.8mF capacitor is given by

WC (t) =12

CvC2( t) =

12

(0.8mF) 10vin ( t)[ ]2 = 0.361− cos(100t)[ ]2 J

SOLUTION 7.19.

Part 1 Since Q = CV, the charge that resides on each plate of the capacitor = 10µF × 100V = 1mC

Part 2 Since V = Q/C, the required voltage = 1mC/5µF = 200V

Part 3 Since V = Q/C, the required voltage = 50µC/1µF = 50V

Part 4 The energy required = 0.5 × 10µF × (100V)2 = 0.05J

SOLUTION 7.20. When 0s≤ t <2s, vC(t) = 25V. Conservation of charge requires that

q1(2−) + q2(2−) = q1(2+ ) + q2(2+ ). Since q2(2− ) = 0 C it follows that

25 V × 150 mF = vC(2+)(150 mF + 100 mF)

Hence

vC (2+ ) =25V ×150mF + 0V ×100mF

150mF +100mF= 15 V

Thus, for t > 2s, vC(t) = 15 V.

SOLUTION 7.21. For this solution consider the figure below in which C1 and C2 are labeled.

There are two cases to consider: (i) t < 2 ms and (ii) t > 2 ms.

Case 1. t < 2 ms. Here, since the current source is zero for t < 0, and C1 is uncharged at t = 0,

vC (t) =1C1

is( )d−∞

t

∫ = vC (0) +1C1

is( )d0

t

∫ =1C1

is( )d0

t

∫

=−12

15 ×10−3 × 500e−500[ ]0

t=1.6 1− e−500t( ) V


Note that vC (2− ms) =1.0114 V. Hence the energy stored over [0, 2 ms] is

WC1(0 < t < 2ms) = 0.0192 1− e−500t( )2 J

andWC2(0 < t < 2ms) = 0

Case 2. 2 ms < t. At t = 2 ms the switch closes, forcing a discontinuity in the capacitor voltages. Tocalculate the capacitor voltages at 2

+ ms, we use conservation of charge. Here, the relevant equation is:

qC1(2+ ms) + qC 2(2+ ms) = qC1(2− ms) + qC2(2− ms) + is( )d

2− ms

2+ ms

∫

Note that since vC 2(2− ms) = 0, qC 2(2− ms) = C2vC 2(2− ms) = 0 and the integral of the boundedcontinuous function is(t) over an infinitesimal interval is zero, this equation reduces to

qC1(2+ ms) + qC 2(2+ ms) = qC1(2− ms)

or equivalently, since for t > 2 ms, vC1(t) = vC 2(t) = vC (t) ,

C1vC (2+ ms) + C2vC (2+ ms) = C1vC (2− ms)Therefore

vC (2+ ms) =C1

C1 + C2vC (2− ms) =

1515 + 25

×1.0114 = 0.37927 V

and it follows that

vC (t) = vC (2+ ms) +1

Ceqis( ) d

2ms

t

∫ = 0.379 +1

C1 + C2is( )d

2ms

t

∫

= 0.379 +−12

40 ×10−3 × 500e−500[ ]0.002

t= 0.379 + 0.6 0.36788 − e−500t( ) V

Hence the energy stored in the two capacitors over the interval [2+ ms, t] is

WCeq(2+ ms < t) =

12

CeqvC2 ( t) −

12

CeqvC2 (2+ ms) J

whereas the instantaneous stored energy, i.e., the energy stored over (-∞,t > 2 ms] is given by

WCeq(t) =

12

CeqvC2 ( t)

What happens between 2-ms and 2

+ms is beyond the scope of the material in this chapter. Please

refer to problem 51 in chapter 8 for an explanation.


SOLUTION 7.22. Part 1 Applying the inductor v-i relationship,

vL (t) = Ld

dtiin (t)

= 2.5mH × −200te−10t + 20e−10 t( )= 0.05e−10 t (−10t +1)

Applying the capacitor v-i relationship, we have:

vC (t) = vC (0) +1C

iin ( )d0

t

∫

=20

1mte−10

0

t

∫

= 20 ×103 −0.1 e−10 − 0.01e−10[ ]0

t

= 2 ×103 − te−10 t − 0.1e−10t + 0.1[ ] V

and

vin (t) = vL ( t) + vC ( t)

= 0.05e−10t (−10t +1)+ 2000 − te−10t − 0.1e−10t + 0.1[ ]= −2000.5 × te−10 t −199.95 × e−10t + 200(V)

The sketches of vL(t), vC(t) and vin(t) are shown below.


Part 2 The energy stored in the inductor is given by

WL (t) =12

L[iin (t)]2 = 0.5t2e−20 t J

The sketch of WL(t) vs. time is shown below.

Part 3 The energy stored in the capacitor is given by

WC (t) =12

C vC (t)[ ]2 = 2 −te−10t = 0.1e−10 t + 0.1[ ]2kJ

The sketch of WC(t) vs time is shown below.


SOLUTION 7.23. Applying the inductor v-i relationship we have that

vL (t) = 0.25d[is( t)]

dt= 4cos(4 t) (V )

Applying the capacitor v-i relationship it follows that

iC ( t) = 0.25d[2v2( t)]

dt= 0.25 × 2 × 4 × 4 × [− sin(4t)] = −8sin(4 t) A

SOLUTION 7.24.

Part 1. By KCL is( t) = iC1(t) + iC 2(t). Applying the v-i relationship for capacitor C1 and C2 we have:

iS (t) = (10mF + 20mF)d

dtvs(t)

= 30 ×10−3 × (12te−5t − 30t2e−5t )

= 0.36te−5t − 0.9t2e−5t A

Part 2. Applying the v-i relationship for the 20mH inductor, it follows that

vout (t) = (20mH )d

dt[12is( t)]

= 0.24 × 4.5t2e−5t − 3.6te−5t + 0.36e−5t( )= 1.08t2e−5t − 0.864te−5t + 0.0864e−5t (V )

Part 3 The energy stored in the 20mH inductor for t > 0 is given by

WL (t) =12

(20mH )[12iS( t)]2

= 1.44 0.36te−5t − 0.9t2e−5t( )2J

= 1.44 × t2e−10 t (0.36 − 0.9t)2 J

SOLUTION 7.25. We denote by vC(t) the voltage across the capacitors C1 and C2 and by iL(t) the currentthrough the inductors L1 and L2. The equivalent capacitance of the parallel combination of C1 and C2 is(C1 + C2) and thus:

iS (t) = (C1 + C2)dvC (t)

dt

Using the v-i relationship for the capacitor C2 it follows that

iC 2(t) = C2dvC (t)

dt

By replacing dvC ( t)

dt, it follows that


iC 2(t) =C2

C1 + C2× iS(t) =

23

is(t)

The equivalent inductance of the series combination of L1 and L2 is (L1 + L2) and thus:

9iC 2( t) = (L1 + L2)diL (t)

dt

Using the v-I relationship for the inductor L2 it follows that

vout (t) = vL2(t) = L2di2( t)

dt

By replacing diL (t)

dt, it follows that

vout (t) =L2

L1 + L2× 9iC 2( t) =

23

9iC 2( t) = 6iC 2(t) = 4is(t)

SOLUTION 7.26. Using the v-i relationship for the capacitor we can write:

vC 2( t) =1

C2iC ( )d

−∞

t

∫

vC1(t) =1C1

iC ( )d−∞

t

∫

where iC(t) is the current through the capacitors C1 and C2. Since C1 and C2 are connected in series:

vin (t) = vC1( t) + vC 2(t) =1C1

+1

C2

iC ( )d

−∞

t

∫

It follows that iC ( )d−∞

t

∫ = vin (t)C1C2

C1 + C2

. Hence

vC 2( t) =1

C2iC ( )d

−∞

t

∫ =1

C2×

C1C2C1 + C2

vin (t) =C1

C1 + C2vin ( t)

Since L1 and L2 are combined in parallel it follows that

vout (t) =1

L1+

1

L2

−1d(AvC 2( t))

dt

=L1L2

L1 + L2× A ×

dvC 2(t)dt

=L1L2

L1 + L2× A ×

C1C1 + C2

dvin ( t)dt


where the last equality follows by replacing vC2(t) with C1

C1 + C2vin (t)

SOLUTION 7.27. Observe first that the 0.3H and 0.6H parallel inductances combine to make a 0.2Hinductance. Also the 0.4H and 1.2H parallel inductances combine to make a 0.3H inductance. Finally, the0.2H and 0.3H inductances are combined in series and the equivalent inductance is 0.5H. Shortly, all theabove steps can be written as: Leq = (0.3H || 0.6H) + (0.4H || 1.2H) = 0.5H.

SOLUTION 7.28. Observe first that the 1mH and 5mH inductors combine to make a 6mH inductance.This inductance combines in parallel with the 3mH inductance to make a 2mH inductance. The next stepis to combine in series the 2mH inductance with the 10mH inductance. The equivalent inductance is12mH. This inductance is combined in parallel with the 36mH inductance and the result is 9mH. Finally,the 9mH inductance is combined in series with the 4mH inductance and the result is 13mH. Shortly, allthe above steps cam be written as: Leq = [(5mH + 1mH) || 3mH + 10mH] || 36mH + 4mH = 13mH.

SOLUTION 7.29. Step 1. The parallel combination of the 0.6mH and 1.2mH inductors is equivalent to a0.4mH inductor.Step 2. The series combination of the 2.4mH and 0.4mH inductors is equivalent to a 2.8mH inductor.Step 3. The parallel combination of the 2.8mH and 7mH inductors is equivalent to a 2mH inductor.

SOLUTION 7.30. The three inductors can be arranged in the seven fashions as shown below.

(a) (b) (c)

(d) (e)

(f) (g)

Leqa = 1mH

Leqb = 1mH + 1mH = 2mH

Leqc = 1mH || 1mH = 0.5mH

Leqd = 1mH + 1mH + 1mH = 3mH

Leqe = 1mH + (1mH || 1mH) = 1.5mH

Leqf = 1mH || (1mH + 1mH) = 0.667mH

Leqg = 1mH || 1mH || 1mH) = 0.333mH


SOLUTION 7.31. For Fig. P7.31a, Leq = 2 + 15 + 10 + 10 + 40 +30 + 20 + 8 = 135mH

The circuit in Fig. P7.31b is equivalent to the following circuit.

Therefore, Leq = 2 + 15 + 10 + 8 = 35mH

The circuit in Fig. P7.31c is equivalent to the following circuit.

Therefore, Leq = 2 + [(15+10) || (10+40) || (20+30)] + 8 = 22.5mH

SOLUTION 7.32. When the switch is open, the circuit in Fig. P7.32 can be rearranged as the following.

Therefore, Leq = 8L || 8L = 4L

When the switch is closed, the circuit in Fig. P7.32 can be rearranged as the following.

Therefore, Leq = 4L || 4L + 4L || 4L = 4L


SOLUTION 7.33. Leq1=Leq2 Without going into a detailed analysis, we present the following intuitiveargument. Note that the points a and b represent points on a balanced bridge circuit meaning that thevoltage between a and b would be zero. Therefore, no current will flow through the additional inductanceL. Therefore the presence of L does not affect the equivalent inductance value.

SOLUTION 7.34. Leq1>Leq2 Without going into a detailed analysis, we present the following intuitiveargument. Note that the points a and b represent points on an unbalanced bridge circuit, meaning that thevoltage between a and b would not be zero. Also note that when two inductors are placed in parallel, theequivalent inductance becomes smaller than either inductance. The addition of the inductor L in circuit 2essentially creates an internal parallel inductance resulting in an Leq2 lower than Leq1.

SOLUTION 7.35. First we add an iin label to the circuit as shown below.

From KVL and the derivative definition of the capacitor

vin (t) = vL1( t) + vL 2(t) = L1diin (t)

dt+ L2

diin ( t)dt

= L1 + L2( ) diin ( t)dt

Equivalently,diin ( t)

dt=

1L1 + L2( ) vin (t)

It follows that

vLk(t) = Lk

diin (t)dt

=Lk

L1 + L2vin ( t)

which is the required voltage division formula.

SOLUTION 7.36. Leq = (11mH) || (19.25mH) + 3mH = 10mH. Applying the voltage division formula,

vL1(t) = vin ( t) ×3mHLeq

= 60te−t mV

and

vL2( t) = vin (t) − vL1( t) =140te−t mV


SOLUTION 7.37. First consider the circuit below which contains the additional label of vin(t).

From KCL and the integral definition of the inductor,

iin (t) = iL1(t) + iL 2(t) =1L1

vin ( )d0

t

∫ +1L2

vin ( ) d0

t

∫

Equivalently

vin ( )d0

t

∫ =1

1

L1+

1

L2

iin (t)

Therefore

iLk(t) =

1Lk

vin ( )d0

t

∫ =

1

Lk1

L1+

1

L2

iin ( t) =

L1L2

LkL1 + L2

iin (t) for k = 1,2

SOLUTION 7.38. In the circuit illustrated in Fig. 7.38,

Leq = (12mH + 27mH) || (130mH) = 30mH

Applying the current division formula,

iL1(t) =12 + 27

12 + 27 +130iin (t) = 0.231e−t 2

mA

iL 2(t) =130

12 + 27 +130iin ( t) = 0.769e−t 2

mA

Also,

vin (t) = Leqddt

iin ( t) = −60te−t 2 V

The instantaneous energy stored in the 130-mH is given by

WL1(t) =12

L1 iL1(t)[ ]2 = 0.8e−2t 2 nJ


SOLUTION 7.39. Using the inductor v-i relationship,

vin (t) = Leqddt

iin ( t) = 30mH ×ddt

iin ( t)

The calculations for vin(t) for t = -2s to 7s are summarized in the following table:

Time Interval d/dt (iin(t)), mAs-1 vin(t), mV-2s< t <=0s 100 30s< t <=2s -200 -62s< t <=3s 0 03s< t <=4s 200 64s< t <=6s 100 36s< t <=7s -200 -6

From the solutions to problem 7.38,

iL1(t) = 0.231 iin ( t)

iL 2(t) = 0.769 iin (t)

Below are the plots for vin(t), iL1(t) and iL2(t).


SOLUTION 7.40. Part 1 Using the current division formula,

iL 2(t) =18

18 + 6iin (t) = 90cos(300 t) mA

Part 2 Leq=1.5mH + 6mH || 18mH = 6mH

vin (t) = Leqddt

iin (t) = -0.679sin(300 t) V

Part 3 Instantaneous power delivered by the source is given by

p( t) = vin (t) × iin(t) = [-0.679sin(300 t)]× [120cos(300 t)] mW

= -40.7sin(600 t) mW

Below is a plot of p(t)

SOLUTION 7.41. Part 1 Leq = 0.3 || 0.9 + 0.4 || 0.4 = 0.425 H

For computing iin(t) we need to apply

iin (t) = iin (t0) +1

Leqvin ( )d

t0

t

∫

to each interval [0, 1], [1, 2], …, [n, n+1], …

The initial condition for the interval [n, n+1] for n even is:

iin (n) = iin (0) +1

Leqvin ( )d

0

n

∫

We assume that iin(0) = 0 and it follows that iin(n) = 0 for n even, since


vin ( )d = 00

n

∫for n even. The initial condition for the interval [n, n+1] for n odd is:

iin (t) = iin (n −1) +1

Leqvin ( )d

n−1

n

∫ =1

0.425(16 ) n−1

n = 37.6 A

For the interval [n, n+1] with n even:

iin (t) = iin (n) +1

Leqvin ( )d

n

t

∫ = 0 +1

0.42516 n

t = 37.6( t − n) A

For the interval [n, n+1] with n odd:


Leqvin ( )d

n

t

∫ = 37.6 +1

0.425(−16 ) n

t = 37.6( n +1− t) A

For the interval [n, n+1] with n even:


Leqvin ( )d

n

t

∫ = 0 +1

0.425(16 ) n

t = 37.6( t − n) A

Hence,

iin (t) =37.6(t − 2n), 2n < t ≤ 2n +1

37.6(2n + 2 − t), 2n +1< t ≤ 2n + 2

where n is a non-negative integer. Below is a sketch for the input current vs. time.

Part 2 The total energy stored in the set of four inductors is given by


W (t) =1

2Leq iin (t)[ ]2

= 300(t − 2n)2 J for 2n < t ≤ 2n +1

300(2n + 2 − t)2 J for 2n +1< t ≤ 2n + 2

where n is a non-negative integer.

Below is a sketch of W(t) vs. t.

Part 3 By the current division formula, the current through the 0.9H inductor is given by

iL 0.9H (t) =0.3

0.3 + 0.9iin (t)

=9.4( t − 2n) A for 2n < t ≤ 2n +1

9.4(2n + 2 − t) A for 2n +1< t ≤ 2n + 2

where n is a non-negative integer.

Below is a sketch of the iL0.9H vs. time.


SOLUTION 7.42.. a)Step 1. The parallel combination of the 1µF and 2µF capacitances is equivalent to a 3µF capacitance.

Step 2. The series capacitors 1.5µF and 3µF combine to make a 1µF capacitance.

Step 3. Finally, the 2µF capacitance is combined in parallel with the 1µF capacitance that was the result of

Step 2 to make a 3µF capacitance.Shortly, the above steps can be written as:

Ceq = [(1µF + 2µF) || 1.5µF] + 2µF] = 3µF

b) Proceeding in a similar fashion as in part a:

Ceq = 30mF + [9mF || (9mF + 18mF) || 5.4mF] = 33mF

SOLUTION 7.43. a)Step 1. Combine the parallel capacitances of 1µF and 2µF to make a 3µF capacitance.

Step 2. Combine the series capacitances of 1.5µF and 3µF to make 1µF capacitance.

Step 3. Combine the parallel capacitances of 2.5µF and 1µF (the result of Step 2) to make a 3.5µFcapacitance.Shortly, the above steps can be written in a condensed form as:

Ceq = 2.5µF + [1.5µF || (1µF + 2µF) = 3.5µF

b) Proceeding in a similar fashion as in part a:

Ceq = (1mF || 2 mF) + (1.2mF || 2mF || 1.5mF) + (4mF || 2.6667mF || 3.2mF || 1.6mF) = 2.1667mF

SOLUTION 7.44.

Part 1 Ceq = ((( 30 || 60 + 40 ) || 30 + 40 ) || 30 + 40 ) || 30 + 40 = 60mF

Part 2 The value of vin(t) is given by

vin (t) = vin (0) +1C

iin ( )d0

t

∫ = 0 +1

0.06200sin(20 )d

0

t

∫ = 166.7 −166.7cos(20t) mV

SOLUTION 7.45. When the switch is open, the circuit in figure P7.45 can be represented by thefollowing.


Therefore, Ceq= 6.4C

When the switch is closed, the circuit in figure P7.45 can be represented by the following.

Therefore, Ceq= ( 4 + 4 ) || ( 16+ 16 ) = 6.4C

SOLUTION 7.46. Ceq1<Ceq2 Without going into a detailed analysis, we present the following intuitiveargument. Note that the points a and b represent points on an unbalanced bridge circuit, meaning that thevoltage between a and b would not be zero. Also note that when two capacitors are placed in parallel, theequivalent capacitance becomes bigger than either capacitance. The addition of the capacitor C in circuit 2essentially creates an internal parallel capacitance resulting in a Ceq2 higher than Ceq1.

SOLUTION 7.47. The three capacitors can be arranged in the seven fashions as shown below.

For configuration (a), Ceq = 1 || 1 || 1 = 0.3333µF

For configuration (b), Ceq = 1 || (1 + 1) = 0.6667µF

For configuration (c), Ceq = 1 || 1 + 1 = 1.5µF

For configuration (d), Ceq = 1 + 1 + 1 = 3µF

For configuration (e), Ceq = 0 + 1 + 1 = 2µF

For configuration (f), Ceq = 1 || 1 + 0 = 0.5µF

For configuration (g), Ceq = 1 + 0 + 0 = 1µF


SOLUTION 7.48. As Q = CV for any capacitor, C=Q/V. The equivalent capacitance of the network isgiven by

Ceq =1C2V

||1C3V

||1C5V

= 0.1 F

SOLUTION 7.49. First consider the circuit below which contains the additional label of iin(t).

From KVL and the integral definition of the capacitor,

vin (t) = vC1( t) + vC 2(t) =1C1

iin ( )d +0

t

∫ 1C2

iin ( )d0

t

∫Equivalently,

iin ( )d0

t

∫ =1

1

C1+

1

C2

vin (t)

Therefore,

vCk (t) =1

Ckiin ( )d

0

t

∫ =

1

Ck1

C1+

1

C2

vin (t) =

C1C2

CkC1 + C2

vin (t)

Hence,

vC1(t) =C2

C1 + C2vin ( t)

vC 2( t) =C1

C1 + C2vin (t)

which is the required voltage division formula.

SOLUTION 7.50. By the voltage division formula,

vC 3(t) =6 ||6

6|| 6 + 9vin (t) = 21(1− e−20t ) V

SOLUTION 7.51. Part 1 By the voltage division formula,


vC1(t) =0.05 + 0.15

0.05 + 0.15 + 0.1vin ( t) = 6.667sin(120 t) V

vC 2( t) = vin (t) − vC1( t) = 3.333sin(120 t) V

Part 2 Let the 0.1µF capacitor be C1 and the 0.15µF capacitor be C2. The energies stored in the 0.1µF

and the 0.15µF capacitors are given by

WC1[0, t] =12

C1 vC12(t) − vC1

2(0)[ ] = 2.222sin 2(120 t) J

WC2[0,t] =12

C2 vC 22( t) − vC 2

2(0)[ ] = 0.8333sin2(120 t) J

SOLUTION 7.52. Part 1 As both the terminals of 0.08F are tied to the voltage source vin(t),

vC1(t) = vin ( t) =100e−2 t V

By the voltage division formula,

vC 2( t) =0.06

0.03 + 0.06vin (t) = 66.67e−2t V

Part 2 Let the 0.03-F capacitor be C2. The energy stored in the 0.03-F capacitor over the interval [0,t] isgiven by

WC2[0,t] =12

C2 vC 22( t) − vC 2

2(0)[ ] = 66.67 e−4 t −1[ ] J

SOLUTION 7.53. First consider the circuit below which contains the additional label of vin(t).

From KCL and the derivative definition of the capacitor

iin (t) = iC1(t) + iC 2(t) = C1dvin (t)

dt+ C2

dvin ( t)dt

= C1 + C2( ) dvin ( t)dt

Equivalently,dvin ( t)

dt=

1C1 + C2( ) iin (t)

It follows that

iCk(t) = Ck

dvin (t)dt

=Ck

C1 + C2iin ( t)

which is the required current division formula.


SOLUTION 7.54. Part 1 By the current division formula,

iC1(t) =0.08

0.08 + 0.03 0.06iin ( t) = 80e−2 t A

iC 2(t) =0.03 0.06

0.08 + 0.03 0.06iin (t) = 20e−2t A

Part 2 Let the 0.03-F capacitor be C2.

vC 2( t) = vC2(0) +1

C2iC 2( )d

0

t

∫ = 333.3(1− e−2t ) V

The energy stored in the 0.03-F capacitor over the interval [0,t] is given by

WC2(0, t) =12

C2 vC 22( t) − vC 2

2(0)[ ] =1666.5(1− e−2t )2 J

SOLUTION 7.55. Part 1 By the current division formula,

iC 2(t) =2

1 + 2is(t) = 20sin(250t) mA

Part 2 By the voltage division formula,

vout (t) =2

1+ 2× 9ic2( t) = 0.12sin(250t) V

Part 3 The current in the 2mH inductor is given by

iL 2(t) = iL2(0) +1L2

vout ( )d0

t

∫ = −0.24 cos(250t) −1[ ] A

The energy stored in the 2-mH inductor is given by

WL2( t) =12

L2iL 22( t) = 57.6 cos(250t) −1[ ]2 J

SOLUTION 7.56. Part 1 The dual network N* is shown below.

Part 2 Since the equivalent conductance seen by Is* is 500S, the equivalent resistance seen by Is* is equalto 0.002Ω.


SOLUTION 7.57. Part 1 The dual network N* is shown below.

Part 2 S*eq = 350S I*out = 8A.

SOLUTION 7.58. First we draw the graph, given below, of the circuit in figure P7.58. We construct thegraph associated with dual network graph for N* from the graph of N. The dual network graph is givenby the dashed lines.

This circuit and its dual have the branch characteristics given in the following table:

ORIGINAL NETWORK DUAL NETWORK

V1 = 40 V I1* = 40 A

V2 = 500 I2 I2* = 500 V2

*

I3 = 2 I2 V3* = 2 V2

*

V4 = 400I4 I4* = 400V4

*

V5 = 200 I5 I5* = 200 V5

*

Finally replace the branches by the elements given in the table above. This produces the dual networkbelow.

SOLUTION 7.59. The dual for the circuit in figure P7.59 is shown below.


SOLUTION 7.60. First we draw the graph, given below, of the circuit in figure P7.60.

This circuit and its dual have the branch characteristics given in the following table:

ORIGINAL NETWORK DUAL NETWORK

V1 = 5 I1 I1* = 5V1

*

I2 = 8 A V2* = 8 V

I3 = 6 A V3* = 6 V

V4 = 4 V I4* = 4 A

V5 = 2 I5 I5* = 2V5

*

I6 = 7 A V6* = 7 V

Now we construct the graph associated with dual network graph for N* from the graph of N. The dualnetwork graph is given by the dashed lines.


The graph of the dual network then is pulled out and flipped vertically to produce the graph topology of thedual network, N*.

Finally replace the branches by the elements given in the table above. This produces the dual networkbelow.

SOLUTION 7.61. Part 1 We first redraw the circuit to eliminate branch crossing. The resultantschematic is shown below.


Then we draw the graph, given below, of the above circuit. We construct the graph associated with dualnetwork graph for N* from the graph of N. The dual network graph is given by the dashed lines.

The resultant dual network given by the above graph is shown below.

Part 2 The equivalent resistance Req* seen by the current source in the dual circuit is equal to 1/Req = 1/αΩ.

SOLUTION 7.62. Part 1 We first redraw the circuit to eliminate branch crossing. The resultantschematic is shown below.


Then we draw the graph, given below, of the above circuit. We construct the graph associated with dualnetwork graph for N* from the graph of N. The dual network graph is given by the dashed lines.

The resultant dual network given by the above graph is shown below.

Part 2 If we label the conductances in the dual network by their corresponding resistance values, we havethe circuit below.


By comparing to the original network, we conclude that Req* = Req.

Part 3 Since the general relationship between Req and Req* is given by Req= 1/Req*. We can write

Req= 1/Req* =1/Req

Solving the above equation, we know that Req = 1 Ω.

Part 4 To retain the special properties of parts (b) and (c), we require one resistance value must be areciprocal of the other one.

SOLUTION 7.63. The dual network is shown below.


From the answer of problem 7.55,

iC2(t) = 20 sin (250t) mA, and vout(t) = 0.12 sin (250t) V

Therefore, vC2*(t) = 20 sin (250t) mV and iout* (t) = 0.12 sin (250t) A

The energy stored in the 2mF capacitor is given by

WC2* ( t) =

12

C2* vout

* (t)[ ]2=

12

L2 iL 2(t)[ ]2 = 57.6[cos(250t) −1]2 J



The equivalent inductance of the dual circuit (Leq*) = (1+2) || 1.5 + 2.5 = 3.5µH.


v1( t) =

1

0.41

0.4+

1

0.2

vs(t) =0.2

0.2 + 0.4vs(t) =

13

vs( t)

Note that the 0.4 F resistor in parallel with the dependent voltage source has no effect on i1(t) and hence isredundant. By the definition of a capacitor,

i1(t) = 0.2dv1( t)

dt=

115

×dvs( t)

dt

Finally, since the last capacitor is initially uncharged,

vout (t) =1

0.5i1( )d

0

t

∫ =215

dvs( )d

d0

t

∫ =2

15vs( t)

A sketch is given below.

SOLUTION 7.67. Following the method outlined in page 269 of the text, we require

vC (t0) −1C

i0( )d ≥140

1/200

∫With i0(τ) = 2A, vC(0) = 20 V, it follows that

20 −14 ≥1C

× 2A ×1

200s

C ≥ 1.667mFTherefore, the standard capacitor value of 1.8mF should be chosen for this application.


SOLUTION 7.68. From the relationships given in Problem 7.67, we have

vL (t) =ddt

L(t)iL( t)[ ] = I0ddt

L(t)[ ] = I0L1 41T1

−t

T12

Below is a plot of vL(t) vs t.

From the plot, we notice that when t=T1, the value of vL(t) changes from positive to negative. This meansthat vL(t) will change sign when a car is inside the loop. Therefore, we can make a circuit to monitor thevoltage vL(t) and whenever a negative voltage is detected, the left turn signal should be initiated during thenext traffic light change.

1st Order Circuit Probs 11/26/01 P8-1 © R. A. DeCarlo, P. M. Lin

SOLUTIONS PROBLEMS CHAPTER 8

SOLUTION 8.1. (a) By KCL, CdvC (t)

dt= −

vC (t)R

or dvC ( t)

dt+

vC (t)RC

= 0 . Using 8.12

vC (t) = vC (0)e−t / =10e−t V where = RC =1s . Plotting this from 0 to 5 sec

»t = 0:.05:5;

»vc = 10*exp(-t);

»plot(t,vc)

»grid

»xlabel('time in s')

»ylabel('vc(t) in V')

0 1 2 3 4 50

2

4

6

8

10

time in s

vc(t)

in V

TextEnd

(b) The solution has the same general form as for (a),iC ( t) = iC (0+)e−t / = −10

5 ×103 e−t = −2e−t mA =

−vC (t)R

.


0 1 2 3 4 5-2

-1.5

-1

-0.5

0

time in s

ic(t)

in m

A

TextEnd

(c) By linearity, if vC(0) is cut in half, all resulting responses are cut in half. If vC(0) is doubled, then all

resulting responses are doubled. Alternately, one view this as a simple change of the initial condition

with the same conclusion reached from linearity.

SOLUTION 8.2. (a) From inspection of the general form, 8.12, 0.1/ = 1= 0.1/ RC ⇒ C = 0.1/R = 5

µF.

(b) Since τ = RC = 0.1, vC (t) = 10e−10 t V.

SOLUTION 8.3. (a) The general solution form is vC (t) = vC (0)e−t / = vC (0)e−t /ReqC

. Using the

given data, take the following ratio, vC (0.001)vC (0.002)

=18.3946.7668

= 2.7183 =vC (0)e−(0.001)/

vC (0)e−(0.002)/ = e0.001/ . Hence,

»K = 18.394/6.7668

K =2.7183

»tau = 1e-3/log(K)

tau =0.0010

»C = 5e-6;

»Req = tau/C

Req =200.0008

»% Req = R*4e3/(R+4e3)


»R = Req*4e3/(4e3-Req)

R =210.5272

»vC0 = 6.7668/exp(-0.002/tau)

vC0= 49.9999 V

(b)

» % vC (t) = 50e−1000 t V

»t = 0:tau/100:5*tau;

»vc = vC0*exp(-t/tau);

»plot(t,vc)

»grid

»xlabel('Time in ms')

»ylabel('vC(t) in V')

SOLUTION 8.4. After one time constant the stored voltage, 8 V, decays to 8/e = 2.943 V. From the

graph, the time at which the output voltage is 2.94 V is approximately 0.19 s. Thus = 0.19 s, and R =

τ/C = 190 Ω.

SOLUTION 8.5. (a) The general form of the inductor current is iL ( t) = iL (0)e−t / = 0.15e−t / A where

= L / R = 2 ×10−3 s. Plotting iL ( t) = 0.15e−500 t A from 0 to 10 msec yields:

»t = 0:.01e-3:10e-3;

»iL = 0.15*exp(-t/2e-3);

»plot(t,iL)

»grid


»xlabel('Time in s')

»ylabel('Inductor Current')

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Time in s

Indu

ctor

Cur

rent

TextEnd

(b) Here vL (t) = −RiL (t) = −22.5e−500 t V.

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-25

-20

-15

-10

-5

0

Time in s

Indu

ctor

Vol

tage

in V

TextEnd

(c) Using linearity, for iL(0) = 50 mA, then vL (t) =−22.5

3e−500t = −7.5e−500t V and for iL(0) = 250

mA, then vL (t) = −22.5250150

e−500t = −37.5e−500t V.

SOLUTION 8.6. Since = L / R, we can solve for L = 5 mH. Then solving iL ( t) = iL (0)e−t / for iL(0)

at t = 4 µsec yields iL(0) =15 mA.


SOLUTION 8.7. (a) We desire to solve iL ( t) = iL (0)e−t / for iL(0) and R in = 0.08 /(R +103).

Using the following ratio, iL (0.05 ms)iL (0.15 ms)

=9.1971.2447

=iL (0)e−(0.05m)/

iL (0)e−(0.15m)/ = e0.1×10−3 / = 7.3889. Hence

»K = 9.197/1.2447

K =7.3889

»tau = 0.1e-3/log(K)

tau =5.0000e-05

»L = 0.08;

»Req = L/tau

Req =1.6000e+03

»R = Req - 1000

R = 599.9862

»iL0 = 9.197e-3/exp(-0.05e-3/tau)

iL0 = 0.0250 A

(b) = 80m /(R +1000) = 50 sec , iL ( t) = 0.025e−t /50×10−6 A.

SOLUTION 8.8. By Ohm’s law, vR(0+) = −(4k ||16k)iL (0+) = −32 V. The time constant

= L /(4 k ||6k) = 25 sec , i.e.,

»Req = (4e3*16e3/(4e3+16e3))

Req =3200

»L = 0.08;


»tau = L/Req

tau =2.5000e-05

Using the general equation, vR( t) = vR (0+)e−t / = −32e−t /25 V. Equivalently,

vR( t) = −ReqiL (t) = −Req 0.01e−4×104 t V.

SOLUTION 8.9. (a) Note that the Thevenin resistance seen by the capacitor is Rth:

»R1 = 360+60*120/(60+120)

R1 =400

»Rth = 400*1200/1600

Rth =300

Hence, vC (t) = vC (0)e−t / = 80e−t /0.15 V where = RTHC = 300 × 0.5 ×10−3 = 0.15 s.

0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

60

70

80

Cap

acito

r V

olta

ge (

V)

TextEnd

Time in s

(b) Here iC (0+) = −vC (0+ ) / RTH = −0.2667 . Therefore iC ( t) = iC (0+)e−t / = −0.2667e−t /0.15 A.

Equivalently, iC ( t) =−vC (t)

Rth= −0.2667e−t /0.15 A.

(c) By voltage division, vR(0+) = vC (0)60||120

(60||120) +160 + 200

= 8 V; for t > 0 vR( t) = 8e−t /0.15 V.


SOLUTION 8.10. First, find the Thevenin equivalent seen at the left of the inductor. Introducing a test

source in place of the inductor we obtain the following KCL equation at that node.

itest = vtest /1k + vtest − 200vtest1k

200 . Let vtest = 1 V. Then

»itest = 1e-3 + (1 - 200*1e-3)/200

itest =0.0050

»Rth = 1/itest

Rth =200

Thus Rth = 200 Ω, = L / Rth = 0.25 ms, and iL ( t) = 0.025e−4000 t A. Next from 8.13b find

vL (t) = −(RTH × iL (0))e−4000 t = −5e−4000 t V, and from Ohm’s law ix (t) = −5e−4000 t mA.

SOLUTION 8.11. For all parts it is necessary to find the Thevenin equivalent resistance seen by the

capacitor. To this end we apply an external test current to the remainder of the circuit to obtain:

vtest = R1itest + R2(itest + itest )

Thus

Rth = vtest / itest = R1 + R2(1+ ) = 120 + 70(1+ )

(a) With α = 4, Rth = 470 Ω, = RthC = 0.1175 s, and vC (t) = vC (0)e−t / = 50e−8.51t V.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

10

20

30

40

50

Time in s

Cap

acito

r V

olta

ge (

V)

TextEnd

(b) With α = -4, Rth = −90 Ω, = RthC = −0.0225 s, and vC (t) = 50e44.44t V.


0 0.02 0.04 0.06 0.08 0.1 0.120

1000

2000

3000

4000

5000

6000

7000

8000

Time in s

Cap

acito

r V

olta

ge (

V)

TextEnd

Note how this is not a stable design as V increases exponentially without bound.

(c) From the general equation developed at the beginning, Rth = 120 + 70(1+ ) > 0 requires that

> −2.7143.

SOLUTION 8.12. Find the Thevenin resistance left of the inductor. Forcing a test current source into

the output node,

vtest = R1itest + R2 itest − R1itest( ) = 100 + 50(1− 100)( )itest = 150 − 5000( )itest

and

Rth = vtest / itest =150 − 5000 Ω

(a) Using the above equation, Rth = −350 Ω and = L / Rth = −7.1429 ×10-5 s. Hence,

iL ( t) = 0.1e14000t A, an unbounded response due to the presence of the negative equivalent resistance.

(b) Rth = 100 Ω, = L / Rth = 2.5 ×10−4 s, vR(0+) = −R2(1− R1)iL(0) = 0 , but more importantly

vR2(t) = −R2(1− R1)iL( t) = 0 × iL ( t) = 0 .

(c) R1 + R2 1− R1( ) = 150 − 5000( ) > 0 implies < 0.03.

SOLUTION 8.13. Over a long period of time the inductor L, is seen as a short circuit. Thus at time 0-,

the current through the inductor is, by current division, Is/2. As such, iL ( t) = 0.5Is e−Rt /2L A. A sketch

will reveal an exponentially decreasing current from an initial 0.5Is A.


SOLUTION 8.14. This is similar to problem 8.13. Here the current turns off at time zero instead of a

switch opening. By current division iL (0+) = Is /2 . The difference between this problem and problem

8.13 is that the Thevenin resistance seen by the inductor is different. Here, RTH = 2R ||0.5R = 0.4R . So

for t > 0, iL = (Is /2)e−R tht /L = 0.5Ise−Rt /2.5L A. A sketch of this function plotted with respect to this

new time constant will be identical to the one in problem 8.13.

Define the time constant of problem 8.13 as τold. The slower decay in the plot below represents

the fall of the inductor current for problem 8.14 relative to that of problem 8.13 which is the faster

decaying curve in the plot below.


SOLUTION 8.15. Over a long period of constant applied voltage, a capacitor looks like an open circuit.

By voltage division, vC (0+ ) =34

Vs and τ = 3RC. Hence

vC (t) = vC (0+)e−t / = 0.75Vse−t /(3RC ) V

SOLUTION 8.16. Same problem as 8.15 for t < 0. For t > 0 the effective resistance changes.

Rth = 3R / /R = 0.75R and τ = 0.75RC. Thus, vC (t) = 0.75Vse−t /(0.75RC) V. Same behavior as in the

previous problem, except for a faster decay than in problem 8.15 due to a smaller effective resistance.

Note how the decreased resistance affects the RC and RL circuit differently.

SOLUTION 8.17. For t < 0 the inductor looks like a short circuit. Let R1 =1333/ /800 = 500 Ω. The

current supplied by the source is Is =12

100 + 500= 0.02 A. By current division,

iL (0+) = 0.02 ⋅1333

800 +1333= 0.0125 A

For t > 0, the switch is opened and the inductor sees only the 800 Ω resistor. Hence, = L / R = 25 µsec

and


iL ( t) = 12.5e−40000 t mA

0 20 40 60 80 1000

2

4

6

8

10

12

14

Time in micro seconds

Indu

ctor

cur

rent

in m

A

TextEnd

SOLUTION 8.18. (a) For t < 0 the applied voltage is constant and at t = 0, the capacitor is like an open

circuit. By voltage division, vC (0+ ) = 3025k ||6.25k

(25k ||6.25k) +1k= 25 V. For 0 < t < 1 ms, the source is off

and the capacitor discharges through three resistors in parallel; thus

= RthC = 833.33 × 0.6 ×10−6 = 0.5 ms and vC (t) = 25e−2000 t V.

(b) From continuity vC (0.001)= 25e−2000(0.001) = 3.383. For t > 1ms, the capacitor keeps on

discharging through only one resistance, the 25 kΩ resistor; thus the new time constant is τnew = 15 ms,

and vC (t) = 3.383e−(t−0.001)/0.015 V.

(c) PlotvC (t) = 25e−2000 t u(t) − u( t − 0.001)[ ] + 3.383e−(t−0.001)/0.015u(t − 0.001) V

»t = 0:.01:12;

»vc = 25*exp(-2*t) .*(u(t)-u(t-1))+3.3834*exp(-(t-1)/15) .*u(t-1);

»plot(t,vc)

»grid

»xlabel('Time in milli-secs')

»ylabel('Capacitor Voltage in V')


0 2 4 6 8 10 120

5

10

15

20

25

Time in milli-secs

Cap

acito

r V

olta

ge in

V

TextEnd

SOLUTION 8.19. (a) From Ohm’s law iL (0+) =54

60 + 30= 0.6 A. For t > 0, the Thevenin resistance

seen by the inductor is Rth = (60 + 30) || 720 = 80 Ω and τ = L/Rth = 1/160 s. Thus

iL ( t) = iL (0+ )e−t / = 0.6e−160t A. From Ohm’s law and current division

v(t) = −60 ×720

90 + 720× iL( t) = −32e−160 t V

(b) From continuity property, iL (0.01±) = 0.6−160(0.01) = 121.14 mA. For t > 10 ms, the Thevenin

resistance seen by the inductor is Rth = (690 + 30) || 720 = 360 Ω and τnew = L/Rth = 1/720 s. Hence,

iL ( t) = 121.14e−720(t−0.01) mA for t > 10 ms. From Ohm’s law and current division

v(t) = −690 ×720

720 + 720× iL (t) = −41.793e−720(t−0.01)u(t − 0.01) V

Therefore

iL ( t) = 0.6e−160 t u( t) − u( t − 0.01)[ ] + 0.12114e−720(t−0.01)u(t − 0.01) A


0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Time in milli-seconds

Indu

ctor

Cur

rent

in A

TextEnd

SOLUTION 8.20. For both circuits we first compute the Thevenin resistance seen to the right of the

capacitor for 0 ≤ t ≤ 60 ms. If we excite the circuit to the right of the capacitor over this time interval,

then itest =vtest200

+(1− 0.25)vtest

100= 0.0125vtest . Let Rth1 =

10.0125

= 80 Ω.

(a) For t < 0, the capacitor acts as an open circuit. Using voltage division, vC (0+ ) =80

80 +133.380 = 30

V. For 0 ≤ t ≤ 60 ms, the time constant is 1 = Rth1C = 40 ms, and

vC (t) = vC (0+ )e−t = 30e−t (Rth1C ) = 30e−25t V

From continuity, vC (60+ ms) = 6.694 V. The new Thevenin resistance is Rth2 = 200 Ω. Thus for t >

60 ms, the time constant is 2 = 200C =100 ms, and vC (t) = 6.694e−10(t−0.06) V. The resulting

capacitor voltage is plotted below.


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Cap

acito

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ge (

V)

TextEnd

(b) It is the same circuit as above for t < 0; thus vC (0+ ) = 30 V. However the Thevenin resistances seen

by the capacitor are different because there is no switch to disconnect the independent voltage source and

its series resistance. First for 0 ≤ t ≤ 60 ms, the Thevenin resistance to the right remains as

Rth1 =1

0.0125= 80 Ω. However, for 0 ≤ t ≤ 60 ms, the Thevenin resistance seen by the capacitor

changes to Rth3 = Rth1 //133.3 = 50 Ω. Then new time constant is 3 = Rth3C = 25 ms and for 0 ≤ t ≤60 ms

vC (t) = 30e−40t V

From continuity, vC (60+ ms) = 2.72 V. The new Thevenin resistance is Rth4 = 200/ /133.3 = 80 Ω.

Thus for t > 60 ms, the time constant is 4 = 80C = 40 ms, and vC (t) = 2.72e−25(t−0.06) V. The

resulting capacitor voltage is plotted below.


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time in ms

Cap

acito

r V

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ge (

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TextEnd

(c) For t < 60 ms, the voltage decays faster in (b) due to the smaller time constant. Similarly, for t > 60

ms.

SOLUTION 8.21. Following, are the switching times with the time constants associated with them.

t = 0 ⇒ Rth = 20 kΩ ⇒ = 20 ms

t = 5 ms ⇒ Rth = 4 kΩ ⇒ = 4 ms

t = 7.5 ms ⇒ Rth = 800 Ω ⇒ = 0.8 ms

It follows that with t in ms,

vC (t) = 10e−0.050t u( t) − u(t − 5)[ ] + 7.788e−0.250( t−5) u(t − 5) − u(t − 7.5)[ ] + 4.169e−1.25(t−7.5)u(t − 7.5) V


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Cap

acito

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TextEnd

*SOLUTION TO 8.22. This solution is done in MATLAB.

% Define switching times, inductance, and Thevenin equivalent resistances.

t1= 12e-6;t2=18e-6;t3=21e-6;L= 0.1;rth1= 800;rth2= 8000;rth3=1600;rth4= 32000;

% Define time constants for each of the four time intervals.

tau1= L/rth1tau2= L/rth2tau3= L/rth3tau4= L/rth4

tau1 = 1.2500e-04tau2 = 1.2500e-05tau3 = 6.2500e-05tau4 = 3.1250e-06

% Compute initial inductor currents for each of the four time intervals.

il1= 100e-3;il2=il1*exp(-t1/tau1)


il3=il2*exp(-(t2-t1)/tau2)il4=il3*exp(-(t3-t2)/tau3)

il2 = 9.0846e-02il3 = 5.6214e-02il4 = 5.3580e-02

% Determine inductor currents for each of the four time intervals. Plot.

t = 0:0.5e-7:36e-6;seg1= il1*exp(-t/tau1) .*(ustep(t)-ustep(t-t1));seg2 =il2*exp(-(t-t1)/tau2) .*(ustep(t-t1)-ustep(t-t2));seg3=il3*exp(-(t-t2)/tau3) .*(ustep(t-t2)-ustep(t-t3));seg4= il4*exp(-(t-t3)/tau4) .* ustep(t-t3);iL=seg1 + seg2 +seg3 + seg4;

plot(t,iL)grid

SOLUTION 8.23.

For circuits with a forced voltage, equation 8.19c is used as a general solution,

vC (t) = vC (∞) + vC to+( ) − vC ∞( )[ ]e−

t

RTHC

.

(a) At time zero the voltage is 0 V. As time approaches infinity, the capacitor looks like an open withvoltage 10 V. The Thevenin resistance is 10 kΩ. Thus for t > 0.


vC (t) = 10 + −10[ ]e−

t

2

=10(1− e−0.5t ) V.

(b) With vin(t) = 0 and vC (0+ ) = 5 V, vC (t) = 5e−

t

2

= 5e−0.5t V.

Part (a) Part (b)

(c) From linearity, vC (t) = 10(1− e−0.5t ) + 5e−0.5t = 10 − 5e−0.5t V. Using Ohm’s law,

iC ( t) =vin (t)

104 −vC ( t)

104 . Thus iC ( t) = 1− 1− 0.5e−0.5t( ) = 0.5e−0.5t mA.

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Cap

acito

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urre

nt (

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)

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(d) This is the same as (a), under the condition that the input is 1.5 times larger. Hence by linearity,

vC (t) = 1.5 ×10(1− e−0.5t ) = 15(1− e−0.5t ) V.

(e) By linearity, ANSWER = –2x(ANSWER to (b)) + 3x(ANSWER to (a)):


vC (t) = −2 × 5e−0.5t + 3×10(1− e−0.5t ) =15(1− e−0.5t ) = 30 − 40e−0.5t V

SOLUTION 8.24.(a) At t = 0-, the capacitor looks like an open circuit; therefore, by voltage division and the continuity

property, vC (0− ) = vC (0+ ) =3R4R

Vs1 = 0.75Vs1 . Similarly, at t = ∞, vC (∞) =3R4R

Vs2 = 0.75Vs2 . The

circuit time constant is = (3R / /R)C = 0.75RC . Hence

vC (t) = 0.75Vs2 + 0.75 Vs1 −Vs2[ ]e−

t

0.75RC

.

(b) A sketch will show an exponentially varying voltage from 0.75Vs1 converging to 0.75Vs2 with thecomputed time constant.(c) The response to the initial condition when the inputs are set to zero, zero-input response, is

vC (t) = Vs1e−

t

0.75RC

. The zero order response, the response with 0V initial condition to a forced

voltage, is vC (t) = Vs2 −Vs2e−

t

0.75RC

.

SOLUTION 8.25.For RL circuits with a forced current, equation 8.19b is used as a general solution:

iL ( t) = iL (∞) + iL to+( ) − iL ∞( )[ ]e−R th

L(t−t o)

.

Since Rth = R = 100 Ω and L = 0.4 H, we have = 4 ms and

iL ( t) = iL (∞) + iL to+( ) − iL ∞( )[ ]e−250(t−to )

(a) Here, iL (0) = 0 and as time approach infinity, the inductor becomes a short and

iL (∞) =10 100 = 0.1 A. Thus iL ( t) = 0.11− e−250 t( ) A.

(b) Here iL (0) = −50 mA and because the input is zero, iL (∞) = 0 . Thus, iL ( t) = −0.05e−250t A. Plotsfor parts (a) and (b) appear below.


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)

TextEnd

(c) By linearity iL ( t) = 0.11− e−250 t( ) − 0.05e−250t = 0.1− 0.15e−250t A. Further, by KVL and Ohm's

law, vL (t) = vin ( t) −100iL ( t) implies vL (t) = 10 −10 +15e−250t = 15e−250t V.

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)

TextEnd

(d) Observe that the new initial condition is –0.5 times the old one and that the new input voltage is 1.5times the old one. Therefore, by linearity,

iL ( t) = 1.5 × 0.11− e−250 t( ) + (−0.5) × −0.05e−250t( ) = 0.15 − 0.125e−250 t

and thus

vL (t) = 15 −15 +12.5e−250t = 12.5e−250t V

The plot is similar to part (c) with initial point 12.5 instead of 15.

SOLUTION 8.26. For this problem Rth = 2R / /0.5R = 0.4R in which case = L Rth = L 0.4R .


(a) At t = 0, the inductor looks like a short circuit. Hence by current division, iL (0−) = iL (0+ ) = 0.5Is1 .A similar argument yields iL (∞) = 0.5Is2 . Using the general form of the solution,

iL ( t) = 0.5Is2 + 0.5 Is1 − Is2[ ]e− 0.4Rt

L .(b) A sketch will show an exponentially varying current from 0.5Is1 A converging to 0.5Is2 .(c) The response to the initial condition when the inputs are set to zero, zero-input response, is

iL ( t) = 0.5Is1e− 0.4Rt

L . The zero state response, the response with no initial condition, to the input

Is2u(t), is iL ( t) = Is2 1− e− 0.4Rt

L

.

SOLUTION 8.27. For this problem, vC (0) = 20 −10e−0.4t( )t=0

=10 V and

vC (∞) =10Is = 20 =t→∞lim 20 −10e−0.4t( ) . Hence Is = 2 A. Further,

iC ( t) = CdvC (t)

dt= 4Ce−0.4t = 0.4e−0.4t which implies that C = 0.1 F. Since

=1/0.4 = 2.5 = (10+ R)C = 0.1(10 + R) , it follows that R = 15 Ω.

SOLUTION 8.28.(a) The Thevenin resistance for this configuration is Rth = 1000/ /1000 = 500 Ω and = RthC = 0.25 s.

Hence vC (t) = vC (0+ )e−t =15e−4 t V is the zero-input response.

(b) Using a source transformation and voltage division, vC (∞) = 3 V. Thus vC (t) = 3 1− e−4 t( ) V.

(c) Here vC (∞) = 4 V, thus vC (t) = 4 1− e−4 t( ) V.

(d) This is the superposition of parts (b) and (c), i.e., vC (t) = 7 1− e−4t( )(e) The complete response is the superposition of parts (d) and (a), i.e., vC (t) = 7 + 8e−4 t V.(f) From linearity,

vC (t) = 0.5 × 7 1− e−4t( ) + 2 ×15e−4 t = 3.5 + 26.5e−4 t

SOLUTION 8.29. Solution done in MATLAB

%Problem 8.29

%RTH= (60||120)+120%tau=L/RTH

RTH=1/(1/60+1/120)+120;tau=0.2/RTH;

%Using superposition at t<0il01=36/120/2;il02=(72/180)*(60/180);il0= il01+il02;


% At t>0 36volts is off thusilinf=il02;

t=0:5*tau/1000:5*tau;ilt=ilinf+(il0-ilinf)*exp(-t/tau);

plot(t,1000*ilt);xlabel('Time in seconds');ylabel('Current in mA');

0 1 2 3 4 5 6 7x 10-3

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Curr

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A

*SOLUTION 8.30.

» % Rth = 120 + 120\\60 = 160 kohm» % tau = Rth*C»tau = 160e3*0.5e-6tau = 0.0800» % Initial capacitor voltage is computed by» % voltage division and superposition»vc0 = 24+24vc0 = 48» % At t = ∞, capacitor looks like an open circuit. Hence»vcinf=24;»t = 0:1e-3:5*tau;»vct = vcinf+(vc0-vcinf)*exp(-t/tau);»plot(t*1000,vct)»grid»xlabel('Time in msec')»ylabel('Capacitor voltage (V)')


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Cap

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)

*SOLUTION 8.31A

»vc0 =0;»% Consider 0 ≤ t ≤ 0.5»vcinf = 50;»% Rth = 600\\300 = 200 ohms»% tau1 = Rth*C»tau1 = 300e-6*200»vc0 = 0;»t = 0:5e-3:1;»vct = (vcinf+(vc0-vcinf)*exp(-t/tau1)) .* (ustep(t)-ustep(t-.5));»»% Consider 0.5 ≤ t ≤ 1»tau2 = tau1»vc5 = (vcinf+(vc0-vcinf)*exp(-.5/tau1))»vcinf2 = 80;»vct2 = (vcinf2+(vc5-vcinf2)*exp(-(t-0.5)/tau1)) .* ustep(t-0.5);»vca = vct+vct2;»plot(t,vca)»grid»xlabel('Time in sec')»ylabel('Capacitor Voltage (V)')


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Cap

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olta

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*SOLUTION 8.31B

»vc0 =0;»% for 0 ≤ t ≤ 0.5»vcinf = 75;»tau1 = 300e-6*300»vc0 = 0;»t = 0:5e-3:1;»vct = (vcinf+(vc0-vcinf)*exp(-t/tau1)) .* (ustep(t)-ustep(t-.5));»% for 0.5 ≤ t ≤ 1»tau2 = 300e-6*200»vc5 = (vcinf+(vc0-vcinf)*exp(-.5/tau1))»vcinf2 = 80;»vct2 = (vcinf2+(vc5-vcinf2)*exp(-(t-0.5)/tau1)) .* ustep(t-0.5);»vcb = vct+vct2;»plot(t,vcb)»grid»xlabel('Time in sec')»ylabel('Capacitor Voltage (V)')»pause»plot(t,vca,t,vcb,'b')»grid


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Cap

acito

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olta

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Comparison of the two responses.

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SOLUTION 8.32.

%Problem 8.32

%Consider t < 0%vin=-20V, thusvc0=(8/10)*(-20);

%For 0<t<20msRTH=1/(1/2e3+1/8e3);tau1=5e-6*RTH;


vcinf=(8/10)*20;t=20e-3;vc20ms=vcinf+(vc0-vcinf)*exp(-t/tau1);

%For 20ms<tRTH=1/(1/2e3+1/8e3+1/1.6e3);tau2=5e-6*RTH;

%By superpositionvcinf2=20*0.5+20*0.4;t=0:(40e-3)/1000:40e-3;

vct= (ustep(t)-ustep(t-20e-3)).*(vcinf+(vc0-vcinf).*exp(-t./tau1)) ...+ ustep(t-20e-3).*(vcinf2+(vc20ms-vcinf2).*exp(-(t-0.02)./tau2));

plot(t,vct);grid;xlabel('time in seconds');ylabel('Volts');

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-15

-10

-5

0

5

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20

time in seconds

Volts

SOLUTION 8.33.(a)%Problem 8.33%(a)

%at t<0 only one source is contributing thus,il0=24/60*0.5;%For t>0RTH=60+1/(1/30+1/60);tau=16e-3/RTH;

%As t goes to infinity, by superposition,ilinf=24/(60)*0.5+24/80*1/3;t=0:5*tau/1000:5*tau;ilt=ilinf+(il0-ilinf).*exp(-t/tau);figure(1);plot(t,1000.*ilt);grid;xlabel('time in seconds');ylabel('Current in mA');


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(b) Using 8.23, The time constant remains the same as in (a). In order to find the inductor voltage, wemust do so indirectly by solving for vL (t) = VA − 60IL . At t=0+,

iL (0+) = 200mA

VA = 246090

+ 243090

− iL (0+ )(60|| 30) = 20V

vL (0+ ) = x( t0+ ) = VA − 60iL (0+ ) = 8V

For t > 0, X e = vL (∞) = 0V , Thus vL (t) = 8e−5000 tV .

(c) By linearity,

iL ( t) = 600 − 200e−5000 tmA

vL (t) = 16e−5000 tV

VA

SOLUTION 8.34.

(a) Since the voltage has been constant for a long time, the capacitor acts as an open circuit. Thus by

voltage division and continuity, vC (0− ) = vC (0+ ) = 0.75Vo .(b) RTH = 6R ||18R || 3R =1.8R .

(c) For that period of time the switch is closed, vC (t) = 0.9Vo − 0.15Voe−t /1.8RC .

(d) Using the previous equation, and by continuity, vC (T− ) = vC (T + ) = 0.9Vo − 0.15Voe−T /1.8RC .(e) The time constant remains the same as the only difference is the source turning off.

(f) For t > T, vC (t) = vC (T + )e−(t−T )/1.8RC .(g)


0 2 4 6 8 10 12 14 16 180

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Normalized time in terms of tau

Norm

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o

SOLUTION 8.35.

(a) For t < 0, The switch is closed, the current source off, and the voltage source has been providing a

constant voltage for a prolonged period of time. Thus, vC (0− ) = −50uV .

(b) Since voltage in continuous across a capacitor, vC (0+ ) = vC (0− ) = −50uV .(c) The thevenin resistance seen by the capacitor is RTH = 200||200 =100Ω , thus = RTHC = 0.2s .

(d) As t goes to infinity the capacitor voltage goes to 16V, thus vC (t) = 16 + (−50u −16)e−5tV .

(e) Again using the continuous property of a capacitor, vC (0.5+ ) = vC (0.5− ) = 14.687V .(f) For t > 0.5s, the switch is open, thus RTH = 200Ω . = RTHC = 0.4s .

(g) As t goes to infinity, the capacitor voltage goes to 32V. vC (t) = 32 + (14.687− 32)e−2.5(t−0.5)V .(h)

0 0.5 1 1.5 2 2.5-5

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Time in seconds

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SOLUTION 8.36.

(a) vC (0+ ) = −5V(b) Doing so in matlab.%Problem 8.36b

%Initial conditionvc0=-5;%From 0<t<80usRTH=300e3;tau=RTH*(1/3)*1e-9;


%So the initial condition on Voutvout0=(10+5)/300e3*60e3-5; %-2Vvoutinf1=10;t=80e-6;vout80us= 10+(vout0-10)*exp(-t/tau);

%For t > 80us%tau stays the samevoutinf2=-5;

t=0:160e-6/1000:160e-6;vout= (ustep(t)-ustep(t-80e-6)).*(10+(vout0-10).*exp(-t./tau)) ...+ustep(t-80e-6).*(-5+(vout80us+5).*exp(-(t-80e-6)./tau));plot(t,vout);xlabel('time in seconds');ylabel('Volts');grid;%Notice that Vout is vin-vct times a constant 60/300 plus vct%Thus Vout=vin(60/300)+vct(1-60/300)

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Vo

lts

(c)


(d)

SOLUTION 8.37.


If the source voltage has been –10 V for a long time then the switch is open and vC (0+ ) = −10V . Thetime constant with the switch open is = 5us . At t > 0, the input voltage changes to 20V. It then followsthat vC (∞) = 20V , and

vC (t) = 20 − 30e−t /5uV . Using the elapsed time formula, 8.24, we wish to find when the switch closes.

ta − 0 = 5us ⋅ln−10 − 200 − 20

= 2.03us . At that time the input voltage is still 20V, and the switch closes.

The time constant is now = 98ns , and vC (∞) = 0.39V . Note that because the voltage converges to avalue greater than zero, this time interval will be from 2.03us to 5us when the input changes back to–10V, thus

vC (t) = 0.39 − 0.39e−t /98nV .vC (5us) = 0.39V . At t > 5us, the voltage changes to –10 V, so

vC (t) = −0.2 + (0.39 + 0.2)e−t /98nV .Using the elapsed formula, we get tb = 0.1us for the voltage to go down to 0V again and cause the switchto open again. At this point the time constant becomes the original value again and

vC (t) = −10 +10e−t /5uV for t > (0.1+5) us.

SOLUTION 8.38.

(a) Introduce a test current source at the output and write KVL,

itest = vtest /200 + (vtest − 6V ) /200 + (6V − vtest ) / 400. Solving for itest = vtest3

400

− 0.015 . This

implies the following,RTH = 400 /3ΩiSC = 15mA

vOC = RTH iSC = 2V

(b) Using the general form, vC (t) = 2 − 8e−15000 t .

SOLUTION 8.39.

(a) Introduce a test voltage and solve for KVL,vtest = 5kitest +101(vtest +1− 40itest) −1 + 40itest

vtest = −10itest −1

vOC = −1V

RTH = −10Ω(b) The complete response is vC (t) = −1 + et V. Note that the voltage goes to infinity as t goes to infinitybecause of the negative time constant.

SOLUTION 8.40.

Compute the thevenin equivalent seen by the inductor at t > 0. Using KCL write,itest = ⋅100 ⋅ itest + (vtest −100itest + 25) /50 . Then one obtains the following,


itest = vtest /125 +1/5 A

RTH = 125ΩiSC = −1/5 A

vOC = −25VA t < 0, the applied voltage has been –25V for a long time. Using the previously obtained thevenin

equivalent and linearity, iL (0−) = 1/5 A . iL (∞) = iSC = −1/5 A , So iL ( t) = 0.2 + 0.4e−6250 tA .

SOLUTION 8.41.(a) Introducing a test source and using KCL,

itest =vtest −1.5vs

40+

vtest

100

0.2 =vs

40+

1.5vs − vtest

40

Solving for the test source current in terms of the test voltage, itest =vtest50

− 0.12 A. Thus the thevenin

equivalent is,Rth = 50Ωvoc = isc * Rth = 6V

(b) vC (t) = 6 − 6e−200 t V.

(c) From linearity all the currents increase by the same ratio, thus vC (t) = 15 −15e−200 t V.(d)

(e)


SOLUTION 8.42.

%Problem 8.42C= 1e-6;vc0=0;

%For 0 < t < 5msRth= 20e3;tau1= Rth*C;vcinf1= 50e-3*Rth;vc5ms= vcinf1+(vc0-vcinf1)*exp(-5e-3/tau1);

%For 5ms < t < 7.5msRth= 4e3;tau2= Rth*C;vcinf2= 50e-3*Rth;vc75ms= vcinf2+(vc5ms-vcinf2)*exp(-(7.5e-3-5e-3)/tau2);

%For t > 7.5msRth=800;tau3= Rth*C;vcinf3= 50e-3*Rth;t=0:12e-3/1000:12e-3;vct= (ustep(t)-ustep(t-5e-3)).*(vcinf1+(vc0-vcinf1).*exp(-t/tau1)) ...+ (ustep(t-5e-3)-ustep(t-7.5e-3)).*(vcinf2+(vc5ms-vcinf2).*exp(-(t-5e-3)/tau2)) ...+ (ustep(t-7.5e-3)).*(vcinf3+(vc75ms-vcinf3).*exp(-(t-7.5e-3)/tau3));plot(t,vct);grid;


xlabel('time in seconds');ylabel('Volts');

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100

150

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250

time in seconds

Vol

ts

SOLUTION 8.43.

(a) From the thevenin resistance RTH = 800Ω , 1 =125 s .(b) From the thevenin resistance RTH = 8kΩ , 2 = 12.5 s .(c) From the thevenin resistance RTH = 1.6kΩ , 3 = 62.5 s .(d) From the thevenin resistance RTH = 32kΩ, 4 = 3.125 s .(e) 0 mA.(f) In MATLAB:%Problem 8.43f

tau1= 125e-6;tau2= 12.5e-6;tau3= 62.5e-6;tau4= 3.125e-6;vs= 100;

il0= 0;ilinf1= vs/800;ilinf2= vs/8e3;ilinf3= vs/1.6e3;ilinf4= vs/32e3;

il12us= ilinf1+(il0-ilinf1)*exp(-12e-6/tau1);il18us= ilinf2+(il12us-ilinf2)*exp(-(18e-6-12e-6)/tau2);il21us= ilinf3+(il18us-ilinf3)*exp(-(21e-6-18e-6)/tau3);

t= 0:36e-6/1000:36e-6;ilt= (ustep(t)-ustep(t-12e-6)).*(ilinf1+(il0-ilinf1).*exp(-t/tau1))+ ...(ustep(t-12e-6)-ustep(t-18e-6)).*(ilinf2+(il12us-ilinf2).*exp(-(t-12e-6)/tau2))+ ...(ustep(t-18e-6)-ustep(t-21e-6)).*(ilinf3+(il18us-ilinf3).*exp(-(t-18e-6)/tau3))+ ...(ustep(t-21e-6)).*(ilinf4+(il21us-ilinf4).*exp(-(t-21e-6)/tau4));

plot(t,1000*ilt);grid;xlabel('time in seconds');ylabel('Current in mA');


0 0.5 1 1.5 2 2.5 3 3.5 4x 10-5

0

5

10

15

time in secondsC

urr

ent i

n m

A

SOLUTION 8.44.

The first stage is a differentiator, and from 8.25, the output of the first op-amp is

= −RCdvin (t)

dt= 0.25RCe−0.25t . The second stage is an integrator and using 8.26,

vout (t) =−1

2RC(0.25RCe−0.25 d )

0

t∫ = 0.5 e−0.25[ ]0

t= 0.5e−0.25t

SOLUTION 8.45.

(a) First note the following relationships,

vout (t) = vC 2(t) =1

C2iC 2( )d

0

t∫

vin (t) = vC1( t)

iC 2(t) = −vin (t) / R − C1dvC1(t)

dt= − vin ( t)

R− C1

dvin (t)dt

Doing the appropriate substitution, and solving,

vout (t) = −1

C2Rvin ( )d

0

t∫ −

C1C2

vin ( t) +C1C2

vin (0).

(b) 4

C2Re−0.25t −1( ) +

C1C2

1− e−0.25t( ) V.

(c) −1C2R

sin( t) +C1C2

1− cos( t)( ) V.

SOLUTION 8.46.

These are two integrator in cascade. Using 8.26, the output of the first stage is

−1

RCvin ( )d = 2 cos(50 )[ ]0

t mV = 2cos(50 t) − 20

t∫ mV. Using the same equation again,

vout (t) = −1

RC(2cos(50 )d = −10 −

250

sin(50 )

0

t

mV = 2cos(50t)mV0

t∫


*SOLUTION 8.47. The following solution is done in MATLAB

c= 1e-6;rf= 10e6;rs=1e6;tau=c*rf;vgain= -10/1;vofinal= -3*vgain;

% Part (a)

voinit= 0;t= 0: tau/100: tau;vout= vofinal + (voinit -vofinal).*exp(-t/tau);

% Time at which output voltage reaches saturation is tsat

tsat= tau*log((0 - vofinal)/(15- vofinal))

plot (t,vout)gridxlabel('Time in secs')ylabel('Output Voltage in volts')

tsat = 6.9315e+00

% Part (b)

voinit=-5;vout= vofinal + (voinit -vofinal).*exp(-t/tau);tsat= tau*log((-5 - vofinal)/(15- vofinal))plot (t,vout)gridxlabel('Time in secs')


ylabel('Output Voltage in volts')

tsat = 8.4730e+00

(c) vC(0-) = 0. Observe saturation at about 6.4 seconds.

V(IVM)

Time (s)Leaky Integrator-Transient-7

(V)

+0.000e+000

+5.000

+10.000

+15.000

+0.000e+000 +2.000 +4.000 +6.000 +8.000 +10.000

vC(0-) = -5 V. Observe that the time of saturation is 8.198 seconds.


V(IVM)

Time (s)Prb Sol 8.47-Transient-16

(V)

-5.000

+0.000e+000

+5.000

+10.000

+15.000

+0.000e+000 +2.000 +4.000 +6.000 +8.000 +10.000

Note that in both cases the time of saturation is much lower than in the MATLAB computations which

assume an ideal op amp. In the Burr Brown model used by the SPICE simulation of this circuit, the

input resistance is 2 MΩ which is comparable with the external input resistance. Hence the assumption

of an infinite input resistance is not valid for the SPICE simulation and causes the discrepancy in the

time of saturation. However, if the external input resistance is changed to 10 kΩ and the feedback

resistance to 100 kΩ with a corresponding change in the capacitor to 100 µF, one obtains results

comparable to the MATLAB computations.

SOLUTION 8.48.Since the op-amps do not load the first stage of the circuit, we can find its transfer function for the op-

amp stage as vout (t)

v + (t)= (1+ K) .

(a) From the problem statement, we know that the overall function, is a scaled integrator. As the op-amp

stage only provides gain it is logical to assume that the R-C stage will perform the integration of the input

times some constant, G. With this in mind we have

vout (t) = G(1+ K) 10sin( )d =−10G(1+ K)

cos( t) −1( )0

t∫ , where G must be negative.

(b) Using the same reasoning, vout (t) = G(1+ K)d(10sin( t))

dt= 10G(1+ K) cos( t)where G is

positive.

For low frequency (a) yields a big output, while (b) a small one. For high frequency the reverse happens.


SOLUTION 8.49.

For (b) the integral i-v relationship is

v(t) = vCeq(0+ ) +1

Ceqi(

0

t∫ )d = vC1(0+ ) + vC 2(0+ ) +

C1 + C2C1C2

i( )d0

t∫ . Repeating the same for (a),

vC1(t) = vC1(0+ ) +1

C1i( )d

0

t∫

vC 2( t) = vC2(0+ ) +1

C2i( )d

0

t∫

By KVL the two capacitor voltage can be added together, thus give the same relationship as for (b).

SOLUTION 8.50.

First calculate Ceq = C1 ||C2 = 0.2F . Then find the initial voltage

vCeq(0+ ) = vC1(0+ ) + vC 2(0+ ) = 30V . The final voltage will be 12V, and the time constant is

= Req ⋅Ceq = 0.4s . Thus vout (t) = 12 + (30 −12)e−2.5tV .

*SOLUTION 8.51. (a) After the switch closes, we have the circuit shown below.

From Chapter 7, Ceq = 0.5 F and vC (0+ ) = vC1(0+ ) − vC 2(0+ ) = 2 − 0 = 2 V. Hence,

iR (t) = iR(0+ )e−t / = iR (0+ )e−t /RC eq =

vC (0+ )R

e−t /RCeq = 4e−4 tu( t) A

(b) For this part we apply the integral definition of the capacitor. Specifically,


vC1(t) = vC1(0+ ) +1

C1iC1( ) d

0

t

∫ = 2 − iR( ) d0

t

∫ = 2 − 4 e−4 d0

t

∫ = 1+ e−4 t V

and

vC 2( t) = vC2(0+ ) +1

C2iC 2( ) d

0

t

∫ = 0 + iR ( )d0

t

∫ = 4 e−4 d0

t

∫ =1− e−4 t V

(c) The energy stored at t = 0+ for each capacitor is:

WC1(0+ ) = 0.5C1vC12 (0+ ) = 2 J

and

WC2(0+ ) = 0.5C2vC 22 (0+ ) = 0 J

Further at t = ∞,

WC1(∞) = 0.5C1vC12 (∞) = 0.5 J

and

WC2(∞) = 0.5C2vC 22 (∞) = 0.5 J

Computing total instantaneous stored energies, we have

WCtot(0+ ) = 2 J and WCtot(∞) = 1 J

Hence the decrease in stored energy from 0+ to ∞ is 1 J.

(d) Computing the energy dissipated in the resistor over [0+, ∞) is

WR(0,∞) = R iR2 ( )d

0

∞

∫ = RvC

2 (0+ )

R2

e−2t /RCeq d

0

∞

∫ =vC

2 (0+ )R

×RCeq

−2× e

−2 t /RCeq ]0

∞

=CeqvC

2 (0+ )

2= 1 J

(e) From the expressions developed in part (d), the dissipated energy is independent of the value of R.

R only affects the rate at which energy is dissipated. Clearly, the energy stored at 0 is 2 J while

the energy dissipated over [0,∞) is 1 J and the remaining energy at t = ∞ is 1 J. Hence

conservation of energy is verified.


SOLUTION 8.52. (a) Using the relations developed in P8.49,

Ceq = 0.2F

vCeq =1− 0.5 = 0.5V

iR (t) = iR(0+ )e−t / = iR (0+ )e−t /RC eq =

vCeq(0+ )

Re

−t /RC eq = e−10 tu(t)

(b) For this part we apply the integral definition of the capacitor. Specifically,

vC1(t) = vC1(0+ ) +1

C1iC1( ) d

0

t

∫ =1− iR( ) d0

t

∫ =1− e−10 d0

t

∫ =1

10(9 + e−10 t ) V

and

vC 2( t) = vC2(0+ ) +1

C2iC 2( ) d

0

t

∫ = 0.5 + 4 iR ( )d0

t

∫ = 0.5 + 4 e−10 d0

t

∫ =1

10(9 − 4e−10t ) V

(c) The energy stored at t = 0+ for each capacitor is:

WC1(0+ ) = 0.5C1vC12 (0+ ) = 0.5 J

and

WC2(0+ ) = 0.5C2vC 22 (0+ ) = 31.25mJ

Further at t = ∞,

WC1(∞) = 0.5C1vC12 (∞) = 405mJ

and

WC2(∞) = 0.5C2vC 22 (∞) =101.25mJ

Computing total instantaneous stored energies, we have

WCtot(0+ ) = 531.25mJ and WCtot(∞) = 506.25mJ

Hence the decrease in stored energy from 0+ to ∞ is 25 mJ.

(d) Computing the energy dissipated in the resistor over [0+, ∞) is


WR(0,∞) = R iR2 ( )d

0

∞

∫ = RvCeq

2 (0+ )

R2

e

−2 t /RCeq d0

∞

∫ =vCeq

2 (0+ )

R×

RCeq

−2× e

−2 t /RCeq ]0

∞

=CeqvCeq

2 (0+ )

2= 25mJ

(e) From the expressions developed in part (d), the dissipated energy is independent of the value of R.

R only affects the rate at which energy is dissipated. Clearly, the energy stored at 0 is 531.25 mJ

while the energy dissipated over [0,∞) is 506.25 mJ and the remaining energy at t = ∞ is 25 mJ.

Hence conservation of energy is verified.

SOLUTION 8.53. As all the switches are open initially, the initial current through the inductors is 0A.

For 0 < t < 50 ms, iL ( t) = 54.54 − 54.54e−20000 t mV. At t > 50 ms, the equivalent inductance is 10 mH,

the initial current through the 110 mH inductance is 54.54 mA, and through the 11 mH inductance 0 A.

So assuming the current splits equally between the two branches in steady state,

iL1 = 27.27 + (54.54 − 27.27)e−220000 t

iL 2 = 27.27 − 27.27e−220000t

SOLUTION 8.54. (a) Charges will distribute in order to achieve equal voltage by KVL. Since q=CV,

vR(0−) = 0V , due to equal capacitance the charges will distribute half and half, vR(0+ ) = 0.5V .

(b) The equivalent capacitance is 2 F, thus vR( t) = 0.5e−0.5t .

*SOLUTION 8.55. (a) Writing a node equation at v we have for all t,

4dvdt

+v4

+ 4ddt

v − vs( ) +v − vs( )

2= 0 (*)

Equivalently,

8dvdt

= −3v4

+ 4dvsdt

+vs2

(**)

Grouping terms and dividing by 8 yields when t > 0,

dvdt

= −332

v +116

(***)


Notice that vs = 1 for t > 0 and for t > 0, dvs( t)

dt=

du(t)dt

= 0.

(b) By inspection v(0-) = 0; both capacitors are uncharged at 0

-. Recall from part (a) that KCL at the

node for v yields (*) which is equivalent to (**). Since conservation of charge follows by

integrating (*) or equivalently integrating (**) we have

8dvd

0−

0+

∫ d = −34

v

0−

0+

∫ d + 4dvsd

0−

0+

∫ d + 0.5 vs

0−

0+

∫ d

Since the integral of a finite integrand over an infinitesimal interval is zero, we have equivalently,

8dvd

0−

0+

∫ d = 0 + 4dvsd

0−

0+

∫ d + 0

Evaluating these integrals we obtain

8 v(0+ ) − v(0−)( ) = 4 vs(0+ ) − vs(0− )( ) = 4u(0+ ) = 4

(c) Since v(0-) = 0, v(0

+) = 0.5 V. Since v satisfies (***), i.e.,

dvdt

= −332

v +116

≡ −1

v + F (***)

from equation 8.17,

v(t) = F + v(0+ ) − F[ ]e−t / =23

−16

e−3t /32

u(t) V

Using v(0-) would have led to an incorrect answer.

SOLUTION 8.56. (a)

x(t) = K1e−t / + K2

dx( t)

dt= −

K1 e−t /

doing the substitution, −K1 e−t / = −

1(K1e−t / + K2) + F . In order to satisfy the equality, K2 = F .


(b) x(t0+ ) = K1e

−t 0 / + F , and K1 = x( t0+ ) − F[ ]et 0 / .

(c) x(t) = K1e−t / + K2 = x( t0+ ) − F[ ]e−(t−t0 )/ + F .

SOLUTION 8.57. (a) From the graph, the initial and final values are 0 and 80 V respectively. That sets

the following constraint, 100R2

R1 + R2= 80V . From vC ( ) = 80 − 80e−1 = 50.57V . Looking at the Graph

= 5ms . Thus R1 ||R2 ⋅C = 5ms . Solving, R1 = 6250Ω and R2 = 25kΩ .

(b) Using the same equalities, R1 = 0.25R2 = 2kΩ , and C =2k || 8K

= 3.125uF .

SOLUTION 8.58. (a) From the graph, the initial and final values are 0 and 100 mA respectively. Thus

R1 = 200 /100m = 2kΩ. From iL ( ) = 100 −100e−1mA = 63.21mA , the graph shows a = 20ms . Thus

L /(R1 || R2) = 20ms, and R2 = 0.25mΩ.

(b) R1 stays the same, L = 20ms(R1 || R2) = 20H .

SOLUTION 8.59. This question is done in matlab

%Problem 8.59

tau1= 20*1;

tau2= 1;

%For 0 < t < ta

vo= 0;

vinfa= 10;

%Using the elapsed time formula,

ta=tau1*log((0-10)/(9-10));

%For ta < t < tb

vinfb=0;

tb=tau2*log((9-0)/(1-0));

%For tb < t < tc

vinfc=vinfa;

tc=tau1*log((1-10)/(9-10));


%Next switching is just a repeat of ta < t < tb

t1=ta;

t2=ta+tb;

t3=t2+tc;

t4=t3+tb;

t5=t4+tc;

t=0:t5/1000:t5-1/1000;

vt= (ustep(t)-ustep(t-t1)).*(10-10.*exp(-t/tau1))+ ...

(ustep(t-t1)-ustep(t-t2)).*(9.*exp(-(t-t1)/tau2))+ ...

(ustep(t-t2)-ustep(t-t3)).*(10-9.*exp(-(t-t2)/tau1))+ ...

(ustep(t-t3)-ustep(t-t4)).*(9.*exp(-(t-t3)/tau2))+ ...

(ustep(t-t4)-ustep(t-t5)).*(10-9.*exp(-(t-t4)/tau1));

Frequency= 1/(tb+tc)

plot(t,vt);

grid;

xlabel('time in seconds');

ylabel('Volts');

(b) The frequency is 0.0217 Hz.

0 20 40 60 80 100 120 1400

1

2

3

4

5

6

7

8

9

time in seconds

Volts

SOLUTION 8.60. When the switch is in position A, =18.18ms . In position B it is = 99.5us . Using

the elapsed time formula, find ta, when the output voltage reaches 90 V.


ta =18.18msln60 −136.3690 −136.36

= 9.07ms . At this point the switch goes to B, and the elapsed time until the

voltage reaches 60 V is tb = 99.5usln90 − 59.4560 − 59.45

= 0.4ms . Adding both time, F =105.6Hz .

SOLUTION 8.61. (a) The circuit can be rearranged in a series of one Vsolar V voltage source, one Lstore

inductor, and one Rstore+Rsolar resistor.

(b) iL ( t) =Vsolar

Rsolar + Rstore1− e−(Rstore+Rsolar )t /L[ ]A .

(c) In this time period the circuit reduces to an Lsotre inductor in series with a Rstore+R1 resistor.

(d)

iL (T1−) =

Vsolar

Rsolar + Rstore

iL ( t) = iT1

− e−(Rstore+ R1)(t−T 1)/L A.

(e) The two elements in series are an Lstore inductor and a resistor Req = Rstore + (R1 || R2).

(f)

iL (T2−) = i

T1− e−(Rstore+ R1)(T2−T 1)/Lsotre A

iL ( t) = iT2

− e−(Re q)(t−T 2)/LstoreA

(g)

PVsolar = VsolariL (t) =V 2

solar

Rsolar + Rstore1− e−(Rstore+Rsolar)t /L[ ]W

PRsolar = RsolariL2( t)W

PRstore = RstoreiL2( t)W

PLstore = LstoreiL ( t)diL (t)

dtW

(h) WL (0,t) =12

LiL2(t)J

SOLUTION 8.62. The light turns off when the current through it goes down to 0.5 mA. This

corresponds to ib = 10uA , and a voltage across the capacitor of vC = ib(R1 + 2k) + 0.5 = 0.7V . The time

constant of this circuit is = (R1 + 2k) || 5k ⋅1000uF = 4s. The final voltage across the capacitor is by

voltage division, 0.1V. Thus using the elapsed time formula t1 = 4ln1.5 − 0.10.7 − 0.1

= 3.39s .


SOLUTION 8.63. (a) Since RC = 10-3

s, from equation 8.25, va( t) = −RCdvs(t)

dt= cos(1000t) V.

Hence vb( t) = −RCdva( t)

dt== − sin(1000t) V, and vout (t) = −

R f

Rvb(t) = sin(1000t) = vs( t) V.

(b) With the switch moved to position B, there is no source in the circuit. But the output at the

switching instant is sin(1000t) V which coincides with vs(t). Hence, the input to the first amp remains

the same and the circuit continues to produce vout (t) = sin(1000t) V, i.e., the circuit becomes an

oscillator.

*SOLUTION 8.64. Before attacking the problem proper, consider driving an ideal unity gain integrator

with the square wave of figure P8.64b. If we start the integration when the square wave goes positive,

then we have a triangular waveform as follows:

On the other hand, if we start the integration when the square wave goes negative, we get the following

waveform

One concludes that without some further physical assumptions, there is no unique solution to this

problem.

Physically speaking all capacitors have a leakage resistance. Hence, in modeling the capacitor we

put a very large resistance in parallel with an ideal C, producing a nearly ideal leaky integrator circuit.

The leaky integrator circuit has a first order response. Hence over time, when the circuit reaches steady

state, the dc level of the resulting output waveform will be proportional to the average level of


the square wave which is zero in this case.

Comments: Actually the proportionality constant in the above statement is the overall dc gain of the

integrator-inverter. See the formulas given in P22.16. Adding up the two formulas, we have

Output(t)max + Output(t)min = K (Vmax + Vmin)

and K = H(0), i.e., the dc gain of the first order low pass system. This leads to

Average of output = (dc gain) (average of input levels)

See the analysis in example 8.7 and later an exact analysis is given in problems 22.15 and 22.16. In

other words, one would expect that the output of our (leaky) integrator in steady state to be given by the

waveform below.

Now we can start to solve the problem. The first part is to design a (leaky) integrator circuit to

produce a triangular waveform of value 2 V peak-to-peak. For this we consider the following figure

which consists of the leaky integrator followed by an inverter.


To handle this analysis, recall that i ≅ ∆q/∆t in which case ∆q = C∆v. Hence, to have a peak-to-peak

voltage at v2(t) and v3(t) of 2 V, we require that

∆v =iin∆t

C=

9R1

×50 ×10−6

C= 2

Hence R1C = 2.25×10-4. If we choose R1 = 10 kΩ, then C = 22.5 nF. At this point the waveform of

v3(t) is given below.

In order to complete the design, we must raise the portion of the curve with positive slope by 1 V and

lower the portion with negative slope by –1 V. This can be done by adding one-ninth of vin(t) to v3(t).

This can be done by using the following circuit. In this circuit, there is a voltage-divider at the non-


inverting terminal of the second op amp. Here V+ equals one-eighteenth of vin(t). However the gain of

the non-inverting portion is 2; therefore one-ninth of the input is added to v3(t) as desired.

CHAPTER 9 PROBLEM SOLUTIONS

SOLUTION TO PROBLEM 9.1. If we can compute expressions for K and q that are real,then these quantities exist by construction. Consider that A, B, K and q must satisfy thefollowing relationship:

K cos(ωt + θ) = K cos(θ)( )cos(ωt) + −K sin(θ)( )sin(ωt) ≡ Acos(ωt) + Bsin(ωt)

Therefore K cos(θ) = A and −K sin(θ) = B . Consequently,

K cos(θ)( )2 + −Ksin(θ)( )2 = K2 = A2 + B2

in which case K = A2 + B2 . Further,

K sin(θ)

K cos(θ)= tan(θ) =

−B

Ain which case

θ = tan−1 −B

A

with due regard to quadrant.

SOLUTION TO PROBLEM 9.2. For the inductor,

WL (t) = 12 L ⋅ iL

2(t) =1

2L Vo

C

Lsin

1

LCt

2

=CV0

2

2sin2 1

LCt

and for the capacitor,

WC (t) = 12 C ⋅vC

2 (t) = 12 C Vo cos

1

LCt

2

=CV0

2

2cos2 1

LCt

.

Hence,

WC + WL = 12 C ⋅ vC

2 (t) + 12 L ⋅ iL

2(t) =CV0

2

2sin2 1

LCt

+ cos2 1

LCt

=

CV02

2

SOLUTION TO PROBLEM 9.3. Since x(t) = (K1 + K2t)e−αt ,

x' (t) = − α ⋅K1e−αt + K2e−αt − α ⋅t ⋅ K2e−αt

and

x' ' (t) = α2 ⋅ K1e−αt − α ⋅K2e−αt − α ⋅K2e−αt + α2 ⋅ t ⋅ K2e−αt

Substituting into the differential equation, we have

α2 ⋅ K1e−αt − 2α ⋅K2e−αt + α2 ⋅ t ⋅ K2e−αt

+ 2α − α ⋅K1e−αt + K2e−α t − α ⋅t ⋅ K2e−αt[ ] + α2 K1e−αt + K2te−αt[ ] = 0

This means that the solution form satisfies the differential equation.

SOLUTION TO PROBLEM 9.4.

(a) Suppose x(T ) = 0 at some T. Then K1es1T = −K2es2T . Since esiT > 0 whenever si

is real and T is finite, K1 & K2 must have opposite signs.

(b) For this we solve for T and show there can only be one solution. Since

K1es1T = −K2es2T and esiT > 0 ,

K1

−K2=

es2T

es1T implies ln

K1

−K2

= ln

es2T

es1T

= s2 − s1( )T

Hence the unique solution is given by

T = ln

K1

−K2

s2 − s1( )

provided s2 ≠ s1 which is the case for distinct roots.

SOLUTION TO PROBLEM 9.5. Suppose x(T ) = 0 at some T >0. This is true if and only if

K1es1T = −K2Tes1T (*)

Since es1T > 0 and T > 0, (*) is true if and only if K1 = −K2T which is true if and only ifK1 & K2 have opposite signs.

SOLUTION 9.6. (a) Denote one period of oscillation by T. Then by definition9950T = 2π . Hence, T = 0.63148 ms. The time constant of decay is 1 ms. Therefore,NT = N × 0 . 6 3 1 4 8 = 1. Hence N =1.5836 cycles.

(b) Here observe that the time constant of decay is 1/σ s. Hence NT = N2πωd

=1

σ. One

concludes that N =ωd

2πσ.

SOLUTION 9.7. The differential equation for the capacitor voltage is

d2vC (t)

dt2 +1

LCvC (t) =

1

LCVs

For t > 0, the characteristic equation is s2 +1

LC= 0 . Hence from table 9.2, the solution

for either the inductor current or capacitor voltage has the general form

vC (t) = Acos(ωdt) + Bsin(ωdt) + XF

where ωd =1

LC. From table 9.2, XF =

Vs

LC

1

LC

= Vs. Further,

vC (0+ ) = A + XF = A + Vs = 0Hence A = −Vs . Also,

vC' (0+ ) = ωd B =

1

LCB =

iL (0+ )

C= 0

Hence B = 0. Therefore

vC (t) = −Vs cos(ωdt) +Vs = Vs 1 − cost

LC

V

To obtain the expression for iL(t) (= iC(t)), we can either repeat the above derivation ordifferentiate and multiply by C. We choose the latter. Therefore

iL (t) =CVs

LCsin

t

LC

=Vs

L

C

sint

LC

A

SOLUTION TO PROBLEM 9.8. Essentially this is example 9.7, case 1, with literals and R =∞. Clearly the circuit is a driven parallel LC circuit having characteristic equation

s2 +1

LC= s +

j

LC

s −j

LC

= s + jωd( ) s − jωd( ) = 0

Thus we obtain

iL (t) = Acos(ωdt) + Bsin(ωdt) + XF = Acos1

LCt

+Bsin1

LCt

+XF

Here XF = IS is the value of the current when the inductor is shorted and the capacitor isopen. Applying the initial conditions,

iL (0+) = A + Is = 0 ⇒ A = − IsFurther

iL' (0+) =

vL (0+)

L=

vC (0+)

L= 0 =

1

LCB

because capacitor voltage is continuous and because

iL' (0+) =

d

dtAcos

1

LCt

+ B sin1

LCt

+ XF

t=0

=B

LC

Hence B = 0. Therefore,

iL (t) = Is 1− cos(1

LCt)

Rather than repeat the above derivation,

vC (t) = vL (t) = LdiL (t)

dt= LIs

d

dt1 − cos

1

LCt

= Is

L

Csin

1

LCt

SOLUTION TO PROBLEM 9.9. Observe that the circuits of figures (a) and (b) are dualcircuits. Hence the numerical value of vout(t) and iout(t) are the same for the sameexcitation. Since the circuit is linear, when the excitation is doubled, the response isdoubled (given zero initial conditions) by linearity. Therefore, iout (t) = 2g(t) .

SOLUTION TO PROBLEM 9.10.(a) The initial conditions are:

vc (0−) = 0 = vc( 0+)iL (0−) =10 / 0.5 = 20 = iL( 0+)

(b)

WL (0)=1

2L ⋅iL

2(0)= 0.5 J

WC (0) = 0

(c) Maximum value of vC when all energy is in capacitor must satisfy 1

2C ⋅ vC,max

2 = 0.5

or equivalently vC,max = 1000 V.

(d) From the text development, the parallel LC circuit has the differential equation,

d2vC

dt2 +1

LCvC = 0

The solution form is:

vC (t) = A ⋅cos1

LCt

+B ⋅sin1

LCt

The initial conditions are:vC (0)= A = 0

and

vC ' ( 0+) =1

LCB =

iC (0+)

C= −

iL( 0+)

CHence

B = −1000Thus

vC (t) = −1000 ⋅ sin1

LCt

V

SOLUTION TO PROBLEM 9.11.The switch has been closed for a long time which means that the inductor acts like a short

and the capacitor like an open. Hence, at t = 0-, vL = 0 and iC = 0. Hence Is divides

equally between the resistors, i.e.,

iL (0−) = iL( 0+) = Is / 2 and vC (0−) = vC ( 0+) = 0

For t ≥ 0, the differential equation is

d2vC

dt2 +1

LCvC = 0

with corresponding response

vC (t) = A ⋅cos1

LCt

+B ⋅sin1

LCt

Evaluating at the initial conditions,

vC (0)= A = 0and

vC ' ( 0 )= B ⋅1

LC=

iC (0+)

C=

−iL (0+)

Cin which case

B = −IS

2

L

CTherefore,

vC (t) = −Is

2

L

Csin

1

LCt


Natural frequency is 1

LC= 5000 ⋅ 2π . Hence,

»C = 100e-9;»L = 1/((10e3 *pi)^2 *C)L = 1.0132e-02

By voltage divider,vC (0−) = vC ( 0+) = 20 mV

Current through L,iL (0−) = iL( 0+) = 0

Voltage across capacitor satisfies,

vC (t) = Acos1

LCt

+ B sin1

LCt

Using the ICs,

vC (0+) = A = 20 mVand

vC ' ( 0 )= B ⋅1

LC=

iC (0+)

C= −

iL( 0+)

C= 0

HencevC (t) = 20 ⋅cos 10,000πt( ) mV

SOLUTION TO PROBLEM 9.13. By definition

ω =1

LC= 2π ⋅40

in which case»C = 0.1e-3;»w = 2*pi*40;»L = 1/(w^2 *C)L = 1.5831e-01 (rad/s).

Observe thatiL (0−) = iL( 0+) = 1A

From the given circuit, the capacitor is never connected to a source. Therefore,vC (0−) = vC ( 0+) = 0. Also, since

d2iLdt 2 +

1

LCiL = 0

it follows thatiL (t) = Acos(ωt) + Bsin(ωt)

From the initial conditionsiL (0)= A = 1

and

iL' (0+ ) = Bω =vC (0+ )

L= 0

iL (t) = cos(80πt) A

SOLUTION TO PROBLEM 9.14.(a) From the continuity property, the capacitor voltage and inductor current remain thesame,

vC (0−) = vC ( 0+) = 5 ViL (0−) = iL( 0+) = 1

However, iC (0−) = 0 but iC (0+ ) = −iL (0+) = −1 A and vL (0−) = 0 but vL (0+) = 5 V.These values change to maintain satisfaction of KVL and KCL.

(b)

vC (t) = Acos1

LCt

+ B sin1

LCt

wherevC (0+) = 5 = A

vC ' ( 0+) = B1

LC=

iC (0+)

C= −

1

C ⇒ B = −

π2

vC (t) = 5cos(2πt) −π2

sin(2πt) V

Alternately, from equation 9.4,

vC (t) = K cos(2πt + θ)

and from equation 9.17b K = A2 + B2 and θ = tan−1 −B

A

in which case

»A = 5; B = -pi/2;»K = sqrt(A^2 + B^2)K = 5.2409e+00»theta = atan2(-B,A)theta = 3.0440e-01»thetadeg = theta*180/pithetadeg = 1.7441e+01

Hence, K = 5.24 and θ = 17o.

(c)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-6

-4

-2

0

2

4

6C

apac

itor

Vol

tage

(V

)

TextEnd

Time in s

SOLUTION TO PROBLEM 9.15. As before, ω =1

LC= 2π ⋅40 in which case L = 0.158

H.Now,

vC (0−) = vc (0+) =100

125⋅ 25 = 20 mV

andiL (0−) = iL( 0+) = 10 mA

The solution form is:iL (t) = Acos(80πt) + Bsin(80πt)

whereiL (0+) = A = 10 mA

and

iL' (0+) =vL( 0+)

L=

vC (0+)

L= 80πB ⇒ B = 0.50265 mA


(a) Note: vC (0−) = vC ( 0+) = −1 V and iL (0)= 5 A. The characteristic equation is:

s2 +1

RCs +

1

LC= s2 + 5s + 4 = (s +1)(s + 4 )= 0

As stated, the circuit is overdamped. Hence,

vC (t) = K1es1t + K2es2 t = K1e−t + K2e−4t

vC (0+) = K1 + K2 = −1

vC ' ( 0+) = s1K1 + s2K2 = −K1 − 4K2 =iC (0+)

C= 0

⇒ K1 =−s2

s2 − s1= −1.3333 and K2 =

s1s2 − s1

= 0.3333 .

Finally,

vC (t) = −1.3333e−t + 0.3333e−4t VUsing MATLAB to plot:

»t = 0:.02:4.5;»vc = -1.33333*exp(-t) + 0.33333*exp(-4*t);»plot(t,vc)»grid»ylabel('Capacitor Voltage in V')»xlabel('Time in s')

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1

-0.9

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

Cap

acito

r V

olta

ge in

V

TextEnd

Time in s

(b)K1 + K2 =1

vC ' ( 0+) = s1K1 + s2K2 = −K1 − 4K2 =iC (0+)

C=

−10

C= −10

Hence,»A = [1 1;-1 -4]A = 1 1 -1 -4»b = [1;-10]b = 1 -10»K = A\bK = -2 3

yielding

vC (t) = −2e−t + 3e−4t V

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Cap

acito

r V

olta

ge in

V

TextEnd

Time in s

Obviously, there is only one zero crossing.

SOLUTION TO PROBLEM 9.17.(a) First,

vC (0−) = vC ( 0+) =100

30030 =10 V

iL (0−) = iL( 0+) = 0.1 A

Clearly, iC (0−) = 0 and vL (0−) = 0 . However,

iL (0+) + iC (0+) +vC (0+)

66.667= 0.1 + iC (0+) + 0.15 = 0

Hence, iC (0+ ) = −0.25 A. Further,vL (0+) = 10 V.

(b) The characteristic equation is:

s2 +1

RCs +

1

LC= 0

From MATLAB,»R = 66.667;C = 25e-6; L = 0.5;»b = 1/(R*C)b = 6.0000e+02»c = 1/(L*C)c = 80000»si = roots([1 b c])si = -3.9999e+02 -2.0000e+02

We take the roots to be: s1 = −400 and s2 = −200 .

(c) Overdamped response implies,

vC (t) = K1es1t + K2es2 t = K1e−400t + K2e−200t

(d)vC (0+) = K1 + K2 =10

vC ' ( 0+) = s1K1 + s2K2 = −400K1 − 200K2 =iC (0+)

C= −104

»A = [1 1;-400 -200];»b = [10;-1e4];»K = A\bK = 40 -30

Finally,

vC (t) = 40e−400 t −30e−200t V(e)»t=0:0.01e-3:25e-3;»vc = 40*exp(-400*t) - 30*exp(-200*t);»plot(t,vc)»grid»ylabel('Capacitor Voltage in V')»xlabel('Time in s')

0 0.005 0.01 0.015 0.02 0.025-6

-4

-2

0

2

4

6

8

10

Cap

acito

r V

olta

ge in

V

TextEnd

Time in s

SOLUTION TO PROBLEM 9.18.(a)

iL (0−) = iL( 0+) = 0vC (0−) = vC (0+) = 5 V

At t = 0+, the circuit is a series RLC with R = 12.5 Ω, L = 2.5 H, and C=0.1 F. Theresulting characteristic polynomial is:

s2 +RL

s +1

LC= s2 + 5s + 4 = (s + 4)(s +1) = 0

Hence, s1,s2 = −4,−1 and the form of the response is:

vC (t) = K1e−4 t + K2e−t

At t = 0+,vC (0+) = K1 + K2 = 5

and

vC '(0) = −4K1 − K2 =iC (0+)

C=

iL (0+)C

= 0

Solve for K1 and K2 we obtain:»A = [1 1; -4 -1];»b = [5;0];

»K = A\bK = -1.6667e+00 6.6667e+00

Hence,

vC (t) = −1.66667e−4 t + 6.66667e−t V

(b)iL (0+) = iL(0−) = 0

Since, the stable (passive) circuit contains no source for t > 0, all initial energy isabsorbed by the resistor. Hence limt→∞ iL ( t) = 0 , i.e., iL (∞) = 0 .

iL ( t) = iC (t) = CdvCdt

=23

e−4 t −23

e−t A

SOLUTION TO PROBLEM 9.19. This circuit is a series RLC in which case the

characteristic equation is always: s2 +RL

s +1

LC= 0

(a) For this time period, R = 2 kΩ in which case the characteristic equation is found andsolved in MATLAB as follows:

»R = 2e3; L = 0.1; C = 0.1e-6;»b = R/Lb = 20000»c = 1/(L*C)c = 100000000»s12 = roots([1 b c])s12 = -10000 -10000

The roots are repeated and: s1, s2 = –10,000. The form of the response is:

iL (t) = (K1 + K2t)e−10000t

To find K1 and K2:

iL (0+) = K1 = 2.5and

iL '(0+) = −25000 + K2 =vL (0+)

L=

vC (0+) − RiL (0+)L

=6 − 2000 × 2.5

0.1= −49940

⇒ K2 = −24940

Thus

iL (t) = (2.5− 24940t)e−10000 t 0 ≤ t ≤ 0.1msec

(b) After the switch closes, R = 1 kΩ,

iL (0.1×10−3) = (2.5 − 24940 × 0.1×10−3)e−1 = 0.0022073and

iL '(10−3) = −25000e−1 − 24940e−1 + 24940e−1 = −9197

The new characteristic equation is computed as follows:

»R = 1e3; L = 0.1; C = 0.1e-6;b = R/Lb = 10000»c = 1/(L*C)c = 100000000»s12new = roots([1 b c])s12new = -5.0000e+03 + 8.6603e+03i -5.0000e+03 - 8.6603e+03i

Hence s1,s2 = −5000 ± j8660.25. The form of the new solution is:

iL ( t') = e−5000 t' Acos( t ') + Bsin( t')[ ] A

where t'= t − 0.1×10−3 and = 8660.25 rad/sec. Observe that

iL ( t'= 0) = 0.0022073 = Aand

iL '( t'= 0) = −5000A + 8660.25B = −9197From MATLAB,»w = imag(s12new(1));»sig = real(s12new(1));»B = (-9197 -sig*2.2073e-3)/wB = -1.0607e+00Hence for t > 0.1 ms,

iL ( t) = e−5000(t−0.1ms) 0.0022073cos ( t − 0.1ms)( ) −1.0607sin ( t − 0.1ms)( )[ ] A

(c)

0 0.5 1 1.5 2 2.5 3 3.5 4x 10-3

-1

-0.5

0

0.5

1

1.5

2

2.5

iL(t

), in

A

TextEnd

time in s

K1 = 2.5;K2 = -24940;t = 0:0.01e-3:4e-3;A = 0.0022073;B = (-9197 -sig*2.2073e-3)/wB = -1.0607e+00iL = (K1 + K2*t) .* exp(-10000*t) .* (u(t)-u(t-1e-4)) ...+ exp(-5000*(t - 1e-4)) .* (A*cos(w*(t - 1e-4))+B*sin(w*(t - 1e-4))) ....* u(t -1e-4);plot(t,iL)gridiL = (K1 + K2*t) .* exp(-10000*t) .* (u(t)-u(t-1e-4))

(d)

»K1 = 2.5;K2 = -24940;»t = 0.1e-3;»iL = (K1 + K2*t) .* exp(-10000*t)iL = 2.2073e-03» % The energy stored in the inductor over [0,0.1ms] is in J:»WL = 0.5*0.1*(iL^2 - 2.5^2)WL = -3.1250e-01» % The energy stored in the capacitor over [0,0.1ms] first» % requires computation of vL and then vC.»vL = 0.1*(K2*exp(-10000*t) - 10000*(K1 + K2*t) .* exp(-10000*t))vL = -9.1970e+02»vC = vL + 2000*iLvC =

-9.1528e+02» % The energy stored in the capacitor over [0,0.1ms] is in J:»WC = 0.5*0.1e-6*(vC^2 - 6^2)WC = 4.1885e-02

» % To compute energy dissipated in resistor, we make» % use of conservation of energy: WR + WC + WL = 0

»WR = -WL - WCWR = 2.7061e-01

SOLUTION TO PROBLEM 9.20. For this problem, iL (0) = 8 A, vC (0) = 20 V. For the

parallel RLC, the characteristic equation s2 +1

RCs +

1LC

= 0 is solved as follows:

»R = 20; C = 0.01e-3; L = 25e-3;»si = roots([1 1/(R*C) 1/(L*C)])si = -4.0000e+03 -1.0000e+03»s1 = si(1); s2 = si(2);

Hence

vC (t) = K1e−4000 t + K2e−1000t

To compute the constants,

vC (0) = K1 + K2 = 20and

vC '(0) = −4000K1 −1000K2 =iC (0+)

C= −

iL (0+) +vC (0+)

RC

»iL0 = 8; vC0 = 20;»vCprime = -(iL0 +vC0/R)/CvCprime = -9.0000e+05»A = [1 1;-4000 -1000];»b = [20;vCprime];»K = A\bK = 2.9333e+02 -2.7333e+02

Hence,

vC (t) = 293.33e−4000t − 273.33e−1000 t VAlso,

iL ( t) = −vC (t)

R− CvC

' ( t) = −2.933e−4000t +10.933e−1000 t A

where»K1 = K(1); K2 = K(2);»KR1 = K1/R; KR2 = K2/R;»KCp1 = -4000*K1*C; KCp2 = -1000*K2*C;»KL1 = -KR1-KCp1KL1 = -2.9333e+00»KL2 = -KR2 - KCp2KL2 = 1.0933e+01»t = 0:0.01e-3:4e-3;»iL = KL1*exp(-4000*t) + KL2*exp(-1000*t);»plot(t,iL)»grid»xlabel('Time in s')»ylabel('Inductor Current in A')»

0 0.5 1 1.5 2 2.5 3 3.5 4x 10-3

0

1

2

3

4

5

6

7

8

9

Time in s

Indu

ctor

Cur

rent

in A

TextEnd

SOLUTION TO PROBLEM 9.21.(a) The circuit is critically damped and the characteristic polynomial is:

s2 +1

RCs +

1LC

= (s + 20)2 = s2 + 40s + 400 = 0

»R = 2; C = 1/80;»L = 1/(400*C)L = 2.0000e-01

(b) vC (0) = 10 V and vC '(0+ ) = −800 =iC (0+ )

C=

−iL (0+ ) − vC (0+ ) / 2C

.

»vC0 = 10; vCp0 = -20*10 -600vCp0 = -800»iL0 = C*800-vC0/2iL0 = 5Hence: iL(0-) = iL(0+) = 5 A.

(c)By simple KCL,

iL ( t) =−180

dvCdt

−vC2

= 5e−20t +150 ⋅ t ⋅e−20t A

SOLUTION TO PROBLEM 9.22.(a) The characteristic equation for the series RLC

s2 +Req

Ls +

1

LC= 0

For critically damped response, want (Req / L)2 − 4 / (LC ) = 0 . Solving yields Req = 4

Ω. Hence,

Req =5R

5 + Rimplies that R = 20 Ω.(b) Solving for the resulting roots implies that»si = roots([1 Req/L 1/(L*C)])si = -50 -50»s1 = 50;

Hence

vC (t) = (K1 + K2t)e−50t

From the initial conditions, vC (0) = K1 = 5 and

vC '(0) = −250 + K2 =iC (0+)

C=

iL (0+)C

=−5

0.01= −500

Hence, K2 = –250 and

vC (t) = (5 − 250t)e−50t V

Set to zero and solve for t = 0.02. See MATLAB plot.

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-1

0

1

2

3

4

5

SOLUTION TO PROBLEM 9.23.(a) The circuit is a series RLC. Hence

»R = 40; C = 0.25e-3; L = 0.1;si = roots([1 R/L 1/(L*C)])si = -200 -200»s1 = si(1);

Since the circuit is critically damped,

iL ( t) = (K1 + K2t)e−200 t AUsing initial conditions,

iL (0) = K1 = 1

iL '(0) = s1K1 + K2 = −200K1 + K2 =vL (0+)

L=

−vC (0+) − 40iL (0+)

L= −350

HenceK2 = −150

Thus

iL ( t) = (1−150t)e−200 t A

Now using MATLAB,»R/Lans = 400»1/(L*C)ans = 40000

»y = dsolve('D2y + 400*Dy + 40000*y = 0, y(0) = 1,Dy(0) = -350')y =exp(-200*t)-150*exp(-200*t)*t

This answer coincides with the analytical solution.

(b) As above, the form of the solution is

vC (t) = (K1 + K2t)e−50t

Applying initial conditions,

vC (0) = K1 = 5

vC '(0) = s1K1 + K2 = −200K1 + K2 =iC (0+)

C=

iL (0+)

C= 4000

Hence, K2 = 5000 and

vC (t) = (5 + 5000t)e−200 t V

In MATLAB,»y = dsolve('D2y + 400*Dy + 40000*y = 0, y(0) = 5,Dy(0) = 4000')y =5*exp(-200*t)+5000*exp(-200*t)*t

which verifies the analytical solution. Plotting we obtain,

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050

2

4

6

8

10

12

Observe, there is no zero crossing as per problem 9.5.

SOLUTION TO PROBLEM 9.24.Source has been on for a long time and turns off at t = 0. Hence

iL (0−) = iL (0+) = 0

vC (0−) = vC (0+) = 0.1× 40 = 4 V

At t ≥ 0, we have a series RLC circuit:

»R = 40; C = 4e-3; L = 2;si = roots([1 R/L 1/(L*C)])si = -1.0000e+01 + 5.0000e+00i -1.0000e+01 - 5.0000e+00i»wd = imag(si(1))wd = 5»sig = -real(si(1))sig = 10

Hence

vC (t) = e−10 t Acos(5 t) + B sin(5t)( )Applying initial conditions,

vC (0) = A = 4

vC '(0) = −10A + 5B =iC (0+)

C=

iL (0+)C

= 0

Hence, B = 8. The final form is:

vC (t) = e−10 t 4cos(5t) + 8sin(5 t)( ) V

Similarly,

iL ( t) = e−10 t A cos(5t) + Bsin(5t)( ) AApplying initial conditions,

iL (0) = A = 0

iL '(0) = −10A + 5B =vL (0+)

L=

−vC (0+) − 40iL (0+)L

= −2

Thus

iL ( t) = −0.4e−10 t sin(5t) A


At t = 0-, the capacitor is an open circuit and the inductor is a short circuit. The

resulting circuit is a simple resistive network. The first step in the solution is to solve thisnetwork for the initial conditions on the capacitor and inductor. Specifically, solve forthe capacitor voltage (i.e. the voltage across the series connection of the 6 Ω resistor andthe independent voltage source) and the inductor current (i.e. the current flowing throughthe 4 Ω resistor).

Verify that iL (0−) = iL (0+) =1 A and vC (0−) =vC (0+) =12 V.

When the switch opens, the branch containing the independent voltage source iseliminated. So, we end up with a series RLC circuit. The equivalent resistance is

Req = 4 + 24 / / 4 8= 20 Ω.

»R = 20; C = 0.01; L = 2;si = roots([1 R/L 1/(L*C)])si = -5.0000e+00 + 5.0000e+00i -5.0000e+00 - 5.0000e+00i

»sig = -real(si(1))sig = 5»wd = imag(si(1))wd = 5

iL ( t) = e−5t A cos(5t) + Bsin(5t)[ ] AApplying ICs,

iL (0) = A =1

iL '(0) = −5A + 5B =vL (0+)

L=

−8

2= −4

Hence B = 0.2. It follows that

iL (t) = e−5t cos(5t) +1

5sin(5t)

A

SOLUTION TO PROBLEM 9.26. The series RLC circuit has characteristic equation

s2 +RL

s +1

LC= s2 +

10L

s +1

LC= s2 + 2 s + 2 + d

2 = 0

From the given response, = 10 =102L

which implies that L = 0.5 H. Further,

1LC

=2C

= 2 + d2 =102 + 10 3( )2

= 400

Hence, C = 5 mF.

Now, from given response, iL (0) = 0 and iL '(0) = 500 3 =vL (0)

L= 2vL(0) .

HencevL (0) = 250 3 V. In addition, vC (0) = −10iL (0) − vL (0) = −250 3 V and

vC '(0) = iL (0) / C = 0

To find the capacitor voltage we have

vC (t) = e−10 t Acos(10 3t) + B sin(10 3t)( )

It follows that vC (0) = −250 3 = A and

vC '(0) = 0 = 2500 3 +10 3 BB = –250. Therefore

vC (t) = −e−10 t 250 3 cos(10 3t) + 250sin(10 3t)( )

SOLUTION TO PROBLEM 9.27. (a) From the given information,

vC (t) = Ae− t cos( t + ) V

where = 2 f = 2 / T =2

0.5 ×10−3 = 4000π rad/s. f = 2 kHz.

The time constant of the exponential term is 1/σ which is 2 ms from the figure.

Hence σ = 500 . The characteristic equation of the parallel RLC is:

s2 +1

RCs +

1LC

= s2 +1

104Cs +

1LC

= s2 + 2 s + 2 + d2 = 0

Since 1

104C= 2 = 1000, C = 0.1 µF. Further,

1LC

=107

L= 2 + d

2 =106 +16π2106

implies that

»L = 1e7/(1e6 + 16*pi^2 *1e6) or L = 63 mH.

(b) From the figure and the above calculations, vC(0) = 10 V and

vC (t) = 10e−500 t cos(4000π t) V,Hence

vC '(t) = −5000e−500t cos(4000π t) − 40000πe−500t sin(4000π t)

implying that vC' (0) = −5000 . Thus

iL (0) = −vC (0) /10 4 − CvC' (0) = −0.001 + 5000 × 0.1×10−6 = −0.5 ×10−3 A

SOLUTION TO PROBLEM 9.28. As given vC (t) = Ae−3t cos(4 t + ) and R = 10 Ω. Thecharacteristic polynomial of the parallel RLC circuit is:

s2 +1

RCs +

1LC

= s2 +1

10Cs +

1LC

= s2 + 2 s + 2 + d2 = 0

Hence 1/(RC) = 6 implies C = 1/60 F. Further d =

4 × 60

L− 36

2= 4 . Hence L = 2.4

H.To change R to obtain a cricially damped circuit,

60R

2

=4

LC=100

Hence R2 = 36 or R = 6 Ω. It follows that 2σ = 10 or σ = 5. The form of the response is:

vC (t) = (K1 + K2t)e−5t

SOLUTION TO PROBLEM 9.29. For all cases, vC(0-)= vC(0+) = 0 and iL(0-)= iL(0+) =20/20 = 1 A. Further for all cases the circuit is a parallel RLC with characteristicequation:

s2 +1

RCs +

1LC

= s2 + 2 s + 2 + d2 = 0

(a)»L = 2e-3; C = 5e-6;»c = 1/(L*C);»R = 10;»b = 1/(R*C);»si = roots([1 b c])si = -10000 -10000»% Solution is cricitally damped.

Thus

vC (t) = (K1 + K2t)es1t = (K1 + K2t)e−104 t V

From ICs,vC (0) = K1 = 0

vC '(0) = s1K1 + K2 = K2 =iC (0+)

C=

−iL (0+)

C= −2 ×105

Hence

vC (t) = −2 ×105te−104 t V

(b)»R = 100;»si = roots([1 1/(R*C) 1/(L*C)])

si = -1.0000e+03 + 9.9499e+03i -1.0000e+03 - 9.9499e+03i

vC (t) = e−1000 t Acos(9950 t) + B sin(9950t)( ) VFrom ICs.

vC (0+) = A = 0

vC '(0) = −1000A + 9950B =iC (0+)

C=

−iL (0+)C

= −2 ×105

in which case B = –20.1. Thus

vC (t) = −20.1e−1000t sin(9950t) V

(c)»R = 87/17;si = roots([1 1/(R*C) 1/(L*C)])si = -3.6328e+04 -2.7527e+03

We define s1,s2 = –2753, –36327. Thus

vC (t) = K1es1t + K2es2t = K1e−2753t + K2e−36327 t VFrom the IC's

vC (0) = K1 + K2 = 0

vC '(0) = s1K1 + s2K2 = −2 ×105

»A = [1 1;si(2) si(1)];»b = [0; -2e5];»K = A\bK = -5.9568e+00 5.9568e+00

Therefore

vC (t) = −5.9568 e−2753t − e−36327t( ) V

SOLUTION TO PROBLEM 9.30. For all cases, vC(0-) = vC(0+) = 0 and iL(0-) = iL(0+) =10/100 = 0.1 A. Further for all cases the circuit is a parallel RLC with characteristicequation:

s2 +1

RCs +

1LC

= s2 + 2 s + 2 + d2 = 0

(a)»R = 50; C = 0.04e-3; L = 0.625;»si = roots([1 1/(R*C) 1/(L*C)])si = -400 -100

Define the two roots as:s1 = –100s2 = –400

Hence,

vC (t) = K1e−100 t + K2e−400t V

From, the initial conditionsvC(0+) = 0 = K1 + K2

and

vC '(0) = s1K1 + s2K2 = −100K1 − 400K2 =−iL (0+)

C= −2500

»A = [1 1;si(2) si(1)];»b = [0;-2500];»K = A\bK = -8.3333e+00 8.3333e+00

Therefore

vC (t) = −8.3333 e−100 t − e−400t( ) V

(b)»L = 0.4;»si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+02 -2.5000e+02

Thus

vC (t) = (K1 + K2t)e−250t VFrom IC's,

vC (0) = K1 = 0

vC '(0) = s1K1 + K2 = K2 =iC (0+)

C=

−iL (0+)

C= −2500

Therefore

vC (t) = −2500te−250 t V

(c)»L = 0.2;»si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+02 + 2.5000e+02i -2.5000e+02 - 2.5000e+02i

vC (t) = e−250 t A cos(250t) + Bsin(250t)[ ] VFrom ICs.

vC (0+) = A = 0

vC '(0) = −250A + 250B = 250B =iC (0+)

C=

−iL(0+)C

= −2500

in which case B = –10. Thus

vC (t) = −10e−250t sin(250t) V

SOLUTION TO PROBLEM 9.31.(a) The indicated behavior occurs when the resistance causes the circuit to be criticallydamped, i.e.,

1RC

2

−4

LC=

4R

2

−16 = 0

Thus R = 1 Ω and

R =1 = R0 + et−5 = 0.8 + et−5

So»t = 5+log(1-0.8)t = 3.3906e+00 (years)

(b) Here, we have a series case: the indicated behavior occurs when the resistancecauses the circuit to be critically damped, i.e.,

RL

2

−4

LC= R2 −144 = 0

Thus R = 12 Ω and

R =12 =R0

1 + et−5 =15

1 + et−5

»t = 5 + log((15-12)/12)t = 3.6137e+00 (years)

SOLUTION TO PROBLEM 9.32. Step 1: Since the step functions are 0 from t = ∞ up to t =0-,

vC(0-) = vC(0+) = 0, iL(0-) = iL(0+) = 0

Step 2: At t = 0+, we have

iC (0+) = 2 −vC (0+)

20− iL (0+) = 2 A

andvL (0+) = vC (0+) − 5iL (0+) − 50 = −50 V

SOLUTION TO PROBLEM 9.33. Since the step function is 0 from t = ∞ up to t = 0-,

vC(0-) = vC(0+) = 0, iL(0-) = iL(0+) =0

Since the circuit is a parallel RLC

s2 +1

RCs +

1LC

= s2 +100s +1600 = 0

»R = 4; L = 0.25; C = 2.5e-3;»si = roots([1 1/(R*C) 1/(L*C)])si = -80 -20

Hence s1,s2 = -20, -80. The general form is:

vC (t) = K1e−80 t + K2e−20t + X f V

When the capacitor is open and the inductor is a short, Xf = 0. Thus,

vC (t) = K1e−80 t + K2e−20t VFrom the ICs

vC(0+) = 0 = K1 + K2

and

vC '(0) = −80K1 − 20K2 =iC (0+)

C=

−iL (0+) − vC (0+) 8 + (20 − vC (0+)) 8

C

=2.5

2.5 ×10−3 =1000

»A = [1 1;-80 -20];»b = [0; 1000];»K = A\bK =

-1.6667e+01 1.6667e+01Hence,

vC (t) = −16.667 e−80 t − e−20t[ ] V

SOLUTION TO PROBLEM 9.34. From the continuity property and the fact that at t = 0-,the capacitor looks like an open and the inductor looks like a short at t = 0-,

iL(0-) = iL(0+) = 1AvC(0-) = vC(0+) = 0

Since the circuit is a parallel RLC

s2 +1

RCs +

1LC

= s2 + 500s + 40000 = 0

»R = 40; C = 0.05e-3;L = 0.5;»si = roots([1 1/(R*C) 1/(L*C)])si = -400 -100

Define the two roots as:s1 = –100s2 = –400

Hence, the general form is:

iL ( t) = K1e−100 t + K2e−400t + X f

When the capacitor is open and the inductor is a short, Xf = –1 A. Thus,

iL ( t) = K1e−100 t + K2e−400t −1

From, the initial conditionsiL (0+) = K1 + K2 −1 =1

and

iL '(0) = −100K1 − 400K2 =vL (0+)

L=

vC (0+)L

= 0

»A = [1 1;-100 -400];»b = [2;0];»K = A\bK = 2.6667e+00

-6.6667e-01

Hence,

iL ( t) =83

e−100 t −23

e−400t −1 A


(a) From the problem specs and the continuity property,

iL(0-) = iL(0+) = 0.008 AvC(0-) = vC(0+) = 2

At t = ∞, the inductor looks like a short and the capacitor looks like an open; hence iL(∞)

= 0 and vC(∞) = 400×0.006 = 2.4 V. The circuit is a series RLC with characteristic

polynomial

s2 +RL

s +1

LC= 0

»R = 400; C = 6.6667e-6; L = 0.2;si = roots([1 R/L 1/(L*C)])s1 = si(1); s2 = si(2);si = -1.5000e+03 -5.0000e+02

Hence

vC (t) = K1e−1500 t + K2e−500t + 2.4 VUsing initial conditions

vC(0) = K1 + K2 + 2.4 = 2

vC '(0) = −1500K1 − 500K2 =iC (0+)

C=

iL (0+)C

=8 ×10−3

6.6667 ×10−6 =1200

»A = [1 1;-1500 -500];»b = [2-2.4;1200]b = -4.0000e-01 1.2000e+03»K = A\bK = -1.0000e+00 6.0000e-01

Hence

vC (t) = −e−1500 t + 0.6e−500t + 2.4 V

(b) vL(t) is going to have the same form as vC(t) above except that vL(∞) = 0 since theinductor is a short at t = ∞. Alternately however, we have

vL (t) = LdiL (t)

dt= L

diC (t)dt

= LCd2vC (t)

dt2 = −3e−1500t + 0.2e−500t V


At t = 0, inductor is a short circuit and the capacitor is an open circuit. Since the currentsource is 0 at t = 0-, and the continuity property,

iL(0-) = iL(0+) = –1 AvC(0-) = vC(0+) = 65 V

For positive time, we have a series RLC circuit with characteristic equation

s2 +RL

s +1

LC= 0

»R = 65; C = 0.1e-3; L = 0.1;si = roots([1 R/L 1/(L*C)])s1 = si(1); s2 = si(2);si = -4.0000e+02 -2.5000e+02

At t = ∞, vC(∞) = 0.6*65 = 39 V. Hence

vC (t) = K1e−400 t + K2e−250t + 39 V

Using initial conditions,vC(0) = K1 + K2 + 39 = 65

vC '(0) = −400K1 − 250K2 =iC (0+)

C=

iL (0+)C

=−1

10−4 = −104

»A = [1 1;s1 s2];»b = [65-39; -1e4];K=A\bK = 2.3333e+01 2.6667e+00

Hence,

vC (t) = 23.333e−400t + 2.6667e−250 t + 39 V

SOLUTION 9.37.

(a) Rth = 200//50 + R = (40 + R). The characteristic equation is:

s2 +1

RthCs +

1

LC

= 0

Critically damped means that both roots are the same, so the discriminant is zero, i.e.

1

RthC

2

−4

LC= 0

Equivalently,

Rth = 40 + R = 0.5L

C= 50 Ω

Thus R = 10 Ω.(b) Short the inductor and open the capacitor. Because the capacitor is in parallel

with the shorted inductor at t = 0-, vC(0+) = vC(0

-) = 0. The Thevenin equivalent

resistance seen by the LC-parallel combination is Rth = 50 Ω from part (a). A simple

calculation indicates that Voc = 0.8*50 = 40 V. Therefore, iL(0+) = iL(0

-) = 40/50 = 0.8

A. To find vR(0+) we use the following equivalent circuit:

Hence,

vR(0+ ) = −40 ×10

10 + 40= −8 V.

To compute the derivative of vR at 0+, consider that

d

dtvR(t)( ) =

d

dt10 × iR(t)( ) = 10

d

dt

−40 − vC (t)

50

= −0.2

d

dtvC (t)( ) = −0.2

iC( t)

C

HencedvR(t)

dt t=0+

= −0.2iC (0+ )

C= −4000iC (0+ )

But

iC (0+ ) =−40

50− iL( 0+ ) = −0.8 − 0.8 = −1.6 A

Therefore,dvR(t)

dt t=0+

= −4000iC (0+ ) = 6400 V/s

(c) Since the circuit is critically damped the roots of the characteristic equation are

s1,2 = −1

2RthC= −200

According to table 9.2 for t ≥ 0,

vR(t) = K1 + K2t( )e−200 t + XF

It follows from the circuit and this equation that

XF = vR(∞) = −8 V

K1 = vR(0+) − XF = 0

dvR(t)

dt t=0+

= −200K1 + K2 = K2 = 6400

Therefore

vR(t) = 6400te−200t − 8( )u(t) V

A plot of the waveform is given below

SOLUTION TO PROBLEM 9.38. For this problem we first compute the Theveninequivalent of the circuit to the left of the capacitor for t > 0. Consider

Now observe

Is = I1 + I2 =Vs −12

500+

Vs − kv1500

=Vs −12

500+

Vs − k 12 −Vs( )500

It follows that

Vs =5002 + k

Is +

1 + k2 + k

12 = Rth Is +Voc

The parallel LC is now driven by this Thevenin equivalent.

»L = 0.01; C = 1e-6;»% Critical damping means (1/(Rth*C))2 - 1/(L*C) = 0»x = sqrt(4/(L*C));»Rthcrit = 1/(C*x)Rthcrit = 50»kcrit = (500 - 2*50)/50kcrit = 8»% For parallel circuit, larger Rth means less damping»% Hence, smaller Rth means overdamped. Smaller Rth»% means larger k. Therefore k > 8 is the ranger for»% overdamped response.

For the critically damped response we have R = Rth = 50 Ω; hence

»R = 50; C = 1e-6; L = 0.01;si = roots([1 1/(R*C) 1/(L*C)])si =

-10000 -10000

in which case

vC (t) = (K1 + K2t)e−50t + X f

At t = ∞, vC (∞) = vL (∞) = 0 in which case X f = 0. From the initial conditions,

vC (0) = K1 = 0

vC '(0) = s1K1 + K2 = K2 =iC (0+)

C=

Voc

Rthcrit− iL (0+)

CHence

K2 =10.8 50

10−6 = 2.16 ×105

Therefore

vC (t) = 2.16 ×105e−50 t V

SOLUTION TO PROBLEM 9.39. (a) The series RLC leads to a characteristic equation ofthe form

s2 +RL

s +1

LC= s2 +

240.2

s +1

0.2C= s2 +120s +

5C

= 0

For a critically damped response, 1202 =20C

. Hence, C = 1.3889 mF.

(b)»C = 20/120^2C = 1.3889e-03»L = 0.2; R = 24;»si = roots([1 R/L 1/(L*C)])si = -60 -60Hence

iL (t) = (K1 + K2t)e−60t + 0.4 A

where iL(∞) = 0.4 because at t = ∞, the capacitor looks like an open and the inductor likea short. Hence all current from the source flows through the inductor.

Using the initial conditions,

iL(0) = K1 + 0.4 = 0 ⇒ K1 = – 0.4

iL '(0) = s1K1 + K2 = 60 × 0.4 + K2 =vL (0+)

L=

−vC (0+) − 24 iL(0+) − 0.4( )L

= 48

Hence, K2 = 24 and

iL ( t) = (−0.4 + 24t)e−60 t + 0.4 A

SOLUTION TO PROBLEM 9.40. This problem differs from 39 in the initial conditioncalculation. Specifically,

iL(0-) = iL(0+) = –0.4vC(0-) = vC(0+) = 0

Again

iL (t) = (K1 + K2t)e−60t + 0.4 Aand

iL(0+) = K1 + 0.4 = – 0.4 ⇒ K1 = – 0.8

iL '(0) = 60 × 0.8 + K2 =vL (0+)

L=

−vC (0+) − 24 iL (0+) − 0.4( )L

= 96

Hence,K2 = 48

and

iL ( t) = (−0.8 + 48t)e−60t + 0.4 A

SOLUTION TO PROBLEM 9.41.(a) At 0-, the capacitor is an open circuit and inductor is a short circuit. So,

vC(0-) = vC(0+) = 5 ViL(0-) = iL(0+) = 5×10-3 – 4×10-3 = 1 mA

Now, at 0+, replace the capacitor by a 5 V voltage source and the inductor by a 1 mAcurrent source. Also, the original independent current source is turned off. Solve theresulting circuit to obtain.

iC + 5×10-3 – 1×10-3 = 0

⇒ iC(0+) = –4 mA

vL(0+) = 0

(b) Since the circuit is a parallel RLC, the characteristic polynomial is

s2 +1

RCs +

1LC

= s2 +1

104Cs +

1LC

= s2 + 2 s + 2 + d2 = 0

»R = 1e3; C = 0.5e-6; L = 0.184;si = roots([1 1/(R*C) 1/(L*C)])si = -1.0000e+03 + 3.1416e+03i -1.0000e+03 - 3.1416e+03i

»wd = imag(si(1))wd = 3.1416e+03

»sig = -real(si(1))sig = 1000

Also, at t = ∞, iL(∞) = 5×10-3 A. Therefore

iL ( t) = e−1000 t Acos(π ×103t) + Bsin(π ×103t)[ ] + 5 ×10−3 A

Using initial conditions

iL (0) = A + 5 ×10−3 =10−3 ⇒ A = −4 ×10−3

and

iL '(0) = −1000A + π ×103B = 4 + π ×103B =vL (0+)

L= 0

Hence,»B = -4e-3/piB = -1.2732e-03

Finally,

iL ( t) = −e−1000 t 4 c o s (π ×103t) +1.2732sin( π ×103t)[ ] + 5 mA

SOLUTION TO PROBLEM 9.42. The response here coincides with that of problem 41 up totime t = 2s. At this point we need the new initial conditions on the circuit at t = 2+.However, at t = 2, e−1000 t = 0 for all practical purposes. Hence, iL (2) = 5 ×10−3 A.

Differentiating the expression for iL(t) and evaluating at t = 2 yields zero by inspection.

This follows because Ke−2000 = 0 for K in the range of 1 to 104. This can also be seen

from the circuit because at t = 2 s, the capacitor has charged to 5 V, making

iL' (2+) =

vL (2+)L

= 0.

To find steady state current, solve the circuit with the new current source value and withthe capacitor and inductor as open and short circuits, respectively:

iL (∞) = 5 ×10−3 + 4 ×10−3 = 9 ×10−3

iL ( t) = −e−1000(t−2) A cos π ×103(t − 2)( ) + B sin π ×103( t − 2)( )[ ] + 9 ×10−3 A

Using the new initial conditions

iL (2+) = A + 9 ×10−3 = 5 ×10−3 ⇒ A = −4 ×10−3

and

iL '(2+) = −1000A + π ×103B = 4 + π ×103B =vL (0+)

L= 0

Hence B = –1.2732×10-3

and for t ≥ 2s,

iL ( t) = −e−1000(t−2) 4cos π ×103( t − 2)( ) +1.2732sin π ×103(t − 2)( )[ ] + 9 mA

SOLUTION TO PROBLEM 9.43. At t = 0-, iL(0-) = 10/20 = 0.5 A and vC(0-) = –5 V by theusual considerations. At t = 0+, we have a parallel RLC circuit. Hence

s2 +1

RCs +

1LC

= s2 +1

104Cs +

1LC

= s2 + 2 s + 2 + d2 = 0

»R = 10; C = 0.05e-3; L = 0.01;si = roots([1 1/(R*C) 1/(L*C)])si = -1.0000e+03 + 1.0000e+03i -1.0000e+03 - 1.0000e+03i»wd = imag(si(1))wd = 1.0000e+03»sig = -real(si(1))sig = 1000

Also, at t = ∞, vC(∞) = –5 V. Therefore

vC (t) = e−1000 t A cos(103t) + Bsin(10 3t)[ ]− 5 V


vC (0) = A − 5 = −5 ⇒ A = 0and

vC '(0) = −1000A +1000B =103B =iC (0+)

C=

−0.5

0.05 ×10−3 = −104

Hence, B =10 and

vC (t) = −10e−1000t sin(103t) − 5 VFurther

vL (t) = −5 − vC ( t) = 10e−1000 t sin(103t) V

SOLUTION TO PROBLEM 9.44. (a) The circuit is a driven series RLC. Hence

s2 +RL

s +1

LC= s2 + 2 s + 2 + d

2 = 0

»R = 400; C = 0.5e-6; L = 0.2;si = roots([1 R/L 1/(L*C)])si = -1.0000e+03 + 3.0000e+03i -1.0000e+03 - 3.0000e+03i»wd = imag(si(1))sig = -real(si(1))wd = 3000sig = 1000

Also, at t = ∞, vC(∞) = 400×0.006 = 2.4 V. Therefore

vC (t) = e−1000 t A cos(3000 t) + B sin(3000t)[ ] + 2.4 V


vC (0) = A + 2.4 = 2 ⇒ A = −0.4and

vC '(0) = −1000A + 3000B = 400 + 3000B =iC (0+)

C=

iL (0+)C

= 1.6 ×104

Hence, B = 5.2 and

vC (t) = e−1000 t −0.4cos(3000t) + 5.2sin(3000t)[ ] + 2.4 V

(b) Consistent with underdamped circuit behavior and because the inductor behaves as ashort and the capacitor as an open at t = ∞ (iL(∞) = 0),

iL ( t) = e−1000 t Acos(3000t) + Bsin(3000t)[ ] V


iL (0+) = A = 0.008 AFurther,

iL '(0) = −1000A + 3000B = −8 + 3000B =vL (0+)

L=

400(0.006 − 0.008) − 20.2

= −14

Hence, B = −0.002 and

iL ( t) = e−1000 t 0.008cos(3000t) − 0.002sin(3000 t)[ ] VFinally

vL (t) = vR(t) − vC (t) = 400 0.006 − iL (t)[ ] − vC (t) = 2.4 − 400iL (t) − vC ( t)

which implies that

vL (t) = e−1000 t 3.6cos(3000t) −13.2sin(3000t)[ ] V

SOLUTION TO PROBLEM 9.45. This circuit is the same series RLC as problem 44. Note

that at t = ∞, vC(∞) = –2.4 V. Hence

vC (t) = e−1000 t A cos(3000 t) + B sin(3000t)[ ]− 2.4 V

Now, the initial conditions are:

vC(0-) = 2.4 = vC(0+) V, iL(0-) = 0 = iL(0+)Thus

vC (0) = A − 2.4 = 2 . 4 ⇒ A = 4.8Further,

vC '(0) = −1000A + 3000B = −4800 + 3000B =iC (0+)

C= 0

Hence, B =1.6 and

vC (t) = e−1000 t 4.8cos(3000t) +1.6sin(3000 t)[ ]− 2.4 V

SOLUTION TO PROBLEM 9.46. For all three cases, assuming iL is pointing downward,

vC (0−) = vC (0+) = 0 and iL (0−) = iL (0+) = 0.1 AAt t = ∞,

vC (∞) = 0 and iL (∞) = −0.2 AFurther

iC (0+) = −iL (0+) +−10 − vC (0+)

50= −0.1− 0.2 = −0.3 A

Lastly, all three cases are for a parallel RLC whose characteristic equation is:

s2 +1

RCs +

1LC

= 0

(a)»R = 50; C = 0.04e-3; L = 0.625;si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -400 -100

Hence,

vC (t) = K1e−100 t + K2e−400t V


and

vC '(0) = s1K1 + s2K2 = −100K1 − 400K2 =iC (0+)

C=

−0.3

0.04 ×10−3 = −7500

»A = [1 1;si(2) si(1)];»b = [0;-7500];»K = A\bK = -25 25

Therefore

vC (t) = −25 e−100t − e−400t( ) V

(b)»R = 50; C = 0.04e-3; L = 0.4;»si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -2.5000e+02 -2.5000e+02

Thus

vC (t) = (K1 + K2t)e−250t VFrom IC's,

vC (0) = K1 = 0

vC '(0) = s1K1 + K2 = K2 =iC (0+)

C= −7500

Therefore

vC (t) = −7500te−250 t V

(c)»L = 0.2;»si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+02 + 2.5000e+02i -2.5000e+02 - 2.5000e+02i

vC (t) = e−250 t A cos(250t) + Bsin(250t)[ ] VFrom ICs.

vC (0+) = A = 0

vC '(0) = −250A + 250B = 250B =iC (0+)

C= −7500

in which case B = –30. Thus

vC (t) = −30e−250t sin(250t) V


At t = 0-, he capacitor is open and the inductor is a short. This together with thecontinuity property implies vC(0-) = vC(0+) = 10 V and iL(0-) = iL(0+) = 1 A byinspection.

Now, for t = 0+, vin = 0, replace capacitor and inductor with a voltage source and acurrent source, respectively (values are those of the initial conditions). Solve for initialcapacitor current and initial inductor voltage to obtain:

vL(0+) = -10 ViC(0+) = iL(0+) – iR1 – iR2 = – 1 A

Notice that the resulting circuit is an undriven parallel RLC circuit with Req = 10//10 = 5Ω.»R = 5; C = 0.01; L = 4/3;si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -15 -5

Hence,

vC (t) = K1e−15t + K2e−5t V


and

vC '(0) = s1K1 + s2K2 = −15K1 − 5K2 =iC (0+)

C=

−10.01

= −100

»A = [1 1;-15 -5];»b= [10; -100];»K = A\bK = 5.0000e+00 5.0000e+00

vC (t) = 5e−15t + 5e−5t V

Verify with dsolve function in matlab: dsolve('D2y+20*Dy+75*y=0,y(0)=10,Dy(0)=-100')

»dsolve('D2y+20*Dy+75*y=0,y(0)=10,Dy(0)=-100')ans =5*exp(-5*t)+5*exp(-15*t)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

3

4

5

6

7

8

9

10


For t = 0-, there is no source present nor has there been a non-zero excitation. Hence,

vC(0-) = vC(0+) = 0 and iL(0-) = iL(0+) = 0

At t = 0+, replace capacitor and inductor by 0-valued voltage and current sources toobtain:

vL(0+) = 10 V, iC(0+) = 1 A

For t > 0, we have a driven parallel RLC circuit with vC(∞) = 10 V. Thus

»R = 5; C = 0.01; L = 4/3;si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -15 -5

and

vC (t) = K1e−15t + K2e−5t +10 V

From, the initial conditionsvC(0+) = 0 = K1 + K2 +10

and

vC '(0) = s1K1 + s2K2 = −15K1 − 5K2 =iC (0+)

C=

10.01

=100

»A = [1 1;-15 -5];»b= [-10; 100];»K = A\bK = -5.0000e+00 -5.0000e+00

in which case

vC (t) = −5e−15t − 5e−5t +10 V

To compute iL, note that iL(∞) = 1 A,

iL ( t) = K1e−15t + K2e−5t +1 A

From, the initial conditionsiL(0+) = 0 = K1 + K2 +1

and

iL '(0) = s1K1 + s2K2 = −15K1 − 5K2 =vL (0+)

L=

104 / 3

= 7.5

»A = [1 1;-15 -5];»b= [-1; 7.5];»K = A\bK = -2.5000e-01 -7.5000e-01

iL ( t) = −0.25e−15t − 0.75e−5t +1 A


Input to this circuit is a superposition of the inputs in problems 9.47 and 9.48. So, theoutput of the circuit here is a superposition of the output of the circuit in problems 9.47and 9.48:

vC (t) = 10e−15t +10e−5t −10 V

For t > 0, by linearity this is the difference of the zero-input circuit response (i.e., due tothe IC's as per problem 47) and the zero-state (zero ICs) as per problem 48.


vC(0-) = –60 V, iL(0-) = –0.1 A(b) By continuity property,

vC(0+) = –60, iL(0+) = –0.1(c)Replace capacitor by voltage source of value –60 V and inductor by current source ofvalue –0.1 A.

vL(0+) + vC(0+) = 60⇒ vL(0+) = 120

andiC(0+) + iR1(0+) – 1 – iL(0+) = 0

⇒ iC(0+) = 1

(d) Req = 120//600 = 100 Ω.»R = 100; C = 1e-3; L = 2;si = roots([1 1/(R*C) 1/(L*C)])si = -5.0000e+00 + 2.1794e+01i -5.0000e+00 - 2.1794e+01i»sig = -real(si(1)); wd = imag(si(1));

(e) Note that if the excitation of 60 V had remained forever, then iL(∞) would be 0.1 A.Therefore for the interval 0 < t < 1,

iL ( t) = e−5t A cos(21.794 t) + B sin(21.794t)[ ] + 0.1 A

(f)iL(0+) = A + 0.1 = –0.1

⇒ A = –0.2

iL' (0+) = −5A + 21.794B =1 + 21.794B =

vL (0+)L

= 60

Hence, B = 2.707 and

iL ( t) = e−5t −0.2cos(21.794t) + 2.707sin(21.794t)[ ] + 0.1 A

(g) For t > 1, the forcing function is zero iL(∞) = 0. Thus,

iL ( t) = e−5(t−1) A cos 21.794(t −1)( ) + Bsin 21.794(t −1)( )[ ] A

From part (f) of the exp(-5t) term, we can guess that iL(1) approximates 0.1 and also that

vC(1) approximates its steady state value of 60 V. These may be off by percent or two,

but are good enough for our engineering calculations. It follows that vL(1+)approximates –60 V.

iL (1+) = A = 0.1and

iL' (1+) = −5A + 21.794B = −0.5 + 21.794B =

vL (1+)L

= −30

Here B = –1.35. Hence

iL ( t) = e−5(t−1) 0.1cos 21.794( t −1)( ) −1.35sin 21.794( t −1)( )[ ] A

*SOLUTION 9.51. To find the initial conditions, use the following equivalent circuit at t =

0–.

By inspection iL (0+ ) = iL( 0− ) = 1 A and vC (0+ ) = vC( 0− ) = 5 V.To find the characteristic roots, set independent source to zero which means open

circuit the independent current source in figure P9.51. This leaves a series RLC with Rth= 10 Ω. Hence

s2 +Rth

Ls +

1

LC= s2 +100s + 2.5 ×104 = 0

Using MATLAB, we find»Rth = 10;C = 0.4e-3; L = 0.1;»s12=roots([1 Rth/L 1/(L*C)])s12 = -50 -50

Since for t > 0, the source is off, we use table 9.1, case 3 to obtain

iL (t) = K1 + K2t( )e−50 t A

It follows that 1 = iL (0+ ) = K1 and

diL

dt(0+ ) = −50K1 + K2 = −50 + K2 =

1

LvL (0+ ) =10vL (0+ )

To find vL (0+) we consider the equivalent circuit valid at 0+:

It follows that

vL (0+) = 5 −10 ×1 = −5 VHence

−50 + K2 = −50or K2 = 0. Finally

iL (t) = e−50tu(t) A

SOLUTION TO PROBLEM 9.52. To find the initial conditions, use the following equivalent

circuit at t = 0–.

By inspection iL (0+ ) = iL( 0− ) = 1 A and vC (0+ ) = vC( 0− ) = 5 V.To find the characteristic roots, set independent source to zero which means open

circuit the independent current source in figure P9.51. This leaves a series RLC with Rth= 10 Ω. Hence

s2 +RthL

s +1

LC= s2 + 20s + 5 ×103 = 0

Using MATLAB, we find»Rth = 10;C = 0.4e-3; L = 0.5;»s12=roots([1 Rth/L 1/(L*C)])s12 = -1.0000e+01 + 7.0000e+01i -1.0000e+01 - 7.0000e+01i

Since for t > 0, the source is off, we use table 9.1, case 2 to obtain

iL ( t) = e−10 t Acos(70t) + Bsin(70t)[ ] A

It follows that 1 = iL (0+) = A and

diLdt

(0+ ) = −10A + 70B = −10 + 70B =1L

vL (0+ ) = 2vL (0+ )

To find vL (0+) we consider the equivalent circuit valid at 0+:

It follows that

vL (0+) = 5 −10 ×1 = −5 VHence −10 + 70B = −10 or B = 0. Finally

iL ( t) = e−10 t cos(70t)u(t) A

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

SOLUTION TO PROBLEM 9.53. (a) vC (0−) = vC (0+) = 150 ×10−3 × 80 =12 V.

(b) iL(0-) = iL(0+) = 150 mA

(c) For these values at 0+, the independent current source becomes an open circuit.Replace the inductor and capacitor with current and voltage sources to represent theinitial conditions. Solve the resulting simple circuit to obtain:

vL(0+) = –6 = 12 – 120*0.15 V, iC(0+) = iL(0+) = –150 mA

(d)»Rth = 120;C = 2/9 * 1e-3; L = 0.6;»s12=roots([1 Rth/L 1/(L*C)])s12 = -1.5000e+02 -5.0000e+01

(e)

vC (t) = K1e−150 t + K2e−50t V


and

vC '(0) = s1K1 + s2K2 = −150K1 − 50K2 =iC (0+)

C= −675

A = [1 1;-150 -50];b= [12; -675];K = A\bK = 7.5000e-01 1.1250e+01

in which case

vC (t) = 0.75e−150 t +11.25e−50t V(f)

iL ( t) = −iC (t) = −CdvC (t)

dt=

29

×10−3 −150 × 0.75( )e−150 t + −50 ×11.25( )e−50 t[ ]= 0.025e−150t + 0.125e−50t A

SOLUTION TO PROBLEM 9.54. (a) At t = 0-, the independent current source is off, theinductor is a short circuit, and the capacitor is an open circuit. By voltage division,

vC(0-) = vC(0+) = (80/100)50 = 40 V(b)

iL(0-) = iL(0+) = 50/100 = 0.5 A

(c) At t = 0+, we have an independent current source. Also, we replace the inductor witha current source and the capacitor with a voltage source to represent the initial conditions.

1.5 – iL(0+) = iC(0+)⇒ iC(0+) = 1 A

Further,vC(0+) + 40 iC(0+) = 80 iL(0+) + vL(0+)

⇒ vL(0+) = 40 V

(d)Rth = 120;C = 2/9 * 1e-3; L = 0.6;s12=roots([1 Rth/L 1/(L*C)])s12 = -1.5000e+02 -5.0000e+01

(e) In steady state, the capacitor is open and the inductor is a short in which case, Xf = 80* 1.5 = 120 V.

vC (t) = K1e−150 t + K2e−50t +120 V

From, the initial conditionsvC(0+) = 40 = K1 + K2 + 120

and

vC '(0+) = −150K1 − 50K2 =iC (0+)

C= 4.5 ×103

A = [1 1;-150 -50];b= [-80; 4.5e3];K = A\bK = -5 -75

in which case

vC (t) = −5e−150t − 75e−50t +120 V

(f)

iL ( t) = 1.5 − iC (t) = 1.5 − CdvC (t)

dt= −0.16667e−150t − 0.83333e−50 t +1.5 A

SOLUTION TO PROBLEM 9.55.(a) At t = 0-, we replace the inductor by a short circuit and the capacitor by an opencircuit; hence

iL(0-) = iL(0+) = 1 Aand

vC(0-) = vC(0+) = (40)(1) – 20 = 20 V

(b) At t = 0+, we replace the capacitor by a voltage source of value 20 V and theinductor by a current source of value 1 A. Since the inductor current is 1 A and theindependent current source outputs 1 A, no current flows through the branch containingthe capacitor. Therefore,

iC(0+) = 0

Also, because of the zero current in the branch containing the capacitor, no voltage dropoccurs across the resistance in series with the capacitor. Therefore, the voltage across theindependent current source is vC(0+). Therefore,

vL(0+) = vC(0+) – 40*iL(0+) = – 20 V

(c) At steady state (large t), the capacitor becomes an open circuit and the inductorbecomes a short circuit. By inspection,

vC(∞) = 40 V

(d)Rth = 80;C = 1/15 * 1e-3; L = 0.1;s12=roots([1 Rth/L 1/(L*C)])s12 = -5.0000e+02 -3.0000e+02

vC (t) = K1e−500 t + K2e−300t + 40 V

(e) From, the initial conditionsvC(0+) = 20 = K1 + K2 + 40

and

vC '(0+) = −500K1 − 300K2 =iC (0+)

C= 0

A = [1 1;-500 -300];b= [-20; 0];K = A\bK = 3.0000e+01 -5.0000e+01

in which case

vC (t) = 30e−500t − 50e−300t + 40 V

SOLUTION TO PROBLEM 9.56.(a) At t = 0-, we replace the inductor by a short circuit and the capacitor by an opencircuit; hence

iL(0-) = iL(0+) = 1 Aand

vC(0-) = vC(0+) = (40)(1) – 20 = 20 V

(b) At t = 0+, we replace the capacitor by a voltage source of value 20 V and theinductor by a current source of value 1 A. Since the inductor current is 1 A and theindependent current source outputs 1 A, no current flows through the branch containingthe capacitor. Therefore,

iC(0+) = 0

Also, because of the zero current in the branch containing the capacitor, no voltage dropoccurs across the resistance in series with the capacitor. Therefore, the voltage across theindependent current source is vC(0+). Therefore,

vL(0+) = vC(0+) – 40*iL(0+) = – 20 V

(c) At steady state (large t), the capacitor becomes an open circuit and the inductorbecomes a short circuit. By inspection,

vC(∞) = 40 V

(d)Rth = 80;C = 62.5e-6; L = 0.1;s12=roots([1 Rth/L 1/(L*C)])s12 = -400 -400

vC (t) = K1 + K2t( )e−400t + 40 V

(e) From, the initial conditionsvC(0+) = 20 = K1 + 40 ⇒ K1 = – 20

and

vC '(0+) = −400K1 + K2 = 8000 + K2 =iC (0+)

C= 0

in which case

vC (t) = − 20 + 8000t( )e−400 t + 40 V

SOLUTION TO PROBLEM 9.57.Step 1:

dvC/dt = iC/C = 2iC

diL/dt = vL/L = 0.5vL

Step 2: iR = vC/2. From KCL, iL – iR – ic = 0. Therefore, ic = iL – vc/2.

Step 3: Similarly, vin – vL – vc = 0, which implies vL = vin – vC. Hence,

dvC/dt = 2 iL – vC

diL/dt = 0.5 vin – 0.5 vC

Step 4. Eliminate terms in vC (see equation 9.47 in text) to obtain:

d2iL/dt2 + diL/dt + iL = 0.5 dvin/dt + 0.5 vin

SOLUTION TO PROBLEM 9.58.(a) At t = 0-,

vC(0-) = vC(0+) = 0iL(0-) = iL(0+) = 0

For t between 0 and 1, we have a parallel RLC circuit, with Req being the parallelcombination of the two 21.1333 Ω resistors.

R = 21.1333/2; C = 15.7729e-3; L = 0.1;si = roots([1 1/(R*C) 1/(L*C)])si = -3.0000e+00 + 2.5000e+01i -3.0000e+00 - 2.5000e+01i

If there were no further switchings, then Xf = 1 A. Hence,

iL ( t) = e−3t A cos(25t) + Bsin(25t)[ ] +1 A

Applying the initial conditions,

iL(0+) = A + 1 = 0 ⇒ A = –1

and

iL' (0+) = −3A + 25B = 3 + 25B =

vL (0+)L

= 0

Hence, for 0 ≤ t < 1,

iL ( t) = −e−3t cos(25 t) + 0.12sin(25 t)[ ] +1 A

Now, for the next interval, we need initial conditions. These are obtained from the aboveequation for iL(t) at t = 1.

iL(1) = e-3[-cos(25) –3/25 sin(25)] + 1 = 0.9514and

iL' (1) = 3e−3 cos(25) + 0.12sin(25)[ ] − e−3 −25sin(25) + 0.12 × 25cos(25)[ ] = −0.1671

The circuit is still a parallel RLC circuit, but now there is no source and R = 1.268:R = 1.268; C = 15.7729e-3; L = 0.1;si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+01 + 3.0002e+00i -2.5000e+01 - 3.0002e+00i

Hence,

iL ( t) = e−25(t−1) A cos(3(t −1)) + B sin(3(t −1))[ ]

Using initial conditionsA = 0.9514

–0.1671 = –25A+ 3B ⇒ B = 7.8726

Thus, for 1 ≤ t < 2

iL ( t) = e−25(t−1) 0.9514cos(3(t −1)) + 7.8726sin(3( t −1))[ ] A

(b)In period between 1 and 2 seconds, the response has a time constant of 1/25. So, when t =2, 25 time constants would have passed from the time the switch is turned (t = 1). Thismeans that the L and C currents and voltages would have settled almost identically totheir values at 0–. A similar argument can be made for the other cycle. Thus the overallresponse effectively becomes a periodic response equal to the response over 0 ≤ t < 2that reflects the periodicity of the switching.

(c)

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Time in s

Indu

ctor

cur

rent

in A

TextEnd


dvc/dt = ic/C = 0.707icdiL/dt = vL/L = 0.707vL

To find expressions for iC and vL we use the following figure.

From this resistive circuit, iC = – vC/1 + iL and vL = vin – 1×iL – vC

˙ v C = −0.707vC + 0.707iL˙ i L = −0.707vC − 0.707iL + 0.707vin

Using equation 9.47,

d2vC

dt2 +1.414dvCdt

+ vC = 0.5vin

»si = roots([1 sqrt(2) 1])si = -7.0711e-01 + 7.0711e-01i -7.0711e-01 - 7.0711e-01i

(b) At steady state (large t), vC = 0.5

vC(t) = e-0.707t[A cos(0.707t) + B sin(0.707t)] + 0.5 V

At t = 0–,vC(0-) = vC(0+) = A + 0.5 = 0 ⇒ A = –0.5

vC' (0) =

iC (0)C

=iL (0) − vC (0)

C= 0 = −0.707A + 0.707B

⇒ B = –0.5

Thusvc(t) = e-0.707t[–0.5 cos(0.707t) – 0.5 sin(0.707t)] + 0.5 V


dvC

dt= 3iC

diLdt

=vL

3

To eliminate iC and vL, consider

HenceiC = iin – vC/2 – iL

vL = vC – 12iL

˙ v C = −1.5vC − 3iL + 3iin

˙ i L =1

3vC − 4iL


d2vC

dt2 + 5.5dvCdt

+ 7vC = 3iin' +12iin

»si = roots([1 5.5 7])si = -3.5000e+00 -2.0000e+00

Note: iin’(t) = 0 for t>0.

At t = 0-, current source is off, inductor is a short circuit, and capacitor is an open circuit.

vc(0-) = vc(0+) = 0iL(0-) = iL(0+) = 0

At t = 0+, the current source is on. Replace the inductor and capacitor by current andvoltage sources to represent the initial conditions. Hence, iC(0+) = 1 A. At t = ∞,vC (∞) =1× 12 / / 2( ) =1.7143. Thus,

vC (t) = K1e−3.5t + K2e−2t +1.7143 VTo find the constants,

vC (0) = K1 + K2 +1.7143 = 0and

vC' (0) =

iC (0)C

= 3 = −3.5K1 − 2K2

»A = [1 1;-3.5 -2];»b = [-1.7143;3];»K = A\bK = 2.8573e-01 -2.0000e+00

Therefore

vC (t) = 0.28573e−3.5t − 2e−2t +1.7143

(b) From equation 9.47b,

d2iLdt2 + 5.5

diLdt

+ 7vC = iin

Since iL (∞) = 0.14286 A,

iL ( t) = K1e−3.5t + K2e−2t + 0.14286 A


iL (0) = K1 + K2 + 0.14286 = 0and

iL' (0) =

vL (0)L

=vC (0) −12iL (0)

L= 0 = −3.5K1 − 2K2

»A = [1 1;-3.5 -2];»b = [-1/7; 0];»K = A\bK = 1.9048e-01 -3.3333e-01

Therefore,

iL ( t) = 0.19048e−3.5t − 0.3333e−2 t + 0.14286 A


dvC1/dt = 0.5iC1

dvC2/dt = 0.5iC2

Writing a node equation,iC1 + vC1/0.5 + (vC1 – vC2)/0.5 = 0

⇒ iC1 = –4vC1 + 2vC2

By symmetry,iC2 = –4vC2 + 2vC1

Hence

˙ v C1 = −2vC1 + vC 2

˙ v C 2 = vC1 − 2vC2

From equation 9.47a,

d2vC

dt2 + 4dvCdt

+ 3vC = 0

(b)»si = roots([1 4 3])si = -3 -1Hence

vC (t) = K1e−3t + K2e−t

(c)

vC1(0) = 2 ⇒ K1 + K2 = 2

and

vC1' (0) =

iC1(0)C

=2(vC 2(0) − vC1(0)) − 2vC1(0)

C=

8 − 8C

= 0 = −3K1 − K2

Hence

vC (t) = −e−3t + 3e−t V

*SOLUTION P9.62. (a) For this problem we need to define a voltage at the output of thefirst op amp as shown below.

Also, let us relate the input and output voltages for an arbitrary leaky integrator as shownbelow.

We now write a node equation at the inverting terminal of the op amp. Here

va

Ra+

vb

Rb+ C

dvb

dt= 0

Equivalentlydvb

dt= −

vb

RbC−

va

RaC(1)

Now we apply the formula of (*) to the second stage of our given op amp circuit toobtain:

dvout

dt= −

vout

RC2−

v1

RC2(2)

where R = 1 MΩ and C2 is to be determined.Applying the formula of (*) to the first stage we obtain:

dv1

dt= −

v1

RC1−

vs

R1C1(3)

The equations (2) and (3) form a coupled set of state equations which we can write as

˙ v out

˙ v 1

=

− 1RC2

− 1RC2

0 − 1RC1

vout

v1

+

0−1

R1C1

vs

Using equation 9.47a of text we can write down the characteristic equation as

s2 +1

RC2+

1

RC1

s +

1

RC1RC2= s +

1

RC2

s +

1

RC1

= 0

We require that the natural frequencies be –4 and –12 in which case

1

RC2= 4,

1

RC1=12

From MATLAB»R = 1e6;»C2 = 1/(R*4)C2 = 2.5000e-07»C1 = 1/(R*12)C1 = 8.3333e-08(b) For this part, the overall dc gain must be 10. The dc gain of the second stage is –1.

The dc gain of the first stage must be −10 = −106

R1; hence R1 =100 kΩ.

(c) Since the roots are distinct and real, for t > 0

vout (t) = K1e−4t + K2e−12t + X f = K1e−4t + K2e−12 t + 10 V

where X f =10 V by part (b). The problem states that the capacitor voltages are initially

zero. Hence

0 = vC2(0)= vout (0) = K1 + K2 +10Equivalently

K1 + K2 = −10 (1)Also,

dvout

dt(0) =

1

C2iC2(0)= −4K1 −12K2 = 0 (2)

because iC2(0)= 0 . This is so because vC1(0)= 0 = v1(0) means no current flowsthrough the 1 MΩ input resistor to stage 2. This fact and the fact that vC2 (0) = 0, means

that no current flows through C2.

From equation 2, K1 = –3 K2. Substituting into equation (1) yields K2 = 5 and

hence K1 = –15. Finally,

vout (t) = −15e−4t + 5e−12 t +10 V

*SOLUTION 9.63. This problem requires the characteristic equation in terms of A. For

this we may set vin = 0 and the circuit becomes the one given below. Note the new label

v1.

The first step is to write a node equation at v1:

v1

100+10−4 d

dtv1 − vout( ) +

1

100v1 −

vout

A

= 0

Equivalentlydv1

dt−

dvout

dt=

100

Avout − 200v1 (*)

Now we write a node equation at vg = vout/A. Here,

10−4

A

dvout

dt+

1

100

vout

A− v1

= 0

Equivalently,dvout

dt= −100vout +100 Av1 (**)

Let us put (*) and (**) in matrix form to obtain:

1 −1

0 1

˙ v 1˙ v out

=

−200 100 / A

100A −100

v1

vout

We can now solve this to obtain the state equations

˙ v 1˙ v out

=

1 1

0 1

−200 100 / A

100A −100

v1

vout

=

−200 +100A −100 +100 / A

100 A −100

v1

vout

Compare these equations with equation 9.37 and use the formula of 9.47b to obtain the

following second order differential equation in vout:

d2vout

dt2 + 300 −100A( ) dvout

dt+104 vout = 0

The discriminant of this characteristic equation is plotted below for 0 < A < 3. For valuesof A > 3, the circuit is unstable. A negative value of the discriminant indicatesunderdamped (1 < A < 3) and a positive value overdamped (0 < A < 1). For A = 1, wehave critical damping.

Solution 9.64 We can write two state equations as follows:(i) From the definition of a capacitor,

dvC1

dt=107iC1

dvC 2

dt= 109iC 2

(ii) From KVL and Ohm's law

dvC1

dt=106 vi − vC1 − vC 2( ) = −106vC1 −106vC 2 +106vi

dvC 2

dt= 108 vi − vC1 − vC 2 − 0.01vC 2( ) = −108vC1 −1.01×108vC2 +108vi

Casting these two equations into a second order differential equation, as described in thetext:

d2vC 2

dt2 +1.02 ×108 dvC 2dt

+1012vC 2 =108 dvidt

The characteristic equation for this differential equation has real roots:»si = roots([1 1.02e8 1e12])si = -1.0199e+08 -9.8049e+03

Since the capacitors become open circuits, vC2(∞) = 0 and vC1(∞) = vi.

vC 2 t( ) = K1e−1.02×108 t + K2e−9.8×103 t V

Applying IC's:

vC 2 0( ) = K1 + K2 = 0Also,

vC 2' (0) =

iC 2(0+)C2

=0.1

10−9 = −1.02 ×108K1 − 9.8 ×103K2

Thus

vC 2' (0) =1 = −1.02K1 − 9.8 ×10−5K2

»b = [0; 1];»A = [1 1; -1.02 -9.8e-5];»K = A\bK = -9.8049e-01 9.8049e-01

vC 2 t( ) = −0.9805e−1.02×108 t + 0.9805e−9.805×103 t V

»t = 0:1/(abs(s1)*100):100/abs(s1);vc = -0.9805*exp(s1*t) + 0.9805*exp(s2*t);»plot(t,vc)»grid»xlabel('Time in s')»ylabel('vout in V')

0 0.2 0.4 0.6 0.8 1x 10-6

0

0.2

0.4

0.6

0.8

1

Time in s

vout

in V

TextEnd

Solution 9.65. First, derive the differential equation by writing state equations:

dvC

dt= 3iC

diLdt

=vL

3

Now, assume that the capacitor is a voltage source and the inductor is a current source,and write by KCL

iC = −vC2

+vCRN

− iL

And by KVL:

vL = vC −12iL

Substitute into the differential equations:

dvCdt

= −32

+3

RN

vC − 3iL

diLdt

=1

3vC − 4iL

Using equation 9.47 we obtain

d2vC

dt2 − −32

+3

RN

− 4

dvCdt

+ −32

+3

RN

−4( ) +1

vC = 0

or equivalently

d2vC

dt2 + 5.5 −3

RN

dvCdt

+ 7 −3

RN

vC = 0

For constant amplitude oscillations, the middle term should be zero, which means that RN

= 3/5.5 = 0.54545 Ω. Thus the negative resistance is – RN = –0.54545 Ω.

SOLUTION 9.66. The problem data is

i1(t) = Im sin(ωt + θ) A

R1 = 500 + 100(Im – 0.01), R2 = 500

Suppose it starts out with exponentially growing amplitude. R1 will increase withincreasing amplitude. This changes the location of the roots until equilibrium is reachedwhere the roots and the amplitude are stable. This is achieved when the roots of thecharacteristic equation describing the output voltage are purely imaginary, i.e.,

(R1 – R2)/(R1R2C) = 0

⇒ R1 = R2 = 500 = 500 + 100(Im – 0.01)

⇒ Im = 0.01

(a)ω0 = 1/[(500)(1µ)] = 2 k-rad/s

(b) Amplitude of i1 is 0.01. Thus i1 = 0.01 sin(ω0t) A. Let v1 = Vm cos(ω0t + φ) V.

Then

dv1/dt = –ω0Vmsin(ω0t + φ) = (0.01/C) sin(ω0t)

⇒ Vm = 0.01/(ω0C) = 5 V and φ = π rad

Now, v2 + v1 + i1R1 – 3v2 = 0 ⇒ v2 = v1/2 + i1R1/2. Finally,

vout = 3v2 = 1.5v1 +1.5R1i1 = 7.5cos( 0t + π) + 7.5sin( 0t) V

Hence the amplitude of vout(t) is: 7.5 2 V.

SOLUTION 9.67. (a)

n =1

RC=104 rad/s

(b) From equation 9.47a,d2v1

dt2 +1

R2C2 v1 = 0

Hence,

v1 t( ) = A cos1

RCt

+ Bsin

1RC

t

V

Using the value of v1(0) we have,

v1 0( ) = A = 5

To compute the second initial condition,

v1' 0 +( ) =

iC1(0+)C

=iR1(0+)

0.1×10−6 = 104 B

But,

iR1(0+) =3v2(0+) − v2(0+) − v1(0+)

R1=

2v2(0+) − v1(0+)

103 = −0.005 Amps

Hence

v1' 0 +( ) =

−5 ×10−3

0.1×10−6 = 104 B

Hence B = –5.

v1 t( ) = 5cos 10,000t( ) − 5sin 10,000 t( ) V

0 0.2 0.4 0.6 0.8 1x 10-3

-8

-6

-4

-2

0

2

4

6

8

Time in s

Vol

tage

v1

in V

TextEnd

*SOLUTION 9.68. For this problem, R2 = 10 kΩ should be Rf = 10 kΩ and R1 = 1 kΩshould be R2 = 1 kΩ.(a) For sustained sinusoidal oscillation, R1 = R2 = 1 kΩ. From equation 9.59,

ω0 =1

R1C=104 rad/s or 1.5915 kHz

(b) From figure P9.68, to obtain an R1 = 1 kΩ, IR1,peak = 0.2 mA. Therefore,

iR1(t) = 0.2sin(ω0t + θ) mA for appropriate θ. Since Cdv1

dt= iC1 = iR1 = 0.2sin(ω0t + θ)

mA, we know that v1 has the following form:v1(t) = V1m cos(ω0t + φ)

In which case

Cdv1

dt= C

d

dtV1m cos(ω0t + φ)( ) = CV1mω0 sin(ω0t + φ+ π) = 0.2sin(ω0t + θ) mA

Therefore CV1mω0 = 0.2 ×10−3. It follows that V1m = 0.2 volts. Here V1m is the peakvalue of the sinusoid. However, the op amp peak output voltage with respect to ground,as shown in problem 66, is 1.5 2V1m . Also, for such a small amplitude, we expect the

output waveform to be quite close to sinusoidal. By choosing a different lamp (R1) witha different characteristic, we can obtain larger peak output voltages.

SOLUTION 9.69. (a) Note that the capacitor is like an open circuit and the inductor islike a short circuit at t=0-. Thus, we can obtain the capacitor voltage by voltage division:

vC 0 −( ) = 10 ×45

= 8 = vC 0 +( )

Similarly, the inductor current is obtained by applying Ohm’s Law:

iL 0 −( ) =105

= 2 = iL 0 +( )

(b) Here, we note that the new initial conditions are just 2.5 times the values that we justobtained in part (a). This can be achieved by simply changing the input voltage source,from 10 to 25 V.

1


SOLUTION 10.1 Using KCL, we can write

CdvCdt

+vCR

= iin t( )

Dividing by C:

dvCdt

+1

RCvC =

iin t( )C

We know that iin t( ) = 20sin 400t( ) mA, which can be represented by a complex exponential,

iin t( ) = Re 20e j400te− jπ/2[ ] mA. For convenience we will simply let iin t( ) = 20e j400 te− jπ/2 mA, knowing

that we must take the real part to complete our solution. The output voltage will also be reparesented as a

complex exponential:

vC t( ) = Vme j 400t+( ) = Vme j400te j

Substituting this expression into the differential equation and canceling e j400 t :

j400Vme j +VmRC

e j =20 ×10−3e− jπ/2

C

Thus

Vme j 1RC

+ j 400

=

20 ×10−3

C ⇒ Vme j =

− j40001000 + j400

= 3.714∠− 111.8o

where the values for R = 100 Ω and C = 5 mF were substituted in. Thus,

Vm = 3.714

= 0 − tan−1 4001000

= −111.8°

Taking into account a 90o phase shift we obtain

vC t( ) = 3.714cos 400t −111.8°( ) = 3.714sin 400t − 21.8°( ) V

and

iout t( ) = 18.57sin 400t − 21.8°( ) mA

SOLUTION 10.2 From KCL and component definitions:

2

iin t( ) −vL25

− iL = 0 ⇒ 0.125

diLdt

+ iL = iin t( ) ⇒ diLdt

+ 250iL = 250iin t( )

We represent the input signal by the complex exponential: iin t( ) = 0.2e j250 t A and the unknown current

can be represented as iL ( t) = IL e j 250t+( ) .

Substituting this into the differential equation and canceling e j250 t :

j250IL e j + 250IL e j = 50

Thus

ILe j j250 + 250( ) = 50 ⇒ IL e j =50

250 + j250= 0.14142∠ − 45o

and

IL = 0.141, = −45° ⇒ iL t( ) = 0.141cos 250t − 45°( ) A

SOLUTION 10.3. Construct differential equation by KVL and device definitions:

vin t( ) − 0.5diLdt

− 200iL = 0 ⇒ diLdt

+ 400iL = 2vin t( )

We represent vin t( ) as the complex exponential function, vin t( ) = 20e j400t V. The current in the inductor

has the form: iL = ILe j 400t+( ) . Substituting into the differential equation and canceling e j400 t :

j400ILe j + 400IL e j = 40

Thus

ILe j j + 400( ) = 40 ⇒ IL e j =40

400 + j 400= 0.070711∠− 45o

and

IL = 0.0707, = −45°, ⇒ iL t( ) = 70.7cos 400t − 45°( ) mA

Hence,

vout t( ) = 14.14cos 400t − 45°( ) V

SOLUTION 10.4. Construct differential equation using KVL and device definitions:

3

vin t( ) − vC − CdvCdt

R = 0 ⇒ RCdvCdt

+ vC = vin t( )

The output voltage is defined as:

vout t( ) = vin t( ) − vC t( )

This means that finding vC is enough to be able to obtain the output voltage. The input voltage is

represented by the complex exponential:

vin t( ) = 20e j250te− jπ/2 V

and vC t( ) = Vme j 250t+( ) . Substituting into the differential equation, dividing by e j250 t , and rearranging:

j250RCVCe j + VCe j = − j20 ⇒ VCe j j250RC +1( ) = − j20

⇒ VCe j =− j201 + j

= 14.142∠− 135o

Now

Voutej = − j20 − VCe j ⇒ Voute

j =10 − j10 =14.142∠ − 45o

Thus, in the time-domain,

vout t( ) = 14.142cos 250t − 45°( ) V

SOLUTION 10.5. The circuit is identical to that of problem 10.1. Thus,

RCdvsdt

+ vs = Ris t( ) ⇒ Ris t( ) − RCdvsdt

= vs

Moreover, the complex exponential solution is given by

vs(t) = Vme j = 223.6e− j63.43o V

Hence

R − jRC223.6e− j63.43o= 223.6e− j63.43o

= 100 − j200

i.e.,

R − jRC(100− j200) = R − 200RC − j100RC =100 − j200

4

Thus RC = 2 s, R =100 + 200RC = 500 Ω, and C = 4 mF.

SOLUTION 10.6. First use KVL

vin − vL − vc − vout = 0 ⇒ vin − LdiLdt

−1C

iLdt−∞

t

∫ − RiL = 0

Differentiating this equation, rearranging, and dividing by L,

d2iLdt2 +

RL

diLdt

+1

LCiL =

1L

dvindt

We represent the input signal with = 104 as vin t( ) =100e j te− jπ/2 V and iL ( t) = IL e j te j A.

Substituting these two expressions into the differential equation and dividing out e j t :

− 2IL e j + jRL

IL e j +1

LCIL e j =

− j100× jL

⇒ ILe j 1LC

− 2 + jRL

=108

⇒ ILe j =108

−75 ×106 + j108= −0.48 − j0.64 = 0.8∠− 126.87o

Solving for the magnitude and angle (by hand or using MATLAB):

iL t( ) = 0.8cos 10,000t −126.87°( ) = 0.8sin 10,000t − 36.87°( ) A

and

vout t( ) = 80sin 10,000t − 36.87°( ) V

SOLUTION 10.7. Using standard reference directions, from KCL and component definitions,

iin = iR + iL + iC =voutR

+ iL (0) +1L

vout ( ) d0

t

∫ + Cdvout

dt

Taking a second derivative and dividing by C yields

d2vout

dt2 +1

RCdvout

dt+

1LC

vout =1C

diindt

5

We now let ω = 2500 rad/s and represent iin(t) by the real part of the complex exponential 0.02e j t .

Further we represent vout(t) as the real part of the complex exponential Vme j( t+ ) = Vme j te j .

Substituting these expressions into the differential equation and taking the indicated derivatives yields

Vm( j )2e j te j +VmRC

j e j te j +1

LCVme j te j =

0.02C

j e j t

Observe that e j t cancels out on both sides of this equation producing

Vme j 1LC

− 2 +jRC

=

j0.02C

Hence

Vme j =j0.02

C1

LC− 2 +

j

RC

=1.28 + j0.96

This was obtained using MATLAB as follows:

»w = 2500;

»L = 40e-3;

»C = 1e-6;

»R = 100;

»Vout = j*0.02*w/(1/L - C*w^2 +j*w/R)

Vout = 1.2800e+00 + 9.6000e-01i

»magVout = abs(Vout)

magVout = 1.6000e+00

»angVout = angle(Vout)*180/pi

angVout = 3.6870e+01

Therefore

vout (t) = 1.6cos(2500t + 36.87o) V

SOLUTION 10.8. (a) From KCL: 3∠15° − 5∠45°− I = 0 ⇒ I = 3∠15°− 5∠45° . In MATLAB,

»Ibar = 3*exp(j*15*pi/180) - 5*exp(j*45*pi/180)

6

Ibar =-6.3776e-01 - 2.7591e+00i

»abs(Ibar)

ans =2.8318e+00

»angle(Ibar)*180/pi

ans =-1.0302e+02

Therefore,

i t( ) = Re Ibar = 2.83cos t −103°( ) A

(b) From KCL,

»Ibar = (1 + 2*j) - (-2 + j*6)

Ibar =

3.0000e+00 - 4.0000e+00i

»abs(Ibar)

ans = 5


ans =-5.3130e+01

Therefore i t( ) = 5cos 50πt − 53°( ) A.

SOLUTION 10.9. We define a Gaussian surface encompassing the three bottom nodes. Thus, KCL

dictates that the sum of 4 currents be zero:

»Ibar = -(-2-j*8) + (3 +j*12) + 10

Ibar =

1.5000e+01 + 2.0000e+01i

»abs(Ibar)

ans =

25


ans =

5.3130e+01

Therefore i t( ) = 25cos 1000t + 53.13°( ) A.

SOLUTION 10.10. First represent the time-domain functions as phasors:

7

V1 = 2∠0°, V2 = 2 2∠− 45°

Then, by KVL

»V1 = 2; V2 = 2*sqrt(2)*exp(-j*pi/4);

»VL = V1 - V2

VL =

0 + 2.0000e+00i

Therefore, VL = 2∠90° V and vL t( ) = 2cos t + 90°( ) = −2sin( t) V

SOLUTION 10.11. Apply KVL by simply following the loop defined by the independent voltage sources:

Vx = 4 j − 2 j −1−1+ (1− j) − (1+ j) = −2

SOLUTION 10.12. First note that Iout =VRR

. VR = − j10 + (5 − j5) −10 ⇒ VR = −5 − j15 V.

Thus, Iout =−5 − j15

5= −1− j 3 = 3.16∠ −108.4° ⇒ iout (t) = 3.16cos 500 t −108.4°( ) A.

SOLUTION 10.13. Using a Gaussian surface,

Iy = −(2 + j3) − (1+ j2) + (1− j5) = −2 − j10 = 10.2∠− 101.3o A.

Therefore, i t( ) =10.2cos 2000 t −101.3°( ) A. Now applying KVL,

Vx = (2 + j2) + (2 + j 3)− (1− 4 j) = 3 + 9 j

»Vx=2+2*j+2+j*3-1+4*j

Vx =

3.0000e+00 + 9.0000e+00i

»abs(Vx)

ans =

9.4868e+00

»angle(Vx)*180/pi

8

ans =

7.1565e+01

Thus, vx t( ) = 9.487cos 2000 t + 71.6°( ) V.

SOLUTION 10.14. (a) At = 1000 rad/s,

YC ( j ) = j C = j4.7 ×10−3 ⇒ C =1.496 F

So, at = 50 , YC ( j ) = j C = j2.3499 ×10−4 S ⇒ ZC ( j ) = − j4255 Ω .

(b) At = 1000 rad/s, ZL = j L = j18.85 ⇒ L = 6 mH. Since impedance is proportional to

frequency, multiplying the frequency by 20 means the impedance is multiplied by 20. Thus, at

= 20 × (103π) rad/s,

ZL = j18.85 × 20 = j 377 ⇒ YL = − j2.652 mS

SOLUTION 10.15. The input voltagephasor as 2∠0° . By inspection:

I1 = j C VC = j10 × 0.1× 2 = j2

From the definition of a dependent V-source and an inductor:

I2 =5I1j L

=j10j2

= 5

Finally,

Vout = 5 × 2 I1 + I2( ) = 10 5 + j2( ) = 50 + j20 = 53.85∠21.8°

vout t( ) = 53.85cos 10t + 21.5°( ) V

SOLUTION 10.16. Using KCL and the definition of a resistor:

V10Ω = 10 6∠0° − 3∠90°( ) = 10(6− 3j) = 60 − j 30 V

Thus,

Z j 0( ) =60 − j 30

j3= −10 − j20 Ω

9

And, the combination of this Z( j 0) with the 10 Ω resistance, at this frequency, is 10 – j5.

»Z=-10-j*20; R = 10;

»ZZ = R*Z/(R+Z)

ZZ =1.0000e+01 - 5.0000e+00i

»C = 1/(-5*j*j*2000*pi)

C =3.1831e-05

This is equivalent to a 10 Ω resistor in series with a 31.83 µF capacitor.

SOLUTION 10.17.

»w = 2*pi*60;

»VL = 3 +12*exp(-j*30*pi/180) + 6 -12*exp(j*30*pi/180)

VL =9.0000e+00 - 1.2000e+01i

»ZL = j*w/60

ZL =0 + 6.2832e+00i

»IL = VL/ZL

IL =-1.9099e+00 - 1.4324e+00i

»abs(IL)

ans =2.3873e+00

»angle(IL)*180/pi

ans =-1.4313e+02

Therefore, iL ( t) = 2.387cos(120πt −143.1o) A.

Solution 10.18 (a)

First represent the inputs with their phasors: Is1 =10∠30° = 8.66 + j5 and Vs2 =100∠0° .

The admittance of the RC combination is: Y1 =1R

+ j C = 0.1+ j0.1. Using superposition and noting that

a 0-volt V-source is a short circuit and a 0-amp current source is an open circuit,

Ix = Y1Vs2 − Is1 =1.3397 + j5 = 5.1764∠75o A

Therefore,

ix t( ) = 5.176cos 100t + 75°( )(b) Let ix1 be the contribution to ix generated by the current source and ix2 the contribution generated by

the voltage source. Then ix1(t) = −10cos(50t + 30o) A, and ix2(t) = −14.142cos(100 t + 45o) A since

10

»Vs2 = 100;

»w = 100;

»R = 10; C = 1e-3;

»Y1 = 1/R + j*w*C

Y1 =1.0000e-01 + 1.0000e-01i

»Ix2 = Y1*Vs2

Ix2 =1.0000e+01 + 1.0000e+01i

»abs(Ix2)

ans =1.4142e+01

»angle(Ix2)*180/pi

ans =45

Therefore ix (t) = −10cos(50t + 30o) −14.142cos(100t + 45o) A.

Solution 10.19 First of all, write out the given values: = 200 rad/s, I1 = 0.5∠90° = 0.5 j A, and

Vs2 = 4∠0° . From KVL Vs1 = 3I1 + j LI1 +I1

j C+ Vs2 which leads to:

»w = 200; I1 = 0.5j; Vs2 = 4;

»Vs1 = 3*I1 + j*w*0.04*I1 + I1/(j*w*1e-3) +Vs2

Vs1 =2.5000e+00 + 1.5000e+00i

»abs(Vs1)

ans =

2.9155e+00

»angle(Vs1)*180/pi

ans =

3.0964e+01

Therefore vs1(t) = 2.9155cos(200t + 30.964o) V.

Solution 10.20 The phasor for the input can be written as = 1000 rad/s and Vin = 10∠60° = 5 + j8.66

V. The currents can be obtained easily by applying Ohm’s law for phasors:

11

IR =10∠60

500= 0.02∠60° A, IL =

10∠60j1000 × 0.25

= 0.04∠− 30° A, and

IC = 10∠60 × j1000 × 2 ×10−6( ) = 0.02∠150° A. Thus

Iin = 0.02∠60° + 0.04∠ − 30° + 0.02∠150° = 0.0283∠15° A

and

iin t( ) = 0.0283cos 1000t +15°( ) A

using the following MATLAB code:

»Vin = 10*exp(j*60*pi/180)

Vin =5.0000e+00 + 8.6603e+00i

»R = 500; L = 0.25; C = 2e-6;

»w = 1e3;

»IR = Vin/500

IR =1.0000e-02 + 1.7321e-02i

»IL = Vin/(j*w*L)

IL =3.4641e-02 - 2.0000e-02i

»IC = j*w*C*Vin

IC =-1.7321e-02 + 1.0000e-02i

»Iin = IR + IL + IC

Iin =2.7321e-02 + 7.3205e-03i

»abs(Iin)

ans =2.8284e-02

»angle(Iin)*180/pi

ans =1.5000e+01

Solution 10.21 In MATLAB

»Iin = -100*j*1e-3; R = 100;

»L = 0.04; C = 2e-6; w = 2500;

»VR = R*Iin

VR =0 - 1.0000e+01i

»VL = j*w*L*Iin

VL =10

»VC = Iin/(j*w*C)

VC =-20

»Vin = VR + VL + VC

12

Vin =-1.0000e+01 - 1.0000e+01i

»abs(Vin)

ans =1.4142e+01

»angle(Vin)*180/pi

ans =-135

Therefore, vin t( ) =14.14cos 2500t −135°( ) V.

Solution 10.22 (a) Here, i1 t( ) = 0.6cos 200t( ) A and vout t( ) = 20sin 200t( ) = 20cos 200t − 90°( ) V.

For = 200 rad/s, the phasors are by inspection: I1 = 0.6∠0° A, Vout = 20∠ − 90° V .

(b) Write down the resistor, inductor, and capacitor current phasors, given Vout:

IR =20∠− 90

1/0.03= 0.6∠ − 90° = − j0.6

IL =20∠− 90j200 × 0.1

=1∠ −180° = −1

IC = 20∠ − 90 × j200 × 0.4 ×10−3( ) = 1.6∠0°

Now, by KCL

I2 = IR + IL + IC − I1 = − j0.6

where we have substituted the above values of branch currents. The time-domain function is:

i2 t( ) = 0.6cos 200t − 90°( ) = 0.6sin 200t( ) A

Solution 10.23 (a) = 400 rad/s and Vin = 20∠0° V. We can easily use the voltage divider formula for

phasors and substitute values to obtain:

Vout =200

200 + j LVin =

201 + j

= 14.14∠ − 45°

in which casevout t( ) = 14.14cos 400t − 45°( ) V

(b) = 250 rad/s and Vin = 20∠− 90° = − j20 V. Again, we can easily use voltage division:

Vout =400

400 +1

j C

Vin =20∠ − 90°

1− j=14.14∠− 45° V

Thus, in the time-domain,

vout t( ) = 14.142cos 250t − 45°( ) V

13

Solution 10.24 (a) = 10,000 rad/s and Vin = 100∠− 90° = − j100 V. Apply the voltage divider

formula:

Vout =100

100 + j L + 1j C

Vin =100

100 + j100 − j0.25Vin = 80∠ −126.87°

The steady-state response is thus,

vout t( ) = 80sin 10,000t − 36.87°( ) V

The phasor method provides for a much easier way of obtaining the steady-state response.

(b) Here, = 2500 rad/s and Iin = 0.02∠0° A. Now, apply current division:

IR =0.01

0.01 + j C + 1j L

Iin =0.01

0.01 + j0.0025 − j0.01Iin = 0.016∠36.86°

By Ohm’s law:

Vout = 100IR =1.6∠36.87° V

Therefore

vout (t) = 1.6cos(2500t + 36.87o) V

Solution 10.25 (a) It is easier to find the admittance first:

Yin ( j100) =1

j L+ j C = − j0.2 + j12.5 = j12.3

⇒ Zin ( j100) = − j0.0813

(b) Yin ( j100) =1

25 j=

1j100 × 0.05

+ j100C . Hence, in MATLAB,

»C = (1/(25*j) - 1/(j*w*L) )/(j*100)

C = 1.6000e-03

Solution 10.26 (a)

Zin ( j ) =1

j C+

R × j LR + j L

=− jC

+j RL

R + j L= 9.975∠− 0.0072°

»R = 10; L = 0.1; C = 1e-3; w = 2e3;

»Zin = -j/(w*C) + j*w*R*L/(R + j*w*L)

Zin = 9.9751e+00 - 1.2469e-03i

14

»abs(Zin)

ans = 9.9751e+00

»angle(Zin)*180/pi

ans = -7.1620e-03

(b) As the frequency increases, the capacitor becomes a short circuit and the inductor becomes an open

circuit. Thus, the impedance approaches R. Analytically,

lim →∞ Zin ( j ) = lim →∞1

j C+ lim →∞

RR

j L+1

= R

(c) Zin ( j ) =− jC

+j RL(R − j L)

R2 + 2L2 = jLR2

R2 + 2L2 −1C

+R 2L2

R2 + 2L2 . It follows that w must satisfy

2LR2C = R2 + 2L2 or equivalently 2 =R2

LR2C − L2 =1

LC 1−L

CR2

. For the given component

values 1−L

CR2

= 0; hence there is no finite value of frequency for which the impedance is real.

SOLUTION 10.27. We note that the input admittance is given by:

Yin ( j ) =1

100 + j 0.1+ j ×10−6 =

100 − j 0.1

104 + 0.01 2 + j ×10−6

=100

104 + 0.01 2+ j ×10−6 − j

0.1

104 + 0.01 2

Thus, ω must satisfy 105 =104 + 0.01 2 or equivalently, = 107 −106 = 3000 rad/s.

SOLUTION 10.28. (a) As usual we will use MATLAB.

»R1 = 20; R2 = 10;

»L = 0.04; C = 0.6e-3;

»w = 250;

15

»Yin = 1/R1 + 1/(R2 + j*w*L) + j*w*C

Yin = 1.0000e-01 + 1.0000e-01i

Hence

Yin ( j250) = 0.1+ j0.1 S

(b) For this part we observe that

Yin ( j250) =1

R1+

1

R2 + jωL+ jωC = 0.1 − 0.05 j + j250C

For this to be real, the imaginary part must be zero, i.e.,

C =0.05

250=

1

5000= 0.2 mF

Solution 10.29 (a) We can derive an expression for the input impedance by noting that it is the series

combination of the resistance and the inductor/capacitor pair connected in parallel. Thus,

Zin ( j ) = R +

1j C

× j L

1

j C+ j L

= R − j

L

C

L −1

C

Equating the real and imaginary parts of the given impedance, R = 4 Ω and

−LLC −1/

=−L

2L − 0.25= 2 ⇒ L = 0.1 H

(b) At zero inductance, the above reactance is zero. Also, at L = 0.125, the denominator of the above

reactance is zero, which means that the reactance is infinite.

Solution 10.30 (a) First derive an expression for the input impedance as a function of frequency:

Zin j( ) = 5 + j L +1

j C= 5 + j L −

1C

We want the imaginary part to be equal to zero. Thus,

L −1C

= 0 ⇒ 2 =1

LC ⇒ = 2500 rad/s

16

where we have substituted the values of L and C.

The magnitude of the impedance is minimum when the imaginary part is zero, and Zin(j2500) = 5.

(b) Derive an expression for the admittance:

Yin ( ) =15

+ j C +1

j L=

15

+ j C −1L

Again, the imaginary part is equal to zero when =1

LC or = 2500, at which point the admittance is

0.2.

Solution 10.31 The input admittance is Yin ( ) =1R

+ j C = 0.008+0.004. Equating this to the given

admittance at = 500yields R = 125 Ω, ωC = 0.004 or C = 8 µF.

Solution 10.32 Derive an expression for the admittance at = 1000:

Yin ( j ) =1

j L+

0.25 × j C0.25 + j C

= − j0.5 +j0.125

0.25 + j0.5= 0.2 − j0.4 S

Note that this is equivalent to a 0.2 S conductance (i.e. 5 Ω resistance) in parallel with a 2.5 mH inductance

(at the given frequency!). Now, the impedance is:

Zin =1

0.2 − j0.4= 1+ j2 Ω

This is equivalent to a resistance of 1 Ω in series with a 2 mH inductance (at the given frequency).

Solution 10.33 The current is zero when the input impedance of the parallel combination of inductor and

capacitor is infinite. The latter is given by:

ZLC( j ) =

1j C

× j L

1

j C+ j L

= − j

L

C

L −1

C

The magnitude of this is infinite when rL −1C

= 0 ⇒ r2 =

1LC

⇒ r =15,811 rad/s. Observe that

Is =Vs

R + ZLC ( j r )= 0

17

Hence is(t) = 0 at r = 15,811 rad/s. At this frequency, the voltage across the LC tank is equal to the input

voltage (since there is no drop across the resistor).

Solution 10.34 (a) By inspection:

Zin =2 × j 0.4

2 + j 0.4=

j 0.8

2 + j 0.4

Yin =1

R+

1

j L= 0.5 − j

2.5

(b) Use the current divider formula, and substitute the given frequency, to obtain:

IL =R

R + j L2 =

2 22 + j2

=1∠ − 45°

and iL ( t) = cos(5t − 45o) A.

Solution 10.35 (a) Using voltage division, Vout =R

R − jC

Vin =j RC

1+ j RCVin which

⇒ ∠Vout = 90o − in − tan−1 RC( ) and Vout =RC Vin

1+ 2R2C2

(b) For this part, we need to make sure that tan−1 RC( ) = 45° ⇒ RC = 1 ⇒ =1/ RC .

(c) At this frequency,

VoutVin

=

1

RCRC

1+1

R2C2 R2C2=

1

2

Solution 10.36 Here, = 1/ RC and VCVin

=

1j C

1

j C+ R

=1

1 + j RC=

11 + j

. Therefore

VC =1

1 + jVin = 0.707Vm∠ − 45°

The time-domain function is

vC t( ) = 0.707Vm cos1

RCt − 45°

V

18

Solution 10.37 (a) The magnitude of the capacitor voltage is 10/14.14 = 0.707 times the magnitude of the

input signal. We just showed in the above problem that

VCVin

=1

1+ j10RC ⇒

VCVin

=1

1 +100R2C2

And we also showed that the ratio is 0.707 when the frequency is 1/RC. So, C = 1/(10R) = 0.01 F.

(b) Again, from the results of the previous problem, the angle is –45 degrees.

Solution 10.38 = 1000 rad/s and Iin = 2∠45° A. The equivalent admittance is

Yeq j( ) =1R

+ j C −1L

= 0.25 + j 0.25 − 0.25( ) = 0.25 S

In MATLAB

»Yeq = 0.25;

»Iin = 2*exp(j*pi/4)

Iin =

1.4142e+00 + 1.4142e+00i

»Vout = Iin/Yeq

Vout =

5.6569e+00 + 5.6569e+00i

»abs(Vout)

ans = 8

»angle(Vout)*180/pi

ans = 4.5000e+01

»% Using Current Division

»IL = (1/(j*1000*4e-3)/Yeq)*Iin

IL =

1.4142e+00 - 1.4142e+00i

»abs(IL)

ans = 2

»angle(IL)*180/pi

ans = -45

19

Therefore, vout (t) = 8cos(1000t + 45o) V and iL ( t) = 2cos(1000t − 45o) A.

(b) If = 618 rad/s, then

Yeq( ) =14

− j0.405 + j0.1545 = 0.25 − j0.25 = 0.3536∠ − 45°

and

»Vout = Iin/Yeq

Vout =

-3.4779e-04 + 5.6565e+00i

»abs(Vout)

ans =

5.6565e+00


ans =

9.0004e+01

»IL = Iin*(1/(j*618*4e-3))/Yeq

IL =

2.2882e+00 + 1.4069e-04i

»abs(IL)

ans =

2.2882e+00

»angle(IL)*180/pi

ans =

3.5228e-03

Therefore, vout (t) = 5.657cos(1000t + 90o) = −5.657sin(1000t) V and iL ( t) = 2.288cos(1000t) A.

Solution 10.39 Write the input phasor: = 1000 rad/s and Iin = 0.01 2∠60°A.

»w =1000;

»Iin = 0.01*sqrt(2)*exp(j*60*pi/180)

Iin =

7.0711e-03 + 1.2247e-02i

»Yeq = 1/500 +1/(j*w*0.25) +j*w*2e-6

Yeq =

2.0000e-03 - 2.0000e-03i

»Vin = Iin/Yeq

20

Vin =

-1.2941e+00 + 4.8296e+00i

»abs(Vin)

ans =

5

»angle(Vin)*180/pi

ans =

1.0500e+02

»IR = Iin*(1/500)/Yeq

IR =

-2.5882e-03 + 9.6593e-03i

»abs(IR)

ans =

1.0000e-02

»angle(IR)*180/pi

ans =

1.0500e+02

»IL = Iin*(1/(j*w*0.25))/Yeq

IL =

1.9319e-02 + 5.1764e-03i

»abs(IL)

ans =

2.0000e-02

»angle(IL)*180/pi

ans =

1.5000e+01

»IC = Iin*j*w*2e-6/Yeq

IC =

-9.6593e-03 - 2.5882e-03i

»abs(IC)

ans =

1.0000e-02

»angle(IC)*180/pi

ans =

-1.6500e+02

21

Therefore, vin t( ) = 5cos 1000t −105°( ) V, iC t( ) = 0.01cos 1000t −165°( ) A, iL t( ) = 0.02cos 1000t +15°( )A, and iR t( ) = 0.01cos 1000t +105°( ) A.

SOLUTION 10.40. = 4 . Using voltage division

V1 =R

R +1

j C+ j L

Vin =2

2 − 4 j + 2 j−8 j( ) = 5.657∠− 45o

Converting back to time:

vout t( ) = 5.657cos 4t − 45°( ) V

SOLUTION 10.41. = 25 rad/s, Vs = 10∠0° V. By voltage division,

VC =− j

0.02 × 25

10 + j0.08 × 25 − j0.02 × 25

10∠0° = −2 j = 2∠ − 90°

Thus, vC t( ) = 2cos 25t − 90°( ) V.

SOLUTION 10.42. VoutVin

=1

3 + j8 + 1j 4C

= 0.2 ⇒ 25 = 3 + j 8 −1

4C

2

.

Thus 8 −1

4C

2

= 25 − 9 ⇒ C =1

16= 0.0625 F.

SOLUTION 10.43. Here, = 3.33 ×103 rad/s and Vin = 50∠0° V. Using phasors,

VR =400

400 −j

3.33×103 ×10−6

× 50 = 40∠36.897° V

and

Vout =100VR j C

100 +1

j C

=100VR

j100 C +1= 3985∠32.1° V

22

Hence, vout (t) = 3985cos(3.33×103t + 32.1°) V.

SOLUTION 10.44. Here, = 104 rad/s, Iin = 0.01∠0° A.

Z1 =100 × j 0.1100 + j 0.1

= 1+ j1000 ⇒ VL = ZLIin = 0.01× ZL = 0.01 + j10

Now, in MATLAB

»w=1e4; R = 100;L = 0.1; C = 0.1e-6;

»Z1 = R*j*w*L/(R+j*w*L)

Z1 = 9.9010e+01 + 9.9010e+00i

»Iin = 0.01;

»VL = Z1*Iin

VL =9.9010e-01 + 9.9010e-02i

»Z2 = 1/(1/R + j*w*C)

Z2 = 9.9010e+01 - 9.9010e+00i

»VC = Z2*VL

VC = 9.9010e+01 + 1.7764e-15i

»abs(VC)

ans = 9.9010e+01

»angle(VC)*180/pi

ans = 1.0280e-15

Thus, vC t( ) = 99cos 10000t( ) V.

SOLUTION 10.45. Here = 40 rad/s and Vin = 120∠0° . This problem is best done in MATLAB usingparallel impedance computation, voltage division, and Ohm's law for phasors:

»R1 = 500; R2 = 80;»C = 0.1e-3; L = 2;»w = 40; Vin = 120;»Z1 = R1/(j*w*C)/(R1 + 1/(j*w*C))Z1 = 1.0000e+02 - 2.0000e+02i»Z2 = R2*j*w*L/(R2 + j*w*L)Z2 = 4.0000e+01 + 4.0000e+01i» Use voltage division»VC = Z1*Vin/(Z1+Z2)VC = 1.2212e+02 - 3.1858e+01i»abs(VC)ans = 1.2621e+02»angle(VC)*180/pians = -1.4621e+01

23

» Use voltage division and Ohm's law for inductors»VL = Z2*Vin/(Z1+Z2)VL =-2.1239e+00 + 3.1858e+01i»IL = VL/(j*w*L)IL =3.9823e-01 + 2.6549e-02i»abs(IL)ans = 3.9911e-01»angle(IL)*180/pians = 3.8141e+00

Therefore, vC (t) = 126.21cos(40 t −14.621o) V and iL ( t) = 0.399cos(40 t + 3.814) A.

SOLUTION 10.46. Here = 40 rad/s and Iin = 0.120∠0° A. This problem is best solved usingMATLAB.

»w = 40; Iin = 0.12;R = 5;»C = 0.004; L = 0.1;»Y1 = 1/R + j*w*CY1 = 2.0000e-01 + 1.6000e-01i

»Y2 = 1/R + 1/(j*w*L)Y2 = 2.0000e-01 - 2.5000e-01i

»% USING CURRENT DIVISION»IL = Iin*Y2/(Y1 + Y2)IL = 7.3171e-02 - 5.8537e-02i»abs(IL)ans = 9.3704e-02»angle(IL)*180/pians = -3.8660e+01

»% AGAIN USING CURRENT DIVISION»IC = Iin*Y1/(Y1 + Y2)IC = 4.6829e-02 + 5.8537e-02i

»% USING OHM'S LAW FOR CAPACITORS»VC = IC/(j*w*C)VC = 3.6585e-01 - 2.9268e-01i»abs(VC)ans = 4.6852e-01»angle(VC)*180/pians = -3.8660e+01

SOLUTION 10.47 Here, = 25 rad/s and IS = 2∠0° A. Now perform a source transformation. The

combination of current source in parallel with resistor is changed into a voltage source in series with the

same resistor. The voltage source value is: VS = 1× IS = 2∠0° V. Apply Ohm’s law to obtain:

I1 =2∠0°

2 + 1j25 × 0.02

= 0.5 + j0.5 = 0.707∠45°

24

Hence, i1(t) = 0.707cos(25 t + 45o) A.

SOLUTION 10.48 Apply a source transformation to obtain:

The impedance of the inductor branch is ZL( j 4) = 2 + j4 × 0.5 = 2 + j2 Ω and

YL( j 4) =1

ZL ( j4)= 0.25 − j0.25 S. Now, the total admittance seen by the source:

Yeq( j 4) =12

+ j0.25 + YL ( j4) = 0.75 S

Thus, by inspection, VC =6∠0°0.75

= 8∠0° V and IC = 0.25∠90°× 8∠0° = 2∠90° A. Therefore

iC ( t) = 2cos(4 t + 90o) = −2sin(4 t) A.

SOLUTION 10.49. Apply a source transformation to obtain:

Then, by inspection, Yeq( j500) =12

+11

+1j2

−1j2

=1.5 S. Thus,

25

VC =2.5∠0°

1.5= 1.667∠0° V = VL ⇒ IL = 1.667/ j2 = 0.833∠− 90° A

In the time-domain:

vC t( ) = 1.667cos 500t( ) V and iL t( ) = 0.833cos 500t − 90°( ) = 0.833sin 500t( ) A.

1


SOLUTION 10.50. The input voltage phasor is = 2000 rad/s and VS = 20∠0° V. Now, do a source

transformation on the phasor circuit:

where

IS =VSZL

=20∠0°

j2000 ×10 ×10−3 = 1∠− 90° = − j A

Now,

Yeq =120

+1

j20+ j2000 × 50 ×10−6 = 0.0707∠45° S

and

VC =ISYeq

=1∠− 90°

0.0707∠45°= 14.14∠ −135° V

SOLUTION 10.51. Use superposition. First, find response to current source using circuit below:

Vx_1 = Is1ZRC = Is13× (− j 3)

3 − j3= 2∠0°× 2.121∠− 45° = 4.242∠− 45°

Now, find the response due to the voltage source using the following circuit:

2

The voltage across the inductor is the same as the input source, and this voltage divides between the series

combination of capacitor and resistor:

Vx_2 =j3

3 − j3Vs2 =

j 33 − j3

3∠90o = 2.121∠ −135o V

Combining the two contributions implies that:

Vx = 4.242∠ − 45o + 2.121∠− 135o = 4.74∠ − 71.6o V

SOLUTION 10.52. (a) As stated, VL = aVs1 + bIs2 . To find a, set Is2 = 0 and use voltage division:

VL =−j30

j30 + j30Vs1 = −0.5Vs1 = −aVs1

To find b, set Vs1 = 0 and use parallel impedance and Ohm's law:

VL = j30/ / j30( )Is2 = j15Is2 = bIs2

Hence

VL = –0.5Vs1 +j15Is2

(b) For this part, Vs1 =10 and Is2 = 0.5∠ − 90o . Hence from the formula,

VL =−0.5Vs1 + j15Is2 = −0.5 ×10 + j15 × 0.5(− j) =−5 + 7.5 = 2.5 V

3

Therefore vL(t) = 2.5 cos(100πt) V.

SOLUTION 10.53. In this problem, we can make use of the linearity property for phasors. Specifically,

from the given information, we can write

V1Iin

=20∠45°10∠0°

= 2∠45° = a and V1Vin

=5∠90°10∠45°

= 0.5∠45° = b

Hence,

V1 = aIin + bVin

Substituting the new values of input current and voltage, we obtain:

V1 = 10∠0° +10∠45° =18.48∠22.5° V

SOLUTION 10.54. (a) For Vout to be zero, we want

RC

RC −j

C

=j L

RL + j L ⇒ RCRL =

LC

(b) Substituting RCC = 2 s and RL = 3 Ω into the above expression gives: L = RCCRL = 6 H.

(c) The bridge circuit can be represented by an impedance Zbridge(jω). The voltage that appears across the

bridge, say Vbridge, is obtained by voltage division. Hence, by the voltage substitution theorem, the problem

may be solved as in part (a) with this new source voltage Vbridge appearing across Zbridge(jω).

SOLUTION 10.55 The input phasor is: = 1000 rad/s and Iin = 2∠45° A, assuming peak value. First

compute

Yth ( j103) = 0.25 −j

1000 × 4 ×10−3 + j1000 × 0.25 ×10−3 = 0.25S ⇒ Zth ( j103) = 4 Ω

Then,

Voc = Iin × 4 = 8∠45° V

4

in which case voc ( t) = 8cos(103t + 45o) V. The final equivalent is a voltage source (having value Voc) in

series with a resistance of 4 Ω.

SOLUTION 10.56. (a) First note that the frequency is given in Hz, so, = 1281.77rad/s and Iin =10∠0°A. Then, in MATLAB,

»R = 0.25; L = 1.17e-3; C = 520e-6;

»w = 2*pi*204;

»Yin = j*w*C +1/(R + j*w*L)

Yin =

1.0815e-01 + 1.7737e-02i

»Zin = 1/Yin

Zin =

9.0039e+00 - 1.4766e+00i

»abs(Zin)

ans =

9.1242e+00

»angle(Zin)*180/pi

ans =

-9.3134e+00

Therefore Zth = 9.1242∠− 9.313° Ω. Finally, Voc = ZthIin = 91.2∠− 9.313° V.

(b) Now, the circuit looks like the following:

Simple voltage division can yield:

VL =ZL

ZL + ZTHVoc = 46.219∠0o ⇒ vL t( ) = 46.2cos 1281.77t( ) V

5

SOLUTION 10.57. For this problem we short the V-source and compute Zth and use voltage division to

find Vo c . Specifically

Zth =1

jωC + 1

0.1+ jωL

= 103 − j10 Ω

and

Voc =

1

jωC1

jωC+ 0.1 + jωL

× 2 = −200 j V

where

»w=1000;

»L = 0.01;

»C = 0.1e-3;

»Zth = 1/(j*w*C + 1/(0.1 + j*w*L))

Zth = 1.0000e+03 - 1.0000e+01i

»Voc = 2*(1/(j*w*C))/(0.1 + j*w*L + 1/(j*w*C))

Voc = 0 - 2.0000e+02i

To compute the load voltage, define

Zload =103 + j10−2 ω = 103 + j10 Ω

Again using voltage division and MATLAB we have

»Zload = 1e3 + j*10;

»Vload = Voc*Zload/(Zload + Zth)

Vload =

1.0000e+00 - 1.0000e+02i

»magVload = abs(Vload)

magVload =

1.0000e+02 (volts)

6

SOLUTION 10.58. Here, = 3.33 ×103 rad/s and Vin = 50∠0° V. Here we note that we already found

Voc in Problem 10.43. Thus, Voc = 3985∠32.1° V. In order to find Zth, we introduce a fictitious 1 A

current source at the A and B terminals:

Now, we note that the VR voltage phasor is zero. Thus, the dependent source has zero volts across it. This

way, the temporary current source sees the parallel combination of a resistor and a capacitor:

VAB =100 × − j

3.33×103× 0.25×10−6

100 − j1200= 99.65∠ − 4.77° V

Thus, Zth = 99.65∠− 4.77° Ω. The Thevenin equivalent consists of Voc in series with

Zth = 99.65∠− 4.77° = 99.3 − j8.2865 Ω. Thus the Norton equivalent is the parallel combination of Zth

and the current source with value

Isc =VocZth

=3985∠32.1°

99.65∠ − 4.77°= 39.99∠36.91o A.

SOLUTION 10.59. = 10,000 rad/s and Vin = 10∠0° V. Again, we have already found Voc in Problem10.44: Voc = 99∠0° V. Now, to compute the impedance, we introduce the temporary current source of 1A:

7

Again, the inductor has no voltage across it. So, the dependent source generates no current. Hence, the

independent source sees the parallel combination of a resistor and a capacitor:

VAB =100 × − j1000( )

100 − j1000= 99.5∠− 5.71° V

Thus Zth ( j104 ) = 99.5∠ − 5.71° Ω. So, the Thevenin equivalent is the series combination of the Voc

source and the above Zth ( j104 ) = 99.5∠ − 5.71° Ω.

SOLUTION 10.60. Inject a current source at terminals A and B. Then, write a KCL equation at node A:

IR = IS − I1 ⇒ VR = IS − I1 ⇒ VC = VAB = VR + I1 = IS − I1 + I1 = IS

Since the voltage across the current source is equal to its current, the equivalent impedance across this

current source is 1 Ω.

SOLUTION 10.61. Inject a current source as usual. Then, write Ohm’s law for phasors for the equivalent

series RLC circuit. Note that the controlling current for the dependent source is the input current:

VAB = j0.01× 200 × IS −j

200 × 0.005IS − 2IS = 2 + j( )IS

Therefore, Zth =VABIs

= 2 + j Ω.

SOLUTION 10.62. Inject a current source Is :

8

Apply KCL: j2I− 6I

6= Is − I ⇒ I =

6j2

Is . Now, VAB is the voltage across the inductor:

VAB =6 × j2

j2Is = 6Is ⇒ Zth = 6 Ω.

Solution 10.63. = 2000 rad/s and VS = 25∠0° V. Now, inject a current source, and express VAB as a

function of this current source and Voc:

Now, write two nodal equations at A and the top of the dependent current source:

V1 − VS

103 + gmVAB + j C V1 − VAB( ) = 0

j C VAB − V1( ) +VABj L

= IS

Substituting values, these can be cast into the following matrix equation:

9

10−3 1 + j 3 − j

− j − j

V1

VAB

=

10-3VS

IS

This can be solved in MATLAB to obtain:

V1

VAB

=100 ×

−2.5 − j2.5 −5 + j10

2.5 + j2.5 5

10-3VS

IS

Thus

VAB = (0.25 + j0.25)Vs + 500Is = 500Is + 6.25 + j6.25

Therefore Zth = 500 Ω and Voc = 6.25 + j6.25 = 8.839∠45o V. For the Norton equivalent we need

Isc =VocZth

=8.839500

∠45o = 0.01768∠45o A.

SOLUTION 10.64. Inject an upward current source, IS2, at VAB. Then, write the following two nodal

equations at Vx and VA: let R1 = R2 = R, then

VxR

+ j C1Vx + j C2 Vx − VAB( ) = IS

j C2 VAB − Vx( ) + gmVx +VAB

R= IS2

which after grouping terms becomes

1

R+ j C1 + j C2

Vx − j C2VAB = IS

− j C2 + gm( )Vx + j C2 +1

R

VAB = IS2

In MATLAB

»R = 100e3; C1 = 1e-9; C2 = 1e-10;

»gm = 0.1e-3; w = 1e3;

»Nodal = [1/R+j*w*C1+j*w*C2 -j*w*C2;-j*w*C2+gm j*w*C2+1/R]

Nodal =

1.0000e-05 + 1.1000e-06i 0 - 1.0000e-07i

10

1.0000e-04 - 1.0000e-07i 1.0000e-05 + 1.0000e-07i

»Nodalinv = inv(Nodal)

Nodalinv =

9.5680e+04 - 2.0070e+04i 2.1024e+02 + 9.5470e+02i

-9.5449e+05 + 2.1120e+05i 9.7783e+04 - 1.0523e+04i

Thus

Vx

VAB

=103 ×

95.68 − j20 0.21+ j0.955

−954.5 + j211.2 97.78 − j10.52

40 ×10-6

IS2

Therefore

VAB = −38.18 + j8.448 + 97.8 − j10.5( ) ×103IS2

from which we identify

Voc = −38.18 + j8.448 = 39.1∠167.5o V

and Zth = 97.8 − j10.5( ) ×103 = 98.35 ×103∠− 6.14o Ω.

SOLUTION 10.65. We solve this problem by the method illustrated in example 6.3 where a fictitious

source is applied and the repsonse is calculated. One can either apply a voltage source or a current source

(see figure 6.10 a and b). Generally speaking, neither choice can be claimed as superior to the other. But

for a specific circuit, one choice can lead to a much simpler solution than the other. To illustrate this point,

we solve the problem with both choices below. Note that although the first method is much simpler than

the second, it lacks the generality. If one more resistor were inserted into the circuit, the simplicity of

solution of solution may disappear totally, whereas the second method will proceed with very few changes.

Method 1. For this solution we apply an arbitrary voltage source, labeled in phasor form as Vs as

indicated in the circuit below.

For this circuit we will compute an equation of the form ofequation 6.11:

11

IA =1

ZthVs − Isc (1)

By inspection of the circuit,

IA = IC + IL −1 = jωCVs +VLjωL

−1 = j2Vs − jVL −1 (2)

ButVL = 0.25IC + Vs = 0.25 jωCVs + Vs = 0.5 j +1( )Vs (3)

Substituting (3) into (2) producesIA = 0.5 + j( )Vs −1 (4)

By comparing (4) with (1), we obtain the answers for the Norton equivlent circuit: Is c = 1 A and Zth =1/(0.5 + j) =

0.4 – j0.8 Ω.

Method 2: For this method we apply an arbitrary current source, labeled in phasor form as Is as indicated

in the circuit below.

Notice that we have added a current label Ix as we plan to use modified nodal analysis method to obtain the

desired answer. For this we will compute an equation of the form

VAB = ZthIs + Voc

Because of the addition of Ix , we can write the modified nodal equations more or less by inspection:

1/ jωL 0 1

0 jωC −1

1 −(1+ j0.25ωC) 0

VL

VAB

Ix

=− j 0 1

0 2 j −1

1 −1 − 0.5 j 0

VL

VAB

Ix

=1

Is

0

where from MATLAB,

12

»w=2000;

»L = 0.5e-3;

»C = 1e-3;

»Y11 = 1/(j*w*L)

Y11 = 0 - 1.0000e+00i

»Y22 = j*w*C

Y22 = 0 + 2.0000e+00i

»Y32 = -(1+j*0.25*w*C)

Y32 = -1.0000e+00 - 5.0000e-01i

Solving the equations in MATLAB produces,

»A = [-j 0 1;0 2*j -1; 1 -1-0.5*j 0];

»Ainv = inv(A)

Ainv =

8.0000e-01 - 6.0000e-01i 8.0000e-01 - 6.0000e-01i 1.6000e+00 + 8.0000e-01i

4.0000e-01 - 8.0000e-01i 4.0000e-01 - 8.0000e-01i 8.0000e-01 + 4.0000e-01i

1.6000e+00 + 8.0000e-01i 6.0000e-01 + 8.0000e-01i -8.0000e-01 + 1.6000e+00i

Multiplying the second row of Ainv times the right-most vector of our equations produces

VAB = (0.4 − j 0.8)Is + (0.4 − 0.8j )

This implies that Zth = 0.4 – j0.8 Ω and Vo c = 0.4–j0.8 V. For the Norton equivalent we need

Isc =VocZth

=1 A

SOLUTION 10.66. (a) = 1000 rad/s and VS = 20∠45° V. By voltage division:

Vout =1

1+ j20∠45° =14.14∠0° V ⇒ IL =14.14∠0° A

(b) At dc, Vout = Vin. We want the frequency at which Vout = 0.1Vin. Thus, we want:

13

VoutVin

=1

1+ j × 0.001= 0.1 ⇒

1

1+ 2 ×10−6 =1

100 ⇒ = 9950 rad/s

SOLUTION 10.67. Z = 25 − j20 Ω. By KCL:

VZ −14∠0°j15

+VZ

25 − j20+

VZ + 8∠90°− j20

= 0

This equation simply needs to be manipulated in order to obtain:

VZ = 41.35∠ − 73.46°

SOLUTION 10.68. In this problem, we note that the impedance, jX, is in series with the parallel RLC

circuit to the right. Thus, all we need to do is to find an expression for the equivalent impedance of the

parallel RLC circuit:

YRLC =130

−j

20+ j0.025 ⇒ ZRLC =19.2 + j14.4

Now, the total impedance seen by the source is 19.2 + j14.4 +jX. Therefore, for this to be real, the

unknown reactance has to be –14.4 Ω. Also, the input current now is I = 96/19.2 = 5 A. Hence

i( t) = 5cos(10 t) A.

SOLUTION 10.69. First, compute the current through the series RC section: IC =1

1− j A.

Now, by KCL, we can write

0 = IC − Ix +1− 2Ix

j=

11− j

− Ix +1− 2Ix

j ⇒ Ix = 0.3 + j0.1= 0.316∠18.44o A

SOLUTION 10.70. = 20 rad/s and Yin = 0.05 + j0.0866 S.

(a) Since Yin =1R

−jL

+ j C , equating the real parts of the above two expressions implies that R =

1/0.05 = 20 Ω.

(b) Similarly, equating the imaginary parts and substituting, we obtain:

14

j C −1L

= j0.0866 ⇒ L = 3.73 H

(c)

VC =IinYin

=20∠30°

0.05 + j0.0866= 200∠− 30° V ⇒ vC t( ) = 200cos 20t − 30o( ) V

(d)

IL =200∠30°j20 × 3.73

= 2.68∠ −120°A ⇒ iL t( ) = 2.68cos 20t −120o( ) A

(e) Voc is just the voltage VC, which was obtained in (c), and Zth =1

Yin=10∠− 60° Ω.

(f) Zth = 5 – j8.66 Ω. This is equivalent to a series combination of a 5 Ω resistance and a 5.77 mF

capacitance at the given frequency ω = 20 rad/s.

SOLUTION 10.71. (a) Vs = 2 + j0V, =1000 rad/s. By KCL

VA− j1.33

+VA

2 + j2+

VA − Vs4

= 0 ⇒ VA = 0.5 − j0.5 = 0.707∠ − 45° V

The time-domain expression is:

vC t( ) = vA t( ) = 0.707cos 1000t − 45o( ) V

(b) IL =0.707∠ − 45°

2 + j2= 0.25∠− 90° A implies that iL t( ) = 0.25cos 1000t − 90o( ) = 0.25sin(1000t) A.

(c) We already determined Voc in part (a). Now, turn off the source to compute the equivalent impedance:

Yth = 0.75 j +1

2 + j2+ 0.25 = 0.5 + j0.5 ⇒ Zth =1− j Ω. This is the series connection of a 1 Ω resistor

with a capacitor of value C =1

=10−3 F. This completes the definition of the Thevenin equivalent.

SOLUTION 10.72. In MATLAB,

»R = 5; L = 1e-3; C = 20e-6;Vs1 = 5; Is2 = 0.5*j;

»G = 1/R; w = 10e3;

»YL = 1/(j*w*L)

YL =

0 - 1.0000e-01i

15

»YC = j*w*C

YC =

0 + 2.0000e-01i

Hence

G(VC − Vs1) + YCVC + YL (VC − Vs1 − 2VR) = Is2

Substituting for VR,

G(VC − Vs1) + YCVC + YL (VC − Vs1 − 2(Vs1 − VC)) = Is2

Therefore

G + YC + 3YL( )VC = G + 3YL( )Vs1 + Is2

Again, using MATLAB,

»a = G + YC + 3*YL

a =

2.0000e-01 - 1.0000e-01i

»b = (G+3*YL)*Vs1 + Is2

b =

1.0000e+00 - 1.0000e+00i

»VC = b/a

VC =

6.0000e+00 - 2.0000e+00i

»abs(VC)

ans =

6.3246e+00

»angle(VC)*180/pi

ans =

-1.8435e+01

Hence, vC (t) = 6.325cos(104 t −18.44o) V.

SOLUTION 10.73. Denote by vC1, the node voltage of the 2.5 mF capacitor . Note that at ω = 800 rad/s,

Vs = 20∠0o V. From this we write a set of nodal equations by inspection after observing the following

from MATLAB:

16

»w = 800;

»C1 = 2.5e-3;

»L = 1.25e-3;

»Y1 = 0.5+j*w*C1-j/(L*w)

Y1 = 5.0000e-01 + 1.0000e+00i

»Yoff=j/(L*w)

Yoff = 0 + 1.0000e+00i

»Y2 = 0.25+j*w*3.75e-3 - j*w*1.25e-3

Y2 = 2.5000e-01 + 2.0000e+00i

This information leads to the following matrix nodal equation:

0.5 + j j

j 0.25 + j2

VC1

Vout

=

0.5Vs

0.25Vs

To solve these equations we again use MATLAB:

»A = [Y1, Yoff;Yoff, Y2]

A =

5.0000e-01 + 1.0000e+00i 0 + 1.0000e+00i

0 + 1.0000e+00i 2.5000e-01 + 2.0000e+00i

»b = [0.5*20;0.25*20]

b =

10

5

»Vnodes = inv(A)*b

Vnodes =

7.1141e+00 - 6.9799e+00i

-3.6242e+00 + 5.3691e-01i

»magVnodes = abs(Vnodes)

magVnodes =

9.9664e+00

17

3.6637e+00

»angVnodes = angle(Vnodes)*180/pi

angVnodes =

-4.4454e+01

1.7157e+02

Therefore

vout (t) = 3.664cos(800t +171.57o) V

SOLUTION 10.74. For this problem we use loop analysis with loops indicated in the figure below.

Since there are no controlled sources, we can write down the loop equations by inspection:

10 + 9 j 6 + 5j

6 + 5 j 12 + 9 j

IA

IC

=

120 −120∠ −120o

120∠120o −120∠− 120o

=180 +103.92

j207.85

The solution of this equation is done in MATLAB as follows:

»b1=120-120*exp(-j*2*pi/3)

b1 = 1.8000e+02 + 1.0392e+02i

»b2=120*exp(j*2*pi/3)-120*exp(-j*2*pi/3)

b2 = 0 + 2.0785e+02i

18

»A = [10+j*9, 6+j*5;6+j*5,12+j*9]

A =

1.0000e+01 + 9.0000e+00i 6.0000e+00 + 5.0000e+00i

6.0000e+00 + 5.0000e+00i 1.2000e+01 + 9.0000e+00i

»I=inv(A)*[b1;b2]

I =

1.4472e+01 - 1.3469e+01i

4.2926e-01 + 1.7703e+01i

»% Please note that using the commands I=inv(A)*[b1,b2]'

»% will lead to the wrong answer because a conjugate is

»% inserted along with the transpose.

»magI = abs(I)

magI =

1.9770e+01

1.7708e+01

»angleI = angle(I)*180/pi

angleI =

-4.2944e+01

8.8611e+01

»IB = -I(1)-I(2)

IB = 1.8288e+00 + 1.7234e+01i

»magIB = abs(IB)

magIB = 1.7331e+01

»angleIB = angle(IB)*180/pi

angleIB = 8.3943e+01

Changing the sign on each source amounts to multiplying its value by "–1". This means that all

magnitudes remain the same, but there is a 180o phase shift for each current, i.e., add 180

o to each current

angle.

19

SOLUTION P10.75. For this problem we have both a transient component to the response and a steady

state component. The steady state component is computed in the usual way because the circuit is stable,

i.e., the time constant is positive. Once the steady state part is computed, we use initial conditions to obtain

the coefficient B in the response.

Part 1: Compute steady state response. For this we use MATLAB,

»R = 0.5; L = 0.866;

»Vs = 10;

»w = 1;

»Zin = R + j*w*L

Zin = 5.0000e-01 + 8.6600e-01i

»IL = Vs/Zin

IL = 5.0002e+00 - 8.6604e+00i

»magIL = abs(IL)

magIL = 1.0000e+01

»angIL = angle(IL)*180/pi

angIL = -5.9999e+01

Hence

iL (t) = 10cos(t − 60o) + Be−0.577t A

Part 2: From the initial conditions we have

iL (0) = 1 = 10cos(t − 60o) + Be−0.577t[ ]t=0= 10cos(−60o ) + B = 5 + B

Hence B = –4. It follows that

iL (t) = 10cos(t − 60o) − 4e−0.577t A

SOLUTION 10.76. KCL dictates that:

IL + IR + IC = IS ⇒ IL = IS – IR – IC

We can perform this sum graphically as follows:

20

j2

-IC

-IR

IL=-j

SOLUTION 10.77. (a) Note that VS = VR + VL, but that VL leads VR by 90 degrees. Similarly, IS = IR +

IC, but IC leads VS by 90 degrees. Also note that the inductor current is also IR, and the capacitor voltage is

VS.

VRIR

VL

VS

ICIS

(b) Using graph paper to construct the phasor diagram to scale, we find the difference between the phase

angles of Is and Vs is zero.

Solution 10.78. First note that VC, the capacitor voltage, will lag IS by 90 degrees. Now, VC plus the

unknown element voltage should result in a vector that runs diagonally between the two vectors. From the

following illustration, it follows that the unknown voltage should have the same phase as the input current:

VC

IS

Vu

VS

VC

21

This means that the unknown element is a resistor. The 45o phase difference implies that VC = Vu or

IsC

= IsR . Therefore R =1C

=1

103 ×10−6 =1000 Ω.

SOLUTION 10.79. The student can construct the phasor diagram using graph paper. The diagram is going

to look like that in the problem statement, except that the proper lengths and angles will be used.

SOLUTION 10.80. As the frequency approaches infinity, the capacitor shorts and the inductor opens. So,

the output voltage is zero. As the frequency approaches zero, the capacitor opens, but the inductor shorts,

so the output is also zero. A plot of the complete response is shown below. (Note that the magnitude

response at 10rad/s is infinite):

»L = 0.04; C = 0.25;

»w = 0: 30/300:30;

»% Vout = Zin * Iin

»Zin = j*w*L ./(j*w*L*j.*w*C + 1);

»plot(w, abs(Zin))

»grid

»ylabel('Magnitude Zin')

»xlabel('Frequency in rad/s')

»plot(w,angle(Zin)*180/pi)

»grid

»xlabel('Phase in degrees')

»ylabel('Phase in degrees')


22

0 5 10 15 20 25 300

5

10

15

20

25M

agni

tude

Zin

TextEnd

Frequency in rad/s0 5 10 15 20 25 30

-100

-80

-60

-40

-20

0

20

40

60

80

100

Frequency in rad/s

Pha

se in

deg

rees

TextEnd

SOLUTION 10.81. At infinite frequency, the resistor current is zero (because the inductor opens). So, theoutput voltage is zero. At DC, the inductor is short, and the output voltage is equal to the input voltage. Theplot of the frequency response is shown below (a logarithmic x-axis is used):

SOLUTION 10.82. The circuit inside the black box is

23

At DC, the capacitor is an open circuit. Thus, the voltage across the resistor is 1mA×R. But we know that

this voltage is 1 from the graph. This means that R = 1 kΩ. Now, in general for the above diagram:

VI

=R

1+ j RC

The magnitude of this function is R/sqrt(2) when = 1/ RC . Substituting the frequency from the graph

(1000 rad/s), we get C = 1 µF.

Solution 10.83. The admittance of the parallel RLC circuit is:

Yin =1R

+1

j L+ j C

Zin =1

Yin=

VsIs

The function we want to find the frequency response for is nothing but the input admittance of the circuit.

Using MATLAB, the following plot can be obtained:

»R = 100; L = 0.1; C = 1e-3;

»w = 0:0.5:300;

»w = 0.01:0.5:300;

»Yin = 1/R + 1. ./(j*w*L) + j*w*C;

»Zin = 1 ./Yin;

»plot(w,abs(Zin))

»grid


»ylabel('Magnitude Zin')

24

0 50 100 150 200 250 3000

10

20

30

40

50

60

70

80

90

100

Frequency in rad/s

Mag

nitu

de Z

in

TextEnd

»plot(w,angle(Zin)*180/pi)

»grid


»ylabel('Phase Zin in degrees')

0 50 100 150 200 250 300-100

-80

-60

-40

-20

0

20

40

60

80

100

Frequency in rad/s

Pha

se Z

in in

deg

rees

TextEnd

25

SOLUTION 10.84. The circuit inside the box is a series RLC circuit. It cannot be a parallel RLC, because

as per problem 83, the admittance of a parallel RLC does not approach zero as w approaches infinity. Thus,

IV

= Yin =1

R +1

j C+ j L

The resonance frequency is 50 rad/s and is determined by 1/sqrt(LC). Given L = 0.4 H, C = 1 mF.

To obtain R, we make use of the fact that, from the given graph at ω = 57 rad/s, the current

magnitude is approximately 0.2 times the peak magnitude. Therefore

1

R2 + 57L −1

57C

2≅

0.2

R2 + 50L −1

50C

2=

0.2R

Hence

R2 + 57 × 0.4 −1

57 ×10−3

2

=R2

0.04= 25R2

R2 + 57x0.4 - 157x0.001

2 ≅ R2

0.04 = 25R2

From which R =5.2561

24=1.0728 Ω.

SOLUTION 10.85. Create three mesh currents in the three planar loops. All currents are clockwise: I1 in

the voltage source loop, I2 in the top bridge loop, and I3 in the bottom one. The three mesh equations are:

V − I1R1 − R2 I1 − I2( ) − R3 I1 − I3( ) = 0

R2 I2 − I1( ) +1

j C1I2 + Rmeter I2 − I3( ) = 0

R3 I3 − I1( ) + Rmeter I3 − I2( ) +1

j C2I3 = 0

The plots of the magnitude and phase of VB − VC = Rmeter I3 − I2( ) are shown in the text.

26

SOLUTION 10.86. We can see all ranges by plotting on a logarithmic scale:

Note that the output will decay when we start to reach the bandwidth of the op-amp. In other words, the

inverting amplifier says that the output is –1 times the input (provided the op amp works properly). Once

the op amp’s gain starts dropping, the output voltage also decays with it.

SOLUTION 10.87. Correction: Change the 0.01 µF capacitor to 1 µF. (a) For this part consider the

diagram below,

From the problem statement, ω = 320π rad/s, and Iin = 0.01∠0o A. Observe that the 50 kΩ resistor input

to the inverting op amp terminal is in parallel with the 100 Ω resistor because of the virtual ground at the op

amp terminals. However, for all practical purposes, this has no effect on the 100 Ω resistor, hence from

Ohm's law

27

VL =100 × j 0.1ω100 + j0.1ω

Iin = 50.265 + j50( ) × 0.01 = 0.50265 + j0.5

From the inverting op amp configuration,

ˆ V =-10VL = −5.0265 − j5

From voltage division,

VC =10−3

10−3 + jω10−9ˆ V = 0.49735 − j0.5( ) × −5.0265 − j5( ) = 5∠179.7o V

Therefore

vC (t) = 5cos(320πt +179.7o )V

Parts (b) and (c). For the SPICE simulation we have the following circuit in B2-SPICE:

which leads to the response below

28

MAG(V(IVM))

Frequency (Hz)Prb10-87-Small Signal AC-2

(V)

+0.000e+000

+100.000m

+200.000m

+300.000m

+400.000m

+500.000m

+1.000 +10.000 +100.000 +1.000k +10.000k

The magnitude at 160 Hz is 0.499 for a 1 mA current input. Thus a 10 mA input current should lead to

4.99 V by linearity which approximates the 5 V computed analytically in part (a). Hence with a 15 V

saturation limit, the input magnitude may increase by a factor of 3 to 30 mA.

SOLUTION 10.88. (a) = 400 rad/s and Vin = 10−3∠ − 90° V. By the virtual short property:

Iin =Vin

1/ j C= j CVin

All this current flows through the 1 MΩ resistor: Vin = − j C106 Vin = 0.4∠ −180° . Thus,

vout(t) = – 0.4cos(400t) V

(b) = 200 rad/s, Vin = 10−3∠0° V, and Vout = 10−3∠− 90° V. Again,

Vout = − j10−3 = − j C × 4 ×105Vin = − j80 ×103C ⇒ C =12.5 nF

29

SOLUTION 10.89. (a) = 800rad/s and Vin = 1∠− 90°V ⇒ Iin =Vin

106 . No current flows into Op-

Amp terminals:

Vout = −Iin1

j C=

j × (− j)800

= 1.25 ×10−3 V

Thus, vout(t) = 1.25cos(800t) mV.

(b) Again, Iin =Vin

200 ×103 A, and Vout = −Iin1

j C ⇒ 10 =

j × (− j)

C2 ×105 ⇒ C = 2.5 nF .

Solution 10.90 (a) = 2 700rad /s and Vin = 1∠0° V ⇒ Iin =1

150 ×103 A. Further,

Vout = −150 ×103 − j

C

150 ×103 − jC

Iin =−1

150 ×103 ×150 ×103 − j

C

150 ×103 − jC

= 0.0015∠90o V

Thus, vout(t) = 1.5cos(2π700t + 90o) mV.

(b)

(c) The output lags the DC response by 45 degrees (note that at DC, the amplifier is inverting, or has a

phase of –180 degrees). Now, the frequency response is really determined by the RC circuit in the

feedback path of the op-amp. The first resistance at the input simply converts the input into a current that

30

drives this RC circuit. It can be shown that a 45 degree phase shift occurs in an RC circuit when the

frequency is 1/RfC (directly from the results of an analysis on an RC circuit). So,

RfC = 1/2000π ⇒ C = 0.016 nF (We know Rf = 10 MΩ.)

Also, at this frequency, the response is 0.707×DC response. The DC response is –107/R. So, the DC gain

is 14.14. Thus, R = 7.07 kΩ.

Solution 10.91 (a) Use the virtual short property: Iin =Vin

103 + 1j C

. All of current flows through the

feedback path: Vout = −104Iin =104 Vin

103 − jC

⇒ VoutVin

= 7.07∠ −135o .

(b)

(c) The spice result looks the same at low frequencies. However, at high frequencies, the response falls

back to zero as the op-amp non-ideal frequency response starts to affect the behavior of the circuit.

SOLUTION 10.92. (a) The negative terminal of the op-amp is at Vs. This implies IR =VSR

. By KVL,

31

VS + ZRCIR = Vo ⇒ VoVS

=1 +1R

− j3RC

3R − jC

=1 +3

j3 RC +1

The MATLAB plot for the given values is:

(c) The spice result looks pretty much the same, especially since the cut-off frequency of this circuit is

much lower than the frequency at which the op-amp ceases to operate as an ideal op-amp.

SOLUTION 10.93. To compute the gain as a function of ω we observe that by the properties of an ideal opamp,

Gain =VoutVin

= −2 × 10−5 + j0.5 ×10−6ω

10−4 + j0.1 ×10−6ω=

20 + j0.5ω100 + j 0.1ω

In MATLAB

»G1 = 1/50e3;»G2 = 1/10e3;»C1 = 0.5e-6;»C2 = 0.1e-6;»w = logspace(-1,5,1500);»Y1 = G1 +j*w*C1;»Y2 = G2 + j*w*C2;»H = Y1 ./ Y2;»semilogx(w,abs(H))»grid

32

SOLUTION 10.94. (a) The equations are:

V1 − Vin1000

+V1 − V2

1000+

V1 − V21/ jwC f

= 0

V2 − V11000

+V2

1/ j C2= 0

V2 = Vout

Substituting values and solving for V1 and V2 ,

V1 = 0.8535∠ − 82oVin

V2 = 0.723∠− 114.7o Vin

The second one is the relation that we are looking for.

(b) At 100 Hz,

V1 = 1.00∠ − 3.68o Vin

V2 = 1.00∠ − 7.3oVin

At 3000 Hz,

V1 = 0.15∠ −102oVin

V2 = 0.0728∠ −164o Vin

33

(c) The response is that of a low-pass filter, as predicted from the results of part (b) above.

SOLUTION 10.95. (a) The two nodal equations:

Vx − Vin200

+ j C1Vx + j C2 Vx − Vout( ) = 0

j C2 0 − Vx( ) −Vout

28 ×103 = 0

In MATLAB,

»w = 2*pi*1.34e3; C = 0.05e-6; R1 = 200; R2 = 28e3;

»A = [1/R1+j*w*C+j*w*C -j*w*C;-j*w*C -1/R2]

A =

5.0000e-03 + 8.4195e-04i 0 - 4.2097e-04i

0 - 4.2097e-04i -3.5714e-05

»b = [1/R1; 0];

»V = A\b

V =

2.6664e-01 - 5.9266e+00i

-6.9859e+01 - 3.1429e+00i

»Vout = V(2)

Vout =

-6.9859e+01 - 3.1429e+00i

»abs(Vout)

ans =

6.9929e+01


ans =

-1.7742e+02

Hence, VoutVin

= 69.93∠− 177.4o .

When the capacitors are shorts, the output is shorted to the virtual ground input, at 0 V. Similarly, when

they are opens, the virtual ground makes sure that vout is zero, since there is no drop across the feedback

resistor.

34

(b) The band-pass response can be computed using any SPICE program.

SOLUTION 10.96. First, note the input-output relationship:

VoutVin

= −Z f

R1

where Zf is the impedance of the parallel RLC circuit. We have studied this circuit extensively earlier, and

we have shown that at “resonance”, the impedance of this circuit is going to be real and equal to the value

of resistance, in this case R2. That’s exactly the requirement of this problem, since we want VoutVin

= −Z f

R1

to simply be equal to R2/R1. It remains to note that this resonance occurs at a frequency ω = 1/sqrt(LC).

SOLUTION 10.97. (a) First, analyze the feedback amplifier circuit. The output of this op-amp circuit is:

Vop2 = −1/ j C

2 ×106 Vout

Also, by voltage division, the voltage at the resistive voltage divider (+ terminal of first op-amp):

VRR =2 ×103

2 ×105 × −1

j C22 ×106 Vout =−0.01

j C22 ×106 Vout

Now, the first op-amp circuit is an inverting amplifier, but it’s + terminal is at VRR now. Thus,

Vout = −10 Vin − VRR( ) + VRR

Substituting the above VRR means that:

Vout =−10

1+1.1

j

Vin

Since the input voltage has unity magnitude and zero phase, the above expression gives the required

magnitude and phase of the output voltage.

(b)

35

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1

2

3

4

5

6

7

8

9

10

Frequency Hz

Mag

nitu

de

TextEnd

(c) As can be seen, the response to zero frequency (i.e. DC) is zero. Also, the circuit goes back very quickly

(less than 2 Hz) to provide the required operation, which is to achieve a gain of 10.

Complex Power Prbs 2/5/00 P11-1 @ DeCarlo & P. M. Lin

CHAPTER 11. PROBLEM SOLUTIONS

SOLUTION 11.1. Using equation 11.3, Pav =12

(et −1)2Rdt =e2 t

2+ t − 2et

0

1

= 0.7580

1

∫ W.

SOLUTION 11.2.

(a) From 11.6, Pav =Vm

2

2R= 50 mW for a sinusoidal input.

(b) From 11.3,

Pav =R2

(10cos(10t))2dt0

/20

∫ + (−10cos(10t))2dt/20

3 /20

∫ + (10cos(10t))2 dt3 /20

2 /10

∫

=Im2 R2

= 50 mW

just as the previous case since the square of the absolute cos(10t) is the same as the square of cos(10t).

(c)

Pav =10R2

0.01cos2(10t)[ ]2dt

0

2 /10

∫ =10−3

2R 0.5cos(20 t) + 0.125cos(40 t) + 0.375[ ]dt

0

2 /10

∫

=12

0.520

sin(20t) +0.125

40sin(40t) + 0.375t

0

2 /10

= 37.5 mW

(d)t=0:1/1000:1;R=1e3;

pta= (0.01*cos(10.*t)).^2.*R;ptb= (0.01*abs(cos(10.*t))).^2.*R;ptc= (0.01*cos(10.*t)).^4.*R;

subplot(3,1,1);plot(t,pta);gridylabel('W');subplot(3,1,2);plot(t,ptb);gridylabel('W');subplot(3,1,3);plot(t,ptc);gridylabel('W');xlabel('time in seconds');


0 0.2 0.4 0.6 0.8 10

0.05

0.1

W TextEnd

0 0.2 0.4 0.6 0.8 10

0.05

0.1

W TextEnd

0 0.2 0.4 0.6 0.8 10

0.5

1x 10-5

W TextEnd

time in seconds

SOLUTION 11.3. (a) For figure a, the period is 2, and Pav =12

v2(t)R0

2∫ dt =

12R

400 +100( ) = 25 W.

In figure b, the period is 1, and Pav =20t( )2

R0

1∫ dt =

400R

t3

3

0

1

=13.3 W.

(b)


0 0.5 1 1.5 2 2.5 3 3.5 410

15

20

25

30

35

40

Time in s

Inst

anta

neou

s P

ower

in W

TextEnd

0 0.5 1 1.5 2 2.5 3 3.5 40

5

10

15

20

25

30

35

40

Time in s

Inst

anta

neou

s P

ower

in W

TextEnd

SOLUTION 11.4. (a) For (a), looking the definition for the effective voltage, one sees graphically thatthe integral over one period, 2, of the squared waveform, is 500. Dividing by the period, and taking the

square root, Veff =15.81 V. For fig. b, Veff = 20t( )2 dt0

1

∫ = 40013

= 11.55 V

(b) Pav = Ieff2 R =

Veff

10

2

8 = 20 W.

(c) Pav =Veff

10

2

8 =10.67 W


SOLUTION 11.5. (a) This can be done graphically quite easily. The period of fig a, is 9s. The total

area of one period of the squared waveform is 75. This yields Ieff =759

= 2.89 A. In fig. b, the area

over one period is 25 which yields Ieff =253

= 2.89 A.

(b) Using current division, Pav = Ieff6090

2

30 =111.36 W.

(c) The same result as (b) is obtained since the effective current is the same.

SOLUTION 11.6. (a) Ieff =12

et −1( )2

0

1∫ dt = 0.615 A.

(b) Pav = Ieff2 R = 0.758 W.

SOLUTION 11.7. (a)

Veff2 =

202

10 + 2cos(20t)( )20

2 /20∫ dt =

202

100t + 2t +240

sin(40t) + 2sin(20t)

0

2 /20

= 102.01

Hence Veff =10.1 V.

(b)

Veff2 =

110cos(2t) + 5cos(4 t)( )2

0∫ dt =1

62.5t[ ]0 = 62.568

Hence Veff = 7.91 V.

(c)

Without going into detailed calculation, note the following fact about v32( t) . Only the product terms that

have the same frequency will produce a non-zero result when integrated. Thus the integral reducesto the following:

Veff2 =

1100cos2(2t) + 25cos 2(4 t) + 25cos2(4 t − π /4) + 50cos(4 t)cos(4t − π /4)( )0∫ dt

Hence

Veff = 50 +12.5 +12.5 + 25cos(−π / 4)[ ] = 9.627 V

SOLUTION 11.8. The voltage is V = 50∠0 V and the impedance Zeq =100 −100 j =141.42∠ − 45o

Ω. Thus I =V

Zeq= 353.6∠45o mA. Hence, Pav = R Ieff

2 =100 ×0.3536

2

2

= 6.2516 W.


SOLUTION 11.9. (a) The equivalent load seen by the source, Zeq =1

j C +1/ R= 2∠ − 36.87o . Thus

VL = IZeq = 10∠− 36.87o V and vL (t) = 10cos(30 t − 36.87o) V.

(b) Pav =VmIm

2cos( v − i ) =

10 × 52

cos(−36.87o) = 20 W, and from 11.4

pL ( t) = 25cos(−36.87o) + 25cos(60t − 36.87o) W

SOLUTION 11.10. (a) First find the impedance Zeq = 3 + j4 = 5∠53.13o Ω, then

Is =VsZeq

=10∠ −143.13o A. The magnitude is 10 A rms or 14.14 A peak-value.

(b) Pav = 10 ⋅ 50cos(53.13o) = 300 W.

(c) 3Is2 = 300 W

SOLUTION 11.11. (a) By KVL, Vs = 10IL + j100IL + 9 ⋅10IL and

IL =Vs

100 + j100=

100∠0

141.42∠45o =1

2∠− 45o A.

(b) PavVs = Vs IL cos(45o) = 50 W and Pav9V1 = 9 ⋅10IL IL cos(180o) = −45 W.

SOLUTION 11.12. (a) By KCL the current through the resistor is 5I. So by KVL,

Vs = j1000I − j500I + 200 ⋅ 5I and hence

I =Vs

1000 + j500= 107.33∠ − 26.57o A

The power absorbed by the resistor is, Pav = 200 ⋅ 5I 5I = 57.6 W.

(b) PavVs = Vs I cos(26.57o) = 11.52 W and Pav4 I = 4I 200 ⋅5I cos(0) = 46.08 W.

SOLUTION 11.13. First find the input impedance, Zeq = 2 − j8 + 6 + j2 = 8 − j6 Ω. Then calculate the

current IL =100

8 − j6= 8 + j6 A. The complex power is then computed:

»ZL = 6 + j*2;»IL = 8 + j*6;»VL = ZL*ILVL = 3.6000e+01 + 5.2000e+01i»SL = VL * conj(IL)SL = 6.0000e+02 + 2.0000e+02i


»abs(SL)ans = 6.3246e+02

Thus the apparent power is 790 VA, the average power 600 W, the reactive power 200 var, and theapparent power is 632.46 VA.

SOLUTION 11.14. (a) Using MATLAB

»Vseff = 100*exp(j*pi/6)Vseff = 8.6603e+01 + 5.0000e+01i

»ZL = 350 +j*1*300;»Zin = 50 + ZLZin = 4.0000e+02 + 3.0000e+02i

»ILeff = Vseff/ZinILeff = 1.9856e-01 - 2.3923e-02i

»ILpk = sqrt(2)*abs(ILeff)ILpk = 2.8284e-01

»ILang = angle(ILeff)*180/piILang = -6.8699e+00

»SL = ZL*ILeff*conj(ILeff)SL = 1.4000e+01 + 1.2000e+01i

Therefore, iL ( t) = 2 0.2( )cos(300 t − 6.87o) = 0.2828cos(300t − 6.87o) A. SL = 14 + j12 VA, and theaverage power is 14 W.

SOLUTION 11.15. First use voltage division and observe that VL =12

Vs = 60∠60° V. Now the

complex power is SL = VLVL

*

(6 + j8)* = 360∠53.13° = 216 + j288 VA. Thus the average power is 216

W.

*SOLUTION 11.16. (a) To find V2 we write a node equation. First we note that Yi = 1/Zi is thecorresponding admittance. Hence

Y1 V2 − Va( ) +Y2V2 + Y3 V2 − Vb( ) = 0Hence

V2 =Y1Va + Y3VbY1 +Y2 +Y3

=100 + j50 =111.8∠26.57o V

obtained using MATLAB as follows

»Z1 = 0.1+j*0.1; Z2 = 0.4+j*2.2;»Z3 = 0.2 + j*0.2; Va = 104 + j*50; Vb = 106 + j*48;»Y1 = 1/Z1


Y1 = 5.0000e+00 - 5.0000e+00i

»Y2 = 1/Z2Y2 = 8.0000e-02 - 4.4000e-01i

»Y3 = 1/Z3Y3 = 2.5000e+00 - 2.5000e+00i

»V2 = (Y1*Va + Y3*Vb)/(Y1 + Y2 + Y3)V2 = 1.0000e+02 + 5.0000e+01i

»magV2 = abs(V2)magV2 = 1.1180e+02»angleV2 = angle(V2)*180/piangleV2 = 2.6565e+01

(b) Again working strictly in MATLAB we have the following complex powers of the loads and the twosources:

»Sz1 = (V2 - Va)*conj((V2-Va)*Y1)Sz1 = 8.0000e+01 + 8.0000e+01i

»Sz2 = V2*conj(V2*Y2)Sz2 = 1.0000e+03 + 5.5000e+03i

»Sz3 = (V2 - Vb)*conj((V2-Vb)*Y3)Sz3 = 1.0000e+02 + 1.0000e+02i

»Sva = Va*conj((Va - V2)*Y1)Sva = 1.0800e+03 + 3.0800e+03i

»Svb = Vb*conj((Vb - V2)*Y3)Svb = 1.0000e+02 + 2.6000e+03i

(c) To verify conservation of power observe that:

»TotSrsPwr = Sva + SvbTotSrsPwr = 1.1800e+03 + 5.6800e+03i

»TotLdPwer = Sz1 + Sz2 + Sz3TotLdPwer = 1.1800e+03 + 5.6800e+03i

which provides the desired verification.

SOLUTION 11.17. Use MATLAB and refer to the following figure:


Z1

Z2 Z3

Z4

%(a)%Bundle the impedances as per the following figure and%obtain the following.Z1=2+2*j;Y1=1/Z1;Y2=2+0.5*j;Z2=1/Y2;ZL3=4*j;YL3=1/ZL3;Y3=4+0.25*j;Z3=1/Y3;Z4=4+4*j;Y4=1/Z4;V1=10+2*j;V2=12+2*j;

%Write out KCL for node 1 and 2%(Va-V1)*Y1=V1*Y2+(V1-V2)*YL3%(Vb-V2)*Y4=V2*Y3+(V2-V1)*YL3Va= (V1*Y2+(V1-V2)*YL3)/Y1+V1Vb= (V2*Y3+(V2-V1)*YL3)/Y4+V2Va = 29.0000 +59.0000iVb = 1.6000e+02+ 2.3400e+02i

(b)Sr3=( (Va-V1)*Y1*2)*conj((Va-V1)*Y1)Sl1= ((Va-V1)*Y1*(2*j))*conj((Va-V1)*Y1)Sc1= V1*conj(V1*0.5*j)Sr1= V1*conj(V1/0.5)Sl3= (V1-V2)*conj((V1-V2)*YL3)Sc2= V2*conj(V2*0.25*j)Sr2= V2*conj(V2/0.25)Sl2= (V2-Vb)*Y4*4*j*conj((V2-Vb)*Y4)Sr4= (V2-Vb)*Y4*4*conj((V2-Vb)*Y4)SVA = Va * conj((Va-V1)*Y1)SVB = Vb * conj((Vb-V2)*Y4)

Sr3 = 9.0250e+02

Sl1 = 0 + 9.0250e+02i

Sc1 = 0 - 5.2000e+01i


Sr1 = 208

Sl3 = 0 + 1.0000e+00i

Sc2 = 0 - 3.7000e+01i

Sr2 = 592

Sl2 = 0 + 9.4660e+03i

Sr4 = 9466

SVA = 1.1115e+03 + 8.4550e+02i

SVB = 1.0057e+04 + 9.4350e+03i

(c) Take the real part of each of the complex power found in (b). The only components with non-zeroaverage power will be the resistors which have 208 W, 592 W, 902.5 W, and 9466 W average powerrespectively.

%(d)Total_passive=Sr1+Sr2+Sr3+Sr4+Sl1+Sl2+Sl3+Sc1+Sc2Total_active= Va*conj((Va-V1)*Y1)+Vb*conj((Vb-V2)*Y4)Total_passive = 1.1168e+04+ 1.0280e+04i

Total_active = 1.1168e+04+ 1.0280e+04i

which verifies the conservation of power.

SOLUTION 11.18. (a) From conservation of energy, the complex power is the sum of the complexpower absorbed by every circuit elements. Thus Ss =1240 + j145 VA, and the apparent power is 1248.4VA. The average power is 1240 W.

(b) From Ss = VsIs* , Is =

SsVs

= 5.428 A.

SOLUTION 11.19. (a) The complex power delivered by the source is the sum of the complex powerconsumed by the circuit elements. Thus Ss = 44 + j28 kVA.

(b) Is =Ss

Vs= 22.675 A

(c) The total power delivered to the three groups of impedance following V1 is S1 = 41.5 + j22. From

the current obtained in (b) V1 =S1 + S2 + S4

Is= 2071.5 V.

(d) From V1, and the total power delivered to Z4 and Z2, I2 =S4 + S2

V1= 12.385 A. Finally

V2 =S2

I2= 1805.5 V.


SOLUTION 11.20. (a) Is =44 + j28

2.3

*

= 22.675∠− (32.47 − 0) = 22.675∠− 32.47° A.

(b) V1 =S1 + S2 + S4

Is*

= 2071.5∠ − 4.54° V

(c) Similarly as before I2 =S1 + S2

V1

*

=12.385∠ − 37.61° A, and V2 = Z2I2 =1805.5∠− 11.05° V.

SOLUTION 11.21. From equation 11.30, we find Q = P1

pf 2 −1 = 455.61 var; thus

S = 1000 + j455.61 =1099∠24.5° VA.

SOLUTION 11.22. (a) S1 =1000 W, and S2 = 800 + j600 VA. Thus the total power delivered by the

source is Stot = 1800 + j600 =1897.37∠18.43° VA, and Is =Stot

*

Vs* = 15.81∠− 18.43° = 15 − j5 A rms.

(b) V1 =S1

Is* = 63.25∠ −18.43° V.

(c) V2 =S2

Is* = 63.25∠18.43° V.

SOLUTION 11.23. Using 11.25, find SL =Pavepf

=30000.75

= 4000 VA, and the load current

IL =SL

VL

=4000120

= 33.33 A. The power absorbed by the transmission line is then from

Pline = R IL2 = 0.2 × 33.33( )2 = 222.2 W.

SOLUTION 11.24. The capacitor must absorb a reactive power of Qnew − Qold = −17.9 kvar. Thus

jQC = − j17.9 = − j CVs2, and C =

−QC

Vs2 = 0.897 mF.

SOLUTION 11.25. From equation 11.30, Qnew = 86.61

(0.94)2 −1 = 31.43 VA. Thus the reactive

power absorbed by the capacitor is –18.57 var. Hence C =−QC

Vs2 = 3.714 µF.


SOLUTION 11.26. Device 1 has a complex power of S1 = P1 + jQ1 = 360 + j 0 VA. Recall equation

11.29, pf =Pave

S, and equation 11.30 Qnew = Pave

1

(pf )2 −1

where with a lagging pf, Q > 0, and with a leading pf for Q < 0. Using equation 11.30, we have for

device 2:

Q2 =14401

(0.8)2 −1 =1080 var

S2 = 1440 + j 1080 VA.S1,2 = S1 + S2 = 1800 + j1080 VA

As an aside we compute the magnitude of the current without the capacitor attached.

Is =S1,2

120= 17.493 A

The capacitor is used to achieve a lower source current with the same average power. The first step is tofind the desired QC. Here

Stot = S1 + S2 + SC =1800 + j1080 + jQCHence

Stot120

=1800 + j1080 + jQC

120= 15

In MATLAB we have:

»QC = sqrt((15*120)^2 - 1800^2) - 1080QC = -1080

From the formula on page 451 of the text,

QC = − C Vsource2

Hence C = 0.2 mF. Finally

pf =Pave

S1 + S2 + SC=

18001800

= 1

SOLUTION 11.27. (a) From equation 11.30, Qold = 71

(0.65)2 −1 = 8.1839 kvar. Therefore, the

power absorbed without the capacitor bank is Sold = 7 + j8.1839 kVA. When the bank is added we

want Qnew = 71

(0.8)2 −1 = 5.25 kvar, and henceSnew = 7 + j5.25 kVA. Thus the reactance that must

be absorbed by the bank is –2.934 kvar, and Ceq =−QC

Vs2 = 0.13511 mF.

(b) As was just determined 5.25 kVA.


(c) Sold =7

0.65=10.77 kVA. Snew =

70.8

= 8.75 kVA. The kVA saving is: 10.77 – 8.75 = 2.02

kVA.»Savings = 20*2.02*12Savings = 4.8480e+02

i.e., $484.80.

SOLUTION 11.28. (a) The apparent power is simply 94kW/0.78=120.51 kVA.

(b) Sm =120.51∠38.74o = 94 + j 75.41 kVA.(c) 75.41 kvar.

(d) Ieff =Sm

*

Veff* = 523.96∠− 38.74 A.

(e) By KVL, Vs = RlineIeff + Veff = 516.08 − j229.52 V.

(f) Ss = VsIeff* = 295.94∠14.76° kVA.

(g) The efficiency is =120.51cos(38.74)295.94cos(14.76)

⋅100 = 32.96% . Note that the line resistance of 0.7 Ω is much

to large for practical usage. This value is chosen for pedagogical reasons.

(h) With a power factor of 0.94, Smnew = 94 + j 34.12 kVA. The average power of the motor must be kept

the same. The reactance that must be provided by the capacitor is, Qnew − Qold = QC = −41.29 kvar. The

proper capacitor current will be IC =j42.29k

Veff* , and ZC =

1j C

=VeffIC

=Veff

2

j 42.29 ×103 . Solving for

C =42.29 ×103

Veff2 = 2.12 mF.

(i) Ieffnew =

Smnew

Veff

*

= 434.79∠ −19.95° A

(j) By KVL, Vs = RlineIeffnew + Veff = 516.1− j103.84 = 526.43∠− 11.38o V.

(k) Ssnew = Vs

new Ieffnew( )*

= 228.89∠8.57° = 226.33 + j 34.12 kVA, and the efficiency is 41.5%.

SOLUTION 11.29. (a) The Thevenin equivalent seen at the output is, Zth = 5Ω− j /(0.1 ). For

maximum power transfer, ZL = Zth* = 5 + j . Note that VOC = 50 V rms, and that Ieff =

VOCZth + ZL

= 5 A

rms. Thus SL = ZLIeffIeff* = 125 + j25 VA, and the average power is 125 W.

SOLUTION 11.30. First find the following Thevenin equivalent,

Zth = 2 + j 4 Ω, Voc =10 2

3 V.

By the maximum transfer property, RL = 2Ω , and C = 1/(4 ) = 0.25 mF.


SOLUTION 11.31. (a) The thevenin equivalent left of the load is by KCL,Itest + Is = VR( j0.001+ 0.001) and Vtest = 3VR + j500Itest V. Substituting for VR,Vtest = Itest 1500 − j1000[ ] + Is 1500 − j1500[ ] V, and Zth = 1500 − j1000 Ω withVoc = Is × 2121.3∠ − 45° V. The value of the load for maximum power transfer is then,ZL = 1500 + j1000 Ω.

(b) The complex power absorbed by the load is, SL = ZLVoc

Zth + ZL

2

= 750 + j500 VA, and the average

power 750 W.

SOLUTION 11.32. Consider

V1 V2

(a) Using nodal analysis get the following equations:

V1 − V2 = 4(V1 /8)

Is + Itest = V1(1/8 + j0.25) + V2(1/16 + j0.5)

Vtest = V2 + 8Itest

Using MATLAB we get the following expression,

Vtest = (8.2847- j0.9110)Itest + (0.2847- j0.9110)Is

From this expression Voc = 28.47 − j91.1 V, and Zth = 8.2847- j0.9110 Ω. The phasor equivalent ofthis circuit is a Vocsource in series with a 8.28 Ω resistor and a 7.32 mF capacitor.(b) ZL = 8.2847+ j0.9110 , which is a 8.28 Ohm resistor in series with a 6.07mH inductor. Sameconfiguration as before.

(c) SL = ZLVoc

Zth + ZL

2

= 275.2 + j30.26 VA and the average power is 275.2 W.

SOLUTION 11.33. (a) The Norton equivalent may be found by inspection as Zth = 10 + j20 Ω andIsc =10 A. Thus for maximum power transfer, ZL = 20 − j20 Ω. This is a 10 Ω resistor and a 0.005 F

capacitor in series. The maximum power is SL = ZLIscZth

Zth + ZL

2

=1250 − j2500 VA. The maximum

average power is 1250 W.(b) If R is set to 20 Ω, the closest that can be achieved to maximum power transfer is ZL = 20 − j20 , orC equal to 0.005 F. With ZL as above, by current division


Iload =10 ×10 + j20

(10 + j20) + (20 − j20)= 3.333 − j6.6667 = 7.45∠63.434o A

The maximum average power then 20 × Iload2 = 1111 W.

(c) Using 11.38, RL = 31.62 Ω. Pav = RLIscZth

Zth + ZL

2

= 600.63 W.

SOLUTION 11.34. (a) From Thevenin Zth = 19.2 − j14.4 Ω, and from 11.38, set RL = 24 Ω.

Voc =Vs(− j 40)30 − j40

= 80∠− 36.87° . The maximum power is Pav =Voc

Zth + RL

2

RL = 74.07 W.

(b) The voltage is V = VocRL

Zth + RL, from this relationship, one sees that as the load resistance increase

to infinity the output voltage goes to Voc, which is the maximum output voltage.

SOLUTION 11.35. Correction: the inductor symbol in the load should be a resistor. Since the sourceresistance is variable, example 6.21 serves as a reference suggesting that R = 0 is the answer. To see thisconsider that

P = 2Iload2 =

2 ×100

(R + 2)2 + L −1C

2 =2 ×100

(R + 2)2 + (2 − 2)2

Hence, decreasing R produces increasing power and the maximum power is transferred when R = 0 withPmax = 50 W assuming that the source voltage is given in rms V.

SOLUTION 11.36. As per problem 35,

P = 10Iload2 =

10 × 50 2( )2

(R +10)2 + L −1C

2 =12.5 ×103

(R +10)2 + L −1C

2

Here, again R = 0 with C chosen to eliminate the reactive term maximizes power transfer. Hence

C =12L

= 0.01 F

with Pmax =125 W.

SOLUTION 11.37. (a) By the maximum power transfer theorem, P1 is maximized when ZL is chosenas the conjugate of Zsource, i.e.

ZL = 10 + j1000 Ω

(b) To find the appropriate values of L and C observe that


ZL( j ) = j L +1

10−4 + j C= j107 L +

10−4 − j C

10−8 +1014C2

=10−4

10−8 +1014 C2 + j 107 L −C

10−8 +1014C2

= 10 + j1000

Equating real parts leads to:

»%10^-4 = 10^-7 + 10^15*C^2»»C = sqrt(1e-4 - 1e-7)/sqrt(1e15)C = 3.1607e-10

Thus C = 0.31607 nF.

Equating imaginary parts using the above value of C leads to:

»w = 1e7;»L = (1e3 + w*C/(1e-8 + 1e14*C^2))/1e7L = 1.3161e-04

Thus L = 0.1316 mH.

(c) In part (b), L and C are chosen to maximize P1, the power delivered to ZL. Since L and C consumeno average power, this maximum power is transferred to the 10 kΩ fixed resistor with thecomputed values of L and C. Thus ZL is the same as in part (b) or ZL = 10 + j1000 Ω.

Since we know ZL,

Pmax =(0.1)2

4 ×10= 0.25 mW

This is the average power consumed by the 10 kΩ resistor. Therefore

V22 = 0.25 ×10−3 ×104 = 2.5

It follows that V2 =1.5811 V. Power to the 10 kΩ fixed resistor is

P10kΩ =V2

2

104

Thus if P10kΩ is maximized, then V2 is maximized.

SOLUTION 11.38. (a) From equation 11.4, if we substitute 2T

⇒ , then the resulting instantaneous

power will be p( t) =VmIm

2cos( v − i ) +

VmIm2

cos(4T

t + v + i ) W where it is clearly seen that the

fundamental period will now be halved. Note that by the same argument the fundamental frequency ofthe instantaneous power is double that of the voltage and current.


(b) As a sinusoid, the fundamental period is 2π/10, any integer multiple of this period will also beperiodic.(c) This is the same as (b) with an offset of 1 V added.

SOLUTION 11.39. First, Feff =1T

f 2(t)dtt0

to +T∫ . Without going into detailed calculation, f 2(t) will

give a summation of two types of products, a product of each element with themselves, and products ofeach element with the other element. In the later case, we know that two cosines multiplied with oneanother and integrated over one period will yield zero if their angular frequency are different. As for theformer case the integral will yields the result we are looking for. For example look at the first two terms,

Feff =1T

f 2(t)dtt0

to +T∫ =

1T

Fo2 + 2F1

2 cos2( 1t + 1) + ...( )dtt0

t0 +T∫

=1T

Fo2 + F1

2 + F12 cos(2 1t + 2 1) + ...( ) dt

t0

t0 +T∫

=1

TFo

2t + F12t + F1

2 sin(2 1t + 2 1)+ ...[ ]0T

= Fo2 + F1

2 ...

SOLUTION 11.40. (a) We are given that vC (t) = Vm sin( t) V. Hence

iC ( t) = CdvCdt

= CVm cos( t) A

It follows that

p( t) = vC ( t)iC (t) = CVm2 sin( t)cos( t) = 0.5 CVm

2 sin(2 t) Watts

Clearly, p(t) has a peak value of 0.5 CVm2 and the integral of the sign over one period is zero implying

that the average value of p(t) is zero.

(b) Here

WC (t) = 0.5CvC2 (t) = 0.5CVm

2 sin2( t) = 0.25CVm2 1− cos(2 t)( ) J

Here the peak value occurs when cos(2ωt) = –1 in which case the peak value is 0.5CVm2 . Further, the

average value of cos(2ωt) over one period, T = π/ω, is zero whereas the average of a constant over the

same period is simply the constant. Hence, WC,ave = 0.25CVm2 J.

(c) From example 11.6,

QC = −IC ,eff VC ,eff = − CVC,eff2 = −0.5 CVm

2 = −2 0.25CVm2( ) = −2 WC,ave .

Therefore, WC,ave = −QC2

.


SOLUTION 11.41. (a) We are given that iL ( t) = Im sin( t) A. Hence

vL (t) = LdiLdt

= LIm cos( t) V

It follows that

p( t) = vL ( t)iL (t) = LIm2 sin( t)cos( t) = 0.5 LIm

2 sin(2 t) watts

Clearly, p(t) has a peak value of 0.5 LIm2 and the integral of the sign over one period is zero implying

that the average value of p(t) is zero.

(b) Here

WL (t) = 0.5LiL2(t) = 0.5LIm

2 sin2( t) = 0.25LIm2 1− cos(2 t)( ) J

Here the peak value occurs when cos(2ωt) = –1 in which case the peak value is 0.5LIm2 . Further, the

average value of cos(2ωt) over one period, T = π/ω, is zero whereas the average of a constant over the

same period is simply the constant. Hence, WL,ave = 0.25LIm2 J.

(c) As an extension to example 11.5,

QL = IL,eff VL,eff = LIL,eff2 = 0.5 LIm

2 = 2 0.25CIm2( ) = 2 WL,ave .

Therefore, WL,ave =QL2

.

SOLUTION 11.42. (a) The complex power absorbed by the load is, SL = VI* = ZII* = Z I2 . Now notethat the average power is the real part of the complex power. Also note that a complex number

multiplied by its complex conjugate will yield a real value. Therefore the real part of ZI 2is just the

real part of Z, R, multiplied by I 2 , Pav = R I2 . With the same reasoning, the reactance is the

imaginary part of Z, X, multiplied by I 2 , Q = X I 2 .

(b) The complex power absorbed by the load is SL = VI* = V(YV)* = Y V 2 . The same reasoning as in

(a) holds thus the real part of the admittance times V 2 , yields Pav = GV 2 . Using the imaginary

part, Q = B V 2 .

SOLUTION 11.43. (a) The equivalent resistance seen by the source is Req = 6 – j9 Ω. So the currentdelivered by the source is:

Is =Vs

6 − j9=

− j1106 − j9

= 10.17∠− 33.69° A

or iL ( t) = 10.17 2 cos(120 t − 33.69o) A. Similarly

VC = − j15 × Is =−15 ×110

6 − j9=152.54∠− 123.69° V


or vC (t) = 152.54 2 cos(120 t −123.69o) V. Further, L =6

=15.9 mH and C =1

15= 176.8 µF.

The instantaneous energy stored in the inductor is

WL (t) = 0.5LiL2(t) = 1.646cos2(120 t − 33.69o) J

and the instantaneous energy stored in the capacitor is

WC (t) = 0.5CvC2 (t) = 4.115cos2(120 t −123.69o) J

The source voltage is zero when t = 0. Therefore

WL (0) = 1.646cos2(−33.69o) = 1.1395 Jand

WC (0) = 4.115cos 2(−123.69o) =1.266 J

(b) and (c) Observe that iL(t) and vC(t) are 90o out of phase. When one is zero, the other has a peakvalue. Therefore whenWC = 0 implies vC(t0) = 0 for appropriate t0; hence WL (t0) = 1.646 J.Similarly, when WL = 0, say at t0, then WC (t0) = 4.115 J.

SOLUTION 11.44. In order to solve this problem, we want to express the power in terms or R's and L'sin both circuits. First, looking at the circuit with just the coil and the 110 V source: I = V / Zcoil ,Zcoil = R + j L , and

Pcoil = I 2R =1102

R2 + 2L2 R = 300 watts (*)

Next, looking at the circuit when a resistance is added in series with the coil, I = V /(8 + Zcoil ),Zcoil = R + j L ,

Pcoil = I 2R =2202

(8 + R)2 + 2L2 R = 300 watts (**)

To find R, solve equation (*) for R2 + 2L2 and substitute into equation (**) to obtain

»R = 300*64/(220^2 - 300*16 -110^2)R = 6.0952e-01

Substituting R into equation (*) yields L = 13.05 mH.

SOLUTION 11.45. The average power consumed by the 2.7 Ω resistor is 250 watts. This allows us tocompute the magnitude of | Icoil |. We know that | Vcoil | is 150 Vrms. Thus we can computethe magnitude of the coil impedance and hence L as follows:

»magIcoil = sqrt(250/2.7)magIcoil = 9.6225e+00

»magZcoil = 150/magIcoilmagZcoil = 1.5588e+01


»% magZcoil^2 = 2.7^2 + (w*L)^2»»w = 2*pi*60;»L = sqrt( magZcoil^2 - 2.7^2)/wL = 4.0725e-02

»% Magnitude of impedance seen by 220 V source is:»»magZin = 220/magIcoilmagZin = 2.2863e+01

»% magZin^2 = (R + 2.7)^2 + (w*L)^2»»Realpart = sqrt(magZin^2 - (w*L)^2)Realpart = 1.6941e+01

»R = Realpart - 2.7R = 1.4241e+01

9/26/01 P12-1 @ DeCarlo & P. M. Lin

CHAPTER 12. PROBLEM SOLUTIONSSOLUTION P12.1. By conservation of energy, the instantaneous power consumed by each load whensummed together is equivalent to the total power consumed by the three phase load. Thus writing out

ptot (t) = pAB (t) + pBC ( t) + pCA ( t) = v AB( t)iAB( t) + vBC (t)iBC ( t) + vCA ( t)iCA (t)

= VL 2 cos( t + v ) ⋅VL 2

Zcos( t + i ) +VL 2 cos( t + v −120o) ⋅

VL 2Z

cos( t + i −120o)

+ VL 2 cos( t + v +120o) ⋅VL 2

Zcos( t + i +120o)

=VL

2

Zcos( v − i ) + cos(2 t + v + i )( ) +

VL2

Zcos( v − i ) + cos(2 t + v + i +120o)( )

+VL

2

Zcos( v − i ) + cos(2 t + v + i −120o)( )

=3VL

2

Zcos( v − i ) =

3VL2

Z× pf

SOLUTION P12.2. To justify the point of this problem we equate the following two equations:(i) For the 3 phase system:

Ploss' = 3× IL

2R ' = 3 ×

39

×PL

2

VL2 R ' =

PL2

VL2 R '

(ii) For the single phase system:

Ploss = 2 × IL2R( ) =

PL2

VL2 2R

It follows that R' = 2R. Since both systems have the same distance of transmission and the resistance ofa wire is inversely proportional to the cross sectional area, the condition R' = 2R implies that the crosssection A' of each wire in the three-phase system need only be half of the area A of the wire in the singlephase system. But there are two wires in the single phase system and three wires in the three-phasesystem. Therefore the ratio of the materials used is:

material in 3− phase systemmaterial in 1− phase system

=3A '

2A=

32

×12

= 75%

SOLUTION P12.3. For row 1 of table 12.1, the impedance in (a) seen between each pair of terminal is

Zik = Z∆ || 2Z∆ = 2Z∆3

. In (b) the impedance seen between any two terminals is

Z jk =Z∆3

+Z∆3

= 2Z∆3

.

9/26/01 P12-2 @ DeCarlo & P. M. Lin

In row 2, the impedance between any two terminals is Z jk = ZY + ZY = 2ZY for (c), andZik = 3ZY || 6ZY = 2ZY for (d).

SOLUTION P12.4. Consider the ∆-Y relationship of the figures below (row 3 of table 12.1):

Let us consider the terminal pair (1,2):

(i) For the Y-connected case, Zth = 2 ×Z∆3

and

Voc =Vp

3∠ − 30o( ) −

Vp

3∠ −150o( ) =

Vp

3∠ 1∠− 30o −1∠ −150o( ) = Vp∠

(ii) For the D-connected case,

Zth = Z∆ / / 2Z∆( ) =2(Z∆)2

3Z∆=

23

Z∆

Now note that there is no load connected to the ∆-configuration. Applying KVL to the indicated loopimplies that:

0 = 3Z∆Iloop +Vp∠ +Vp∠ −120o + Vp∠ +120o = 3Z∆Iloop

Hence Iloop = 0 . Finally, Voc = V12 = Z∆Iloop +Vp∠ = Vp∠ . Therefore, looking into terminals 1-2,

both the ∆-configuration and the Y-configuration have the same Thevenin equivalent. For terminal pairs(1-3) and (2-3), the proof is virtually the same. Hence this establishes the equivalence in row 3 of table12.1.

To establish the equivalence in row 4 of table 12.1, we do all the same computations with theslightly different labeling to obtain the same result, i.e., the circuits are equivalent.

SOLUTION 12.5. For (a) first note the following relationship VN =V1Z1

+V2Z2

Z1 || Z2 || Z3( ) , which is

obtained by KCL at the center node, N, with node 3 as the reference node.. Write out KCL at terminal 1,

I1 =V1Z1

−V1

Z1Z1+

V2Z2Z1

Z1 || Z2 || Z3( ) = V1

Z3 + Z2Z2Z3 + Z1Z3 + Z1Z2

− V2

Z3

Z1Z2 + Z2Z3 + Z3Z1

. Do the

9/26/01 P12-3 @ DeCarlo & P. M. Lin

same for terminal 1 in (b), I1 = V11

Z31+

1Z12

−

V2Z12

. Now substitute the formulas in the problem

statement into the later equation, I1 = V1Z2 + Z3

Z1Z2 + Z2Z3 + Z3Z1

− V2

Z3

Z1Z2 + Z2Z3 + Z3Z1

, which is

the same as the equation for (a).Using the same method, the same result is obtained for node 2. So the fact that the substitution

of the equivalence in (b) yields the same equation as in (a) proves that the equivalences are accurate.

SOLUTION 12.6. By Ohm’s law IA =Vp

Z=

200∠0°10 + j5

= 17.89∠− 26.57° A. By the same relationship

IB = 17.89∠ −146.57° A, and IC = 17.89∠93.43° A. The neutral line current is the sum of the otherthree and is zero. The power of each phase is the same, as they have the same load, and current

magnitude. Using the current at terminal A, the total power is, P = 3R IA2 = 9602 W.

SOLUTION 12.7. For this balanced Y-Y connection,we follow the method of example 12.6. UsingMATLAB:

»Z = 20 + j*10'Z = 2.0000e+01 + 1.0000e+01i

»VAN = 200;»IA = VAN/ZIA = 8.0000e+00 - 4.0000e+00i

»IAmag = abs(IA)IAmag = 8.9443e+00

»IAangle = angle(IA)*180/piIAangle = -2.6565e+01

»% Use phase inference to obtain:»IBmag = IAmagIBmag = 8.9443e+00

»IBangle = IAangle - 120IBangle = -1.4657e+02

»ICmag = IAmagICmag = 8.9443e+00

»ICangle = IAangle + 120ICangle = 9.3435e+01

»Ptotave = 3*real(Z)*IAmag^2Ptotave = 4.8000e+03

9/26/01 P12-4 @ DeCarlo & P. M. Lin

SOLUTION 12.8. Using the same approach as in the example 12.6 we compute

VAN = VsA ×Z

Z + Zg= 120 ×

4 + j34 + j3 + 0.1 + j0.2

= 116.94 V

Thus the drop is: 120 −116.94

120×100 = 2.55 % .

SOLUTION 12.9. A 5.5 % drop corresponds to VAN = 113.4 V. This means

Z

Z + Z1 + Zg=

VAN120

=0.945 ×120

120= 0.945

From this equation,

Z + Z1 + Zg =Z

0.945= 5.291.

So 4.05 + 0.1( )2 + 3.15 + 0.2( )2 = 27.99 . Solving the quadratic equation results in = 0.7893.

SOLUTION 12.10. Solving the single phase equivalent circuit for phase A,

IA =Vp −Vp∠120°

10 + j5= 58.87∠− 56.57°

Using phase inference,

IB = IA × ∠− 120° = 58.87∠ −176.57°IC = IA × ∠120° = 58.87∠64.43°

Alternately, using much more work,

IB =Vp∠ −120°− Vp

10 + j5= 58.87∠ −176.57°

IC =Vp∠120° −Vp∠ −120°

10 + j5= 58.87∠64.43°

The total power, using the previously derived relationship, is

P = 3VL IL pf = 3 × 380 × 58.87 × pf = 34.655 kWwhere

pf = cos tan−1 510

= 0.8944

9/26/01 P12-5 @ DeCarlo & P. M. Lin

SOLUTION 12.11. (a) By KVL and Ohm's law,

IAB =Vp

30 + j15= 6.57∠− 26.57°

By phase inference,

IBC = IAB × ∠ −120° = 6.57∠ −146.57°ICA = IAB × ∠120° = 6.57∠93.43°

Alternately,

IBC =Vp∠ −120°

30 + j15= 6.57∠ −146.57°

ICA =Vp∠120°30 + j15

= 6.57∠93.43°

To compute the line currents we have in Amps:

IA = IAB − ICA =11.36∠ − 56.57°IB = IBC − IAB = 11.36∠− 176.57°IC = ICA − IBC = 11.36∠63.43°

(b) The average power is P = 3 × 30 IAB2 = 3.88 kW, and the total reactive power is

Q = 3 ×15 IAB2 = 1.94 kvar.

SOLUTION 12.12. From the efficiency we know that 25 ⋅ 746Pdeliverd

= 0.85 , so the total power being

delivered to the load is 21941 W. Using 12.4, IL =21941

3 ⋅VL ⋅ pf= 63.31 A.

SOLUTION 12.13. (a) Using the power efficiency relationship, Ptot =300 ⋅ 746

0.935= 239.36 kW.

(b) Using 12.4, IL =239.36 ×103

3 ⋅VL ⋅ pf= 68.28 A.

SOLUTION 12.14. The magnitude of the power delivered to each delta connected load is 79.787 kW.We can now perform the analysis on a single phase. Using equation 11.30, we obtain

9/26/01 P12-6 @ DeCarlo & P. M. Lin

Qold = 797871

0.88

2

−1 = 43064 vars

and

Qnew = 797871

0.95

2

−1 = 26225 vars

Hence

Sold = 79787 + j 43064

Snew = 79787 + j26225

So the reactive power to be supplied by the capacitor is QC = 16839 = 43064 – 26225. Thecapacitance

C =QC

Vl2 = 8.44 uF.

SOLUTION 12.15. Replacing the delta configuration of the source by its Y-equivalent, the followingvoltages (in V) are obtained,

VAN =Vp

3∠ − 30°, VBN =

Vp

3∠− 150°, VCN =

Vp

3∠90°

From the single phase equivalent,

IA =VAN

Z=19.63∠− 56.57°

From phase inferenceIB = IA × ∠− 120° = 19.63∠− 176.57°IC = IA × ∠120° =19.63∠63.43°

The power factor of the load is


= 0.8944

Hence, the total power in the balanced system is

P = 3VL IL ⋅ pf = 3 × 380 ×19.63 × 0.8944 =11.556 kW.

SOLUTION 12.16. First replace the delta connected load by it Y-equivalent. Then analyzing a singlephase and using phase inference, we obtain in amps,

9/26/01 P12-7 @ DeCarlo & P. M. Lin

IA =Vp

(Z / 3)= 34.08∠− 26.56°

IB = IA × ∠− 120° = 34.08∠− 146.56°IC = IA × ∠120° = 34.08∠93.43°

The power factor of the load is


= 0.8944

The total power is P = 3VL IL ⋅ pf = 3 ⋅127 ⋅ 34.08 ⋅ pf = 11.61 kW. The voltage across each load is

3 ⋅Vp = VL = 219.97 V.

SOLUTION 12.17. (a) Referring to figure 12.2, the following relationships may be pointed out:

VAB = VAN + VNB = VAN − VBN

VBC = VBN + VNC = VBN − VCN

VCA = VCN + VNA = VCN − VAN

(b) Perform a delta to Y-transformation and use Ohm’s law and phase inference to obtain in amps:

IA =220∠− 30°

3 ⋅ Z=11.36∠ − 56.57°

IB = 11.36∠ −176.57°IC = 11.36∠63.43°

(c) First,


= 0.8944

The average power, Pav = 3VL IL pf = 3 × 220 × IA × 0.8944 = 3.87 kW.

The reactive power may be found as follows, Q = 3 × 5IL2 =1.936 kvar.

SOLUTION 12.18. Part (a) Apply ∆−Y transformation to the source.Apply the transformation of row 3, Table 12.1 to the source. The result is the circuit shown in figure12.19 of example 12.8. Therefore we may use many of the calculated values in example 12.9. Inparticular, the line currents are

IA = 19.49/-68.9o A, IB = 19.49/-188.9o A, and IC= 19.49/51.09o A

Part (b). Compute line-to-neutral voltage. From figure 12.21 and by Ohm's law

9/26/01 P12-8 @ DeCarlo & P. M. Lin

VAN = IA ZY = 19.49/-68.9o × 5∠36.87o = 97.45∠−32.03o VBy inference on phase angles

VBN = 97.45∠−152.03o V and VCN = 97.45∠87.97o V

Part (c). Compute line voltages VAB, VBC, and VCA. From the voltage phasor diagram of figure 12.2.

VAB = 3 VANe j30o= 168.8e− j2.03o

V

By inference on phase angles,

VBC = VABe− j120o=168.8e− j122.03o

V

VCA = VABe j120o=168.8e− j117.97o

V

Part (d). Compute the total power. From equation 12.4

Ptotal = 3 ×VL × IL × pf = 3 ×168.8 ×19.49 × 0.8 = 4558 W

SOLUTION 12.19. Perform a delta to Y transformation on the source. This yields

V1N =Vp

3∠− 30o =

380

3∠ − 30o V. Perform a delta to Y-transformation on the load. This yields a

new impedance Znew =Z3

= 4 + j 3 Ω. Now analyzing a single phase and finding the others by

inference yields, in amps,

IA =V1N

Z1 + Znew= 42.18∠− 67.97°

IB = 42.18∠− 187.97°IC = 42.18∠52.03°

The average power delivered to the load is Pav = 3 × IA2 × 4 = 21.35 kW.

SOLUTION 12.20. Performing a delta-to-Y-transformation on the source and the load, the new sourcemagnitudes are multiplied by 1 3 , and the loads are divided by 3. Observe that Zg = 0.15 + j0.45 Ω,Z1 = 0.1 + j0.2 Ω, and Z =12 + j9 Ω. Using Ohm’s law on phase one, and then inference on thephase angles for the other currents, yields in amps,

IA =Vp∠− 30°

3Zg

3+ Z1 +

Z

3

= 41.14∠− 68.91°

IB = 41.14∠ −188.91°IC = 41.14∠51.09°

9/26/01 P12-9 @ DeCarlo & P. M. Lin

The total power may be calculated as follows, Pav = 3123

IL

2 = 12 × (41.14)2 = 20.310 kW.

SOLUTION 12.21. Perform delta-to-Y-transformation on the delta load. Because of the property of theneutral to be ground in balance circuit, the two loads in the same phase combine in parallel. Thus the newimpedance seen in one branch is

ZYnew = ZY ||Z∆3

=(5 + j5) × (5 − j 3)(5 + j5) + (5 − j3)

= 4.044 + j0.1923 Ω

Now looking at a single phase and inferring for the others,

IA =Vp∠ − 30°

3 ⋅ ZYnew=

230∠ − 30°3 4.044 + j0.1923( ) = 32.84∠ − 32.73° A

IB = 32.84∠ −152.73° AIC = 32.84∠87.27° A

The complex power is S = 3 ⋅ ZYnew IL2 =13.07 + j0.615 kVA.

SOLUTION 12.22. For this problem we do not need to use any 3-phase knowledge if we are clever.We will provide a clever solution here. VDN = VDB – VNB. Using B as the reference we write a nodeequation at N to obtain:

0 =VNBZ1

+VNAZ1

+VNCZ1

=VNBZ1

+VNB − VAB

Z1+

VNB − VCBZ1

Therefore

3VNB = VAB + VCB = 115 −115∠− 120o =172.5 + j99.59

From voltage division

VDB =Z3

Z3 + Z2VAB =

2 + j42 + j4 + 4 − j2

115 = 57.5 + j57.5 V

Now using MATLAB we have

»115-115*exp(-j*2*pi/3)ans = 1.7250e+02 + 9.9593e+01i»VNB =ans/3VNB = 5.7500e+01 + 3.3198e+01i»VDB = 115*(2+j*4)/(6+j*2)VDB = 5.7500e+01 + 5.7500e+01i

»VDN = VDB - VNBVDN = 7.1054e-15 + 2.4302e+01i

»magVDN = abs(VDN)magVDN = 2.4302e+01

»angVDN = angle(VDN)*180/piangVDN = 9.0000e+01

9/26/01 P12-10 @ DeCarlo & P. M. Lin

SOLUTION P12.23. First we write the following two loop equations,

VsA − VsA∠− 120 = IA Zg + Z1 + ZA( ) + IC + IA( ) Z + Z1 + Zg( )VsA∠120 − VsA∠ −120 = IA + 2IC( ) Zg + Z1 + Z( )

and solve in MATLAB:(a)%Problem 12.23aZg=0.05+0.15i;Z=4+3i;Z1=0.1+0.2i;ZA=5+2i;

VA= 120;VB= 120*exp(-120*i*pi/180);VC= 120*exp(120*i*pi/180);

%Write mesh equations in matrix form and solve:% i.e. [IA;IC] = A - 1*CA=[ 2*(Z1+Zg)+ZA+Z Z+Z1+Zg; Zg+Z1+Z 2*(Zg+Z1+Z)];Ainv= A^(-1);C=[VA-VB;VC-VB];B=Ainv*C;IA= B(1);IC= B(2);IB= -IA-IC;IAabs(IA)180*angle(IA)/piIBabs(IB)180*angle(IB)/piICabs(IC)180*angle(IC)/pi

%And By Ohm's law,VAN= IA*ZAabs(VAN)180*angle(VAN)/piVBN= IB*Zabs(VBN)180*angle(VBN)/piVCN= IC*Zabs(VCN)180*angle(VCN)/pi

IA = 21.76∠–29.12° A

IB = 23.88∠–155.60° A

IC = 20.64∠ 82.39° A

9/26/01 P12-11 @ DeCarlo & P. M. Lin

VAN = 117.20∠-7.32° V

VBN = 119.4∠–118.73° V

VCN = 103.2∠119.26° V(b)% From phasor diagramVAB = VAN - VBNabs(VAB)180*angle(VAB)/piVBC = VBN - VCNabs(VBC)180*angle(VBC)/piVCA = VCN-VANabs(VCA)180*angle(VCA)/pi

VAB = 195.5 ∠27.34°VBC = 194.9 ∠–92.04°VCA = 196.99 ∠147.8°(c) Calculating the average power delivered to each load using the general formula

Ptot = R I2∑ = 5 × 21.76( )2 + 4 × 23.88( )2 + 4 × 20.64( )2 = 6.353 kW

SOLUTION P12.24. Doing this problem in MATLAB,%Problem 12.24ZDg=0.15+0.45i;ZD=12+9i;Z1=0.1+0.2i;ZDA=15+6i;%Apply the delta to Y transformation to the load as follows,ZA=(ZD*ZDA)/(2*ZD+ZDA);ZB=(ZDA*ZD)/(2*ZD+ZDA);ZC=(ZD*ZD)/(2*ZD+ZDA);

%Apply delta to Y trans. to the source,V1= 180/sqrt(3)*exp(-30*i*pi/180);V2= 180/sqrt(3)*exp(-150*i*pi/180);V3= 180/sqrt(3)*exp(90*i*pi/180);Zg= ZDg/3;

%Write nodal equations in matrix form and solve%i.e. [IA;IC]=A-1*CA=[ 2*(Z1+Zg)+ZA+ZB ZB+Z1+Zg; Zg+Z1+ZB 2*(Zg+Z1)+ZC+ZB];Ainv= A^(-1);C=[V1-V2;V3-V2];B=Ainv*C;IA= B(1);IC= B(2);IB= -IA-IC;IAabs(IA)

9/26/01 P12-12 @ DeCarlo & P. M. Lin

180*angle(IA)/piIBabs(IB)180*angle(IB)/piICabs(IC)180*angle(IC)/pi

IA = 17.4 ∠–63.1° A

IB = 20.1 ∠178.9° A

IC = 19.5 ∠51° A(b)% By Ohm's law,VAN= IA*ZA;VBN= IB*ZB;VCN= IC*ZC;%From phasor diagramVAB=VAN - VBNabs(VAB)180*angle(VAB)/piVBC = VBN-VCNabs(VBC)180*angle(VBC)/piVCA =VCN - VANabs(VCA)180*angle(VCA)/pi

VAB = 170.40 ∠–2.5° V

VBC = 168.58 ∠–122.4° V

VCA = 169.75 ∠118.1° V(c)% by Ohm's lawIAB=VAB/ZDAabs(IAB)180*angle(IAB)/piIBC=VBC/ZDabs(IBC)180*angle(IBC)/piICA=VCA/ZDabs(ICA)180*angle(ICA)/pi

IAB = 10.55 ∠–24.3o A

IBC = 11.24 ∠–159.3° A

ICA = 11.32 ∠81.2° A(d)

Ptot = R I2∑ =15 × 10.55( )2 +12 × 11.24( )2 +12 × 11.32( )2 = 4.721 kW

9/26/01 P12-13 @ DeCarlo & P. M. Lin

SOLUTION P12.25. The average power that the load draws is P = 440 ⋅ 40 ⋅ pf = 8.8 kW. From this weobtain the old and new complex powers:

Sold = 8.8 + j15.242 kVA and Snew = 8.8 + j15.242 + jQC kVA

Here the reactance supplied by the capacitor is QC = − CVl2 = −7299 var. So the new power factor is

pf = cos tan−1 Qnew8800

= 0.742 . From S = Vl Il = 11.855 kVA, we solve for

Il = IA = IB = 26.94 A.

SOLUTION P12.26. Using Ohm’s law and the power factor of the load,

IC = −440 ⋅ C∠90° = 16.59∠ − 90° A, IA = 40∠60° A, and IB = −IC − IA = 26.94∠ −137.9° A.

SOLUTION P12.27. By simply applying Ohm’s law to each load and then KCL at node N, we get thefollowing:

IA =VA50

= 4.4 A, IB =VB50

= 4.4∠ −120° A, IC =VC250

= 0.88∠120° A, and

IN = IA + IB + IC = 3.52∠− 60° A.

Finally,

PA = 50 × 4.4( )2 = 968 W, PB = 50 × 4.4( )2 = 968 W, PC = 250 × 0.88( )2 = 193.6 W.

SOLUTION P12.28. Using MATLAB,

VA = 220VB = 220*exp(-120*pi*i/180);VC = 220*exp(120*pi*i/180);

%Use KVL to solve for the different currentsIA = (VA - VB)/50 + (VA - VC)/50abs(IA)180*angle(IA)/pi

IB = (VB - VC)/100 + (VB - VA)/50abs(IB)180*angle(IB)/pi

IC = (VC - VB)/100 + (VC - VA)/50abs(IC)180*angle(IC)/pi

IA = 13.2 AIB = 10.1 ∠–130.9° A

9/26/01 P12-14 @ DeCarlo & P. M. Lin

IC = 10.1 ∠130.9° A

The total power delivered to each resistor may be calculated using the following relationship P =Vl

2

R,

where Vl2 = 3Vp

2 using Y to Delta source transformation. Thus PAB = 2904 W, PBC =1452 W, and

PCA = 2904 W.

SOLUTION P12.29. Consider the circuit of figure 12.29 with two additional node labeling:

Choose N as the reference node and apply KCL to node M. This yields the following node equation:

0.02 VM − 220( ) + 0.02 VM − 220e− j120o

+ 0.004 VM − 220e j120o

= 0

Equivalently2.2VM = 88 − j152.42 or VM = 40 − j69.282 V

The remaining calculations proceed in MATLAB as follows:»% Compute node voltage VM»X = +220 +220*exp(-j*2*pi/3) + 0.2*220*exp(j*2*pi/3)X = 8.8000e+01 - 1.5242e+02i»VM = X/2.2VM = 4.0000e+01 - 6.9282e+01i

»% Compute line currents»IA = 0.02*(220 - VM)IA = 3.6000e+00 + 1.3856e+00i»magIA = abs(IA)magIA = 3.8575e+00»angIA = angle(IA)*180/piangIA = 2.1052e+01»IB = 0.02*(220*exp(-j*2*pi/3) - VM)IB = -3.0000e+00 - 2.4249e+00i»magIB = abs(IB)magIB = 3.8575e+00»angIB = angle(IB)*180/piangIB = -1.4105e+02»IC = 0.004*(220*exp(j*2*pi/3) - VM)IC = -6.0000e-01 + 1.0392e+00i»magIC = abs(IC)

9/26/01 P12-15 @ DeCarlo & P. M. Lin

magIC = 1.2000e+00»angIC = angle(IC)*180/piangIC = 1.2000e+02

»% Compute total power»Ptot = 50*magIA^2 + 50*magIB^2 + 250*magIC^2Ptot = 1.8480e+03

SOLUTION P12.30. (a) The equivalent resistance of a 40 ft size #14 wire is

Req1 = 40 ⋅2.5751000

= 0.103Ω . The voltage at the light and appliance is then

Vapp = 115 − 2 ⋅ Req110 = 112.94 V, which is equivalent to a 1.8% drop, thus the wire gauze isappropriate.

(b) Repeating the previous calculations, Req2 = 50 ⋅1.6191000

= 0.08095 Ω.

Vapp = 115 − 2 ⋅ Req214 = 112.7334 V, which corresponds to 1.97 %. So the wire gauze is appropriate.

(c) Under normal operating condition the current in the live wire should equal the current in the neutralwire; thus no current should be in the ground wire.

(d) Note the following relationship, VAB = VAN + VNB = VAN − VBN . This means thatVAN =115 V and VBN = 115∠180° V.

SOLUTION P12.31. Assuming a very large resistance for the person who touches the prong. For prongA, the voltage is approximately V = 115 V. For prong N, the voltage is approximately V = 0. For theground prong G, no current flows through the wire, so the voltage should be ground or zero.

SOLUTION P12.32. For an approximate analysis, we use the circuit models below given the followingassumptions:

(i) The resistance from body to ground is very large (possibly due to rubber shoes);(ii) The resistances of all connecting wires are negligible.(iii) The motor winding is represented by R1 in series with R2. Here we can further estimate

that. R1 + R2 = 115/3 = 38.3 Ω, and R1 = R2 = 19.15 Ω.(iv) The trigger switch of the drill has been depressed.

(a) The point P is connected to the metal case as shown below. It is obvious that Vcase = Vp = 0.

9/26/01 P12-16 @ DeCarlo & P. M. Lin

(b) The point M is connected to the metal case as shown below. The current through the hot wire is115/19.15 = 6 A. If the fuse capacity is smaller than 6 A, then it will blow and Vcase = VM = 0. On the

other hand , if the fuse capacity is greater then 6 A, then R2 is nearly shorted and Vcase = VM = 0.

(c) The point Q is connected to the metal case as shown below.

9/26/01 P12-17 @ DeCarlo & P. M. Lin

Both R1 and R2 are nearly shorted. The line current is so large that it will blow the fuse. Hence Vcase= VQ = 0.

CONCLUSION: In all three cases, the person touching the metal case of the defective appliance willexperience zero or a very low voltage. The circuit is safe.

SOLUTION P12.33. For an approximate analysis, we use the circuit models below given the followingassumptions:

(i) The resistance from body to ground is very large (possibly due to rubber shoes);(ii) The resistances of all connecting wires are negligible.(iii) The motor winding is represented by R1 in series with R2. Here we can further estimate

that. R1 + R2 = 115/3 = 38.3 Ω, and R1 = R2 = 19.15 Ω.(iv) The trigger switch of the drill has been depressed.

(a) The point P is connected to the metal case as shown below. It is obvious Vcase = Vp = 0.

(b) The point M is connected to the metal case as shown below. Simple voltage divider action leads to

9/26/01 P12-18 @ DeCarlo & P. M. Lin

Vcase = VM = 0.5×115 = 57.5 V

(c) The point Q is connected to the metal case as shown below.

It is obvious that Vcase = VQ = 115 V.

CONCLUSION: In all three cases, the fuse will not blow. For cases 2 and 3, the voltage appearing onthe metal case may cause serious injury to the person.

Prbs Ch 13 March 18, 2002 P13-1 ©R. A. DeCarlo, P. M. Lin

1


SOLUTION 13 .1 . Given

′ ′ i ( t) +16 ′ i (t) + 4Bi( t) = ′ v (t) + 8v(t)

(a) with v(t) = ′ v ( t) = 0 and i1(0) + i2(0) = 0

(i) the characteristic equation is

s2 +16s + 48 = 0(ii) the characteristic equation has factors

(s + 4) (s + 12) = 0

and hence

s1,s2 = −4,−12

(iii) Equivalent circuit at t = 0

(iv) Here by KCL

i1(0) = i2(0) = 6A

At point (1)

i1(0) + i2Ω(0) = i2(0)

6A + i2Ω(0) = 6A

i2Ω(0) = 0

and

vL1 = vR2(0) = 2i2Ω (0) = 0V

Then by KVL at t = 0

vL1(0) − v6Ω (0) + vL 2(0) = 0 ⇒ −vL 2 = −36V

L2di2dt

= −36V

′ i (0) = ′ i 2(0) = −36A / s

(v) from part ii

i( t) = Ae−4 t + Be−12 t

and then

′ i (t) = −4 Ae−4 t −12Be−12 t

Then using the initial conditions

A + B = 6

−4A −12B = −36


2

solving yields B = 1.5 and A = 4.5. Then

i( t) = 4.5e−4 t +1.5e−12t A

(b) If v(t) = 12V and i(0) = ′ i (0) = 0 , then ′ v ( t) = 0 and

′ ′ i ( t) +16 ′ i (t) + 48i(t) = 8v(t)

with

i( t) = Ae−4 t + Be−12 t

′ i ( t) = 4Ae−45 −12Be−12t

′ ′ i ( t) =16Ae−4 t +144Be−12t

Then at t = ∞ when v(∞) = 12V

i(∞) = C, ′ i (∞) = 0 and ′ ′ i (∞) = 0

Thus

48C = 96 ⇒ C = 2Then

i( t) = Ae−4 t + Be−12 t + 2

and

i(0) = 0 = A + B + 2

′ i (0) = 0 = −4 A −12B

Multiply the first of these by 4 yields

4 A + 4B = −8

−4A −12B = 0

Solving yields B = 1 and A = -3. Thus for t > 0

i( t) = −3e−4 t + e−12 t + 2 A

(c) If v(t) is as in fig. 13.1(b) then

v(t) =2t 0 < t ≤ 2

0 t ≥ 2

The value of C in part (b) will change and at t = 2s, a new set of initial conditions will be required

(obtainable from the solution at t = 2s) and these would be used in the decay portion described by

′ ′ i ( t) +16 ′ i (t) + 48i(t) = 0 t ≥ 2s


3

SOLUTION 13 .2 .

(a) Use the figure with the currents i1through i5 designated in the circuit below.

Then work from v0 to vin using repeated applications of KVL, KCL and the elemental equations:

i1 = 2vo

i2 = 2 ′ ′ v oi3 = i1 + i2 = 2 ′ v o + vo( )

with

v2 = vo

v1 = 0.5i3 + v2 = 0.5 2 ′ v o + vo( ) + vo[ ] = ′ v o + 2vo

then

i4 = 2 ′ v 1 = 2d

dtvo1 + 2vo( ) = 2 ′ ′ v o + 2 ′ v o( )

i5 = i3 + i4 = 2 ′ v o + vo( ) + 2 ′ ′ v o + 2 ′ v o( ) = 2 ′ ′ v o + 6 ′ v o + 2vo

Finally

vin = 0.5i5 + v1 = 0.5 2 ′ ′ v o + 6 ′ v o + 2vo( ) + ′ v o + 2vo = ′ ′ v o + 4 ′ v o + 3vo

Hence

˙ v out (t) + 4 ˙ v out (t) + 3vout ( t) = vin (t)

(b) Note from part (a) that

vout (t) = v2(t)

and

v1( t) = ˙ v out (t) + vout ( t)

Hence

vout (o) = v2(o) =1V

˙ v out (o) = v1(o) − vout (o) = 7 −1 = 6V

(c) The characteristic equation

s2 + 4s + 3 = 0has factors

s +1( ) s + 3( ) = 0

and roots

s1,v2 = −1,−3


4

Thus, because of the input, vin (t) = 6V

vout (t) = Ae−t + Be−3t + C

˙ v out (t) = −Ae−t − 3Be−3t

˙ v out (t) = Ae−t + 9Be−3t

and at t = ∞

˙ v out (∞) + 4vout (∞) + 3vout (∞) = 6V

0 + 0 + 3C = 6V

C = 2V

and

vout (t) = Ae−t + Be−3t + 2V

SOLUTION 13 .3 . (a) From the given differential equation, the characteristic equation is

s3 +14s2 + 52s + 24 = (s + 6)(s2 + 8s + 4) = 0

Therefore the roots a = −6, b = −4 + 2 3 = −0.5359,c = −4 − 2 3 = −7.4641.

(b) (i) v(0) = vC3(0) = 6 V, as given.

(ii) To compute v'(0) we write a nodal equation at node 3. In particular

(v − vC1) + (v − vC2) + 0.5v '+v = 0 which implies that

v '(t) = 2vC1(t) + 2vC 2( t) − 6v(t) (*)

Hence v '(0) = 2vC1(0) + 2vC 2(0) − 6v(0) = 24 +18 − 36 = 6 V/s.

(iii) To compute v' '(0) we first differentiate equation (*). This yields

v ''(t) = 2v 'C1( t) + 2v 'C 2 ( t) − 6 2vC1(t) + 2vC 2( t) − 6v( t)( ) (**)

To express v 'C1( t) and v 'C 2 (t) in terms of the node voltages we write node equations at nodes 1and

2 respectively. At node 1

(vC1(t) − vC 2(t)) + (vC1(t) − v(t)) + 0.5vC1' (t) = i(t)

and at node 2

(vC 2( t) − vC1(t)) + (vC 2( t) − v( t)) + 0.5vC 2' (t) = 0

Hence

v 'C1( t) = −4vC1(t) + 2vC 2(t)) + 2v(t) + 2i(t)


5

and

vC 2' ( t) = 2vC1(t)) − 4vC 2(t) + 2v( t)

Substituting these two equations into (**) yields the desired result when t is set to 0. However, this

quantity has no direct physical meaning.

v ''(t) = 2 −4vC1( t) + 2vC 2(t)) + 2v( t) + 2i(t)( ) + 2 2vC1(t)) − 4vC 2(t) + 2v( t)( ) − 6 2vC1(t) + 2vC 2( t) − 6v(t)( )

Hence, v' '(0) = 2(–48 + 18 + 12 + 2i(0)) + 2(24 – 36 + 12) – 6(24 + 18 – 36) = –72 + 4i(0) V/s2.

Finally, using the characteristic roots found in part (a) and assuming a constant input, the form

of the solution is

v(t) = Ae−6t + Be−0.5359t + Ce−7.4641t + D

Following the methods of example 13.2,

A = 0, B = (7.3301 – 1.0774×i(0)), C = (–1.3301 + 0.0774×i(0)), and D = i(0).

(c) (i) Not proportional to a single voltage but it is proportional to iC3(0).

(ii) Much more complex.

(iii) No.

(iv) No. This is why we use the Laplace transform approach.

SOLUTION 13 .4 .

(a) f (q + T0)]q=t = f ( t + T0)

(b) e−5q cos 0.5πq + 0.25π( )]q=2 t= e−10t cos πt + 0.25π( )

SOLUTION 13 .5 .

(a) Let T = 1.


6

-1 -0.5 0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

(b) Again let T = 1.

-1 -0.5 0 0.5 1 1.5 2 2.5 3-2

-1.5

-1

-0.5

0

(c)

-1 -0.5 0 0.5 1 1.5 2 2.5 30

0.05

0.1

0.15

0.2

0.25

(d)


7

-1 -0.5 0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

(e)

-1 -0.5 0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

(f) i ( t − i)i=0

r

∑(g)

-1 -0.5 0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

(h) Pulses of height 1 and width T.


8

SOLUTION 13 .6 .

(a) L[ f1( t)] = F1(s) = f1(t)e−stdt0−∞

∫ == e−stdtT1

T2∫ =e−st

−s

T1

T2

=1s

e−sT1 − e−sT2( )(b) f2( t) = f1( t) . Hence, the answer is the same as in (a).

(c) L[ f3( t)] = F3(s) = f3(t)e−stdt0−∞

∫ = −2 ( t)cos(4πt − 0.25π)e−stdt0−∞

∫ = −222

= − 2

(d) F4(s) = f4(t)e−stdt0−∞

∫ = −2 (t − T )cos(4πt − 0.25π)e−stdt0−∞

∫ = −2cos(4πT − 0.25π)e−sT

(e) F5(s) = f5(t)e−stdt0−∞

∫ = (t) − ( t − T )[ ]e−stdt0−∞

∫ = 1− e−sT

SOLUTION 13 .7 .

(a) F(s) = f (t)e−stdt0−∞

∫ = 5e−4 te−stdt0−∞

∫ = −5e−(s+ 4)t

s + 4

0−

∞

=5

s + 4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

t

f(t)

TextEnd


9

(b) F(s) = f (t)e−stdt0−∞

∫ = 5e−4 te−stdt1−∞

∫ = −5e−(s+4) t

s + 4

1−

∞

=5e−(s+4)

s + 4= e−s 5e−4

s + 4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

t

f(t)

TextEnd

(c) F(s) = f (t)e−stdt0−∞

∫ = 5e−4(t−1)e−stdt1

∞∫ = −5e4 e−(s+ 4)t

s + 4

1−

∞

= e−s 5s + 4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

t

f(t)

TextEnd

(d) F(s) = f (t)e−stdt0−∞

∫ = 5e−4( t−1) (t)e−stdt0−∞

∫ = 5e4

(e) F(s) = f (t)e−stdt0−∞

∫ = 5e−4( t−1) (t −1)e−stdt0−∞

∫ = 5e−s


10

(f)

F(s) = f (t)e−stdt0−∞

∫ = 2 u(t)u(1− t)[ ]e−0.25te−stdt0−∞

∫ = 2 e−0.25te−stdt0−1

∫ =2

s + 0.251− e−(s+0.25)( )

= 2 e−0.25te−stdt0−∞

∫ − 2 e−0.25te−stdt1−∞

∫ =2

s + 0.251− e−s( )

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

f(t)

TextEnd

F(s) = f (t)e−stdt0−∞

∫ =

SOLUTION 13 .8 .

(a) F(s) = f (t)e−stdt0−∞

∫ = Ae t e−stdt0−∞

∫ = Ae−(s− )tdt0−∞

∫ =A

s−

(b) F(s) = f (t)e−stdt0−∞

∫ = Ae (t−1) u( t −1)e−stdt0−∞

∫ = Ae− e−(s− )tdt1−∞

∫ = e−s As−

(c) F(s) = f (t)e−stdt0−∞

∫ = Ae (t−1) ( t)e−stdt0−∞

∫ = Ae−

(d) F(s) = f (t)e−stdt0−∞

∫ = Ae te−stdt0−1

∫ = −Ae−(s− )t

(s − )

0−

1

=A

s −1− e−(s− )( )


11

SOLUTION 13 .9 . Consider the following in which q = at and t = q/a.

L[ f (at)] = f (at)e−st dt

0−

∞

∫ =1a

f (q)e−(s/a )q dq

0−

∞

∫ =1a

Fsa

SOLUTION 13 .10 .

(a)

e j t = cos t + j sin t

e− j t = cos t − j sin t

Add these equations and divide by 2 to obtain

cos( t) =e j t + e− j t

2

Similarly, subtract the equations and divide by 2j to obtain

sin( t) =e j t − e− j t

2 j

Note that

L e j t[ ] =1

s − j, L e− j t[ ] =

1s + j

(b)

L cos t[ ] = Le j t + e− j t

2

=0.5

s− j+

0.5s + j

=s

s2 + 2

and

L sin t[ ] =−0.5 js − j

+0.5 j

s + j=

s2 + 2

SOLUTION 13.11 . From the time frequency scaling property,


12

L sin( t) u(t)[ ] =1 1

s

2

+1

=s2 + 2

Using the time differentiation property,

L[ cos( t)u( t)] = Lddt

sin( t)u(t)( )

= sL[sin( t) u(t)]− sin(0) =

s

s2 + 2

Finally using the multiplication by t property

L[− t cos( t)u(t)] =dds

s

s2 + 2

=

s2 + 2 −2 s2

s2 + 2( )2

Hence

L[sin( t)u(t) − t cos( t)u( t)] =2

s2 + 2 −2 s

s2 + 2( )2 =2 s2 + 2( ) − 2 s2

s2 + 2( )2 =2 3

s2 + 2( )2

SOLUTION 13.12. (a)

g1(t) = Atsin( t), f ( t) = sin(t)u(t), F (s) =1

s2 +1

By the frequency scaling property

L f ( t)[ ] =1 1

s

2

+1

=1 2

s2 + 2

=s2 + 2

By the multiplication by t property

L t sin( t)u( t)[ ] = −dds s2 + 2

= −

−2 s

s2 + 2( )2

=2 s

s2 + 2( )2

By the linearity property

L Atsin( t)[ ] = A2 s

s2 + 2( )2


13

(b)

g2( t) = Aeat sin( t)u( t), f ( t) = sin(t)u(t), F(s) =1

s2 +1

By the frequency scaling property,

L sin( t)[ ] =s2 + 2

By the damping property

L eat sin( t)[ ] =s − a( )2 + 2

and by the linearity property

L g2( t)[ ] = As − a( )2 + 2

(c) Before beginning, note that

sin t +( ) = cos( )sin( t) + sin( )cos( t)

Here

g3(t) = Aeat sin t +( )u(t) = cos( )Aeat sin( t)u( t) + sin( )Aeat cos( t)u( t)

f (t) = sin(t)u(t), F(s) =1

s2 +1

From the linearity property and part (b),

G3(s) = Acos( )s − a( )2 + 2 + Lsin( )Aeat cos( t)u(t)

By the differentiation in the time domain property cos(t) =ddt

sin(t) which implies

L cos(t)[ ] = s L sin(t)[ ] − f (0) =s

s2 +1− 0 =

s

s2 +1

By the frequency scaling property

L cos t[ ] =1 s

s( )2+1

=12

s 2

s2 + 2

=s

s2 + 2

Then by the linearity property


14

L Asin( )cos( t)[ ] = A sin( )s

s2 + 2

Then by the frequency-shift property

L Asin( )eat cos( t)[ ] = Asin( )s − a

(s− a)2 + 2

G3(s) = L Aeat sin t +( )[ ] = Acos( )s − a( )2 + 2

+ Asin( )s − a

(s− a)2 + 2

=A

s − a( )2 + 2cos( ) + sin( )(s − a)( )

SOLUTION 13.13. (a) We are given that

f t( ) = sin t( ) ⇒ F(s) =1

s2 +1

And must find the transform of

g1(t) = Atcos( )tu( t)

By the time differentation property

cos t( ) =ddt

sin t( ) ⇒ L cos t( )[ ] =s

s2 +1− sin 0−( ) =

s

s2 +1

By the frequency scaling property with

L cos t( )[ ] =1

Fs

=

1 s

s( )2+1

=s 2

2 s2 + 2( ) =s

s2 + 2

Using the multiplication-by-t property

L t cos t( )[ ] = −dds

s

s2 + 2

= −

s2 + 2 − s(2s)

s2 + 2( )2 =s2 − 2

s2 + 2( )2

Finally, by the linearity property

L Atcos t( )u(t)[ ] =A 2 − s2( )s2 + 2( )2


15

(b) Let g2( t) = Aeat cos t( )u( t) . Recall from part (a) that

L Acos t( )[ ] = As

s2 + 2

By the Frequency-shift property

L Aeat cos( t)[ ] = As− a

s − a( )2 + 2

(c) From Trig identities,

cos t +( ) = cos( t)cos( ) − sin( t)sin( )

Recall from part (b) that

L Aeat cos( t)[ ] = As− a

s − a( )2 + 2

Hence

L Aeat cos( )cos( t)[ ] = Acos( )s − a

s− a( )2 + 2

It follows that

L Aeat cos t +( )[ ] = Acos( )s− a

s − a( )2 + 2− A sin( )L eat sin t( )[ ]

= A cos( )s− a

s − a( )2 + 2− A sin( )

(s− a)2 + 2 = Acos( )(s− a) − sin( )

s − a( )2 + 2

SOLUTION 13 .14. (a) We are given that

L cos t( )[ ] =s

s2 +1

which implies by the frequency scaling property that

L cos t( )[ ] =s

s2 + 2

Using the multiplication-by-t property

L t cos t( )[ ] = −dds

s

s2 + 2

= −

s2 + 2 − 2s2

s2 + 2( )2 =s2 − 2

s2 + 2( )2


16

(b) If g( t) = teat cos( t), then using part (a) and frequency shift property,

L teat cos( t)[ ] =s − a( )2 − 2

s− a( )2 + 2[ ]2

SOLUTION 13 .15 . (a) With

sinh(at) =eat − e−at

2

L sinh at( )[ ] =12

1s− a

−1

s + a

=

12

s + a − s− a( )s2 − a2

=

12

2a

s2 − a2

=

a

s2 − a2

(b) With

cosh(at) =eat + e−at

2

L cosh(at)[ ] =12

1s + a

+1

s − a

=

12

s − a + s + a

s2 − a2

=

12

2s

s2 − a2

=

s

s2 − a2

SOLUTION 13 .16 . (a) From Problem 15

L sinh at( )[ ] =a

s2 − a2

So that by the multiplication by t property

L t sinh at( )[ ] = −dds

a

s2 − a2

= −

−a(2s)

s2 − a2( )2 =2as

s2 − a2( )2

(b) From Problem 15

L cosh(at)[ ] =s

s2 − a2

So that by the multiplication-by-t property


17

L t cosh(at)[ ] = −dds

s

s2 − a2

= −

s2 − a2( ) − s(2s)

s2 − a2( )2 =−s2 + a2 + 2s2

s2 − a2( )2 =s2 + a2

s2 − a2( )2

SOLUTION 13.17 . Here

F(s) =s + 2s +1

(a) Since g1(t) = 5 f ( t − 2), use the time shift and linearity properties to obtain

L g1(t)[ ] = 5e−2s s + 2s +1

(b) Since g2( t) = 5e−2t f (t), use the frequency shift and linearity properties,

L g2( t)[ ] = 5s + 2( ) + 2s + 2( ) +1

= 5

s + 4s + 3

(c) From part (a),

L 5 f ( t − 2)[ ] = 5e−2s s + 2s +1

Therefore, since g3(t) = 5e−2 t f ( t − 2) , by the frequency shift property

L g3(t)[ ] = G1(s + 2) = 5e−2(s+ 2) (s + 2) + 2(s + 2) +1

= 5e−2(s+2) s + 4

s + 3

(d) Since g4 (t) = 5tf ( t) , use the multiplication-by-t and linearity principle to obtain

L g4 (t)[ ] = −dds

5 s + 2( )s +1

= −

5 (s +1)− (s + 2)[ ](s +1)2 =

5

(s +1)2

SOLUTION 13 .18 . In all parts

L f ( t)u(t)[ ] = F (s) =s

s2 + 4


18

(a) L Af ( t − T )u( t − T )[ ] = Ae−Tss

s2 + 4

(b) L Atf ( t)u(t)[ ] = −dds

Af (s) = −Adds

s

s2 + 4

= −A

s2 + 4( ) − s(2s)

s2 + 4( )2

= As2 − 4( )

s2 + 4( )2

(c) Note that the answer is simply a time shift of the function given in (b).

L A( t − T ) f (t − T)u(t − T )[ ] = Ae−sTs2 − 4( )

s2 + 4( )2

(d) This function is that of part (a) multiplied by t. Hence, by the multiplication by t property,

L Atf ( t − T )u(t − T )[ ] = −Adds

e−Tss

s2 + 4

= −Ae−Ts − Tse−Ts( )(s2 + 4) − 2s × se−Ts

s2 + 4( )2

= Ae−Ts 2s2 − 1− Ts( )(s2 + 4)

s2 + 4( )2

= Ae−Ts Ts3 + s2 + 4Ts − 4

s2 + 4( )2

SOLUTION 13 .19 . In all parts

F(s) = L f ( t)u( t)[ ] =s2 + 2

(a) By the time shift property, L Af ( t − T )u( t − T )[ ] = Ae−Ts

s2 + 2( )

Using the multiplication-by-t property,

L Atf ( t)u(t)[ ] = −Adds

F (s)[ ] = −Adds s2 + 2

= A

(2s)

s2 + 2( )2 = A2 s

s2 + 2( )2


19

© The answer here is an application of the time shift property to the answer of part (b).

L A( t − T ) f (t − T)u(t − T )[ ] = Ae−sT 2 s

s2 + 2( )2

(d) The answer here uses the multiplication-by-t property applied to the answer of part (a).

L Atf ( t − T )u(t − T )[ ] = −Adds

e−Ts

s2 + 2( )

= ATe−Ts(s2 + 2) + 2se−Ts

s2 + 2( )2

= A e−Ts Ts2 + 2s + T 2

s2 + 2( )2

Solution 13.20. (a)

f (2 t) = (2t) + (2 t −1)

For the first term on the right, the peak occurs at t = 0 and

(2 t)dt

0−

0+

∫ = 0.5 ( )d = 0.5

0−

0+

∫

under the transformation τ = 2t. For the second term, the function peaks at t = 0.5 and

(2 t −1)dt

0.5−

0.5+

∫ = 0.5 ( )d = 0.5

0−

0+

∫

under the transformation τ = 2t – 1. Hence

f (2 t) = (2t) + (2 t −1) = 0.5 ( t) + (t − 0.5)[ ]Therefore, a = 0.5 = b.


20

(b) (i) Here F(s) = L f ( t)[ ] =1 + e−s . By the time scaling property

L f (2 t)[ ] =12

F (s /2) = 0.5 1+ e−0.5s( )(b)-(ii) For this part,

L f (2 t)[ ] = 0.5L (t) + (t − 0.5)[ ] = 0.5 1+ e−0.5s( )

SOLUTION 13.21. (a)

L v(t)[ ] = 2L g"( t)[ ] − L g'(t)[ ] = 2s2F (s) − 2sg(0−) − 2g'(0− ) − sF(s) + g(0−)

= 2s2 − s( )F (s) − 2sg(0−) − 2g'(0− ) + g(0− ) =(2s−1)(s +1)

s2

(b)

L v(t)[ ] = 2L f "( t)[ ] − L f '(t)[ ] = 2s2F (s) − sF(s) − 2sf (0−) − 2 f '(0 −) + f (0−)

= 2s2 − s( )F (s) − 2s − 2 +1 =2s−1( )(s +1)

s2 − 2s − 2 +1

(c)

L v(t)[ ] = L g'( t)[ ] − L g(q)dq−∞

t

∫

= sG(s) − g(0− ) −G(s)

s−

1s

g(q)dq−∞

0−

∫ = s−1s

G(s)

= s −1s

s +1

s3 = s +1( ) 1

s2 −1

s4

(d)

L v(t)[ ] = L f '( t)[ ] − L f (q)dq−∞

t

∫

= sF(s) − f (0−) −F(s)

s−

1s

f (q)dq−∞

0−

∫

= s +1( ) 1

s2 −1

s4

−1−

−1

s

assuming λ > 0. The expression is ill-defined if λ ≤ 0.


21

SOLUTION 13.22 .

(a) (i) If

F(s) = L f ( t)u( t)[ ] = ln

s2 + 4

s2

= ln s2 + 4( ) − ln s2( )Then by the multiplication-by-t property

L −tf (t)u(t)[ ] = +dds

ln s2 + 4( ) − ln s2( )[ ]=

2s

s2 + 4−

2s

s2 =2s

s2 + 4−

2

s=

2s2 − 2s2 − 8

s s2 + 4( ) =−8

s s2 + 4( ) (ii) Using the solution to (a)-(i), by the frequency shift property

L −te−2 t f ( t)u(t)[ ] =−8

s + 2( ) s + 2( )2 + 4( ) =−8

s + 2( ) s2 + 4s + 8( )=

−8

s3 + 6s +16s +16

(b) If

G(s) =−8

s s2 + 4( )

a partial fraction expansion may be employed

G(s) =−8

s s2 + 4( ) =K1s

+As + B

s2 + 4=

−2s

+2

s2 + 4

Hence,

g( t) = 2cos(2t) − 2[ ]u( t)

and

f ( t) = −g(t)

t=

2t

−2t

cos(2t)

SOLUTION 13.23: Part (a)-(i): From table 13.2, the multiplication by t property implies that


22

L[−tf (t)u(t)] =

dds

F(s) =dds

lns + as − a

=

dds

ln s + a[ ] − ln s − a[ ]( ) =1

s + a−

1s − a

=−2a

s2 − a2

Part (a)-(ii): Let us make use of the answer to part (a)-1. Let G(s) = L[−tf ( t)u(t)]=−2a

s2 − a2 .

Then by the frequency shift property in table 13.2,

L[−te−at f (t)u(t)] = L[e−at − tf (t)u( t)( )] = G(s + a) =−2a

(s + a)2 − a2 =−2a

s(s + 2a)

Part (b): g( t) = L–1[G(s)]= L–1 −2a

s2 − a2

= L–1 1

s + a

− L–1 1

s − a

= e−at − eat( )u(t)

More specifically,

g( t) = e−at − eat( )u(t) = −2eat − e−at

2

u( t) = −2sinh(at) u(t)

Hence f ( t) =g(t)−t

=2sinh(at)

t.

SOLUTION 13 .24 .

(a)-(i) If

F(s) = L f ( t)u( t)[ ] = ln

s + as + b

= ln(s + a) − ln(s + b)

Then by the multiplication-by-t property

L −tf (t)u(t)[ ] = +

dds

ln s + a( ) − ln s + b( )[ ] =1

s + a−

1s + b

=s + b − (s + a)(s + a)(s + b)

=b − a

(s + a)(s + b)

(ii) By the frequency-shift property

L −te−at f (t)u(t)[ ] =b − a

(s + 2a)(s + a + b)=

b − a

s2 + s(3a + b) + 2a(a + b)

(b) If


23

G(s) = L −tf ( t)u(t)[ ] =b − a

(s + a)(s + b)

a partial fraction expansion may be employed

G(s) =b − a

(s + a)(s + b)=

K1s + a

+K2

s + b=

1s + a

+−1

s + b

Hence, g( t) = e−at − e−bt( )u(t)

and

f ( t) =g(t)−t

=e−bt

t−

e−at

t

SOLUTION 13.25. The relationship is f ( t) =ddt

g(t) or equivalently, g( t) = f (q)dq−∞

t

∫ .

Now we have that f ( t) = 6 (t) −12 (t − 2) + 6 (t − 4) . Therefore,

F(s) = 6 −12e−2s + 6e−4s

From the time integration property,

G(s) =F (s)

s=

6s

−12e−2s

s+

6e−4s

s

SOLUTION 13 .26 . For 0 ≤ t < T1, we see that g( t) = f (q)dq−∞

t

∫ . Thus one presupposes here that

the relationship is f ( t) =ddt

g(t) or equivalently, g( t) = f (q)dq−∞

t

∫ . As such E = A – B and D = A –

B + C.

Further, f ( t) = A (t) − B ( t − T1) + C (t − T2) which implies that

F(s) = A − Be−T1s + Ce−T2s

Thus


24

G(s) =F (s)

s=

As

−Be−T1s

s+

Ce−T2s

s

SOLUTION 13 .27 .

(a) f ( t) = u(t) + u(t −1) ⇒ f (s) =1s

+e−s

s=

1s

1− e−s( )(b)

f ( t) = u(t) + u(t −1)− u(t − 3) ⇒ f (s) =1s

+e−s

s−

e−3s

s=

1s

1+ e−s − e−3s( )(c)

f ( t) = u(t) + u(t −1)− 2u(t − 3) ⇒ F (s) =1s

+e−s

s−

2e−3s

s=

1s

1 + e−s − 2e−3s( )(d)

f ( t) = 2u(t) − u(t − 2) − u( t − 3) ⇒ F(s) =2s

−e−2s

s−

e−3s

s=

1s

2 − e−2s − e−3s( )

SOLUTION 13 .28 .

(a) f ( t) = 2r(t) − 2r(t −1) ⇒ F (s) =2

s2 −2e−s

s2 =2

s2 1− e−s( )

(b) f ( t) = 2r(t) − 2r(t −1)+ r(t − 2) ⇒ F (t) =2

s2 −2e−s

s2 +e−2s

s2 =1

s2 2 − 2e−s + e−2s( )

(c) f ( t) = 2r(t) − 2r(t −1)− 2r( t − 2) + 2r(t − 3). It follows that

F(s) =2

s2 −2e−s

s2 −2e−2s

s2 +2e−3s

s2 =2

s2 1− e−s − e−2s + e−3s( )


25

SOLUTION 13 .29 . (a) Here f ( t) =32

r(t) − 3r(t − 2) +32

r(t − 4) . Thus,

F(s) =3

2s2 −3e−2s

s2 +3e−4s

2s2 =3

2s2 1− 2e−2s + e−4s( )

(b) Here f ( t) =VoT

r(t) −2VoT

r(t − T ) +VoT

r(t − 2T ). Thus

F(s) =VoT

1

s2 −2e−T

s2 +e−2T

s2

(c) Here f ( t) = 2r(t −1)− 4r(t − 2) + 4r(t − 4) − 2r(t − 5)

F(s) =1

s2 2e−s − 4e−2s + 4e−4s − 2e−5s( )

SOLUTION 13 .30 .

(a) Here f ( t) = 2r(t) − 2r(t −1)− 2u( t − 4) implies

F(s) =2

s2 −2e−s

s2 −2e−4s

s=

2

s2 1− e−s − se−4s( )

(b) Here f ( t) = 2u(t) − r(t − 2) + r(t − 4) implies

F(s) =2s

−e−2s

s2 +e−4s

s2 =2

s2 s− e−2s − e−4s( )

SOLUTION 13.31. (a) Here f ( t) = 2u(t) − r(t) + 2r(t − 2) − 2r(t − 4) − 2u( t − 4)

Thus F(s) =2s

−1

s2 +2e−2s

s2 −2e−4s

s2 −2e−4s

s.

(b) f ( t) = −u(t) + r(t) − r(t − 2) − u( t − 2). Hence F(s) =−1s

+1

s2 −e−2s

s2 −e−2s

s.


26

(c) f ( t) = 2r(t) − 2r(t −1)− 2u( t −1) + 2u(t − 2) − 2r( t − 2) + 2r(t − 3).

Hence F(s) =2

s2 −2e−s

s2 −2e−s

s+

2e−2s

s−

2e−2s

s2 +2e−3s

s2 .

2/23/02 page P14.1 © R. A. DeCarlo, P. M. Lin


SOLUTION 14 .1 .(a)

Z(s) =R(Ls +

1

Cs)

R + Ls +1

Cs

=Cs(RLCs2 + R)

Cs(RCs + LCs2 +1)=

RLC(s2 +1 LC)

LC(s2 +R

Ls +

1

LC)

=R(s2 +1 LC)

s2 +R

Ls +

1

LC(b)

Z(s) = R +(Ls)(

1

Cs)

Ls +1

Cs

= R +LCs

C(LCs2 +1)= R +

Ls

LCs2 +1

= RLCs2 + Ls + R

LCs2 +1=

RLC(s2 +1

RCs +

1

LC)

LC(s2 +1

LC)

Hence,

Z(s) =R s2 +

1

RCs +

1

LC

s2 +1

LC

SOLUTION 14.2.(a)

Zin (s) =Vs(s)Is(s)

=(10 + 0.2s)(

80

s)

10 + 0.2s +80

s

=s(800 +16s)

s(0.2s2 +10s + 80)=

800s + 4000

s2 + 50s + 400

(b) If is( t) = 3e−20 tu(t) A then

Is(s) =3

s + 20and

Vs(s) = Zin (s)Is(s) =800s + 4000

s2 + 50s + 400−

3

s + 20=

2400s +12,000

(s +10)(s + 20)(s + 40)

=K1

s +10+

K2

s + 20+

K3

s + 40Here

K1 =2400s +12,000(s + 20)(s + 40) s=−10

=+12,000 − 24,000

(10)(30)= −40

K2 =2400s +12,000(s +10)(s + 40) s=−20

=12,000 − 48,000

(−10)(20)=180


K3 =2400s +12,000(s +10)(s + 20) s=−40

=12,000 − 96,000

(−30)(−20)= −140

Vs(s) =180

s + 20−

40s +10

−140

s + 40and for t > 0,

vs(t) = 180e−20 t − 40e−10t −140e−40 t V

SOLUTION 14.3.(a)

Yp(s) = Cs +1R

= 2 ×10−3s +1

0.10= 2 ×10−3(s + 50)

Then

Zp(s) =500

s + 50and

Zin (s) =1.25s +500

s + 50=

1.25s2 + 62.5s + 500s + 50

and

Yin (s) =Is(s)Vs(s)

=1

Zin (s)=

s + 50

1.25s2 + 62.5s + 500=

0.80s + 40

s2 + 50s + 400

With vs(t) = 90e−40 tu( t) , then

Vs(s) =90

s + 40and

Is(s) = Vs(s)Yin (s) =90

s + 40⋅

0.80s + 40(s +10)(s + 40)

=72s + 3600

(s +10)(s + 40)2 =K1

s +10+

C1s + 40

+C2

(s + 40)2

Here

K1 =72s + 3600

(s + 40)2s=−10

=3600 − 720

(30)2 =2880900

= 3.20

and with

p(s) =72s + 3600

s +10→ p(−40) =

720−30

= −24

′ p (s) =(s +10)(72) − (72s + 3600)

(s +10)2

′ p (−40) =−30(72) − (−2880 + 3600)

(−30)2 =−2160 − 720

900= −3.20

Then

C2 =p(−40)

0!= −24 , C1 =

′ p (−40)1!

= −3.20 ,


Is(s) =3.20s +10

−3.20

s + 40+

24

(s + 40)2

and for t > 0

is( t) = 3.20e−10t − 3.20e−40t + 24te−40t A

SOLUTION 14.4.(a) Find Zin (s) vis Yin (s)

Yin (s) = Cs +1

Ls + 20+

1

10=

200Cs +10LCs2 +10 + 20 + Ls

10(Ls + 20)

=10LCs2 + (200C + L)s + 30

10Ls + 200and

Zin (s) =1

Yin (s)=

10Ls + 200

10LCs2 + (200C + L)s + 30Ω

With C = 10−3F and L = 0.05 H

Zin (s) =0.50s + 200

0.0005s2 + 0.25s + 30=

1000s + 4 ×105

s2 + 500s + 60,000Ω

(b) If is( t) = 0.3u(t) A, then

Is(s) =0.30

sand

Vin (s) = Zin (s)Is(s) = 300s + 400

s(s2 + 500s + 60,000)

= 300s + 400

s(s + 2000)(s + 300)

= 300K1s

+K2200

+K3300

It follows that

K1 =s + 400

(s + 200)(s + 300) s=0=

400200(300)

=1

150

K2 =s + 400

s(s + 300) s=−200=

200(−200)(100)

= −1

100and

K3 =s + 400

s(s + 200) s=−300=

100(−300)(−100)

=1

300Thus


Vin (s) = 300

1

150s

−

1

100s + 200

+

1

300s + 300

=2

s−

3

s + 200+

1

s + 300and for t > 0

vin (t) = 2 − 3e−200t + e−300tV

SOLUTION 14.5.

Z(s) =s + 20s + 40

and the network is “at rest”(a) If

vin (t) = 20u(t) → Vin (s) =20s

then

Iin (s) =Vin (s)Z(s)

=20s

s + 40s + 20

= 20

s + 40s(s + 20)

Using a partial function expansion

Iin (s) = 20s + 40

s(s + 20)

= 20

K1s

+K2

s + 20

in which case

K1 =s + 40s + 20 s=0

=4020

= 2, K2 =s + 40

s s=−20=

20−20

= −1

Thus

Iin (s) = 202s

−1

s + 20

and

iin (t) = 20(2 − e−20t )u( t) A(b) Note that

vin (t) = 20e−40t → Vin (s) =20

s + 40Then

Iin (s) =Vin (s)Z(s)

=20

s + 40s + 40s + 20

=

20s + 20

in which case

iin (t) = 20e−20 tu( t) A(c) Note that

vin (t) = 20e−20t → Vin (s) =20

s + 20Then


Iin (s) =Vin(s)Z (s)

=20

s + 20s + 40s + 20

= 20

s + 40

(s + 20)2

Using a partial fraction expansion

Iin (s) = 20s + 40

(s + 20)2

= 20C1

s + 20+

C2

(s + 20)2

Herep(s) = s + 40 → p(−20) = 20

p'( s) =1 → p'(−20) =1and

C1 =p'(−20)

1!=1, C2 =

p(−20)0!

=201

= 20

in which case

Iin (s) = 201

s + 20+

20

(s + 20)2

so that

iin (t) = 20e−20t + 400te−20 t( )u( t) A

SOLUTION 14.6.(a) Apply an arbitrary Iin (s) to the upper terminal of Fig. P14.6a. Assuming branch currents Ia(s) andIb(s), it follows by KCL that

Iin (s) = Ia(s) + Ib(s) = 0.020sVa(s) + 0.005s Va (s) − sVa (s)[ ]= (0.020 + 0.005 − 0.015)sVa(s) = 0.010sVa (s)

Hence,

Zin (s) =Va(s)Iin (s)

=1

0.010s=

100s

Ω

(b) Similarly apply an arbitrary Iin (s) to Fig P14.6b to obtain, in the s-domain, by KCL

Iin (s) = 10sVin (s) +1

50Vin (s) + 0.10s Vin (s) −

Vin (s)

4

= 10s +

1

50+

30

4s

Vin (s)

=2000s2 + 4s +1500

200s

Vin (s)

Hence

Yin (s) =Iin (s)Vin (s)

=10s2 + 0.02s + 7.50

s

(c) Here we apply Vin (s) to the input terminals of figure P14.6c. By KCL

Vin (s) =10Iin (s) + 0.2sIin (s) +400

sIin (s) −

12

Iin (s)

= 10 + 0.2s +

200s

Iin

in which case


Zin (s) =Vin (s)Iin (s)

=0.2s2 +10s + 200

sΩ

SOLUTION 14 .7 . Writing two loop equations we obtain:

Vin (s) =100I1(s) + 200I2(s)and

100I1(s) + 100 +100

s

I2(s) = 0

In matrix form (dropping the s-dependence)

100 200

100 100 +100

s

I1I2

=

Vin

0

Using Cramer's rule,

I1 =

detVin 200

0 100 +100

s

det100 200

100 100 +100

s

=100s +100

s×

1

100100s +100

s

− 200

Vin

Hence

Zin (s) =VinI1

= −100s −1s +1

SOLUTION 14.8.Working in the s-domain, apply KVL to the left side of the circuit to obtain

Vin (s) =100

sIin (s) +

s100

Iin (s) + V2(s)

Now apply KCL to the right side to obtain

Iin (s) =s

100V2(s) +

100s

V2(s)

Thus

Vin (s) =s2 +104

100s

Iin (s) +V2(s)

To find V2(s) note that

Iin (s) =s2 +104

100sV2(s)

implying that


V2(s) =100s

s2 +104

Iin (s)

Thus

Vin (s) =s2 +104

100s+

100s

s2 +104

Iin (s) =

(s2 +104 )2 +104 s2

100s(s2 +104)Iin (s)

implying that

Zin (s) =Vin (s)Iin (s)

=s4 + 3 ×104 s2 +108

100s(s2 +104 ) Ω

and

Yin (s) =1

Zin (s)=

100s(s2 +104 )

s4 + 3 ×104 s2 +108 S

SOLUTION 14.9.Three mesh equations for the circuit

R + Z2(s) −Z2(s) −R

−Z2(s) 2R + Z2(s) −R

−R −R 2R + Z1(s)

I1(s)

I2(s)

I3(s)

=Vin (s)

0

0

Solve for I1(s) via Cramer’s rule

I1 =

det

Vin −Z2(s) −R

0 2R + Z2(s) −R

0 −R 2R + Z1(s)

det

R + Z2(s) −Z2(s) −R

−Z2(s) 2R + Z2(s) −R

−R −R 2R + Z1(s)

=Vin 3R2 + 2R Z1 + Z2( ) + Z1Z2( )

d(s)

where

d(s) = R + Z2( ) 3R2 + 2R Z1 + Z2( ) + Z1Z2( ) + Z2 −Z2(2R + Z1) − R2( ) − 2R2Z2 + 2R3( )= R3 + 2R2Z1 + 2R2Z2 + 3RZ1Z2

Under the condition that Z1(s)Z2(s) = R2, we have

Zin (s) =VinI1

=R3 + 2R2Z1 + 2R2Z2 + 3RZ1Z2

3R2 + 2R Z1 + Z2( ) + Z1Z2=

4R3 + 2R2 Z1 + Z2( )4R2 + 2R Z1 + Z2( ) = R

SOLUTION 14.10.

(a) Yin (s) = Cs +1R

implies a parallel RC circuit with values R and C respectively.


(b) Yin (s) =1

Zin (s)= 2s +

14

which is a parallel RC circuit of values 4 Ω and 2 F respectively.

(c) Zin (s) =1 +1

2s + 0.25=1 +

1Yb(s)

. Using the result of part (b), this circuit is a 1 Ω resistor in series

with the parallel RC of part (b).

(d) Zin (s) =2s + 8s + 2

= 2 +4

s + 2= 2 +

1

0.25s +1

2

. Using the results of parts (b) and (c), this circuit is a 2 Ω

resistor in series with a parallel combination of a 0.25 F capacitor and a 2 Ω resistor.

(e) Zin (s) =s + 3s +1

+s + 6s + 4

= 2 +2

s +1+

2s + 4

= 2 +1

0.5s +1 2+

10.5s +1 0.5

. Using the above results, this

circuit is a 2 Ω resistor in series with a parallel combination of a 0.5 F capacitor and a 2 Ω resistor which isin series with another parallel combination of a 0.5 F capacitor and a 0.5 Ω resistor.

SOLUTION 14.11.(a) Clearly this is an inductor of value L in series with a resistor of value R.(b) Inverting the admittance we have Zin(s) of the form of part (a). Hence the circuit is a 0.5 H inductorin series with a 10 Ω resistor.

(c) Yin (s) = 0.2 +1

0.5s +10= 0.2 +

1Zb(s)

. Using the result of part (b), the circuit is 0.2 S resistor in

parallel with a series connection of a 0.5 H inductor and a 10 Ω resistor.

(d) Yin (s) =10s + 50

s +1= 10 +

40s +1

= 10 +1

0.025s + 0.025. This is similar to part (c). Hence the circuit is a

10 S resistor in parallel with a series connection of a 25 mH inductor and a 0.025 Ω resistor.

(e) Yin (s) =s + 3s +1

+s + 6s + 4

= 2 +2

s +1+

2s + 4

= 2 +1

0.5s + 0.5+

10.5s + 2

. Hence, the circuit is a 2 S

resistor in parallel with the series connection of a 0.5 H inductor and a 0.5 Ω resistor which in turn is inparallel with a 0.5 H inductor and 2 Ω resistor.

SOLTUION 14.12.

(a) Zin (s) = Ls +1Cs

represents a series connection of an inductance L and a capacitance C.

(b) Yin (s) = Cs +1Ls

represents a parallel connection of an inductance L and a capacitance C.

(c) Zin (s) =0.125s2 +1

0.25s= 0.5s +

10.25s

which is a series connection of a 0.5 H inductor and a 0.25 F

capacitor.

(d) Yin (s) =0.125s2 +1

0.25s= 0.5s +

10.25s

which is a parallel connection of a 0.5 F capacitor and 0.25 H

inductor.


(e) Zin (s) =s2 +1

s+

0.25s2 +10.25s

= 2s +1s

+4s

= 2s +1

0.2s which is a 2 H inductor in series with a 0.2 F

capacitor.

(f) Zin (s) =s2 +1

s+

0.25s

0.25s2 +1= s +

1s

+1

s +1

0.25s

. This circuit is a 1 H inductor in series with a 1 F

capacitor which is in series with a parallel connection of a 1 F capacitor and a 0.25 H inductor.

(g) Yin (s) =s2 +1

s+

0.25s

0.25s2 +1= s +

1s

+1

s +1

0.25s

. This circuit is a 1 F capacitor in parallel with a 1 H

inductor which is in parallel with a series connection of a 1 H inductor and a 0.25 F capacitor.

SOLUTION 14.13.With L[vout ( t)] = Vo(s) and L[vin (t)] = Vi (s) and vout (0− ) = 0,

sV0(s) + 25V0(s) +100

sV0(s) = 5Vi (s) −

10s

Vi (s)

which implies that

s2 + 25s +100s

V0(s) =

5s −10s

Vi (s)

The transfer function is

H(s) =V0(s)Vi (s)

=5s−10

s2 + 25s +100=

5s −10(s + 5)(s + 20)

(a) If vin (t) = te−5tu(t) V, then

Vi (s) =1

(s + 5)2

and

Vout (s) =5s−10

(s + 20)(s + 5)3 =K1

s + 20+

C1s + 5

+C2

(s + 5)2 +C3

(s + 5)3

K1 =5s −10

(s + 5)3s=−20

=−110

(−15)3 =−110−3375

=22675

p(s) =5s−10s + 20

′ p (s) =(s + 20)(5) − (5s = 10)

(s + 20)2 =110

(s + 20)2 = 110(s + 20)−2

′ ′ p (s) =−220

(s + 20)3

p(−5) =−25 −10

15=

−3515

= −73

= −1575675

′ p (−5) =110

(−15)2 =110225

=330675


′ ′ p (−5) = −220

(−15)3 = −2203375

= −44675

Then

C3 =p(−5)

01= −

1575675

, C2 =′ p (−5)1!

=330675

, C1 =′ ′ p (−5)2!

=12

−44675

= −

22675

and

Vout (s) =1

67522

5 + 20−

22s + 5

+330

(s + 5)2 −1575

(s + 5)3

This yields

vout (t) =1

67522e−20t − 22e−5t + 330te−5t −

15752

t2e−5t

u( t) V

(b) If vin (t) = u(t) V,

Vin (s) =1s

and

Vout (s) =5s −10

s(s + 5)(s + 20)=

K1s

+K2s + 5

+K3

s + 20

K1 =5s =10

(s + 5)(s + 20) s=0=

−10100

= −110

K2 =5s −10

s(s + 20) s=−5=

−25 −10(−5)(+15)

=−35−75

=7

15and

K3 =5s−10s(s + 5) s=−20

=−100 −10(−20)(−15)

=−110300

=−1130

Thus

vout (t) =715

e−5t −1130

e−20t −1

10

u(t) V

By virtue of linearity and time invariance, if vin (t) = [u( t) − u(t − 0.5)] V,

vout (t) =7

15e−5t −

11

30e−20t −

1

10

u(t)

−7

15e−5(t−0.5) −

11

30e−20(t−0.5) −

1

10

u(t - 0.5) V

SOLUTION 14.14.Here vin (t) = cos( t) u(t) V and iout ( t) = 2sin(t)u( t) A, in which case


H(s) =Iout (s)Vin (s)

=

2

s2 +1s

s2 +1

=2s

SOLUTION 14.15.

Here vin (t) = te−tu( t) V which implies that Vin (s) =1

(s +1)2 . Further,

vout (t) = (1+ t − 0.5t2)e−tu( t) + sin(t)u(t) − cos(t)u(t) V in which case

Vout (s) =1

s +1+

1

(s +1)2+

1

(s +1)3 +1

s2 +1−

s

s2 +1(a) Hence

H(s) =Vout (s)Vin (s)

=1

s +1+

1

(s +1)2+

1

(s +1)3 −s −1

s2 +1

(s +1)2

= (s +1)+1 +1

(s +1)−

(s−1)(s +1)2

s2 +1Simplifying

H(s) =s3 + 2s2 + 5s + 2

(s +1)(s2 +1)

(b) If vin (t) = (1+ t)u( t) V, then Vin (s) =1s

+1

s2 =s +1

s2 . Hence

Vout (s) = H (s)Vin (s) =s3 + 2s2 + 5s + 2

s2(s2 +1)=

5s

+2

s2 −4s

s2 +1implying that

vout (t) = 5 + 2t − 4cos(t)[ ]u( t) V

SOLTUION 14.16.(a) By a voltage divider (Fig. P14.16a)

Vout (s) =Z4 (s)

Z3(s) + Z4 (s)Vin (s)

and

H(s) =Z4 (s)

Z3(s) + Z4 (s)(b) In Fig. P14.16b,

Yin (s) = Y1(s) + Y2(s)and

Vout (s) =1

Yin (s)Iin (s) =

1Y1(s) + Y2(s)

Iin (s)


Hence

H(s) =Vout (s)Iin (s)

=1

Y1(s) +Y2(s)(c) By current division,

Iout (s) =

1

Z3(s) + Z4(s)

Y1(s) + Y2(s) +1

Z3(s) + Z4(s)

Iin (s) =1

Y1(s) + Y2(s)[ ] Z3(s) + Z4 (s)[ ] +1Iin (s)

Hence

Vout (s) = Z4(s)Iout (s) =Z4(s)

Y1(s) +Y2(s)[ ] Z3(s) + Z4(s)[ ] +1Iin (s) .

and

H(s) =VoutIin

=Z4 (s)

[Y1(s) +Y2(s)] [Z3(s) + Z4 (s)] +1

SOLUTION 14.17. With Vin (s) = Vi and Vout (s) = V0 , H(s) =V0Vi

. By voltage division,

H(s) =V0Vi

=

1

10−4 s +10−3

103 +1

10−4 s +10−3

=1

0.1s + 2=

10s + 20

(a) Vout (s) =400

s(s + 20)=

20s

−20

s + 20 ⇒ vout (t) = 20 − 20e−20 t( )u( t) V. Plot omitted.

(b) If vin (t) = 40 u(t) − u( t − 0.2)[ ]V , then by linearity and time invariance

vout (t) = 20(1− e−20t )u( t) − 20 1− e−20(t−0.2)[ ]u(t − 0.2) V

(c) If vin (t) = 40 u(t) + u(t − 0.2)[ ]V, then by linearity and time invariance

vout (t) = 20(1− e−20t )u( t) + 201 − e−20(t−0.2)[ ]u( t − 0.2) V

(d) If vin (t) = 40e−20 tu(t) V, Vi (s) =40

(s + 20). Hence,

V0(s) = H (s)Vi (s) =400

(s + 20)2 ⇒ vout (t) = 400te−20 tu(t) V

(e) If vin = 40te−20 tu(t) V , then Vi (s) =40

(s + 20)2 . Hence,

V0(s) = H (s)Vi (s) =400

(s + 20)3 ⇒ vout ( t) = 200t2e−20tu( t) V


SOLUTION 14.18.(a) By voltage division

Vout (s) =

2s +1

s

2 +2

s+

2s +1

s

Vin (s) =

2s +1

s2s + 2 + 2s +1

s

Vin (s) =2s +14s + 3

Vin (s)

Hence


=2s +14s + 3

(b) With vin (t) = 8u(t) then

Vout (s) = H (s)Vin (s) =2s +14s + 3

8s

=

16s + 8s(4s + 3)

Using MATLAB

»n = [16 8]; d = [4 3 0];»[r,p,k] = residue(n,d)r = 1.3333e+00 2.6667e+00p = -7.5000e-01 0k = []Then

vout (t) =83

+43

e−0.75t

u( t) V

(c) If vin (t) = 8sin(2 t)u(t) , then

Vout (s) = H (s)Vin (s) =2s +14s + 3

16

s2 + 4

Using MATLAB,

»ilaplace( (32*s+16)/((4*s+3)*(s^2+4)) )ans =-32/73*exp(-3/4*t)+32/73*cos(2*t)+280/73*sin(2*t)

Hence,

vout (t) = −0.43836e−0.75t + 0.43836cos(2t) + 3.8356sin(2t)( )u(t) V

(d) With vin (t) = 8sin(8 t)u(t)

Vout (s) = H (s)Vin (s) =2s +1

4s + 3

64

s2 + 64

=

128s + 64

(4 s + 3)(s2 + 64)

=0.12391s + 31.907

s2 + 64−

0.12391

s + 0.75Using MATLAB

»ilaplace((128*s + 64)/((4*s+3)*(s^2+64)))ans =-128/1033*exp(-3/4*t)+128/1033*cos(8*t)+4120/1033*sin(8*t)


»»128/1033ans = 1.2391e-01»4120/1033ans = 3.9884e+00

vout (t) = 0.12391cos(8t) + 3.9884sin(8t) − 0.12391e−0.75t( )u( t) V

SOLUTION 14.19.

With a source transformation Iin (s) =Vin (s)

R.

(a) By current division,

IC (s) =Cs

1

R+ Cs +

1

Ls

Vin (s)R

=LCs2

LCs2 +L

Rs +1

Vin (s)R

=s2

s2 +1

RCs +

1

LC

Vin (s)R

Here

H(s) =IC (s)Vin (s)

=1R

s2

s2 +1

RCs +

1

LC

(b) With R =23

Ω, C = 0.5F and L = 1H ,

H(s) =32

s2

s2 + 3s + 2

If vin (t) = e−tu( t) V, then Vin (s) =1

s +1. Hence

IC (s) = H (s)Iin (s) =32

s2

s2 + 3s + 2

1

s +1

=

3s2

2(s +1)2(s + 2)=

K1s + 2

+C1s +1

+C2

(s +1)2

Using MATLAB,»n = [3 0 0]; d = conv([2 4],[1 2 1]);»[r,p,k] = residue(n,d)r = 6.0000e+00 -4.5000e+00 1.5000e+00p = -2.0000e+00 -1.0000e+00 -1.0000e+00k = []»


IC (s) =6

s + 2−

4.5s +1

+1.5

(s +1)2

and

iC ( t) = 6e−2t −92

e−t +32

te−t

u(t) A

SOLUTION 14.20.(a) Make a source transformation:

Vin (s) =1Cs

Iin (s) =250

sIin (s)

By voltage division

Vout (s) =10

250s

+250

s+

120

s +10

250

sIin (s)

=

2500

s120

s2 +10s + 500

Iin (s)

=50,000

s s2 + 200s +10,000( ) Iin (s)

and


=50,000

s(s2 + 200s +10,000)(b) If iin (t) = ( t) implies Iin (s) = 1. Using MATLAB»n = 50e3; d = [1 200 10e3 0];»[r,p,k] = residue(n,d)r = -5 -500 5p = -100 -100 0k = []Hence

Vout (s) =50,000

s(s +100)2 =5s

−5

s +100−

500

(s +100)2

and

vout (t) = 5 − 5e−100t − 500te−100 t( )u( t) V

This is the impulse response

(c) If Lin (t) = 100u(t) mA so that Iin (s) =0.1s

. Therefore

Vout (s) =5000

s2(s +100)2

In MATLAB,


»n = 5000; d = conv([1 0 0],[1 200 1e4])d = 1 200 10000 0 0»[r,p,k] = residue(n,d)r = 1.0000e-02 5.0000e-01 -1.0000e-02 5.0000e-01p = -100 -100 0 0k = []Hence

Vout (s) =−0.01

s+

0.5

s2 +0.01

s +100+

0.5

(s +100)2

and

vout (t) = 0.01e−100t + 0.5te−100 t − 0.01 + 0.5t[ ]u( t)

(d) By superposition and time invariance, ifiin (t) = 100 u(t) + u( t −1)[ ] mA

then the result of part (c) can be adjusted to

vout (t) = 0.01e−100t + 0.5te−100 t − 0.01 + 0.5t[ ]u( t)

− 0.01e−100(t−1) + 0.5te−100(t−1) − 0.01 + 0.5(t −1)[ ]u( t −1)

V

SOLUTION 14.21. For this problem change the 20 mH inductor to one of 0.3 H. (a)

Yin = 115

+ 10.3s + 90

+ 10.1s + 10

= ( s + 200) (s + 400)

15( s+ 100) (s + 300)

and

H(s) = IoutIin

= 1/15Yin

= ( s + 100) (s + 300)( s+ 200) (s + 400)

(b) If iin(t) = δ(t), then Iin(s) = 1 and

Iout(s) = H(s) = ( s + 100) (s + 300)( s+ 200) (s + 400)

= 1 - 50s + 200

- 150 s + 400

Henceiout(t) = δ(t) + (- 50 e-200t - 150 e-400 t) u(t) A

(c) We first find the response to iin(t) = 16u(t) mA. Here Iin(s) = 0.016/s and


Iout(s) = H(s)Iin(s) = 0.016( s + 100) (s + 300)

s( s+ 200) (s + 400) = 0.006

s + 0.004s + 200

+ 0.006 s + 400

Henceiout(t) = (6 + 4 e-200t + 6 e-400 t)u(t) mA

By linearity and time invariance, the response to iin(t) = 16[ut) – u(t – 0.01)] mA is

iout(t) = (6 + 4 e-200t + 6 e-400 t)u(t) - (6 + 4 e-200(t - 0.01) + 6 e-400 (t - 0.01))u(t - 0.01) mA

A plot of iout(t) using MATLAB is given below.

t= 0: 0.0005: 0.05;f1= (6 + 4*exp(-200*t) + 6*exp(-400*t)).*u(t);f2= (6 + 4*exp(-200*(t-0.01)) + 6*exp(-400*(t-0.01))).*u(t-0.01);iout= f1 - f2;plot(t, iout)gridylabel('iout in mA')xlabel(' time in second')

-10

-5

0

5

10

15

20

0 0.01 0.02 0.03 0.04 0.05

iout

in m

A

time in second

REMARK: Notice that the resistor current is not continuous.

SOLUTION 14.22 . (a) First observe that the admittance of a parallel LC is

YLC(s) = Cs +1Ls

Using voltage division,


Vout (s) =C1s +

1

Ls

C1s +1

Ls+ C2s +

1

Ls

Vin (s) =C1s +

1

Ls

C1 + C2( )s +2

Ls

Vin (s) =LC1s2 +1

L C1 + C2( )s2 + 2Vin (s)

Finally


=C1

C1 + C2( )s2 +

1

LC1

s2 +2

L C1 + C2( )= 0.2

s2 + 4 ×106

s2 +1.6 ×106

(b) Using MATLAB,»syms s t»ilaplace(0.2*(s^2+4e6)/(s^2+1.6e6))ans =1/5*Dirac(t)+120*10^(1/2)*sin(400*10^(1/2)*t)»120*10^(1/2)ans = 3.7947e+02

h(t) = 0.2L−1 s2 + 4 ×106

s2 +1.6 ×106

= 0.2 (t) + 379.47sin(1264.9t)u(t) V

SOLUTION 14.23.

Y1(s) = C1s +1R1

=R1C1s +1

R1

Y2(s) = C2s +1

R2=

R2C2s +1R2

Then Z1(s) =R1

R1C1s +1 and Z2(s) =

R2R2C2s +1

. By voltage division,

Vout (s) =

R2

R2C2S +1R1

R1C1s +1+

R2

R2C1s +1

Vin (s) =

R2

R2C2s +1

Vin (s)

R1R2C2s + R1 + R1R2C1s + R2

(R1C1s +1)(R2C2s +1)

=R2(R1C1s +1)

(C1 + C2)R1R2s + R1 + R2Vin (s)

Thus the transfer function is:


=R2(R1C1s +1)

(C1 + C2)R1R2s + R1 + R2

(b) If C1 = 0.5 F , C2 =1.0 F and vin (t) =10 u(t) V, then


Vout (s) = H (s)Vin (s) =0.5R1R2s + R2

1.5R1R2s + R1 + R2

10s

Moreover, with R1R2

= 4 so that R1 = 4R2

Vout (s) =2R2

2s + R2

6R22s + 5R2

10s

=

20R22(s +

1

2R2)

6R22 s(s +

5

6R2)

=103

s +1

2R2

s(s +5

6R2)

The partial fraction expansion is

Vout (s) =103

s +1

2R2

s s +5

6R2

=103

K1s

+K2

s +5

6R2

Observe that

K2

s +5

6R2

→L−1

K2e−

5

6R2t

and that it is required that

−5

6R2= −

53

ThusR2 = 0.5 Ω, R1 = 4R1 = 2 Ω

and

Vout (s) =103

s +1

s s +5

3

=103

0.6s

+0.4

s +5

3

=2s

+4 / 3

s +5

3Thus,

vout (t) = 2 +43

e− 5

3t

u(t)

(c) If R1C1 = R2C2 = , then the transfer function is

H(s) =R2(R1C1s +1)

R1R2C1s + R1R2C2s + R1 + R2=

R2 s + R2

R2(R1C1) + R1(R2C2)[ ]s + R1 + R2

= R2( s +1)(R1R2)( s +1)

= R2R2 + R1

The zero-state response is


Vout (s) =R2

R2 + R110u(t)

(d) Using H(s) from part (c) with the requirement that R1C1 = R2C2 , then

H(s) =R2

R1 + R2=

110

With R2 = 106 Ω, then 10R2 = R1 + R2 ⇒ R1 = 9R2 = 9 MΩ. Since C2 = 5 ×10−12 F, then

C1 =R2C2

R1= 0.556 ×10−12 F

SOLUTION 14.24.

(a) H(s) =Vout (s)Vin (s)

. Here Ib(s) =Vin (s)2000

. The parallel admittance at the right is

YR(s) = Cs +1R

=RCs +1

Rso that

ZR(s) =R

RCs +1Then

Vout (s) = −ZR(s) Ib(s) = −R

2000(RCs +1)

(b) With V1(s) = Vin (s) −Vout (s) , then

Cs Vout (s) −Vin (s)( ) +1sVout (s) +

12

Vout (s) − 3Vin (s) + 3Vout (s)[ ] = 0

Hence

Cs +32

Vin (s) = Cs +

1s

+ 2

Vout (s)

and


=s Cs +1.5( )Cs2 + 2s +1

=s s + 0.75( )s2 + s + 0.5

(c) Transform the current source iin (t) into a voltage source. In the s-domain with Iin (s) = 2Vin (s)

Here ZC (s) =1

Cs=

12s

which implies YC (s) = 2s . A single node equation yields

23

Vout (s) −Vin (s)[ ]−23

Vout (s) + 2sVout (s) = −23

Vin (s) +23

−23

+ 2s

Vout (s) = 0


23

Vin (s) = 2sVout (s) implies Vout (s) =Vin (s)

3s

But Vin (s) =Iin (s)

2 in which case

Vout (s) =Iin (s)

6sand


=16s

SOLUTION 14.25. Here in the s-domain

Iin (s) =Vin (s) −VC (s)

0.5s +10and with a node at VC (s)

−Iin (s) +sVC (s)500

+12

Iin (s) = 0 implies sVC (s)

50=

12

Iin (s) =Vin (s) −VC (s)

s + 20Hence,

s2 + 20s + 500500

VC (s) = Vin (s)

and the transfer function is

H(s) =VC (s)Vin (s)

=500

s2 + 20s + 500

With vin (s) = 4 5 u(t) V which implies Vin (s) =4 5

s, and

VC (s) =2000 5

s(s +10 − j20)(s +10 + j20)=

K1s

+K

s +10 − j20+

K*

s +10 + j20

where K* designates the complex conjugate of K

K1 =2000 5

s2 + 20s + 500 s=0

=2000 5

500= 8.944

K =2000 5

s(s +10 + j20)s=− (10− j20)

=2000 5

(−10 + j20)( j40)=

2000 5

−(800 + j400)= −4.472 + j2236 = Jeij

where = 153.44o . Then with

K* = −4.472 − j2.236 = 5e− j

VC (s) =8.944

s+

A + jBs +10 + j20

+A − jB

s +10 − j20Here


A = −4.472, B = 2.236, A2 + B2 = 5and

arc tanBA

= arc tan2.236−4.472

= arc tan1

−2= −153.44o

With the help of Table 13.1.

vC (t) = 8.944 +10e−10t cos(20 t +153.44o)[ ]u(t) V

SOLUTION 14.26. In the s-domain we first find Vx (s) in terms of Vin (s) via voltage division:

Vx (s) =Zp(s)

40 + Zp(s)Vin (s)

where

Zp(s) =(0.40s)(40)0.40s + 40

=40s

s +100Hence

Vx (s) =

40s

s +100

40 +40s

s +100

Vin (s) =40s

80s + 4000=

0.5ss + 50

Vin (s)

and

IL (s) =Vx (s)0.4s

=2.5s

Vx (s)

Then from the right hand side by another voltage division

Vout (s) =10

1000

s+10

0.25Vx (s) =10s

10s +10000.25Vx (s) =

0.25ss +100

Vx (s)

(a) If vin (s) = 20(1− e−40t )u(t), then Vin (s) =20s

−20

s + 40. Hence

Vx (s) =0.5s

s + 50

20s

−20

s + 40

=

ss + 50

10s

−10

s + 40

and

IL (s) =2.5s

Vx (s) =2.5

s + 50

1s

−1

s + 40

=

0.05s

+−0.25s + 40

+0.2

s + 50Hence

iL ( t) = 0.05 + 0.2e−50 t − 0.25e−40t[ ]u( t) A

(b) Vin (s) =20s

−20

s + 40. From, part (a), it was found that

Vout (s) =0.25ss +100

Vx (s) =0.25ss +100

×0.5ss + 50

Vin (s) =s

s +100×

ss + 50

×2.5s

−2.5

s + 40


=100s

s + 40( ) s + 50( ) s +100( ) =−20/ 3s + 40( ) +

10s + 50( ) +

−10 /3s +100( )

Thus,

Vout ( t) = 10e−50t −103

e−100 t −203

e−40 t

u(t) V

SOLUTION 14.27. In both parts (a) and (b), the op-ampis ideal. It will not draw current and thevirtual groundprincipal requires that

v + = v− = 0

(a) For a note at the inverting terminal with mode voltage v1( t) = 0 , KCL gives in the s-domain

Vin (s)R1

= −CsVout (s) −Vout (s)

R2 implies

Vin (s)R1

= − Cs +1R2

Vout (s)

in which caseVin (s)

R1= −

R2Cs +1R2

Vout (s)

Then,


= −R2R1

1R2Cs +1

= −

1R1C

1

s +1

R2C

To make

H(s) = −20

s + 4make

1R1C

= 20 and 1

R2C=

14

If

C = 1 F = 10−6F

1R2C

= 4 or R2 =1

4C=

106

4= 250 kΩ

and1

R1C= 20 or R1 =

220C

=10620

= 50 kΩ

(b) With Fig. 14.27b in the s-domain and v1( t) at the inverting terminal, KCL gives

C1sVin (s) +Vin (s)

R1= −C2sVout (s) −

Vout (s)R2

R1C1s +1R1

Vin (s) = −

R2C2s +1R2

Vout (s)


Then


= −R2R1

R1C1s +1R2C2s +1

= −

R1R2C1R1RC2

s +1

R1C1

s +1

R2C2

= −C1C2

s +1

R1C1

s +1

R2C2

(c) IfH(s) = −5, C2 =1 F and R2 =1MΩ

thenC1C2

= 5 and C1 = 5 F

The bracketed term must cancel and with

R2C2 = 106(10−6) = 1Then with C1 = 5 F

R1C1 = 1

R1 =1C1

=1

5 ×10−6 = 200kΩ

(d) Using H(s) in part (b)

H(s) = −C1C2

s +1

R1C1

s +1

R2C2

to obtain

H(s) = −5s +1s + 2

with C2 =1 F

C1 = 5C2 = 5 ×106F (5 F)

1R1C1

= 1 or R1 =1C1

=106

5= 200kΩ

1R2C2

= 2 or R2 =1

2C2= −

106

2= 500kΩ

SOLUTION 14.28. Here, the op-amp will not draw current at the non-inverting terminal and theprincipal of the virtual ground demand that

v1 = v2 = Vin (s)For Fig. 14.28 in the s-domain with a node V1(s)taken at the inverting terminal

Vin (s)R1

+Vin (s) −Vout (s)

Zp(s)= 0

Here


Zp(s) =

R2

Cs1

Cs+ R2

=R2

R2Cs +1=

1

C(s +1

R2C)

and1R1

+ C(s +1

R2C)

Vin (s) = C(s +

1R2C

)Vout (s)

R1C s +1

R2C

+1

Vin (s) = R1C s +

1R2C

Vout (s)

and


=R1C(s +

1

R2C) +1

R1C(s +1

R2C)

=s +

1

R2C+

1

R1C

s +1

R1C

Here 1

R2C+

1R1C

= 4

and1

R1C= 2

If C = 1 F then R1 =1

2C=

106

2= 500 kΩ

and 1

R2C= 4 − 2 = 2 implies R2 = 500 kΩ .

SOLUTION 14.29. For the non-inverting configuration in the s-domain, each of the two op-amps incascade have a transfer function

H(s) = −Z f (s)

Zin (s)Then for the two op-amps

H0(s) = −Z f 1(s)

Zin1(s)

−

Z f 2(s)

Zin,2

=Z f ,1(s)Z f ,2(s)

Zin,1(s)Zin,2(s)

For Fig. P14.29a in the s-domain

Zin,1 = 25kΩ, Zin,2 = 50kΩ , Z f ,1 =1

C s +1

RC

=250,000

s + 5, and Z f ,2 =

1

C s +1

RC

=250,000s + 2.5

Hence for Fig. P14.29a


Ha(s) =

250,000

s + 2.525,000

250,000

s + 5.050,000

=50

s + 2.5( )(s + 5)

If vin (t) = u(t), then Vin (s) =1s

and

Vout (s) =50

s s + 2.5( )(s + 5)=

4s

−8

s + 2.5+

4s + 5

Hence,

vout (t) = 4 − 8e−2.5t + 4e−5t[ ]u(t) V

(b) For Fig. P14.29b in the s-domain, Zin,1, Z f1 , and Z f 2 are in part (a). However,

Zin,2 =1Cs

=250,000

sThus

Hb(s) =

250,000

s + 2.525,000

250,000

s + 5250,000

s

=10

s + 2.5

ss + 5

=

10s

s +5

2

(s + 5)

With Vin (s) =1s

,

Vout (s) =10s

(s + 2.5)(s + 5)

1s

=

20(s + 2.5)(s + 5)

=4

s + 2.5−

4s + 5

and

vout (t) = 4e−2.5t − 4e−5t( )u( t) V

SOLUTION 14.30.(a) -(b). The subcircuit is an integrator, with

Vout(t)V1 (s)

= - 1s

(c) This subcircuit is again an integrator, with

V1(t)V2 (s)

= - 1s

(d) Applying KCL to the inverting input terminal of the top left op amp, we have

G2Vout(s) + V3(s) + V2(s) = 0 or

V2(s) = - G2Vout(s) - V3(s)


(e) Applying KCL to the inverting input terminal of the bottom op amp, we have

G3V1(s) + G1 Vin(s) + V3(s) = 0 or

V3(s) = - G3V1(s) - G1 Vin(s)

(f) The results of parts (b), (c) and (d) do not involve Vin. Therefore, , we can solve for V1, V2 and V3in terms of Vout from these threes equations:

V1(s) = - sV out(s)

V2(s) = - sV 1(s) = s2Vout(s) and

V3(s) = - V 2(s) - G2Vout(s) = - s2 + G2 Vout(s)

Substituting these relationships into the result of part (e), we obtain

- sG3Vout(s) + G1 Vin(s) - s2 + G2 Vout(s) = 0

Therefore

H(s) = Vout(s)Vin (s)

= G1

s2 + G3s + G2

SOLUTION 14.31. Use the parallel equivalent circuit model for the capacitor with the standard

directions for voltage and current as given in figure 14.16. For the single node with vC (0− ) = 20 V,

VC (s)10

s

+VC (s)40 +10

−110

vC (0− ) = 0 implies s

10+

150

VC (s) =

110

(20) = 2

Equivalently, (50s +10)VC (s) = (s + 0.2)VC (s) = 20 or VC (s) =20

s + 0.2. Therefore,

vC (t) = 20e−0.2tu(t) V

SOLUTION 14.32. Using the equivalent model for the inductor in figure 14.19, we can compute thetotal admittance as

Y(s) =52s

+140

+110

=52s

+18

=2s + 40

16s=

s + 208s

Using current division,


Iout (s) =0.1

s + 20

8s

×−iL (0)

s= −

0.8iL (0−)s + 20

= −1.6

s + 20

Thus

iout ( t) = −1.6e−20t u( t) A

SOLUTION 14.33. Using the equivalent model for the inductor in figure 14.19 and for the capacitorusing figure 14.16, we may combine the current sources to form an equivalent source (with C = 0.1 F) toobtain

Ieq(s) =110

vC (0−) −iL (0−)

s= 0.2 −

1s

Note that

Y(s) = Cs +1LS

=LCs2 +1

Ls= C

s2 +1

LCs

With C = 0.1 F and L = 0.4 H, 1

LC= 25 and

Z(s) =1C

s

s2 + 25

=

10s

s2 + 25Thus

VC (s) = Z(s)Ieq(s) =10s

s2 + 25

s− 55s

=

2(s− 5)

s2 + 25=

2s

s2 + 25−

10

s2 + 25and

vC (t) = 2cos(5t) − 2sin(5 t)( ) u(t) V

SOLUTION 14.34. Consider the equivalent circuit below:

Writing a single node equation we have,


0 = 0.5Vout − 0.2VR1 +1

5 + 0.4sVout + 0.8( )

= 0.5Vout + 5 × 0.2 × 15 + 0.4s

Vout + 0.8( ) + 15 + 0.4s

Vout + 0.8( )

= 0.5Vout +2

5 + 0.4sVout + 0.8( )

Therefore

Vout =−8

s + 22.5and

vout (t) = −8e−22.5tu(t) V

SOLUTION 14.35. Redraw the circuit in the s-domain and use an equivalent circuit for the capacitor(figure 14.16) that accounts for the initial condition. By KCL

VC (s)R

+ CsVC (s) = Cv(0−) + Iin (s)

With R = 50 Ω and C = 0.02 F,VC (s)

50+ 0.02sVC (s) = 0.02vC (0−) + Iin (s)

or

VC (s) + sVC (s) = vC (0−) + 50Iin (s)which is equivalent to

(s +1)VC (s) = vC (0−) + 50Iin (s)

(a) With vC (0− ) = 8 V and iin (t) = 40 (t) mA so that Iin (s) = 0.04 ,

(s +1)VC (s) = 8 + 2 implies VC (s) =10s +1

and

vC (t) = 10e−10 tu( t) V

(b) With vC (0− ) = 1 V and iin (t) = 200e−tu(t) mA we have that Iin (s) =0.2s +1

. Thus

VC (s) =vC (0−)(s +1)

+50

(s +1)Iin (s) =

1(s +1)

+10

(s +1)2

and

vC (t) = e−t +10te−t( )u( t) V

SOLUTION 14.36.


(a) By current division

H(s) = IL(s) Iin (s)

= 1Ls + 1

(b) The given data in Laplace transforms are:

Iin(s) = I0

s2 and IL (s) = 15

s2 - 3

s + 3

s+ 5 = 75

s2 s+ 5

Under the assumption of zero initial inductor current,

IL (s) = H(s) Iin (s) = H(s) = IL(s) Iin (s)

= I0

Ls + 1 s2 = I0/L

s + 1/L s2 = 75

s2 s+ 5

Equating coefficients, we obtain the answers

L = 1/5 = 0.2 H and I0 = 75L =15 A

(c) The s-domain equivalent is shown below.

Applying KVL to the right mesh, we have

LsIL(s) - L iL(0-) + 1 ×[IL(s) - 1] = 0

Solving for IL(s) from this equation, and equating it to 10/(s+ 5), we have

IL(s) = L iL(0-) + 1L s + 1

= iL(0-) + 1/L s + 1/L

= 10 s + 5

from which L= 0.2 H and iL (0-) = 5 A.

SOLUTION 14.37.

(a) By inspection,


H(s) = IL(s) Vin (s)

= 12s + 200

= 0.5s + 100

Given vin(t) = 2u(t) V, then Vin(s) = 2/s , and

IL(s) = 1s s + 100

= 0.01s

- 0.01s + 100

in which case iL(t) = 0.01 ( 1 - e-100t) A

Plots are omitted.

(b) By linearity and time invariance,

iL(t) = 0.01 ( 1 - e-100t)u(t) - 0.01( 1 - e-100(t- 0.05))u(t - 0.05) A

(c) Correction: (a) should be (c). With nonzero initial inducto current, the s-domain equivalentbecomes:

Given iL(0-) = 0.01 A and vin(t) = 2e-200tu(t) V, then Vin(s) = 2/(s + 200) and

IL(s) = 2

s + 200 + 0.02

2s + 200 = 0.01s + 3

( s + 100)( s + 200)

= 0.02 s + 100

- 0.01 s + 200

in which caseiL( t) = 0.02e-100t - 0.01e-200t u(t) A

(d) Correction . (b) should be (d).Given iL(0-) = 0.01 A and vin(t)= 2cos(200t)u(t) V, then

Vin(s) = 2ss2 + 40000

and


IL(s) =

2ss2 + 40000

+ 0.02

2s + 200 = 0.01s2 + s + 400

( s + 100)( s2 + 40000)

We use MATLAB to do the partial fraction expansion.

n= [ 0.01 1 400];d= conv([ 1 100], [ 1 0 40000]);[ r p k ] = residue (n,d)r = 0.0010 - 0.0020i 0.0010 + 0.0020i 0.0080p = 1.0e+02 * 0.0000 + 2.0000i 0.0000 - 2.0000i -1.0000

From the above MATLAB output,

IL(s) = 0.008 s + 100

+ 0.001-j0.002 s - j200

+ 0.001+j0.002 s + j200

= 0.008 s + 100

+ 0.002s + 0.8 s2 + 2002

From table 13.1, item 18,

iL( t) = 0.008e-100t + 0.002 cos(200t) + 0.004 sin(200t) u(t) A


SOLUTION 14.38. In the s-domain, we break the response up into the part due to the initial conditionand the part due to the source with the initial condition set to zero. The transfer function with the initialcondition set to zero is

H(s) =VC (s)Vin (s)

=1 Cs

R +1 Cs=

1 RCs +1 RC

=0.25

s + 0.25

Using the parallel equivalent circuit for the charged capacitor while setting the source voltage to zero,the capacitor voltage due only the initial condition is:

VC,IC (s) =1

1

R+ Cs

CvC (0− )[ ] =vC (0−)s + 0.25

Hence,

VC (s) =0.25

s + 0.25Vin (s) +

vC (0−)s + 0.25

and

IC (s) =Vin (s) −VC (s)

20= 0.05 1−

0.25s + 0.25

Vin (s) −

0.05vC (0−)s + 0.25

=0.05s

s + 0.25Vin (s) −

0.05vC (0−)s + 0.25

for all inputs and initial conditions.

(a) If vin (t) = 20u(t) and vC (0− ) = 10 V, then Vin (s) =20s

and

VC (s) =5

s(s + 0.25)+

10s + 0.25

=20s

−10

s + 0.25 ⇒ vC (t) = 20 −10e−0.25t( )u( t) V

and

IC (s) =1

s + 0.25−

0.5s + 0.25

=0.5

s + 0.25 ⇒ iC ( t) = 0.5e−0.25tu(t) A

(b) If vin (t) = 5e−0.25tu( t) V, then Vin (s) =5

s + 0.25. Hence,

VC (s) =1.25

(s + 0.25)2 +10

s + 0.25 ⇒ vC (t) = 10 +1.25t( )e−0.25tu(t) V

and

IC (s) =0.25s

(s + 0.25)2 −0.5

s + 0.25=

−0.25s + 0.25

−0.0625

(s + 0.25)2

Hence

iC ( t) = − 0.25 + 0.0625t( )e−0.25tu(t) A


0 5 10 15 200

1

2

3

4

5

6

7

8

9

10

Time in s

Inpu

t and

Cap

acito

r vo

ltage

s, V

TextEnd

SOLUTION 14.39. The figure which accounts for the initial conditions is given below.

(a) For the zero-input response, the above circuit reduces to a parallel RLC driven by two currentsources.

Hence VC(s) equals the total current divided by the total admittance, i.e.,


VC (s) =CvC (0−) +

iL (0−)

s

Cs +1

R+

1

Ls

=svC (0−) +

iL (0− )

C

s2 +1

RCs +

1

LC

=20s +10

s2 + 250s +104 =26.6

s + 200−

6.6s + 50

Hence

vC(t) = [26.6e–200t

– 6.6e–50t

]u(t) V

(b) For the zero-state response, the current sources disappear. Executing a source transformation on

the remaining voltage source, we obtain a current, I(s) = Vin(s)/(Ls), driving a parallel RLC circuit.Hence, the zero input response is

VC (s) =

Vin (s)

Ls

Cs +1

R+

1

Ls

=1

LC

Vin (s)

s2 +1

RCs +

1

LC

=20000

s3 + 250s2 +104 s=

2s

+0.6667s + 200

−2.6667s + 50

Hence

vC(t) = [2 + 0.6667e–200t

– 2.6667e–50t

]u(t) V(c) By superposition, the complete response is the sum of the answers to (a) and (b). Hence

vC(t) = [2 + 27.267e–200t

– 9.2667e–50t

]u(t) V(d) By linearity and time-invariance,

vC(t) = [2 + 0.6667e–200t

– 2.6667e–50t

]u(t)

+ [4 + 1.3334e–200(t–0.01)

– 5.3334e–50(t–0.01)

]u(t – 0.01) V


SOLUTION 14.40. f ( t) = Lin (t) = e−2 tu(t)A [part (c)]

for the zero-input response, f ( t) = Lin (t) =L2(0− )

s [part (d)]

for the zero-input response, f ( t) = Lin (t) = CvC (0)A [part (e)]

(a) Yin (s) =1 +1s

+ s +1=s2 + 2s +1

s=

(s +1)2

s

(b) Iout (s) =Y1(s)Yin (s)

Iin (s) =s +1

s2 + 2s +1

s

=s(s +1)

(s +1)2 =s

s +1Iin (s)

and

H(s) =Iout (s)Iin (s)

=s

s +1

(c) If iin (t) = e−2 tu(t) A, then Iin (s) =1

s + 2, then

Iout (s) = H (s)Iin (s) =s

(s +1)(s + 2)=

−1s +1

+2

s + 2

which implies that the zero-state response is

iout ( t) = 2e−2 t − e−t( )u(t) A

(d) If iL (0−) = 2 A, vC (0− ) = 0, and iin (t) = 0. Using the parallel equivalent circuit for the inductor,figure 14.19, we have

Iout (s) = H (s)i(0−)

s=

ss +1

−

2s

= −

2s +1

⇒ iout ( t) = −2e−tu( t) A

(e) Use the parallel equivalent circuit for the capacitor, figure 14.16, to obtain by current division,

Iout (s) = −

1

s+1

1

s+1 + s +1

[CvC (0−)] = 4s +1

s2 + 2s +1=

4s +1

⇒ iout (t) = 4e−tu(t) A

(f) By superposition, the complete response is the sum of the answers to parts (c), (d) and (e).

SOLUTION 14.41. With vin (t) = 4u(t)V and vC (0− ) = 1 V, a single node equation at the front half of

the circuit yields with CvC (0− ) = 1×1= 1:


−245

+ 2 +

4s

+ s

VC1(s) −1= 0 ⇒

s2 + 2s + 4s

VC1(s) =8s

+1 =s + 8

s

or

VC1(s) =s + 8

s2 + 2s + 4

For the rear-half, represent the capacitor by a series equivalent circuit. Thus we can obtain an equivalentvoltage source with value:

Veq(s) = 2VC1(s) −VC (0− )

s=

2(s + 8)

s2 + 2s + 4−

1s

=2s(s + 8)− (s2 + 2s + 4)

s s2 + 2s + 4( )or equivalently

Veq(s) =s2 +14s − 4

s s2 + 2s + 4( )By a voltage division,

Vout (s) =

1

s1

s+ 0.5

Veq(s) +vC (0−)

s=

2s + 2

×s2 +14s− 4

s s2 + 2s + 4( ) +1s

=2s2 + 28s− 8 + s + 2( ) s2 + 2s + 4( )

s s + 2( ) s2 + 2s + 4( ) =s2 + 6s + 36( )

s + 2( ) s2 + 2s + 4( )Using MATLAB»num = [1 6 36];»den = conv([1 2],[1 2 4])den = 1 4 8 8»[r,p,k] = residue(num,den)r = 7.0000e+00 -3.0000e+00 - 2.8868e+00i -3.0000e+00 + 2.8868e+00ip = -2.0000e+00 -1.0000e+00 + 1.7321e+00i -1.0000e+00 - 1.7321e+00ik =

However, it would appear easier here to use ilaplace:»syms t s»ilaplace((s^2+6*s+36)/((s+2)*(s^2+2*s+4)))ans =7*exp(-2*t)-6*exp(-t)*cos(3^(1/2)*t)+10/3*exp(-t)*3^(1/2)*sin(3^(1/2)*t)


»Hence

vout (t) = 7e−2t + e−t 3.334 3 sin( 3t( ) − 6cos( 3t)[ ]u(t) V

SOLUTION 14.42. Using the series equivalent circuit (figure 14.17) for C1, we have

I1K(s) = vC1(0-)/s

R + 1sC1

= - 0.25/s1000 + 50/s

= -0.251000s + 50

Next, since vC 2(0−) = 0 , we have

VC 2(s) =I1k (s)C2s

=−0.25 × 500s 1000s + 50( ) =

−0 / 1 2 5s(s + 0.05)

Finally,

Vout(s) = -VC2(s) = 0.125s(s + 0.05)

= 2.5( 1s - 1s + 0.05

)

andvout(t) = 2.5( 1 - e-0.05t )u(t) V

SOLUTION 14.43. (a) It is preferable to use the series equivalent circuit (figure 14.17) for C1, and theparallel equivalent circuit (figure 14.16) for C2.(b) The current through the 2.5 kΩ resistor is given by

I2.5K (s) = vC1(0-)/s

R1 + 1sC1

= - 2/s2500 + 5000/s

= - 22500s + 5000

=-8×10-4

s + 2

Next,

Vout(s) =VC2(s) = I2.5K (s) 1sC2 + 1

R2

= -8×10-4

s + 2 × 1

0.0002s + 0.0002 = - 4

(s + 2) (s + 1)

(c) Hence

Vout(s) = - 4(s + 2) (s + 1)

= - 4 1s + 1

- 1s +2

and

vout (t) = 4 e−2t − e−t( )u(t) V

(d) SPICE plot omitted.

SOLUTION 14.44. (a) From voltage division,


H(s) = VC2Vin

= Z2Z1+ Z2

=

R2R2C2s + 1

1C1s

+ R1 + R2R2C2s + 1

= R2C1sR1C1R2C2s2 + ( R1C1 +R2C2 + R2C1)s + 1

(b) If vin(t) = 15u(t) V, then Vin(s) = 15/s and

VC2 = H(s)Vin = 1.75ss2 + 4.25 s + 1

×15s = 7

s+ 0.25 - 7

s+ 4Hence

vC 2( t) = 7 e−0.25t − e−4 t( )u( t) V

0 1 2 3 4 5 6 7 80

5

10

15

Time in s

vin

and

vc2

TextEnd

(c) Using the series equivalent for C1, we have

VC2 = H(s)vC1(0-)

s = 1.75ss2 + 4.25 s + 1

×15s

which is the same as result in part (b). Therefore

vC 2( t) = 7 e−0.25t − e−4 t( )u( t) V

(d) Using the parallel equivalent for C2, we have


VC2(s) = C2vC2(0-) 127

s + 47

+ 12 + 1/s

=27

×15× 127

s + 47

+ s2s + 1

= 1s + 0.25

+ 14s + 4

Hence

vC 2( t) = e−0.25t +14e−4 t( )u(t) V

(e) By linearity, the answer is the sum of parts (b), (c) and (d).

SOLUTION 14.45. (a) Z1 = R1 + L1s and Z2 = R2L2sR2 + L2s

. From Ohm's law

IL1 = VinZ1+ Z2

= Vin

R1 + L1s + R2L2sR2 + L2s

= (R2 + L2s)Vin

(R1 + L1s) (R2 + L2s) + + R2L2s

Using current division, we have

IL2 = R2R2 + L2s

IL1= R2Vin(R1 + L1s) (R2 + L2s) + + R2L2s

Therefore

H(s) = IL2Vin

= R2(R1 + L1s) (R2 + L2s) + + R2L2s

= G1

G1L1G2L2s2 + (G1L1+ G2L2 + G1L2)s + 1

With the given element values,

H(s) = 22×1×4

7×7

8s2 + (2×1+ 4

7×7

8 + 2×7

8)s + 1

= 2s2 + 4.25s + 1

(b) If vin(t) = 15u(t) V, then Vin(s) = 15/s and

IL2 = H(s)Vin = 2s2 + 4.25 s + 1

×15s = 30

s - 32s+ 0.25

- 2s+ 4

Hence

iL 2(t) = 30 − 32e−0.25t + e−4 t( )u( t) A

Plot omitted.

(c) Using the series equivalent for L1, we have

IL2 = H(s) L1iL1(0-) = 2s2 + 4.25 s + 1

×1×15 = 8s + 0.25

- 8s +4

Therefore

iL 2(t) = 8 e−0.25t − e−4 t( )u(t) A


(d) Using the parallel equivalent for L2, we have

IL2(s) = L2iL2(0-)

L2s + R2(R1 + L1s)

R2 + (R1 + L1s)

= 78

×15

78

s + 1.75(0.5 + s)

1.75 + (0.5 + s)

= 15( s + 2.25)

s2 +4.25s + 1 = 8

s + 0.25 + 7

s + 4

HenceiL2(t) = (8e-0.25t + 7 e-4t) u(t) A

(e) By linearity, the answer is the sum of parts (b), (c) and (d).

SOLUTION 14.46. (a) Using the result of problem 14.44(a)

VC22Iin

= R2C1sR1C1R2C2s2 + ( R1C1 +R2C2 + R2C1)s + 1

= 1.75ss2 + 4.25 s + 1

Therefore

H1(s) = VC2Iin

= = 3.5ss2 + 4.25 s + 1

(b) Using the result of problem 14.45(a),

H2(s) = IL2VC2

= R2(R1 + L1s) (R2 + L2s) + + R2L2s

= 2s2 + 4.25s + 1

(c)

H(s) = IL2Iin

=H1(s)H2(s) = 3.5ss2 + 4.25s + 1

× 2s2 + 4.25s + 1

= 7s(s2 + 4.25s + 1)2

(d) We first represent the initialized capacitor by the series equivalent (figure 14.17), and then apply asource transformation. From this circuit, by utilizing the expression derived in part (c), we have

IL2vC1(0)

2 s

=sIL27.5

H(s) =H 1(s)H2(s) = 7s(s2 + 4.25s + 1)2

ThereforeIL2(s) = 52.5

(s2 + 4.25s + 1)2 = 1.9911

s + 4 + 3.7333

( s + 4)2 - 1.9911

s + 0.25 + 3.7333

( s + 0.25)2

and

iL2(t) = [(1.9911 + 3.7333t) e-4t +(- 1.9911 + 3.7333t) e-0.25t ] u(t) A

Note: the book answer for part (d) should be divided by 2.

(e) Since Iin(s) = 15/s, we have


IL2(s) = H(s)Iin(s) = 105

(s2 + 4.25s + 1)2 = 3.9822

s + 4 + 7.4666

( s + 4)2 - 3.9822

s + 0.25 + 7.4666

( s + 0.25)2

and

iL2(t) = [(3.9822 + 7.4666t) e-4t +(- 3.9822 + 7.4666t) e-0.25t ] u(t) A

SOLUTION 14.47. (a) For this passive circuit, we may write the nodal equations by inspection.

0.8s + 2 + 10s - 10

s

- 10s 1 + 10

s

VCVR

= 2Vs1-Is2

(b) Vs1= 3/s and Is2 = 3/s. We solve for VR by Cramer's rule to obtain

VR =

0.8s + 2 + 10s

6s

- 10s - 3s

0.8s + 2 + 10s - 10

s

- 10s 1 + 10

s

= - 2.4s2 - 6s + 30s(0.8s2 + 10s + 30)

= - 4s + 7.5

+ 1s

and

vR( t) = 1− 4e−7.5t( )u(t) V

(c) We represent the initialized capacitor by the parallel equivalent circuit ( figures 14.16) In this casethe nodal equations becomes

0.8s + 2 + 10s - 10

s

- 10s 1 + 10

s

VCVR

=

6s + 2.4

- 3s

Solve for VR by Cramer's rule to obtain

VR =

0.8s + 2 + 10s

6s + 2.4

- 10s - 3s

0.8s + 2 + 10s - 10

s

- 10s 1 + 10

s

= - 2.4s2 +18s + 30s(0.8s2 + 10s + 30)

= 1s - 16

s + 7.5 + 12

s + 5


vR(t) = (1 - 16e-7.5t + 12e- 5t) u(t) V

SOLUTION 14.48. (a) After performing the suggested source transformation, and representing theinitialized capacitor and inductor by their series equivalent circuits, we can write two mesh equations byinspection:

0.5 + 1.25s - 1.25

s

- 1.25 s 1 + 0.1s + 1.25

s

Is1IL

= Vs1 -

vC(0)s

vC(0)s + LiL(0) + Is2

(b) With Vs1= 3/s, Is2 = 3/s, vC(0) = 0, and iL(0) = 3 A, the above mesh equation becomes

0.5 + 1.25s - 1.25

s

- 1.25 s 1 + 0.1s + 1.25

s

Is1IL

= Vs1 -

vC(0)s

vC(0)s + LiL(0) + Is2

=

3s

0.3 + 3s

Solve for IL(s) by Cramer's rule to obtain

IL(s) = 0.15s2 +1.875s + 7.50.05s 3 + 0.625s2 + 1.875s

= 2s + 7.5

+ -3s + 5

+ 4 s

Therefore

iL(t) = (4 + 2e-7.5t -3 e-5t) u(t) A

SOLUTION 14.49. (a) Represent the initialized capacitors by their parallel equivalent circuits.(b) Write two nodals equation by inspection

0.001s + 0.4 - 0.2- 0.2 0.001s + 0.4

VC1VC2

= 0.2Vin + 0.001vC1(0-)0.001vC2(0-)

=

2.4s + 0.006

0.002

(c) Solve for VC2(s) by Cramer's rule to obtain

VC2 =

0.001s + 0.4 2.4/s + 0.006- 0.2 0.002

0.001s + 0.4 - 0.2- 0.2 0.001s + 0.4

= 2s2 + 2×103s + 48×104

s(s2 + 8×102s + 12×104) = 0

s + 600 + -2

s + 200 + + 4

s

(d)


vC2(t) = (4 - 2e-200t ) u(t) V

SOLUTION 14.50. (a) Let VC denote the node voltage across the capacitor. By inspection the nodalequations in matrix form are:

1+1 /R + 4s −1 /R

−1 /R 1 +1/ R +1 / ( 4s)

VC

Vout

=

Vin

Vin / ( 4s)

(b) By Cramer's rule,


=det

1 +1/ R + 4s 1

−1 /R 1 / ( 4s)

det1+1 /R + 4s −1 /R

−1 /R 1 +1/ R +1 / ( 4s)

=

( 4s +1)(1 +1/ R)

4s (1+1/ R + 4s)(1+ 4s / R + 4s) −1/ R2( )

=(4s +1)(1 +1/ R)

(1 + 8s +16s2)(1+1 /R)=

1(1+ 4s)

Clearly, R does not affect the transfer function. The question is why? Note that the circuit can beredrawn as a balanced Wheatstone bridge circuit in which there is no voltage across R and no currentthrough R. Hence R has no effect on the transfer function and on the impedance at the input. Hence Rcan be removed in the analysis of the circuit. In this case, the transfer function follows trivially byvoltage division.(c) In view of the answer to (b), the impedance can be calculated with R removed. Hence

Zin (s) =1+

1

4s

1+ 4s( )

1+1

4s

+ 1+ 4s( )

=1 + 4s( )2

1 + 4s( )2 =1

Hence, the input impedance is a constant resistance and the network is called a constant resistancenetwork.

(d) The input is vin(t) = 10e–atu(t) V and R = 5 Ω. Find vout(t) for t ≥ 0 for the three cases, a = 0,

0.5, 0.25.(d) From part (b), for s ≠ 0.25,

Vout (s) =0.25

(s + 0.25)×

10s + a

=2 . 5 / ( 0 . 2 5− a)

s + a−

2 . 5 / ( 0 . 2 5− a)(s + 0.25)

which leads to

vout (t) =2.5

0.25 − a

e−at − e−0.25t( )u( t) V

For a = 0.25,


Vout (s) =2.5

(s + 0.25)2 implying that vout (t) = 2.5te−0.25tu( t) V

SOLUTION 14.51.(a) This bridged-T circuit was analyzed in problem 14.9. Here R = 1 Ω, Z1(s) = 0.25s and Z2(s) = 4/s.

Since the condition Z1(s) Z2(s) = R2 is met, we have Zin(s)= 1Ω.

(b) The s-domain equivalent circuit accounting for initial conditions is given below.

(c) Two nodal equations at Vc and Vout are:

VC - Vin + VC - Vout + 0.25sVC = 0.25 vC(0-)and

Vout - VC + Vout - Vin

0.25s + Vout = iL(0-)

s

Writing these in matrix form, we have

0.25s + 2 -1

-1 4s

+ 2

VC

Vout

= Vin + 0.25vC(0-)

4Vins

+ iL(0-)s

Solving for Vout by Cramer's rule yields


Vout(s) = 4s+4

Vin(s) + 0.5(s+ 8)

(s+4)2 iL(0-) + 0.5s

(s+4)2 vC(0-)

(d) Given vin(t) = 4u(t) - 3e-t u(t) V, then

Vin(s) = 4s - 3s+1

= s + 4s( s + 1)

and

Vout(s) = 4s(s + 1)

+ 0.25(s+ 8)

(s+4)2 + 0.75s

(s+4)2

Taking the inverse Laplace transform, we obtain, for t ≥ 0,

vout(t) = (4 - 4 e-t) + (0.25 e-4t + te-4t) + (0.75 e-4t - 3 te-4t) = 4 - 4 e-t + e-4t - 2te-4t V

SOLUTION 14.52. A supernode is defined by drawing a curve to enclose the controlled voltage source.One node within the supernode has voltage Vout and the other has voltage V1Ω that is equal to

V1Ω = -2I1 - Vout = -2 Vin - VC 2

= - Vin + VC - Vout

Next, we write nodal equations at VC and the supernode: At node VC

0.5 (VC - Vin ) + 0.5sVC + 0.5s (VC- Vout ) = 0

At the supernode

12s

(Vout- VC ) + 12

Vout + - Vin + VC + Vout 1

= 0

In matrix form, the nodal equations are:

0.5(s + 1 + 1/s) -0.5/s1 - 0.5/s 1.5 +0.5/s

VC Vout

= 0.5 Vin Vin

Solving by Cramer's rule yields

H(s) = VoutVin

= 0.5s + 0.75/s0.75s + 1+ 1.5/s

= 2( s2 + 1.5)

3s2 + 4s + 6

SOLUTION 14.53. A supernode is defined by drawing a curve to enclose the controlled voltage source.One node within the supernode has voltage Vout and the other has voltage V1Ω which is equal to


V1Ω = -2I1 - Vout = -2 Vin - VC 2

= - Vin + VC - Vout

Next, write nodal equations at VC and the supernode: At node VC

0.5 (VC - Vin ) + (0.25s + 4s )VC

+ ss2 + 16

(VC- Vout ) = 0

At the supernode

ss2 + 16

(Vout- VC) + 12

Vout + - Vin + VC + Vout 1

= 0

In matrix form, the nodal equation are:

0.25s + 0.5 +s/( s2 + 16) -s/(s2 + 16)

1 - s/(s2 + 16) 1.5 + s/(s2 + 16) VC

Vout = 0.5 Vin

Vin


H(s) = VoutVin

= 0.25s 3 +5.5s0.375s3 + s2+ 9s + 12

SOLUTION 14.54. Write nodal equations at V1 and V2:

0.5 (V1 - Vin ) + 0.125( V1 + 0.2V2 ) + 0.1s(V1- V2 ) = 0 and

0.1s(V2- V1) + 14

V2 + V21

- 5V1 = 0

In matrix form, the nodal equations are:

(0.1s + 0.625) (-0.1s + 0.025)(-0.1s - 5) (0.1s + 1.25)

VC Vout

= 0.5 Vin 0


H(s) = IoutVin

= V2Vin

= 0.05s + 2.5- 0.31s + 0.9062

SOLUTION 14.55.(a) Simply replace each capacitor by the parallel form circuit model given in figure 14.16.(b) For this passive circuit, we can write the nodal equation by inspection.


0.5s +2 -1 0-1 0.5s +2 -10 -1 0.5s +2

VC1VC2VC3

= Vin + 0.5vC1(0)

0.5vC2(0)0.5vC3(0)

Solving for VC3 by Cramer's rule yields

VC3 (s) = Vin(s) + 0.5vC1(0) + 0.25s + 1 vC2(0) +( 0.125s2 + s +1.5 )vC3(0)

0.125s3 + 1.5s2 + 5s + 4

(c) Substituting Vin(s) = 12/s, vC1(0) = 0, vC2(0)= 6, and vC3(0) = 2 into the above expression, weobtain

VC3 (s) = 12/s + 6 0.25s + 1 + 2( 0.125s2 + s +1.5 )

0.125s 3 + 1.5s2 + 5s + 4 = 0.25s 3 + 3.5s2 + 9s +12

s 0.125s 3 + 1.5s2 + 5s + 4

Now use MATLAB to do the partial fraction expansion.n= [ 0.25 3.5 9 12];d= [ 0.125 1.5 5 4 0];[ r p k ] = residue (n,d)r = -2.5000 4.0000 -2.5000 3.0000p = -6.8284 -4.0000 -1.1716 0

From the MATLAB output, we have, for t ≥ 0,

vC3(t) = 3 - 2.5 e-6.828t + 4e-4t - 2.5e-1.1716t V

SOLUTION 14.56. For this problem we utilize loop analysis with loops as indicated below.


In doing the following loop analysis, note that we will use gmVout = I3 and that due to our judiciouschoice of loops

Vout =1Cs

+ Ls

I1

or equivalently,

0 =1

Cs+ Ls

I1 − Vout

For loop 1,

V1 = 2R +1Cs

+ Ls

I1 + 2R I2 + RgmVout

For loop 2,V1 − V2 = 2R I1 + 7R I2 + 5RgmVout

In Matrix form1000

s+ 0.016s 0 −1

2 +1000

s+ 0.016s 2 2

2 7 10

I1I2

Vout

=0

V1

V1 − V2

By Cramer's rule


Vout =

det

1000s

+ 0.016s 0 0

2 +1000

s+ 0.016s 2 V1

2 7 V1 − V2

det

1000

s+ 0.016s 0 −1

2 +1000

s+ 0.016s 2 2

2 7 10

=−

1000

s+ 0.016s

(5V1 + 2V2)

6000

s+ 0.096s − 10 +

7000

s+ 0.112s

Hence

Vout =

1000

s+ 0.016s

(5V1 + 2V2)

10 +1000

s+ 0.016s

=s2 + 62500

s2 + 625s + 62500(5V1 + 2V2)

The answers to (a) and (b) are clear at this point.(c) Using MATLAB»n = [21 0 21*62500];»d = [1 625 62500 0];»[r,p,k] = residue(n,d)r = 35 -35 21p = -500 -125 0k = []Hence

vout(t) = 21− 35e−125t + 35e−500t( )u t( ) V.

SOLUTION 14.57. (a) Replace the LC combination by a 1 V source after setting V1 and V2 to zero.

We need to compute the current leaving the 1 V source which will be 1/Rth. Let the left node be denoted

by Va and the right node by Vb. Also let G = 1/R. The nodal equations are by inspection

2G 0

0 1.25G

Va

Vb

=

gm + G

−gm + 0.25G

⇒

Va

Vb

=

gm + G( ) 2G

−gm + 0.25G( ) 1.25G


Thus, the current leaving the 1 V source isI1V = G 1−Va( ) + 0.25G 1− Vb( ) = G 1− gm + G( ) 2G( ) + 0.25G 1− −gm + 0.25G( ) 1.25G( ) = 0.5G − 0.5gm + 0.25G + 0.2gm − 0.05G = 0.7G − 0.3gm

Substituting G = 1/R we obtain

Rth =1

I1V=

10.7G − 0.3gm

=R

0.7 − 0.3gmR=

10.7 − 0.3 × 2

= 10 Ω

(b) Replace the LC combination by a short circuit and compute Isc. This makes the controlled sourcezero. By inspection

Isc =V12R

+V25R

Thus

Voc = RthIsc = RthV12R

+V25R

= 10 0.5V1 + 0.2V2( )

(c) By voltage division

Vout =ZLC

Rth + ZLCVoc =

1

Cs+ Ls

1

Cs+ Ls + Rth

Voc =s2 +1 LC

s2 +Rth

Ls +1 LC

5V1 + 2V2( )

=s2 + 62500

s2 + 625s + 625005V1 + 2V2( )

SOLUTION 14.58. (a) The last equation is the constraint equation for the controlled floating voltagesource. Hence, we have

V1 − V2 − z0(s)I0 = 0(b) By Cramer's rule,

V2 =

det

1

R+ Cs

Iin 1

0 0 −1

1 0 − z0(s)

det

1

R+ Cs

1 R 1

0 Cs −1

1 −1 −z0(s)

=− Iin

−21

R+ Cs

− z0(s)Cs

1

R+ Cs

=Iin

1

R+ Cs

2 + z0(s)Cs( )

(c) Here


V2 =Iin

1

R+ Cs

2 + LCs2( )

in which case =2

LC.

SOLUTION 14.59. (a) Since the switch has been at position A for a very long time, the inductor lookslike a short and iL(5-) = iL(5+) = 10/4 = 2.5 A. For t > 5, the switch moves to position B and theinductor current decays according to

iL ( t) = iL (5+ )e−(t−5)/ = 2.5e−(t−5)/0.1 = 2.5e−10(t−5) A

(b) Note that iL(0-) = iL(0+) = 0. Hence

IL (s) =1

10s + 4Vin (s) =

0.1s + 0.4

50s

−50

s + 0.5

=

12.5s

−62.5

s + 0.4+

50s + 0.5

Hence for 0 ≤ t ≤ 5s, iL ( t) = 12.5 − 62.5e−0.4 t + 50e−0.5t A. Here iL(5-) = iL(5+) = 8.1458 A.For t > 5, the inductor decays with a time constant of 0.1 s. Thus

iL ( t) = 8.1458e−10(t−5)

SOLUTION 14.60. (a) Since the switch has been at position A for a very long time, the capacitorlooks like an open and vC(5-) = vC(5+) = 40 V. For t > 5, the switch moves to position B and thecapacitor voltage decays according to

vC (t) = vC (5+ )e−(t−5)/ = 40e−(t−5)/2 = 40e−0.5(t−5) V

(b) Note that vC(0-) = vC(0+) = 0. Hence

VC (s) =1 /Cs

1 /Cs + 40Vin (s) =

12.5s +12.5

50s

−50

s +12.5

=

7812.5

s s +12.5( )2

In MATLAB,»syms t s»ilaplace(7.8125e3/(s*(s+12.5)^2))ans =50-625*t*exp(-25/2*t)-50*exp(-25/2*t)Hence for 0 ≤ t ≤ 5s, vC (t) = 50 − 625te−12.5t − 50e−12.5t V. Here vC(5-) = vC(5+) = 50 V.For t > 5, the capacitor voltage decays with a time constant of 0.08 s. Thus

vC (t) = 50e−12.5(t−5) V


SOLUTION 14.61. (a) Since the switch has been closed for a very long time, the capacitor looks likean open and vC(5-) = vC(5+) = 32 V. For t > 5, the switch opens and the capacitor voltage decaysaccording to

vC (t) = vC (5+ )e−(t−5)/ = 32e−(t−5)/0.4 = 32e−2.5(t−5) V

(b) Note that vC(0-) = vC(0+) = 0 and vout = vC. Hence for 0 ≤ t ≤ 5,

VC (s) =1

1

50+

1

200+ 0.002s

Vin (s)50

=10

s +12.5Vin (s) =

10s +12.5

50s

−50

s +12.5

=6250

s(s +12.5)2

In MATLAB,»syms t s»ilaplace(6250/(s*(s+12.5)^2))ans =40-500*t*exp(-25/2*t)-40*exp(-25/2*t)

Hence for 0 ≤ t ≤ 5s, vC (t) = 40 − 500te−12.5t − 40e−12.5t V. Here vC(5-) = vC(5+) = 40 V.For t > 5, the capacitor voltage decays with a time constant of 0.4 s. Thus

vC (t) = 40e−2.5(t−5) V

SOLUTION 14.62. (a) At 0-, vC( 0-) = 0 and iL(0

-) = 50/10 = 5 A.

(b) For this part, consider the equivalent circuit below.

By inspection,

VC =5

s 1+ 0.5s +1

0.184s

=5

0.5s2 + s +1

0.184

=10

s +1( )2 + π2

From table 13.1,


vC (t) = 3.1831e−t sin(πt)u( t) V(c) Using MATLAB,

SOLUTION 14.63. (a) Since the switch has been closed for a long time, iL(1-) = iL(1

+) = 30/0.8 = 37.5 A

and vC(1-) = vC(1

+) = 0. Represent the initialized inductor by its parallel equivalent circuit. Then

VC (s) =−37.5

s×

1

Cs +1

Ls

=−150

s2 + 4

Hence from table 13.1,

vC (t ') = −75sin(2 t') V ⇒ vC (t) = −75sin(2( t −1 ) ) V for t > 1s.

(b) All initial conditions at t = 0 are zero. For 0 ≤ t ≤ 1s,

VC (s) =1

1

0.8+ 0.25s +

1

s

Vin (s)0.8

=5s

s2 + 5s + 4Vin (s) =

150s

s2 + 5s + 4

1s

−1

s + 2

=300s

s2 + 5s + 4

1s(s + 2)

=

100s +1

−150s + 2

+50

s + 4Hence, for 0 ≤ t ≤ 1s,

vC (t) = 100e−t −150e−2 t + 50e−4 t V

Here, vC(1-) = vC(1+) = 17.403 V. Note, vin(1-) = 25.94 V. Next,


iC (1−) = CdvCdt

t=1

= 0.25 −100e−1 + 300e−2 − 200e−4[ ] = 0.037378

Thus in MATLAB»vin1 = 30*(1 - exp(-2))vin1 = 2.5940e+01»vc1 = 17.403vc1 = 1.7403e+01»ic1 = 0.25*(-100*exp(-1) + 300*exp(-2)-200*exp(-4))ic1 = 3.7378e-02»iL1 = (vin1 - vc1)/0.8 - ic1iL1 = 1.0634e+01

Therefore, iL (1−) = iL (1+ ) =10.634 A. For t ≥ 1, we use the parallel equivalent circuit for both theinductor and the capacitor:

VC (s) = e−s 4s

s2 + 4

−iL (1+ )s

+ CvC (1+ )

= e−s 4s

s2 + 4

−10.634s

+ 4.3507

Therefore from table 13.1, for t > 1,

vC (t) = −21.268sin 2(t −1)( ) +17.403cos 2(t −1)( ) V

Plots omitted.

SOLUTION 14.64. Here vC(0) = 0 for both capacitors.Part 1: 0 ≤ t ≤ 1s.

Vout (s) =

10

s

20 +10

s

20s

−20

s + 2

=

20s(s + 0.5)(s + 2)

From MATLAB»num = 20;»den = [1 2.5 1 0];»[r,p,k] = residue(num,den)r = 6.6667e+00 -2.6667e+01 2.0000e+01p = -2.0000e+00 -5.0000e-01 0k = []


Therefore for 0 ≤ t ≤ 1,

vout (t) = 20 − 26.667e−0.5t + 6.667e−2t( ) u( t) − u( t −1)( )

Part 2. 1 ≤ t. Here the initial condition for the right-most capacitor is vout (1− ) = vout (1 + ) = 4.7281 V.As above, the left-most capacitor has zero value at t = 1s. Let us use the series equivalent circuit for theright capacitor. Then,

IC (s) = e−s −4.7281s

×1

10 +20

s

= e−s −0.47281s + 2

Therefore,

Vout (s) =10s

IC (s) + e−s 4.7281s

= e−s −4.7281

s(s + 2)+

4.7281s

= e−s 2.3641

s+

2.3641s + 2

and for t ≥ 1,

vout (t) = 2.3641+ 2.3641e−2(t−1)( )u(t −1) V

SOLUTION 14.65. Assume the switch has been in position A for a long time. Both capacitors behave asopen circuits and both capacitors have initial voltages at t = 0 of 10 V. For t ≥ 0, use the parallelequivalent circuits for both capacitors and write nodal equations. Let the left capacitor have voltageVCa.

0.005s + 0.03 −0.01

−0.01 0.0025s + 0.01

VCa

Vout

=

0.005 ×10

0.0025 ×10

=

0.05

0.025

By Cramer's rule,

Vout (s) =det

0.005s + 0.03 0.05

−0.01 0.025

det0.005s + 0.03 −0.01

−0.01 0.0025s + 0.01

= 10s +10

s2 +10s +16=

4 0 /3s + 2

−10 / 3s + 8

Therefore for t ≥ 0,

vout (t) =403

e−2t −103

e−8 t V

SOLUTION 14.66. (a) Zin (s) = 2 +1

0.5s +2

s

= 2 +2s

s2 + 4.

(b) Here, the initial condition is zero and


VC (s) =

2s

s2 + 4

2 +2s

s2 + 4

×10s

=10

s2 + s + 4

»syms t s»ilaplace(10/(s^2+s+4))ans =4/3*exp(-1/2*t)*15^(1/2)*sin(1/2*15^(1/2)*t)Hence using MATLAB above or table 13.1 we have for 0 ≥ t ≤ 1.5s,

vC (t) = 5.164e−0.5t sin(1.9365t) Vand

vC (1.5− ) = vC (1.5 + ) = 0.57237 V(c) Use the parallel equivalent circuit for the left capacitor. The right capacitor has a zero initial voltageat t = 1.5. Hence, we do not use an equivalent circuit for the right capacitor.

(d) Therefore

e1.5sVC (s) =1

0.5s + 2 +s

s + 4

× 0.5 × 0.57237 = 0.57237s + 4

s2 +10s +16

(e) In MATLAB»[r,p,k] = residue(0.57237*[1 4],[1 10 16])r =V 3.8158e-01 1.9079e-01p = -8 -2k = []

Hence

VC (s) = e−1.5s0.57237s + 4

s2 +10s +16= e−1.5s 0.19079

s + 2+

0.38158s + 8

(f) Finally

vC (t) = 0.19079e−2(t−1.5) + 0.38158e−8(t−1.5)[ ]u(t −1.5) V

SOLUTION 14.67. (a) v1(0−) = v1(0+ ) = v2(0− ) = v2(0+ ) =R2R

16 = 8 V.

(b) For 0 ≤ t ≤ 1, v1( t) = 8 V and v2(t) = 8e−t /RC = 8e−0.6931t V.

(c) v1(1−) = 8 V and v2(1− ) = 8e−0.6931 = 4 V.


(d) From KVL, v1(1+ ) = v2(1 + ). From conservation of charge,

1× v1(0−) +1× v2(0−) =12 = 2 × v1(0+ ) = 2 × v2(0+ ) . Therefore v1(1+ ) = v2(1 + ) = 6 V.(e) This response represents a decay with time constant τ = 2R = 2.8854 s. Hence

v1( t) = v2(t) = 6e−0.34657( t−1)u(t −1) VIt follows that

v1(3) = v2(3) = 6e−0.34657×2 = 3 V

(f) Both capacitor voltages change abruptly at t = 1.

SOLUTION 14.68. Label the current down through the first inductor as i1(t).

(a) i1(0−) = i1(0+ ) =1 A a n d iout (0− ) = iout (0+ ) = 0 .(b) For 0 ≤ t we use a parallel equivalent for the first inductor. By current division

Iout (s) =

1

5 + 0.1s

0.057143 +1

0.35s+

1

5 + 0.1s

×−1s

=−175

s2 + 275s + 2500

Use MATLAB to do the partial fraction expansion

num = -175; den = [ 1 275 2500]; [ r, p, k] = residue (num, den)

r = 0.6831 -0.6831p = -265.5869 -9.4131

From the MATLAB output

Iout (s) = - 0.68313s+ 9.4131

+ 0.68313s+ 265.59

Therefore ,

iout ( t) = 0.68313 e−265.59t − e−9.413t( )u(t) A

SOLUTION 14.69. (a) Here we use voltage division:


V1(s) =

1

4 ×10−6s1

4 ×10−6s+

1

4 ×10−6 s+

1

1×10−6s

×30s

=306s

=5s

Therefore, v1(0+ ) = 5 V.(b) Again use voltage division:

V1(s) =

1

5 ×10−6 s1

5 ×10−6s+

1

1×10−6 s+

1

2 ×10−6s

×40s

=8017s

Therefore, v1(0+ ) = 4.7059 V.

SOLUTION 14.70. (a) Consider a mesh current I(s) in the usual direction and use the series equivalentcircuit for each capacitor. Thus

I(s) =1

1

4 ×10−6 s+

1

4 ×10−6s+

1

1×10−6 s

3s

−0.3s

−0.9s

−0.6s

=

1.21.5

×10−6 = 0.8 ×10−6

Therefore for t > 0,

V1(s) =1

4 ×10−6sI(s) +

0.3s

=0.8 ×10−6

4 ×10−6s+

0.3s

=0.5s

⇒ v1( t) = 0.5 V

Similarly for t > 0

V2(s) =0.8 ×10−6

4 ×10−6s+

0.9s

=1.1s

⇒ v2(t) =1.1 V

and

V3(s) =0.8 ×10−6

1×10−6s+

0.6s

=1.4

s ⇒ v3( t) = 1.4 V

(b) Consider a mesh current I(s) in the usual direction and use the series equivalent circuit for eachcapacitor. Thus

I(s) =1

1

5 ×10−6 s+

1

1×10−6s+

1

2 ×10−6 s

4s

−0.3s

−0.9s

−0.6s

=

2.21.7

×10−6 =1.2941×10−6

Therefore for t > 0,

V1(s) =1

5 ×10−6sI(s) +

0.3s

=1.2941×10−6

5 ×10−6s+

0.3s

=0.55882

s ⇒ v1( t) = 0.55882 V

Similarly for t > 0


V2(s) =1.2941×10−6

1×10−6 s+

0.9s

=2.1941

s ⇒ v2(t) = 2.1941 V

and

V3(s) =1.2941×10−6

2 ×10−6s+

0.6s

=1.2471

s ⇒ v3(t) =1.2471 V

SOLUTION 14.71. (a) For 0 < t < 2, the 150 mF capacitor is charged to 25 V. From conservation ofcharge,

0.15 × 25 = 0.15vC (2+ ) + 0.1vC (2+ ) = 0.25vC (2+ )Therefore

vC (2+ ) =0.15 × 25

0.25=15 V

This voltage remains constant for t > 2 s.(b) For 0 < t < 2, the 150 mF capacitor is charged to 25 V. From conservation of charge,

0.15 × 25 + 0.1×10 = 0.15vC (2+ ) + 0.1vC (2+ ) = 0.25vC (2+ )Therefore

vC (2+ ) =0.15 × 25 + 0.1×10

0.25= 19 V

This voltage remains constant for t > 2 s.

SOLUTION 14.72. (a) Let the middle node have voltage Va(s). Then writing node equations

8s −2s

−2s 6s

Va

Vout

=

1.144

0

⇒

Va

Vout

=

144

6 2

2 8

1.144 s

0

=

0.156 / s

0.052 / s

V

Thus for t > 0, vout (t) = 0.052 V.

(b) Again define Va(s) as the middle node. Then

Va(s) =

3

10s1

10s+

1

4s+

3

10s

×0.286

s=

613

×0.286

s=

0.132s

Hence for t > 0,

Vout (s) =

1

4s1

2s+

1

4s

×0.132

s=

0.1323s

=0.044

s ⇒ vout (t) = 0.044 V


SOLUTION 14.73. With switches in position A, the equivalent capacitance to the right of v2 is 4 mF.Therefore at t = 0+, by voltage division

V1(s) = V2(s) =10s

⇒ v1( t) = v2( t) =10 V for 0 < t < 1.

Hence with the switches in position B, let us write a single node equation using the parallel equivalentcircuit the initialized capacitors:

0.002sV2(s) − 0.002 ×10 + 0.004sV2(s) + 0.004 ×10 = 0.004s×11s

Equivalently0.006sV2(s) = 0.044 − 0.02 = 0.024 ⇒ V2(s) = 4 / s

Hence, for t > 1 s, v2(t) = 4 V and v1( t) = 7 V.

SOLUTION 14.74. (a) At t = 0+, the frequency domain equivalent circuit is given below.

To compute Ceq, we observe

Ceq = 1+1

1

3+

1

2+

1

6

= 2 mF

By voltage division

V1 =Ceqs

Ceqs + 0.002s×

20s

=10s

and V2 =0.002s

Ceqs + 0.002s×

20s

=10s

Hence, for 0 < t <1, v1(t) = v2(t) = 10 V.(b) When the switch moves to B, the pertinent part of the equivalent frequency domain circuit is givenbelow.


By superposition,

V1 =1

1 + 2×

8s

+1

0.003s× 0.02 −

10.003s

× 0.01=6s

Hence, for 1 < t < 2, v1(t) = 6 V and by KVL, v2(t) = 2 V.

SOLUTION 14.75. When the switch in position A, the 2 µF capacitor is charged to –2 V. Hence, the

charge on the top plate is C*vC = –4 µC. When the switch is moved to position B, due to the virtual

ground, the 2 µF capacitor voltage is zero meaning it cannot retain any charge. Hence, assuming an

ideal op amp, all charge moves to the 1 µF capacitor with –4 µC on the left plate. Hence, vout = −(–4

µC)/1 µF = 4 V.

SOLUTION 14.76. (a) vout(t) = 0 for 0 < t < 1 ms. Every time the switch moves to position A, thecapacitor, C, charges to 8 V. When the switch moves to position B, because of the virtual ground, allcharges moves to the kC capacitor. Hence with k = 1, vout(t) = – 8 V for 1 ms < t < 3 ms. For 3 ms < t <

5 ms, vout(t) = – 16 V. Repeating the pattern implies that for 5 ms < t < 7 ms, vout(t) = – 24 V, etc. Seefor example figure 14.51.(b) With k = 0.5, the voltages computed in part (a) double.(c) With k = 2, the voltages computed in part (a) are halved.

SOLUTION 14.77.(a) For this part consider the circuit below.


For the normalized design C = 1 F, G1 = 0.5 S, G2 = 2 S, and G3 = 1.5 S. After magnitude scaling with

Km = 106, then C = 1 F , R1 = 2 MΩ, R2 = 5 0 0 kΩ, R3 = 6 6 6 . 7 kΩ.

(b) For this part, consider the circuit below.

For the normalized design C = 1 F, G1 = 1 S, G3 = 2 S, G2 = 0.5 S, and G4 = 2.5 S. After magnitude

scaling with Km = 106, then C = 1 F , R1 =1 MΩ, R3 = 5 0 0 kΩ, R2 = 2 MΩ, R4 = 4 0 0 kΩ.

(c) Now consider the circuit below.


For the normalized design C = 1 F, G1 = 0.25 S, G3 = 0.5 S, G2 = 0.75 S, G4 = 1 S, and G5 = 1 S. After

magnitude scaling with Km = 106, then C = 1 F , R1 = 4 MΩ, R3 = 2 MΩ, R2 = 4 / 3 MΩ,R4 =1 MΩ, R5 = 1 MΩ.

(d) Finally, we consider the following circuit.

For the normalized design C = 1 F, G1 = 2 S, G3 = 1.5 S, G6 = 1 S, G2 = 0.5 S, G4 = 2 S, and G5 = 2 S.

After magnitude scaling with Km = 106, then C = 1 F , R1 = 0 . 5 MΩ, R3 = 2 /3 MΩ ,R6 = 1 MΩ, R2 = 2 MΩ , R4 = 0 . 5 MΩ, R5 = 0 . 5 MΩ.


SOLUTION 14.78. With V1 = Vout a prototype design is given by the topology below.

Using MATLAB»Km = 1e7;»Gin = 1; G1 = 1.405; G2 = 0.402; G3 = 0.942;»DG = 1.865;»Rinnew = Km/GinRinnew = 10000000»R1new = Km/G1R1new = 7.1174e+06»R2new = Km/G2R2new = 2.4876e+07»R3new = Km/G3R3new = 1.0616e+07»DRnew = Km/DGDRnew = 5.3619e+06

Later, when we study frequency scaling, Km will be smaller and the filter will have a cutoff frequency ina more reasonable range.

SOLUTION TO 14.79. Note corrections to problem statement. W(0+) should be W(∞) in part (b) and inpart (c) one should calculate W(0-) – W(∞). The frequency domain equivalent circuit is given by thefigure below.


(a) To find the required time functions, we first find their Laplace equivalents.

I(s) =

a

s−

b

s

R +1

C1s+

1

C2s

=

a − b

R

s +1

RC1+

1

RC2

≡K1

s + p1

in which case i( t) = K1e− p1tu(t) where

K1 =a − b

R and p1 =

1RC1

+1

RC2

.

Further,

VC1(s) =− I(s)C1s

+as

=as

−

K1

C1s s + p1( ) = a −

K1C1p1

×

1s

+K1

C1p1×

1s + p1( )

in which case

vC1(t) = a −K1

C1p1

+

K1C1p1

e− p1t

u( t)

Also, by symmetry,

VC 2(s) =I(s)C2s

+bs

=bs

+

K1

C2s s + p1( ) = b +

K1C2 p1

×

1s

−K1

C2p1×

1s + p1( )

in which case

vC 2( t) = b +K1

C2 p1

−

K1C2 p1

e− p1t

u( t)

(b) The total energy stored in the capacitors at time 0- is

W (0− ) =12

C1a2 +12

C2b2

Also at t = ∞,

vC1(∞) = a −K1

C1p1

and vC 2(∞) = b +

K1C2p1

.

Hence


W (∞) =12

C1 a −K1

C1p1

2

+12

C2 b +K1

C2 p1

2

(c)

Ri2( t)dt0

∞

∫ = RK12e−2 p1tdt

0

∞

∫ =RK1

2

2p1=

(a − b)2

2Rp1=

(a − b)2

21

C1+

1

C2

Observe that

W (0− ) −W (∞) =12

C1a2 +12

C2b2 −12

C1 a −K1

C1p1

2

−12

C2 b +K1

C2 p1

2

=aK1p1

−bK1p1

−12

K12

C1p12 −

12

K12

C2p12 =

(a − b)2

Rp1−

12

×(a − b)2

Rp1=

(a − b)2

2Rp1=

(a − b)2

21

C1+

1

C2

This indicates that the total energy lost between 0- and infinity is the energy dissipated in the resistorand the result is independent of the value of R.(d) When R → 0,

limR→0

I(s) =

a

s−

b

s1

C1s+

1

C2s

=a − b

1

C1+

1

C2

=C1C2

(C1 +C2)(a − b)

Therefore

i( t) =C1C2

(C1 +C2)(a − b) (t)

Further

VC1(s) =−C2

s(C1 +C2)(a − b) +

as

=C1

s(C1 +C2)a +

C2s(C1 +C2)

b

and

VC 2(s) =C1

s(C1 +C2)(a − b) +

bs

=C1

s(C1 +C2)a +

C2s(C1 +C2)

b

Therefore

vC1(t) = vC 2(t) =C1

(C1 +C2)a +

C2(C1 +C2)

b

u(t)

SOLUTION TO 14.80. (a) From conservation of charge

C1vC1(0−) + C2vC 2(0− ) = C1vC1(0+ ) + C2vC 2(0+ ) = C1 + C2( )vC1(0+) = C1 + C2( )vC 2(0+ )Therefore


vC1(0+ ) = vC2(0+ ) =C1vC1(0− ) + C2vC 2(0− )

C1 + C2( )(b) Inserting values into out answer for part (a) yields

vC1(0+ ) = vC2(0+ ) =C1

C1 + C2( ) = 0.5 V

and the voltage remains the same for t > 0.(c) Before the switch is closed, the energy in C2 is zero and the energy in C1 is the total stored energy:

Wtot (0−) = WC1(0− ) = 0.5C1vC1

2 (0− ) = 0.5 JAfter the switch is closed,

Wtot (0+ ) = WC1(0+ ) + WC 2(0+ ) = 0.5C1vC1

2 (0+ ) + 0.5C2vC 22 (0+ ) = 0.25 J

(d) (i) Using the series equivalent circuit for C1, we have

I(s) =1

R +2

s

×1s

=1 /R

s + 2 / R ⇒ i( t) =

1R

e−2 t Ru(t) A

Thus

VC 2(s) =I(s)

s=

1/ Rs(s + 2 / R)

=0.5s

−0.5

s + 2 / R ⇒ vC 2( t) = 0.5 1− e−2 t R( )u(t) V

and

VC1(s) = −I(s)s

+1s

=−1/ R

s(s + 2 / R)+

1s

=0.5s

+0.5

s + 2 / R ⇒ vC1(t) = 0 . 5 1+ e−2 t R( )u(t) V

(ii) The energy dissipated in the resistor is given by

WR(0,∞) = R i2( )d0

∞

∫ =1R

e−4 t Rd0

∞

∫ = −e−4 t R

4

0

∞

=14

J

(iii) For all R,

Area under i(t) = i( )d0

∞

∫ =1R

e−2 t Rd0

∞

∫ = −e−2 t R

2

0

∞

=12

Further, as R → 0, i( t) =1R

e−2 t Ru(t) has a decay that becomes infinitely fast and its magnitude (1/R)

approaches ∞. Thus we have infinite height, zero-width, but a finite area of 0.5. Thus as R → 0,i( t) → 0.5 (t) A. (We have avoided a more rigorous explanation as the above argument is moreplausible to sophomores.). As R → 0, the exponential terms in the expressions for vC1(t) and vC 2( t)have infinitely fast decays and hence disappear from the expressions yielding the stated result.


SOLUTION TO 14.81.(a)

Zin (s) = 2s + 4.5(s+ 0.5)(s + 4)

= 1s+ 0.5

+ 1 s + 4

= Za (s) + Zb (s)

Ya (s) = 1Za (s)

= s+ 0.5

Yb (s) = 1Zb (s)

= s + 4

From the above expressions, the RC circuit consists of a series connection of (a 1 farad capacitor inparallel with a 2 Ω resistor) and (a 1 F capacitor in parallel with a 0.25 Ω resistor).

(b)Yin (s) = 12s + 440

(s+ 120)(s + 20) = 10

s+ 120 + 2

s + 20 = Ya (s) + Yb (s)

Za (s) = 1Ya (s)

= 0.1s + 12

Zb (s) = 1Yb (s)

= 0.5s + 10

From the above expressions, the RL circuit consists of a parallel connection of (0.1 H inductor in serieswith a 12 Ω resistor) and (a 0.5 H inductor in series with a 10 Ω resistor).

(c) Following the hint, we have

Yin (s)s = 0.225s +0.075

(s+ 0.2)(s + 0.5) = 0.1

s+ 0.2 + 0.125

s + 0.5

Hence

Yin (s) = 0.1ss+ 0.2

+ 0.125s s + 0.5

= Ya (s) + Yb (s)

Za (s) = 1Ya (s)

= s + 0.20.1s

= 10 + 10.5s

Zb (s) = 1Yb (s)

= s + 0.50.125s

= 8 + 10.25s

From the above expressions, we see that each term in Yin (s) represents a series RC circuit. The RCcircuit for Yin(s) consists of a parallel connection of (a 0.5 F capacitor in series with a 10 Ω resistor)and (a 0.25 F capacitor in series with a 8 Ω resistor).

(d) CORRECTION: for part (d),change the second term to 2s/(s2 + 2).


Yin (s) = 0.5ss2+ 1

+ 2s s2+ 2

= Ya (s) + Yb (s)

Za (s) = 1Ya (s)

= s2 + 1 0.5s

= 2s + 1 0.5s

Zb (s) = 1Yb (s)

= s2 +22s

= 0.5s + 1s

From the above expressions, we see that each term in Yin (s) represents a series LC circuit. The LCcircuit for Yin(s) consists of a parallel connection of (a 0.5 F capacitor in series with a 2 H inductor)and (a 1 F capacitor in series with a 0.5 H inductor).

SOLUTION 14.82. CORRECTIONS TO PROBLEM STATEMENT: (i) v0(t), should read vout(t)and (ii) there should be a connection from the circuit inside the shaded box to the bottom line orreference node.(a) (i) 0 ≤ t < 1 ms. Since the capacitor voltage is initially zero and the switch is in position (a), asimple source transformation yields a Norton equivalent (seen by the capacitor) consisting of a 20 mAcurrent source in parallel with 9.8039 kΩ resistor. Hence

Vout (s) =1

1

98039+10−6s

×.02s

=20 ×103

s s +102( )

Using MATLABn = .02*1e6;p1 = 1e6/Rthp1 = 102»d = [1 p1 0];»[r,p,k] = residue(n,d)r = -1.9608e+02 1.9608e+02p = -102 0k = []

Hence, for 0 ≤ t < 1 ms, vout (t) = 196.08 1− e−102t( )u( t) V. It follows that vout (1 ms) = 19.014 V.

(ii) 1 ms ≤ t < 1.05 ms. The frequency domain equivalent circuit is given below.


Writing a single node equation yields

19.014 ×10−6 =Vout −

200

s10000

+Vout −

1

s10

+10−6 sVout

Equivalently 19.014 +1.2 ×105

s= s +100100( )Vout or

ˆ V out =19.014s +1.2 ×105

s s +100100( )Therefore, ˆ v out (t ') = 1.1988 +17.815e−100100t'( )u(t ') , and for 1 ms ≤ t < 1.05 ms,

vout (t) = ˆ v out (t − 0.001) in which case

vout (t) = 1.1988 +17.815e−100100(t−0.001)( )u( t − 0.001)

Again using MATLAB,

(b) Part 1: for t > 0 up to t1 which denotes the time when vout(t) reaches 80 V, i.e., the capacitor ischarging. The frequency domain equivalent circuit is


Using our knowledge of part (a), this circuit simplifies to

Hence

Vout (s) =5 ×10−6 +

0.02

s

10−6 s +1

9803.9

=5s + 20 ×103

s s +102( )

and from MATLAB»syms s t»ilaplace((5*s+20e3)/(s^2+102*s))ans =10000/51-9745/51*exp(-102*t)

in which case

vout (t) = 196.08 −191.08e−102t( )u( t) V

From this expression,

vout (t1) = 80 = 196.08 −191.08e−102 t1 Vandt1 = log((80-10000/51)/(-9745/51))/(-102)t1 = 4.8864e-03

This part of the problem considers t1 ≤ t < t2, i.e., the capacitor is discharging where vout(t2) = 5.The equivalent frequency domain circuit is given below which is a slight modification of the circuit of(a)-(ii):


Making use of our knowledge of part (a)-(ii), we have

ˆ V out =80s +1.2 ×105

s s +100100( )

in which case ˆ v out (t ') = 1.1988 + 78.801e−100100t '( )u( t') V, and

vout (t) = 1.1988 + 78.801e−100100(t−t1)( )u( t − t1)

Here t2' = 3.0286 ×10−5s and t2 = t1 + t2

' = 4.9167 ms where t2' is the duration of the discharge cycle.

As a final point, note that the frequency of the sawtooth is 1/t2 = 203.39 Hz. Finally a plot is givenbelow.


SOLUTION 14.83. CORRECTION: In example 14.10, page 560, delete the four minus signs in the

equation for VC(s) and one more for vC(t).

We use MATLAB instead of SPICE to solve this problem. Applying voltage division to the circuit offigure P14.83, we have

VC (s) = 1

Cs1

Cs + Ls + R

LiL(0-) = iL(0-)

C 1s2 + R

Ls + 1

LC

= 108

s2 + 125s + 1.25 ×109

From table 13.1, item 18

vC (t) = 2828e-62.5t sin(35,355t) u(t) V

A plot of vC(t) is given below with the vertical axis in V and the horizontal axis in seconds.

-3000

-2000

-1000

0

1000

2000

3000

0 0.2 0.4 0.6 0.8 1

x10-3

The waveform for the first few cycles is essentially the same as the example 14.10. Thus for the firs fewcycles, the lossless circuit of example 14.10 is a good approximation to the more accurate circuit modelof this problem. The effect of the presence of 100 Ω resistance is a slow decay (with respect to msintervals) of the peak values.

1/25/02 P16-1 © R. A. DeCarlo, P. M. Lin

PROBLEM SOLUTIONS

Solution 16.1.(a) By the definition of the convolution integral

f2( t) ∗ f2( t) = f2(t − ) f2( )d =−∞

∞

∫ 2u( t − )2u( )d = 4 u( t − )d0

∞

∫−∞

∞

∫The integrand, u( t − ) , is nonzero only when ≤ t. This suggests that there are two regions ofconsideration: t < 0 and t ≥ 0.Case 1: t < 0. Here u( t − ) = 0 since is restricted to the interval [0,∞). Hence

f2( t) ∗ f2( t) = 0 , for t < 0.Case 2: t ≥ 0.

f2( t) ∗ f2( t) = 4 u(t − )d = 4 d0

t

∫−∞

∞

∫ = 4t , for t ≥ 0.

In sum,

f2( t) ∗ f2( t) =0, t < 0

4t, t ≥ 0

(b) By the definition of the convolution integral

f2( t) ∗ f3(t) = f2(t − ) f3( )d =−∞

∞

∫ 2u( t − )4e−2 u( )d = 8 e−2 u( t − )d0

∞

∫−∞

∞

∫

The integrand, u( t − ) , is nonzero only when ≤ t. This suggests that there are two cases toconsider: t < 0 and t ≥ 0.Case 1: t < 0. Here u( t − ) = 0 since is restricted to the interval [0,∞). Hence

f2( t) ∗ f3(t) = 0, for t < 0.

Case 2: t ≥ 0.

f2( t) ∗ f3(t) = 8 e−2 d = −4e−2 ]0

t

0

t

∫ = 4(1− e−2 t )

In sum,

f2( t) ∗ f2( t) =0, t < 0

4(1− e−2t ), t ≥ 0

(c) By the definition of the convolution integral and the sifting property of the delta function

f1(t) ∗ f3( t) = f1(t − ) f3( )d =−∞

∞

∫ 5 ( t − )4e−2 u( )d =−∞

∞

∫= 20e−2 u( )] =t

= 20e−2tu( t)

(d) By the definition of the convolution integral


f3(t) ∗ f3( t) = f3(t − ) f3( )d =−∞

∞

∫ 4e−2(t− )u( t − )4e−2 u( )d =16e−2 t u( t − )d0

∞

∫−∞

∞

∫

The integrand, u( t − ) , is nonzero only when ≤ t. This suggests that there are two cases toconsider: t < 0 and t ≥ 0.Case 1: t < 0. Here u( t − ) = 0 since is restricted to the interval [0,∞). Hence

f3(t) ∗ f3( t) = 0 , for t < 0.Case 2: t ≥ 0.

f3(t) ∗ f3( t) =16e−2t d = 16te−2 t

0

t

∫In sum,

f3(t) ∗ f3( t) =0, t < 0

16te−2t , t ≥ 0

(e) By the definition of the convolution integral and the sifting property of the delta function

f1(t + 2) ∗ f2( t + 4) = f1(t + 2 − ) f2( + 4)d = 5 (t + 2 − )2u( + 4)d =−∞

∞

∫−∞

∞

∫= 10u( + 4)] =t+2 =10u(t + 6)

(f) By the distributive property of convolution

f2( t) ∗ f2(t) + f3(t)[ ] = f2(t) ∗ f2(t) + f2( t) ∗ f3(t)

Using the results of parts (a) and (b) the result follows immediately

f2( t) ∗ f2(t) + f3(t)[ ] =0, t < 0

4(1+ t − e−2t ), t ≥ 0

Solution 16.2.(a) By definition

f3(t) = f1( ) f2( t − )d =−∞

∞

∫ K1u( − T1)K2u(t − − T2)d = K1K2 u( t − − T2)dT1

∞

∫−∞

∞

∫Here observe that u( t − − T2) = 0 for > t − T2 . Hence there are two cases to consider:t − T2 < T1 and t − T2 ≥ T1 .

Case 1: t − T2 < T1 . Here u( t − − T2) = 0 , since is restricted to the interval [T1,∞) .f3(t) = 0

Case 2: t − T2 ≥ T1 .


f3(t) = K1K2 d = K1K2( t − T2 − T1)T1

t−T2

∫In sum,

f3(t) =0, t < T1 + T2

K1K2(t − T2 − T1), t ≥ T1 + T2

(b) By definition

f3(t) = f1( t − ) f2( )d =−∞

∞

∫ K1u(t − + T1)K2u( + T2)d = K1K2 u(t − + T1)d−T2

∞

∫−∞

∞

∫

Here observe that u( t − + T1) = 0 for > t + T1. Hence there are two cases to consider:t + T1 < −T2 and t + T1 ≥ −T2 .

Case 1: t + T1 < −T2 . Here u( t − + T1) = 0 , since is restricted to the interval [−T2 ,∞): f3(t) = 0.

Case 2: t + T1 ≥ −T2 .

f3(t) = K1K2 d = K1K2(t + T1 + T2)−T2

t+T1

∫In sum,

f3(t) =0, t < −T1 − T2

K1K2(t − T2 − T1), t ≥ −T1 − T2

(c) By definition

f3(t) = f1( t − ) f2( )d =−∞

∞

∫ K1u(t − )K2e−a u( )d = K1K2 e−a u( t − )d0

∞

∫−∞

∞

∫

The integrand is nonzero only when ≤ t. Hence, there are two cases to consider: t < 0 and t ≥ 0.

Case 1: t < 0. f3(t) = 0, for t < 0.Case 2: t ≥ 0.

f3(t) = K1K2 e−a

0

t

∫ d = K1K2e−a

−a

0

t

=K1K2

a(1− e−at ).

Therefore

f3(t) =K1K2

a(1− e−at )u(t). for t ≥ 0.

(d) By definition

f3(t) = f1( t − ) f2( )d =−∞

∞

∫ K1u(t − + T1)K2e−a u( )d = K1K2 e−a u( t − + T1)d0

∞

∫−∞

∞

∫


The integrand is nonzero only when ≤ t + T1 . Hence, there are two cases to consider: t + T1 < 0and t + T1 ≥ 0.

Case 1: t + T1 < 0. Here u( t − + T1) = 0 , since is restricted to the interval [0,∞). Hencef3(t) = 0.

Case 2: t + T1 ≥ 0.

f3(t) = K1K2 e−a

0

t+T1

∫ d = K1K2e−a

−a

0

t+T1

=K1K2

a1− e−a (t+T1)[ ].

Therefore

f3(t) =K1K2

a1− e−a (t+T1 )[ ]u(t + T1).

(e) By definition

f3(t) = f1( t − ) f2( )d =−∞

∞

∫ K1u(−t + )K2e−a u( )d = K1K2 e−a u(−t + )d0

∞

∫−∞

∞

∫

The integrand is nonzero only when ≥ t. Hence, there are two cases to consider: t ≤ 0 and t > 0.Case 1: t ≤ 0. Here u(− t + ) = 1, since ≥ 0. Hence

f3(t) = K1K2 e−a

0

∞

∫ d = K1K2e−a

−a

0

∞

=K1K2

a, for t ≤ 0.

Case 2: t > 0.

f3(t) = K1K2 e−a

t

∞

∫ d = K1K2e−a

−a

t

∞

=K1K2

ae−at , for t > 0.

In sum,

f3(t) =

K1K2

a, t ≤ 0

K1K2

ae−at , t > 0

Solution 16.3.(a) By definition

f3(t) = f1( ) f2( t − )d =−∞

∞

∫ K1e−a u( )K2e−a (t− )u( t − )d

−∞

∞

∫ = K1K2 e−atu( t − )d0

∞

∫

The integrand, u( t − ) , is nonzero only when ≥ t. Hence, there are two cases to consider: t < 0and t ≥ 0.

Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Hence f3(t) = 0, for t < 0.Case 2: t ≥ 0.


f3(t) = K1K2e−at d0

t

∫ = K1K2e−at t , for t ≥ 0.

In sum,

Ic(s) =Cs

Cs +1

R

Iin (s)

(b) By definitiona = 1K =1

The integrand, u( t − ) , is nonzero only when ≥ t. Hence, there are two cases to consider: t < 0and t ≥ 0.

Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Hence f3(t) = 0, for t < 0.Case 2: t ≥ 0.

f3(t) = K1K2e−bt e(b−a ) d0

t

∫ =t if a = b

1b − a

e(b−a )t −1[ ] if a ≠ b

Therefore, for t ≥ 0,

f3(t) =K1K2e−bt tu( t) if a = bK1K2

b − ae−at − e−bt[ ] if a ≠ b

(c) By replacing K1 = 50, K2 = 20 and a = 10 in the formula of f3(t) in part (a) the answer forpart (i) is easily obtained as

f3(t) =0, t < 0

1000e−10t t, t ≥ 0

For part (ii) the parameters have the following values: vc ( t) = e−(t− )

−∞

t

∫ d = e −t ]−∞

t=1,

K2 = 0.2, a = 10 and b = 0.2. Using these values in the formula developed in part (b) for f3(t) theanswer follows immediately

f3(t) = 0.102 e−0.2t − e−10 t( )u( t)

SOLUTION 16.4.(a) Using the impulse response theorem and the definition of the convolution integral the responseof the system, y(t), can be computed as follows

y(t) = h( t − )v( )d−∞

∞

∫ = 2e−2(t− )u( t − ) u( −1)− u( − 3)[ ]d−∞

∞

∫

Observing that u( −1)− u( − 3) is nonzero only when 1 ≤ < 3 yields


y(t) = 2 e2( −t)u(t − )d1

3

∫

The integrand in the above equation is nonzero only when ≤ t. This suggests three regions ofconsideration: t <1, 1 ≤ t ≤ 3, and 3 < t.Case 1: t <1. Here u( t − ) = 0 , since is restricted to the interval [1,3]. Hence y(t) = 0, for t <1.Case 2: 1 ≤ t ≤ 3.

y(t) = 2 e2( −t)

1

t

∫ d = e2( −t) ]1t

=1− e2(1−t ) , for 1 ≤ t ≤ 3

Case 3: 3 < t.

y(t) = 2 e2( −t)

1

3

∫ d = e2( −t) ]13

= e2(1−t )(e4 −1), for 3 < t

In sum,

y(t) =0, t < 1

1− e2(1−t ), 1≤ t ≤ 3

e2(1−t )(e4 −1), 3 < t

A picture of y(t) is sketched in the next figure.

(b) Using the impulse response theorem and the definition of the convolution integral the responseof the system, y(t), can be computed as follows

y(t) = h( t − )v( )d−∞

∞

∫ = 2e−2(t− )u( t − )u(2 − t + ) u( −1)− u( − 3)[ ]d−∞

∞

∫

Observing that u( −1)− u( − 3) is nonzero only when 1 ≤ < 3 yields

y(t) = 2 e2( −t)u(t − )u(2 − t + )d1

3

∫The integrand in the above equation is nonzero only when t − 2 ≤ ≤ t. This suggests four regionsof consideration: t <1, 1 ≤ t ≤ 3, 3 < t ≤ 5and 5 < t.Case 1: t <1. Here u( t − ) = 0 , since is restricted to the interval [1,3]. Hence y(t) = 0, for t <1.Case 2: 1 ≤ t ≤ 3. Here u( t − )u(2 − t + ) is nonzero only when 1 ≤ ≤ t. Therefore,


y(t) = 2 e2( −t)

1

t

∫ d = e2( −t) ]1t

=1− e2(1−t ) , for 1 ≤ t ≤ 3.

Case 3: 3 < t ≤ 5. Here u( t − )u(2 − t + ) is nonzero only when t − 2 ≤ ≤ 3. Therefore,

y(t) = 2 e2( −t)

t−2

3

∫ d = e2( −t) ]t−2

3= e2(3−t) − e−4 , for 3 < t ≤ 5.

Case 4: 5 < t. Here u( t − )u(2 − t + ) = 0 , since is restricted to the interval [1,3]. Therefore,y(t) = 0, for 5 < t.


(c) By the impulse response theorem, the zero-state response of the circuit y(t) isy(t) = h(t) ∗v( t)

Using the definition of the convolution integral and the sifting property of delta function it followsthat

y(t) = 2h(t) − 2h( t −1) + h(t − 2) =

0, t < 0

2, 0 ≤ t < 1

−2, 1≤ t < 2

0, 2 ≤ t < 3

1, 3 ≤ t < 4

0, 4 ≤ t

Using the waveform of h(t) given in figure P16.4, y(t) is sketched in the next picture.


SOLUTION 16.5.(a) By definition

f4(t) = f1( t − ) f2( )d = ( t − − 2) ⋅2 ⋅ u( +1)d−∞

∞

∫−∞

∞

∫

By the sifting property of delta function it follows that f4(t) = 2u( +1)] =t−2 = 2u( t −1).

(b) By the definition of convolution and the sifting property of delta function

f5(t) = f1( t − ) f3( )d = (t − − 2)e−2 u( )d−∞

∞

∫−∞

∞

∫= e−2 u( )] =t−2

= e−2(t−2)u( t − 2)

(c) By definition

f6( t) = f2( t − ) f3( )d = 2u(t − +1)e−2 u( )d−∞

∞

∫−∞

∞

∫ = 2 e−2 u(t − +1)d0

∞

∫

The integrand in the above equation is nonzero only when ≤ t +1. This suggests two regions ofconsideration: t < −1and −1≤ t .Case 1: t < −1. Here u( t − +1)= 0, since 0 ≤ . Hence f6( t) = 0 , for t < −1.Case 2: −1≤ t . Here u( t − +1) is nonzero only when ≤ t +1. Therefore,

f6( t) = 2 e−2 d = e−2 ]0

t+1= 1− e−2(t+1)

0

t+1

∫It follows that

f6( t) = 1− e−2(t+1)[ ]u( t +1)

(d) By the definition of convolution and the sifting property of delta function we have


f7( t) = f1(t − ) f3( − 2)d = ( t − − 2)e−2( −2)u( − 2)d−∞

∞

∫−∞

∞

∫ =

= e−2( −2)u( − 2)] =t−2= e−2(t−4)u( t − 4) .


f6( t) = 1− e−2(t+1)[ ]u( t +1)

Here observe that u(t – τ) = 0 for τ > t. Hence, there are two cases to consider: t ≤ 0 and t > 0.

Case 1: t ≤ 0 y ( t ) = K e a τ d τ

−∞

t

∫ = K e a τ

a

−∞

t = K

e at

a

Case 2: t > 0

y(t) = K eaτdτ−∞

t

∫ = K eaτdτ−∞

0

∫ = Keaτ

a

−∞

0

=K

a

(b) By definition

y(t) = K u(τ − t)−∞

∞

∫ e−aτ

u(τ)dτ = K u(τ)t

∞

∫ e− aτ

dτ

Here observe that u(τ – t) = 0 for τ < t; hence the lower limit of integration is t. Also, because of the

presence of u(τ) in the integrand, there are two cases to consider: t < 0 and t ≥ 0.

Case 1: t < 0

y(t) = K e−aτdτ0

∞

∫ = Ke−aτ

−a

0

∞

=K

a

Case 2: t ≥ 0

y(t) = K e−aτdτt

∞

∫ = Ke− aτ

−a

t

∞

=K

ae−at

SOLUTION 16.7.(a) Using the definition of the convolution integral and the sifting property of delta function,f5(t)can be computed as below

f5(t) = f2(t − ) f4( )d = e−a (t− )u(t − ) ( − 4)d−∞

∞

∫−∞

∞

∫ =

= e−a (t− )u(t − )] =4= e−a (t−4)u( t − 4)

A picture of f5(t), for a = 1, is sketched in the next figure.


(b) By definition

f6( t) = f1(t − ) f1( )d = K 2 u(t − )u( )d−∞

∞

∫−∞

∞

∫ = K2 u( t − )d0

∞

∫

Since u( t − ) is nonzero only when ≤ t, there are two regions of consideration: t < 0and 0 ≤ t.Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Hence

f6( t) = 0 , for t < 0.Case 2: 0 ≤ t.

f6( t) = K 2 d0

t

∫ = K 2t , for 0 ≤ t.

A picture of f6( t) , for K =1, is sketched in the next figure.

(c) By definition

f7( t) = f1(t − ) f2( )d = Ku( t − )e−a u( )d−∞

∞

∫−∞

∞

∫ = K e−a u(t − )d0

∞

∫

Since u( t − ) is nonzero only when ≤ t, there are two regions of consideration: t < 0and 0 ≤ t.Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Therefore, f7( t) = 0 , for t < 0.Case 2: 0 ≤ t.

f7( t) = K e−a d0

t

∫ =−Ka

e−a ]0

t=

Ka

1− e−at( ) , for 0 ≤ t.


A picture of f7( t) , for t < 0 and a = 1, is sketched in the next figure.

(d) By definition

f8(t) = f1( t − ) f3( )d = Ku( t − )ea u(− )d−∞

∞

∫−∞

∞

∫ = K ea u(t − )d−∞

0

∫

The integrand, )( −tu , is nonzero only when t≤ . This suggests two regions of consideration:0≤t and t<0 .

Case 1: 0≤t .

f8(t) = K ea u(t − )d−∞

t

∫ =Ka

ea ]−∞

t=

Ka

eat , for t ≤ 0.

Case 2: 0 < t.

f8(t) = K ea u(t − )d−∞

0

∫ =Ka

ea ]−∞

0=

Ka

, for 0 < t.

In sum,

f8(t) =

K

aeat , t ≤ 0

K

a, 0 < t

A picture of f8(t), for K =1 and a = 1, is sketched in the next figure.


SOLUTION 16.8.(a) Using the current division formula

Ic(s) =Cs

Cs +1

R

Iin (s)

By Ohm’s law the Laplace transform of capacitor’s voltage

Vc (s) =1Cs

Ic (s)

Therefore the transfer function of the circuit

H(s) =Vc (s)Iin (s)

=1

Cs +1

R

=1

s + 4

Taking the inverse Laplace transform of H(s) yields the impulse response h(t) = e−4 tu( t) .

(b) By the impulse response theorem

vc ( t) = iin (t) ∗ h( t) = iin ( t −−∞

∞

∫ )h( )d = 3e−(t− )

−∞

∞

∫ u(t − )e−4 u( )d =

= 3 e−(t+ 3 )

0

∞

∫ u(t − )d

The integrand is nonzero only when ≤ t. Therefore there are two regions of consideration:t < 0and 0 ≤ t.Case 1: t < 0. Here u( t − ) = 0 , since 0 ≤ . Hence vc ( t) = 0 , for t < 0.Case 2: 0 ≤ t.


vc ( t) = 3 e−(t+ 3 )

0

t

∫ d = −e−(t+ 3 )]0

t= e−t − e−4 t , for 0 ≤ t.

In sum,

vc ( t) = e−t − e−4 t( )u(t)V.

SOLUTION 16.9.(a) By voltage division formula

Vout (s) =

1

Cs

R +1

Cs

Vin (s)

Therefore the transfer function


=

1

Cs

R +1

Cs

=1

s +1

Taking the inverse Laplace transform of H(s) yields yields the impulse response h(t) = e−tu(t).By the impulse response theorem and the convolution definition

vout (t) = h( t − )vin ( )−∞

∞

∫ d = e−(t− )

−∞

∞

∫ u( t − ) u(− ) + 2e−2 u( )[ ]d =

= e−(t− )

−∞

0

∫ u( t − )d + 2 e−(t+ )

0

∞

∫ u(t − )d

For both integrals the integrand is nonzero only when ≤ t. This suggests two regions ofconsideration: t < 0and 0 ≤ t. Case 1: t < 0. Here the second integral is zero since, for this integral, is restricted to [0,∞).

vc ( t) = e−(t− )

−∞

t

∫ d = e −t ]−∞

t=1, for t < 0.

Case 2: 0 ≤ t.

vc ( t) = e−(t− )

−∞

0

∫ d + 2 e−(t+ )

0

t

∫ d

= e −t ]−∞

0− 2e−(t− )]0

t= e−t − 2 e−2t − e−t( ) = 3e−t − 2e−2t , for 0 ≤ t

(c) By the impulse response theorem and the definition of convolution


vout (t) = h( t − )vin ( )−∞

∞

∫ d = e−(t− )

−∞

∞

∫ u( t − )e−a| |d

= e−(t− )

−∞

0

∫ u( t − )ea d + e−(t− )

0

∞

∫ u( t − )e−a d

= e−t e (a+1)

−∞

0

∫ u( t − )d + e−t e (1−a )

0

∞

∫ u(t − )d

For both integrals the integrand is nonzero only when ≤ t. This suggests two regions ofconsideration: t < 0and 0 ≤ t. Case 1: t < 0. Here the second integral is zero since, for this integral, is restricted to [0,∞).

vout (t) = e−t e (a+1)

−∞

t

∫ d = e−t e (a+1)

a +1

−∞

t

=eat

a +1, for t < 0.

Case 2: 0 ≤ t.

vout (t) = e−t e (a+1)

−∞

0

∫ d + e−t e (1−a )

0

t

∫ d =

= e−t e (a+1)

a +1

−∞

0

+ e−t e (1−a)

0

t

∫ d =e−t

a +1+ e−t e (1−a)

0

t

∫ d

Here observe that a +1 is nonzero since a > 0.For computing the second integral, in case 2, we need to distinguish two subcases: a = 1

and a ≠ 1.

e−t e (1−a)

0

t

∫ d =e−t t if a =1

11− a

e−at − e−t( ) if a ≠1

Therefore, for 0 ≤ t,

vout (t) =

e−t

a +1+ e−t t if a = 1

e−t

a +1+

1

1− ae−at − e−t( ) if a ≠ 1

SOLUTION 16.10.(a) The impulse response is obtained by taking the inverse Laplace transform of the transferfunction

h(t) = −2e−0.2tu(t)

By the impulse response theorem the response y(t)equals


y(t) = h(t) ∗v( t) = h( t − )v( )d−∞

∞

∫Substituting v(t) = u( t +1)− u( t −1) in the above integral yields

y(t) = h( t − ) u( +1)− u( −1)[ ]d−∞

∞

∫

Here observe that u( +1)− u( −1) is nonzero only when −1≤ ≤ 1. Hence

y(t) = h( t − )d = −2 e−0.2( t− )

−1

1

∫ u(t − )d−1

1

∫

The integrand is nonzero only when ≤ t. This suggests three regions of consideration: t < −1,−1≤ t < 1 and 1 ≤ t.Case 1: t < −1. y(t) = 0Case 2: −1≤ t < 1.

y(t) = −2 e−0.2( t− )

−1

t

∫ d =10 e−0.2(t+1) −1[ ] , for −1≤ t < 1.

Case 3: 1 ≤ t.

y(t) = −2 e−0.2( t− )

−1

1

∫ d =10e−0.2t e−0.2 − e0.2( ) , for 1 ≤ t.

(b) The transfer function of the leaky integrator (see equation 14.14 in the textbook) is given by

H(s) =−

1

R1

Cs +1

R2

where R2 is the leakage resistance of the capacitor C and R1 is the resistance connected at theinverting input of the op amp. Equating the two expressions of H(s) we obtain that

−1

R1

Cs +1

R2

=−2

s + 0.2

Matching the coefficients and taking into account that the smallest resistor is 10kΩ the following

values are obtained: R1 =10kΩ, R2 = 100kΩ and C = 5 ⋅10−5F .

(c) The impulse response is obtained by taking the inverse Laplace transform of the transferfunction

h(t) = Ke−atu(t)

By the impulse response theorem the response y(t)equals


y(t) = h(t) ∗v( t) = h( t − )v( )d−∞

∞

∫

Substituting v(t) = u( t + T ) − u( t − T ) in the above integral yields

y(t) = h( t − ) u( + T ) − u( − T )[ ]d−∞

∞

∫

Here observe that u( + T ) − u( − T ) is nonzero only when −T ≤ ≤ T . Hence

y(t) = h(t − )d = K e−a (t− )

−T

T

∫ u(t − )d−T

T

∫

The integrand is nonzero only when ≤ t. This suggests three regions of consideration: t < −T ,−T ≤ t < T and T ≤ t.Case 1: t < −T . y(t) = 0Case 2: −T ≤ t < T .

y(t) = K e−a (t− )

−T

t

∫ d =Ka

1− e−a (t+T )[ ] , for −T ≤ t < T .

Case 3: T ≤ t.

y(t) = K e−a (t− )

−T

T

∫ d =Ka

e−at eaT − e−aT( ) , for T ≤ t.

In sum,

y(t) =

0, t < −TKa

1− e−a (t+T )[ ], −T ≤ t < T

K

ae−at eaT − e−aT( ), T ≤ t

SOLUTION 16.11.(a) First observe, from figure P16.11(a), that

f2( t) = (−2t + 4) u(t) − u( t − 2)[ ]By definition

f3(t) = f1( t − ) f2( )d = 4u( t − )(−2 + 4) u( ) − u( − 2)[ ]d = −8 ( − 2)u( t − )d0

2

∫−∞

∞

∫−∞

∞

∫

The integrand is nonzero only when ≤ t. This suggests three regions of consideration: t < 0,0 ≤ t < 2 and 2 ≤ t.Case 1: t < 0. Here u( t − ) = 0 due to the fact that is restricted to the interval [0,2]. Hence

f3(t) = 0, for t < 0.


Case 2: 0 ≤ t < 2.

f3(t) = −8 ( − 2)d = −82

2− 2

0

t

∫0

t

= −4 t2 − 4t( ) , for 0 ≤ t < 2.

Case 3: 2 ≤ t.

f3(t) = −8 ( − 2)d = −82

2− 2

0

2

∫0

2

= 16, for 2 ≤ t.

In sum,

f3(t) =0, t < 0

−4 t2 − 4t( ), 0 ≤ t < 2

16, 2 ≤ t

A picture of f3(t) is sketched in the next figure.

(b) First observe, from figure P16.11(b), that

f2( t) = t u( t) − u( t − 2)[ ] + (4 − t) u(t − 2) − u( t − 4)[ ]By definition

f3(t) = f1( t − ) f2( )d−∞

∞

∫By replacing f1(t) and f2( t) with their expressions we have

f3(t) = 4u(t − ) u( ) − u( − 2)[ ] + (4 − ) u( − 2) − u( − 4)[ ] −∞

∞

∫ d =

= 4u( t − ) u( ) − u( − 2)[ ]d + 4u(t − )−∞

∞

∫ (4 − ) u( − 2) − u( − 4)[ ]−∞

∞

∫ d =

= 4 u(t − )d + 4 (4 − )u(t − )2

4

∫0

2

∫ d


The integrands are nonzero only when ≤ t. This suggests four regions of consideration: t < 0,0 ≤ t < 2, 2 ≤ t < 4, and 4 ≤ t.Case 1: t < 0. Here u( t − ) = 0 due to the fact that is restricted to the interval [0,4]. Hence

f3(t) = 0, for t < 0.

Case 2: 0 ≤ t < 2. here observe that the second integral is zero since, for this integral, is restrictedto the interval [2,4]. Therefore

f3(t) = 4 d = 2t2

0

t

∫ , for 0 ≤ t < 2.

Case 3: 2 ≤ t < 4.

f3(t) = 4 d + 4 (4 − )2

t

∫0

2

∫ d = 8 + 4 4 −2

2

2

t

= −2t2 +16t −16, for 2 ≤ t < 4.

Case 4: 4 ≤ t.

f3(t) = 4 d + 4 (4 − )2

4

∫0

2

∫ d = 8 + 4 4 −2

2

2

4

=16, for 4 ≤ t.

In sum,

f3(t) =

0, t < 0

2t2, 0 ≤ t < 2

−2t2 +16t −16, 2 ≤ t < 4

16, 4 ≤ t


SOLUTION 16.12. (a) By voltage division,


H(s) =Vout

Vin=

2

s

2s + 5 + 2

s

=1

s2 + 2.5s +1=

2 / 3

(s + 0.5)−

2 / 3

(s + 2)

Hence, the impulse response is

h(t) =2

3e−0.5tu(t) −

2

3e−2 tu(t)

(b) By definition

h(t) * vin (t) = h(t −τ )vin (τ)−∞

∞

∫ dτ =20

3e−0.5(t−τ)u(t −τ )

−∞

∞

∫ eτu(−τ) dτ

−20

3e−2(t−τ)u(t −τ )

−∞

∞

∫ eτu(−τ) dτ

Case 1: t ≤ 0.

vout (t) =20

3e−0.5(t−τ)

−∞

t

∫ eτ dτ −20

3e−2(t−τ)

−∞

t

∫ eτdτ =20

3e−0.5t

−∞

t

∫ e1.5τ dτ −20

3e−2t

−∞

t

∫ e3τdτ

=20e−0.5t

4.5e1.5τ[ ]−∞

t−

20e−2t

9e3τ[ ]−∞

t= 4.444et − 2.222et = 2.222et

Case 2: t > 0.

vout (t) =20e−0.5t

4.5e1.5τ[ ]−∞

0−

20e−2t

9e3τ[ ]−∞

0=

20e−0.5t

4.5−

20e−2t

9

SOLUTION 16.13.(a) The impulse response of the circuit has been computed in problem 16.12

h(t) =23

e−0.5t ( t) −23

e−2tu(t)

By the impulse response theorem and the convolution definition

vout (t) = h(t) ∗vin (t) = h(t − )vin ( )d =−∞

∞

∫


=203

e−0.5(t− ) − e−2(t− )[ ]u(t − )e−| |d−∞

∞

∫

The integrand is nonzero only when ≤ t. Hence

vout (t) =203

e−0.5(t− ) − e−2(t− )[ ]e−| |d−∞

t

∫The existence of the function e−| | under the integral suggests two regions of consideration: t ≤ 0and 0 < t.Case 1: t ≤ 0.

vout (t) =203

e−0.5(t− ) − e−2(t− )[ ]e d−∞

t

∫ =

=203

e−0.5t e1.5

−∞

t

∫ d −203

e−2 t e3 d =−∞

t

∫

=204.5

e−0.5t e1.5[ ]−∞

t−

209

e−2t e3[ ]−∞

t=

= 4.444et − 2.222et = 2.222et , for t ≤ 0.Case 2 : 0 < t.

vout (t) =203

e−0.5(t− ) − e−2(t− )[ ]e d−∞

0

∫ +203

e−0.5(t− ) − e−2(t− )[ ]e− d0

t

∫ =

=203

e−0.5t+1.5 − e−2t+ 3[ ]d−∞

0

∫ +203

e−0.5t−0.5 − e−2 t+[ ]d0

t

∫ =

=203

e−0.5t+1.5

1.5−

e−2t+ 3

3

−∞

0

+203

e−0.5t−0.5

−0.5− e−2 t+

0

t

=

= 17.778e−0.5t − 20e−t + 4.444e−2 t , for 0 < t.

In sum,

vout (t) =2.222et , 0 ≤ t

17.778e−0.5t − 20e−t + 4.444e−2t , 0 < t

SOLUTION 16.14.(a) The impulse response of the circuit is obtained by taking the inverse Laplace transform of H(s)

h(t) = 2e−t − 2e−2t + 4e−4 t( )u( t)

(b) The result follows from the following MATLAB code:


>> p = [-1,-2,-4];>> r = [2,-2,4];>> k = 0;>> [n,d] = residue(r,p,k)n =

4 14 16

d =

1 7 14 8

Therefore,

H(s) =4s2 +14s +16

s3 + 7s2 +14s + 8(c) By the impulse response theorem

y(t) = u( t) ∗ h(t) = u(t)* 2e−tu(t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]Using the distributive property of convolution we have

y(t) = u( t) ∗ 2e−tu(t)[ ] + u( t) ∗ −2e−2tu( t)[ ] + u(t) ∗ 4e−4 tu(t)[ ]In problem P16.2(c) the following equation has been obtained

K1u(t)[ ]∗ K2e−atu(t)[ ] =K1K2

a(1− e−at )u(t)

Using the above equation y(t)is immediately obtained

y(t) = 2 1− e−t( )u( t) − 1− e−2t( )u(t) + 1− e−4 t( )u(t) =

= 2 − 2e−t + e−2t − e−4 t( )u(t)

(d) By the impulse response theorem

y(t) = f ( t) ∗ h(t) = 8u(− t) − 8u(− t − T )[ ]* 2e−tu(t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]Using the distributive property of convolution we have

y(t) = 8u(− t) ∗ 2e−tu(t) − 2e−2tu( t) + 4e−4 tu( t)[ ]−

−8u(−t − T )* 2e−tu(t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]We denote


y1( t) = 8u(−t) ∗ 2e−tu( t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]By the time invariance property it follows that

y(t) = y1( t) − y1( t − T )

In order to compute y1( t) we will use the following equation which has been obtained in problem16.2(e)

K1u(−t)[ ]∗ K2e−atu( t)[ ] =

K1K2

a, t ≤ 0

K1K2

ae−at , t > 0

Therefore,

y1( t) =16, t ≤ 0

16e−t − 8e−2 t + 8e−4 t , 0 < t

The zero-state response to the input f ( t) can now be computed

y(t) =0, t ≤ 0

16e−t − 8e−2t + 8e−4 t −16, 0 < t ≤ T

16 e−t − e−(t−T )[ ] − 8 e−2t − e−2(t−T )[ ] + 8 e−4 t − e−4(t−T )[ ], T < t

SOLUTION 16.15.(a) Using the convolution theorem the transfer function of the cascade is

H(s) = L h(t)[ ] = L h1( t) ∗ h2( t) ∗ h3( t)[ ] = L h1( t)[ ]⋅ L h2( t)[ ]⋅ L h3( t)[ ] = H1(s) ⋅ H2(s) ⋅ H3(s)

From table 13.1

H1(s) =1s

H2(s) =10

s + 2

H3(s) =2

s2

Therefore,

H(s) =20

s3(s + 2)

A partial fraction expansion of H(s) can be obtained using the residue command in MATLAB:

>> num = [20];>> den = [1 2 0 0 0];>> [r,p,k] = residue(num,den)


r =

-2.5000 2.5000 -5.0000 10.0000

p =

-2 0 0 0

k =

[]

Hence

H(s) =−2.5s + 2

+2.5s

+−5

s2 +10

s3

Taking the inverse Laplace transform yields the impulse response of the cascade

h(t) = −2.5e−2tu( t) + 2.5u(t) − 5tu( t) + 5t2u(t)

(b) By the impulse response theorem and the convolution theorem, the Laplace transform of thestep response of the cascade equals

Y(s) = H (s) ⋅U (s) =20

s3(s + 2)⋅1s

=20

s4 (s + 2)

A partial fraction expansion of H(s) can be obtained using the residue command in MATLAB:

>> num = [20];>> den = [1 2 0 0 0 0];>> [r,p,k] = residue(num,den)

r =

1.2500 -1.2500 2.5000 -5.0000 10.0000

p =

-2 0 0


0 0

k =

[]

Hence

Y(s) =1.25s + 2

+−1.25

s+

2.5

s2 +−5

s3 +10

s4

Taking the inverse Laplace transform yields the step response of the cascade

y(t) = 1.25e−2 tu(t) −1.25u(t) + 2.5tu(t) − 2.5t2u(t) +1.667t3u( t) .

SOLUTION 16.16.(a) By the voltage division formula

Vout (s) =

1

Cs

R +1

Cs

Vi (s) =1

CRs +1Vin (s)

Therefore, the transfer function of the circuit is


=1

CRs +1=

1s +1

Taking the inverse Laplace transform yields the impulse response h(t) = e−tu(t).(b) From table 13.1 the Laplace transform of vin (t) is

Vin (s) =1s

+1

(s +1)2

By the impulse response theorem and the convolution theorem it follows that

Vout (s) = H (s) ⋅Vin (s) =1

s +11s

+1

(s +1)2

=1

s(s +1)+

1

(s +1)3

A partial fraction expansion of Vout (s) is

Vout (s) =1s

−1

s +1+

1

(s +1)3

Taking the inverse Laplace transform yields


vout (t) = u( t) − e−tu(t) − 0.5t2e−tu( t).

Using the time domain convolution method vout (t) can be computed as follows

vout (t) = h( t − )vin ( )d−∞

∞

∫ =

= e−(t− )u( t − ) u( ) + e− u( )[ ]d−∞

∞

∫From the experience earned by computing convolution integrals we know that the computation of theabove integral requires more computational work than the Laplace transform method. Morecomputations imply, of course, more sources of errors.From the solution of this problem we have seen that, in the case of the Laplace transform method,the computational burden consists in computing Laplace and inverse Laplace transforms. For alarge class of functions these transforms can be found in tables(for example table 13.1). The onlycomputation that we did, in the solution of this problem, was the partial fraction expansion ofVout (s) .

(c) In this case the (unilateral) transform method cannot be used because vin (t) ≠ 0 for t < 0.

SOLUTION 16.17.(a) The impulse response can be obtained by taking the inverse Laplace transform of H(s) .Therefore

h(t) = 8e−10 tu(t)

(b) From table 13.1

Vin (s) =8

s2 +16


Vout (s) = H (s) ⋅Vin (s) =64

(s +10)(s2 +16)

Vout (s) can be further written as

Vout (s) =−0.5517s + 5.517

s2 +16+

0.5517s +10

= −0.5517s

s2 +16+1.379

4

s2 +16+

0.5517s +10

The above expansion of Vout (s) can be obtained by using the technique of example 13.14, page

514. Taking the inverse Laplace transform yields

vout (t) = −0.5517cos(4 t)u( t) +1.379sin(4 t)u(t) + 0.5517e−10tu( t).


(c) In this case

Vin (s) = L 2e−2t sin(4t)u(t)[ ] =8

(s + 2)2 +16Therefore

Vout (s) = H (s) ⋅Vin (s) =64

(s + 2)2 +16[ ](s +10)

Using again the technique of example 13.14, page 514, Vout (s) can be written as

Vout (s) =−0.8s + 4.8

(s + 2)2 +16+

0.8s +10

= −0.8s + 2

(s + 2)2 +16+1.6

4

(s + 2)2 +16+

0.8s +10


vout (t) = −0.8e−2t cos(4 t)u(t) +1.6e−2t sin(4 t)u(t) + 0.8e−10 tu(t) V.

In this context the Laplace transform method is faster than the time domain convolution. This isdue to the fact that vin (t) has a relatively complicated expression and the convolution integral willrequire more computational work than the Laplace transform method.

(d) In this context the Laplace transform method cannot be used because vin (t) ≠ 0 for t < 0. Thetime domain convolution method will be used to compute the response vout (t). By the impulseresponse thorem

vout (t) = h(t) ∗vin (t) = 8e−10 tu(t)[ ] ∗ u(− t)[ ]

Using the result of problem 16.2, part (e), it follows that

vout (t) =0.8, t ≤ 0

0.8e−10t , 0 < t

SOLUTION 16.18.(a) Replacing R1, R2 , C1 and C2 with their values the transfer function can be obtained

H(s) =s

s2+5s + 2

The only zero of H(s) is 0 and the poles of H(s) are –0.5 and –2. A partial fraction expansion ofH(s) is:

H(s) =−0.167s + 0.5

+0.667s + 2

The impulse response can be obtained by taking the inverse Laplace transform of H(s)

h(t) = −0.167e−0.5t + 0.667e−2t


(b) vout (t) will be computed using the Laplace transform method. This approach is valid becauseh(t) and vin (t) are zero for t < 0.From table 13.1 the Lapace transform of vin (t) is

Vin (s) =1

(s + 2)2


Vout (s) = H (s) ⋅V (s) =s

2s2 + 5s + 2

1

(s + 2)2 =s

2s4 +13s3 + 30s2 + 28s + 8

A partial fraction expansion of Vout (s) can be obtained using the residue command in MATLAB:>> a = [1 0];>> b = [2 13 30 28 8];>> [r,p,k] = residue(a,b)

r = 0.0741 0.1111 0.6667 -0.0741

p = -2.0000 -2.0000 -2.0000 -0.5000

k = []Therefore,

Vout (s) =−0.0741s + 0.5

+0.0741s + 2

+0.1111

(s + 2)2 +0.6667

(s + 2)3


vout (t) = −0.0741e−0.5t + 0.0741e−2t + 0.1111te−2t + 0.3333t2e−2 t[ ]u( t) V.

One would prefer the time domain convolution method to compute vout (t), but the computationsmay require more work relatively to the Laplace transform method.

(c) In this part vin (t) ≠ 0 for t < 0. Therefore the time domain convolution method will be used tocompute vout (t).By the impulse response theorem

vout (t) = h(t) ∗vin (t) =

= −0.1667e−0.5(t− ) + 0.6667e−2(t− )[ ]u−∞

∞

∫ (t − )e2 u(− )d =


= −0.1667e−0.5(t− ) + 0.6667e−2(t− )[ ]−∞

0

∫ u(t − )e2 d

The integrand of the previous integral is nonzero only when ≤ t. This suggests two regions ofconsideration: t < 0 and 0 ≤ t.Case 1: t < 0.

vout (t) = −0.1667e−0.5t e2.5

−∞

t

∫ d + 0.6667e−2 t e4

−∞

t

∫ d =

= −0.1667e−0.5t e2.5

2.5

−∞

t

+ 0.6667e−2t e4

4

−∞

t

=

= 0.1e−2t , for t < 0.Case 2: 0 ≤ t.

vout (t) = −0.1667e−0.5t e2.5

−∞

0

∫ d + 0.6667e−2 t e4

−∞

0

∫ d =

= −0.1667e−0.5t e2.5

2.5

−∞

0

+ 0.6667e−2t e4

4

−∞

0

=

= −0.1667e−0.5t + 0.6667e−2 t , for 0 ≤ t.

SOLUTION 16.19.Replacing Rand C with their values

H(s) =s − 5s + 5

= 1−10

s + 5

The zero-state response vout (t) will be computed using the time domain convolution methodbecause vin (t) ≠ 0 for t < 0.The impulse response of the circuit is

h(t) = ( t) −10e−5tu( t)By the impulse response theorem

vout (t) = h(t) ∗vin (t) = h( )vin ( t − )d =−∞

∞

∫

= 10 ( ) −10e−5 u( )[ ]co−∞

∞

∫ s10( t − )[ ]d =

= 10 ( )co−∞

∞

∫ s( t − )d −10 10e−5 u( )co−∞

∞

∫ s 10(t − )[ ]d

Using the sifting property of the delta function and expanding cos(t − )it follows that


vout (t) = 10cos( t − ) −100cos(10t) e−5 co0

∞

∫ s(10 )d −100sin(10t) e−5

0

∞

∫ sin(10 )d

Using the definition of the Laplace transform we observe that

e−5 cos0

∞

∫ (10 )d = L cos(10t)u(t)[ ]s=5 =s

s2 +100

s=5

= 0.04

and

e−5 sin0

∞

∫ (10 )d = L sin(10t)u( t)[ ]s=5 =10

s2 +100

s=5

= 0.08

Thereforevout (t) = 10cos(10t) − 4cos(10 t) − 8sin(10t)

= 6cos( t) − 8sin(10t) =10cos 10t + tan−1(43)[ ]

Notice that vout (t) and vin (t) have the same frequency and magnitude.

SOLUTION 16.20.(a) First notice that vin (t − T) = u(t). Therefore w(t) = u( t) and

W (s) =1s

(b) Using the properties of the Laplace transform it follows that

Voutw (s) = H (s) ⋅W (s) =

2s(s + 2)

A partial fraction expansion of Voutw (s) is

Voutw (s) =

1s

−1

s + 2


voutw (t) = u( t) − e−2 tu(t)

(b) Sincevin (t) = w( t + T )

it follows, by the time invariance property, that

voutv (t) = vout

w (t + T ).Therefore,

voutv (t) = u( t + T ) − e−2(t+T )u(t + T ) V.

SOLUTION 16.21.(a) First observe from figure P16.21 that

vin (t) = u(t + T ) − u(t − T )From the definition of w(t) it follows that


w(t) = vin ( t − T ) = u(t) − u(t − 2T )Therefore

W (s) =1s

−1s

e−2sT

(a) By the impulse response theorem and the convolution theorem it follows that

Voutw (s) = H (s) ⋅W (s) =

2s(s + 2)

1− e−2sT( ) =1s

−1

s + 2

1− e−2sT( ) =

=1s

−1

s + 2−

1s

−1

s + 2

e

−2sT

Taking the inverse Laplace transform and using the time shift property of the Laplace transformyields

voutw (t) = 1− e−2 t( )u( t) − 1− e−2(t−2T )[ ]u( t − 2T )

Becausevin (t) = w( t + T )

it follows, by the time invariance property, that

voutv (t) = vout

w (t + T ) == 1− e−2(t+T )( )u(t + T ) − 1− e−2(t−T )[ ]u( t − T ) V.

SOLUTION 16.22. (a) The use of t = t + T1 in the problem statement means replace t by t + T1 .

However, strictly speaking we should have used a statement of the form t = t ' +T1 which is done in

the proof below. By definition of the convolution and the property of commutivity,

f (t − T1) * g(t) = f (t − T1 −τ )g(τ) dτ−∞

∞

∫ = f (t' −τ)g(τ) dτ−∞

∞

∫

t'=t−T1

= f (t' ) * g(t' )[ ]t'=t−T1

Observe that t = t ' +T1 . Hence

f (t − T1)* g( t)[ ]t=t'+T1= f (t' ) * g(t' )

Realizing that t and t' are simply dummy variables, we immediately obtain the result. From a

systems perspective, this corresponds to the property of time-invariance where a shift of an input

function by T1 yields a corresponding shift of the output function by T1.

(b) The steps in this part are similar to those of part (a).


f (t − T1) * g(t − T2) = f (t − T1 −τ )g(τ − T2) dτ−∞

∞

∫

Let λ = τ − T2 in which case τ =λ + T2 and dλ = dτ . Hence

f (t − T1) * g(t − T2) = f (t − T1 − T2 −λ )g(λ)dλ−∞

∞

∫ = f (t' −λ)g(λ )dλ−∞

∞

∫

t'=t−T1−T2

= f (t' )* g( t')[ ]t'=t−T1−T2

Since t = t ' +T1 + T2 , and t and t' are dummy variables, we have

f (t − T1)* g( t − T2 )[ ]t=t'+T1+T2= f (t' )* g(t')

and the result follows.

SOLUTION 16.23. From table 13.1, G(s) =1

s + 2, H(s) =

1

(s + 2)2 .

(a) Define p(t) = f(t – 2) = u(t).

Then, P(s) =1

s. Consider

P(s)G(s) =1

s(s + 2)=

0.5

s−

0.5

s + 2

Hence

p(t) * g(t) = 0.5u(t) − 0.5e−2 tu(t)

From problem 16.22 part (a),

f (t)* g(t) = p(t) * g(t)[ ]t=t+2 = 0.5u(t) − 0.5e−2tu(t)[ ]t=t+2= 0.5 − 0.5e−2(t+2)[ ]u(t + 2)

(b) Define p(t) = f(t – 2) = u(t).

Then, P(s) =1

s. Consider

P(s)H(s) =1

s(s + 2)2 =0.25

s−

0.25

s + 2−

0.5

(s + 2)2

Hence


p(t) * h(t) = 0.25 − 0.25e−2t − 0.5te−2t( )u(t)

From problem 16.22 part (a),

f (t)* h( t) = p(t)* h( t)[ ]t=t+2 = 0.25 − 0.25e−2 t − 0.5te−2t( )u(t)[ ]t=t+2

= 0.25 − 0.25e−2(t+2) − 0.5(t + 2)e−2(t+2)[ ]u(t + 2)

SOLUTION 16.24.(a) The pictures of f ( t) and g( t) are sketched in the next figures

(b) Using the convolution theorem it follows thatL f (t − 2 ) ∗ g( t − )[ ] = L f (t − 2 )[ ]⋅L g( t − )[ ]

From table 13.1

L f (t − 2 )[ ] = L u( t)[ ] =1s

L g(t − )[ ] = L sin(t)u( t)[ ] =1

s2 +1Therefore

L f (t − 2 ) ∗ g( t − )[ ] =1

s(s2 +1)=

1s

−s

s2 +1Taking the inverse Laplace transform yields

f ( t − 2 ) ∗ g(t − ) = u(t) − cos( t)u(t) = 1− cos(t)[ ]u(t)

Using the property give in problem 16.22(b) it follows that

f ( t) ∗ g( t) = f (t − 2 ) ∗ g( t − )[ ]t=t+2 + = 1− cos(t + 3 )[ ]u(t + 3 )


SOLUTION 16.25.Define

w(t) = vin ( t − 2)Hence

w(t) = u( t)and, from table 13.1,

W (s) =1s

From table 13.1 we also have that

H1(s) =1

s +1 and H2(s) =

1

(s +1)2

The impulse response of the cascade ish(t) = h1(t) ∗ h2(t)

Hence the transfer function of the cascade is

H(s) = H1(s)H2(s) =1

(s +1)3

We denote by voutw (t)the zero state response due to the input w(t). Hence,

Voutw (s) = H (s) ⋅W (s) =

1

s(s +1)3

A partial fraction expansion of Voutw (s) is obtained using the residue command in MATLAB:

>> a = [1];>> b = [1 3 3 1 0];>> [r,p,k] = residue(a,b)

r = -1.0000 -1.0000 -1.0000 1.0000

p = -1.0000 -1.0000 -1.0000 0

k = []

Therefore

Voutw (s) =

1s

+−1

s +1+

−1

(s +1)2 +−1

(s +1)3



voutw (t) = 1− e−t − te−t − 0.5t2e−t[ ]u(t) V

Due to the fact thatvin (t) = w( t + 2)

the time invariance property implies that

vout (t) = voutw (t + 2) =

= 1− e−(t+ 2) − ( t + 2)e−(t+2) − 0.5(t + 2)2e−(t+ 2)[ ]u(t + 2) V.

SOLUTION 16.26.(a) Using the sifting property of the delta function it follows that

f4(t) = (t) + (t − 4)[ ] ∗ f2(t) = f2(t) + f2(t − 4)

The right-hand side of the above equation interprets as a graphical sum of shifted pictures of f2( t) .A picture of f4(t) is sketched in the next figure.

0 1 2 3 4 5 6 70

2

4

(b) In order to compute the area beneath f2( t − ) ⋅ f2( ) four regions will be considered: t < 0,0 ≤ t <1, 1 ≤ t < 2 and 2 ≤ t.Step 1: t < 0. In this case f2( t − ) ⋅ f2( ) = 0 for all . Therefore

f2( t) ∗ f2( t) = 0 for t < 0.

Step 2: 0 ≤ t <1. In this case f2( t − ) ⋅ f2( ) = 16 for 0 ≤ ≤ t and is zero elsewhere. The areabeneath f2( t − ) ⋅ f2( ) equals 16t . Therefore

f2( t) ∗ f2( t) =16t for 0 ≤ t <1.

Step 3: 1 ≤ t < 2. In this case f2( t − ) ⋅ f2( ) = 16 for t −1 < ≤1 and is zero elsewhere. Hencethe area beneath f2( t − ) ⋅ f2( )equals

f2( t) ∗ f2( t) =16(2 − t) for 1 ≤ t < 2.Step 4: 2 ≤ t. In this case f2( t − ) ⋅ f2( ) = 0 for all . Therefore

f2( t) ∗ f2( t) = 0 for2 ≤ t.

A picture of f5(t)is sketched in the next figure.


(c) In order to compute the area beneath f2( t − ) ⋅ f3( ) five regions will be considered: t < 0,0 ≤ t <1, 1 ≤ t < 2, 2 ≤ t < 3 and 3 ≤ t.Step 1: t < 0. In this case f2( t − ) ⋅ f3( ) = 0 for all . Therefore f2( t) ∗ f3(t) = 0 for t < 0.Step 2 : 0 ≤ t <1. In this case f2( t − ) ⋅ f3( ) = 8 for 0 ≤ ≤ t and is zero elsewhere. Thereforethe area beneath f2( t − ) ⋅ f3( ) equals

f2( t) ∗ f3( ) = 8t for 0 ≤ t <1.Step 3: 1 ≤ t < 2. Here

f2( t − ) ⋅ f3( ) =8, t −1 < <1

24, 1 ≤ < t

0, otherwise

Hence, the area beneath f2( t − ) ⋅ f3( ) equals

f2( t) ∗ f3( ) = 81 − (t −1)[ ] + 24( t −1) = 8(2t −1) for 1 ≤ t < 2.

Step 4: 2 ≤ t < 3. In this case f2( t − ) ⋅ f3( ) = 24 for t −1 < < 2 and is zero otherwise. Hence,the area beneath f2( t − ) ⋅ f3( ) equals

f2( t) ∗ f3(t) = 24 2 − (t −1)[ ] = 24(3− t) for 2 ≤ t < 3.

Step 5: 3 ≤ t. Here f2( t − ) ⋅ f3( ) = 0 for all . Therefore f2( t) ∗ f3(t) = 0 for t < 0.A picture of f6( t) is sketched in the next figure.


SOLUTION 16.27.By the impulse response theorem, it follows that the response is

y(t) = h(t) ∗ f (t) == h( t) ∗ (t) − (t −1)[ ]

Using the distributive property of convolution and the sifting property of delta function y(t) can bewritten as

y(t) = h(t) − h(t −1)The right-hand side of the above equation interprets as a graphical sum of (shifted) pictures of h(t). The pictures of h(t), h(t −1) and y(t) are sketched in the next figures.

SOLUTION 16.28.(a) From the picture of f ( t) and h(t) in figure P16.28 we observe that, in order to compute thearea beneath h(t − ) f ( ) , we need to consider four cases: t < 0, 0 ≤ t < 4, 4 ≤ t < 8 and 8 ≤ t.Step 1: t < 0. Here h(t − ) f ( ) = 0 for all . Therefore the area beneath h(t − ) f ( ) equals zeroand

h(t) ∗ f (t) = 0 for t < 0.Step 2: 0 ≤ t < 4. In this case h(t − ) f ( ) =1 for 0 ≤ ≤ t and is zero otherwise. Hence the areabeneath h(t − ) f ( ) equals

h(t) ∗ f (t) = t for 0 ≤ t < 4.Step 3: 4 ≤ t < 8. In this case

h(t − ) f ( ) =1, t − 4 < ≤ 4

2, 4 ≤ ≤ t

0, otherwise

Therefore the area beneath h(t − ) f ( ) equalsh(t) ∗ f (t) = 4 − ( t − 4)[ ] + 2(t − 4) = t for 4 ≤ t < 8.


Step 4: 8 ≤ t. Here h(t − ) f ( ) = 2 for t − 4 < ≤ t and is zero otherwise. Henceh(t) ∗ f (t) = 2 t − (t − 4) = 8[ ] for 8 ≤ t.


(b) The impulse response ish(t) = u( t) − u(t − 4)

By the impulse response theorem

y(t) = x(t) ∗ h( t) = x( )h( t − )d−∞

∞

∫ = x( ) u(t − ) − u(t − − 4)[ ]−∞

∞

∫ d

Here observe that u( t − ) − u(t − − 4) is nonzero only when t ≤ < t − 4.Therefore

y(t) = x( )dt

t−4

∫which interprets as the running area under x(t) over the interval [t − 4,t].

SOLUTION 16.29.The response, y(t), is obtained as indicated in the statement of the problem, by using the followingMATLAB code:

>> tstep = 1;>> vin = [1];>> h = [0, 2, 3, 1, 1];>> y = tstep*conv(vin, h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(vin)+length(h));>> plot(t,y)>> grid

The response is plotted in the next figure.


SOLUTION 16.30.A picture of vin (t) sketched in the next figure.

In order to plot the response, y(t), the MATLAB code of problem 16.29 will be used with only onemodification. Namely

vin = [1, 1, 2, 2]

as it can be observed from the picture of vin (t) with the time step tstep = 1.Therefore the MATLAB code is:

>> tstep = 1;>> vin = [1, 1, 2, 2];>> h = [0, 2, 3, 1, 1];>> y = tstep*conv(vin,h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(vin)+length(h));>> plot(t,y)>> grid

The response is plotted in the next figure.


SOLUTION 16.31.In order to compute the area beneath v(t − )h( ) seven regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, 2 ≤ t < 3, 3 ≤ t < 4, 4 ≤ t < 5 and 5 ≤ t.

Step 1: t < 0. For t in this region v(t − )h( ) = 0 for all . Hence

y(t) = v(t) ∗ h( t) = 0 for t < 0.

Step 2: 0 ≤ t <1. In this case v(t − )h( ) = v0 × h0 for 0 ≤ ≤ t and is zero otherwise. Therefore thearea beneath v(t − )h( ) equals

y(t) = v(t) ∗ h( t) = v0 × h0 × t for 0 ≤ t <1.

Step 3: 1 ≤ t < 2. For t in this region we have

v(t − )h( ) =

v1× h0, 0 ≤ ≤ t −1

v0 × h0, t −1< < 1

v0 × h1, 1≤ < t

0, otherwise

Therefore the area beneath v(t − )h( ) equals

y(t) = v(t) ∗ h( t) = v1× h0 × ( t −1)− 0[ ] + v0 × h0 × 1− ( t −1)[ ] + v0 × h1× ( t −1)== t × (v1× h0 − v0 × h0 + v0 × h1)− v1× h0 + 2 × v0 × h0 − v0 × h1, for 1 ≤ t < 2

Step 4: 2 ≤ t < 3. In this case

v(t − )h( ) =

v1× h0, t − 2 < < 1

v1× h1, 1≤ ≤ t −1

v0 × h1, t −1< < 2

v0 × h2, 2 ≤ < t

0, otherwise

Hence, for 2 ≤ t < 3,y(t) = v(t) ∗ h( t) =

= v1× h0 × 1− (t − 2)[ ] + v1× h1× (t −1)−1[ ] + v0 × h1× 2 − ( t −1)[ ] + v0 × h2 × (t − 2) == t × (−v1× h0 + v1× h1− v0 × h1+ v0 × h2) + 3× v1× h0 − 2 × v1× h1 + 3 × v0 × h1− 2 × v0 × h2

Step 5: 3 ≤ t < 4. In this case

v(t − )h( ) =

v1× h1, t − 2 < < 2

v1× h2, 2 ≤ ≤ t −1

v0 × h2, t −1< < 3

0, otherwise

Hence, for 3 ≤ t < 4,y(t) = v(t) ∗ h( t) =

= v1× h1× 2 − (t − 2)[ ] + v1× h2 × (t −1)− 2[ ] + v0 × h2 × 3− ( t −1)[ ] == t × (−v1× h1 + v1× h2 − v0 × h2) + 4 × v1× h1− 3 × v1× h2 + 4 × v0 × h2

Step 6: 4 ≤ t < 5. In this case v(t − )h( ) = v1× h2 for t − 2 < < 3 and is zero otherwise. Thereforey(t) = v(t) ∗ h( t) = v1× h2 × 3− ( t − 2)[ ] = v1× h2 × (5 − t) for 4 ≤ t < 5.


Step 7: 5 ≤ t. For t in this region v(t − )h( ) = 0 for all . Hence

y(t) = v(t) ∗ h( t) = 0 for 5 ≤ t.In sum,

y( t) =

v0 × h0 × t , 0 ≤ t < 1

t × (v1 × h0 − v0 × h0 + v0 × h1) − v1 × h0 + 2 × v0 × h0 − v0 × h1, 1 ≤ t < 2

t × (−v1 × h0 + v1 × h1 − v0 × h1 + v0 × h2) + 3 × v1 × h0 − 2 × v1 × h1 + 3 × v0 × h1 − 2 × v0 × h2, 2 ≤ t < 3

t × (−v1 × h1 + v1 × h2 − v0 × h2) + 4 × v1 × h1 − 3 × v1 × h2 + 4 × v0 × h2, 3 ≤ t < 4

v1 × h2 × (5 − t), 4 ≤ t < 5

0, otherwise

Hence,

y1 = y(1) = v0 × h0 = 6y2 = y(2) = v0 × h1+ v1× h0 = 8y3 = y(3) = v0 × h2 + v1× h1 = −6y4 = y(4) = v1× h2 = 4

(b) Using the expressions of p(x) and q(x) it follows that

p(x) × q(x) = x 3 × (v0 × h0) + x 2 × (v0 × h1+ v1× h0) + x × (v0 × h0 + v1× h1)+ v1× h2We observe that the coefficients of p(x) × q(x) are exactly y1 , y2 , y3 and y4 , respectively. Therefore

r(x) = p(x) × q(x) .

SOLUTION 16.32.(a) This part will be solved using the techniques of convolution algebra. Therefore we can write f3(t) as

f3(t) = f1(−1)(t) ∗ f2

(1)( t)Where the superscript (-1) means integration and the superscript (1) means differentiation. From figureP16.32 we observe that

f1(t) = 4 u(t) − u(t − 4)[ ]Hence

f1(−1)(t) = 4 tu(t) − (t − 4)u(t − 4)[ ] =

= 4 r( t) − r(t − 4)[ ]

By inspection, from the same figure, we have

f2(1)( t) = 4 (t) − 2 ( t − 2) + 2 (t − 4) − 2 ( t − 6) + ( t − 8)[ ]

Using the sifting property of the delta function f3(t) can be computed as follows

f3(t) = 4 r(t) − r(t − 4)[ ] ∗ 4 ( t) − 2 (t − 2) + 2 ( t − 4) − 2 (t − 6) + ( t − 8)[ ] == 16 r( t) − 2r(t − 2) + r( t − 4) − r(t − 8) + 2r( t −10) − r(t −12)[ ]



(b) Using the techniques of problem 16.31 and considering the time step tstep = 2, the polynomials p(x) ,q(x) and r(x) can be associated with the functions f1(t), f2( t) andf3(t), respectively, as below:

p(x) = 4x + 4

q(x) = 4 x3 − 4 x2 + 4 x − 4

r(x) = 32x4 − 32

We need to verify that the equalityp(x) ⋅q(x) ⋅tstep = r(x)

holds. The equality indeed holds because

p(x) ⋅q(x) ⋅tstep = 32(x +1)(x 3 − x 2 + x −1) = 32(x4 −1)= r(x).

The results obtained in part (a) and part (b) coincide.

SOLUTION 16.33.(a) Using the techniques of convolution algebra f3(t) can be written as

f3(t) = f1(−1)(t) ∗ f2

(1)( t)

Where the superscript (-1) means integration and the superscript (1) means differentiation. From figureP16.33 we observe that

f1(t) = 2 u(t +1)− u( t − 4)[ ]Therefore

f1(−1)(t) = 4 (t +1)u(t +1)− ( t − 4)u( t − 4)[ ] =

= 4 r( t +1)− r( t − 4)[ ] = g(t)

By inspection, from the same figure, we have

f2(1)( t) = 4 (t) − 8 (t − 2) + 6 (t − 4) − 2 (t − 6)

Using the sifting property of the delta function f3(t) can be computed as follows

f3(t) = 4g( t) − 8g( t − 2) + 6g( t − 4) − 2g( t − 6)


This is plotted in MATLAB as follows:

>> t = -2:0.01:14;>> g = 2*(t+1).*u(t+1)-2*(t-4).*u(t-4);>> g1 = 4*g;>> g2 = -8*( 2*(t-1).*u(t-1)-2*(t-6).*u(t-6) );>> g3 = 6*( 2*(t-3).*u(t-3)-2*(t-8).*u(t-8) );>> g4 = -2*( 2*(t-5).*u(t-5)-2*(t-10).*u(t-10) );>> f3 = g1+g2+g3+g4;>> plot(t,f3);>> grid;


(b) To account for the fact that f1(t)is nonzero for negative t the following formula (see problem 16.22)f1(t) ∗ f2(t) = f1(t −1)∗ f2(t)[ ]t=t+1

will be used to compute f3(t). Using a slightly modified version of the code of problem 16.31, we have

>> f1 = [2, 2, 2, 2, 2];>> f2 = [4, 4, -4, -4, 2, 2];>> T = 1;>> tstep = T;>> f3 = tstep*conv(f1,f2);>> f3 = [0 f3 0];>> t = -1:tstep:tstep*(length(f1)+length(f2))-1;>> plot(t,f3)>> grid


The results of parts (a) and (b) coincide.

SOLUTION 16.34. This problem is solved using the techniques of the convolution algebra with the

graphical method left to the student.

f3(t) = f1(t) * f2(t) = f1(t)[ ](−1)* f2(t)[ ](1)

where the superscript (-1) means integration and the superscript (1) means differentiation. By inspection,

f1(t )[ ](−1) = 4tu(t) − 4(t − 6)u(t − 6) = g(t)

and

f2(t)[ ](1) = 4δ(t) − 8δ( t − 2) + 8δ(t − 6) − 4δ(t − 8)

Hence the response say y(t) satisfies

f3(t) = 4g(t) −8g( t − 2) +8g(t − 6) − 4g(t −8)

This is plotted in MATLAB as follows:

»t=0:.05:20;

»g = 4*t .* u(t) - 4*(t-6) .*u(t-6);

»g1=4*g;

»g2 = -8*(4*(t-2) .* u(t-2) - 4*(t-8) .*u(t-8));

»g3 = 8*(4*(t-6) .* u(t-6) - 4*(t-12) .*u(t-12));

»g4 = -4*(4*(t-8) .* u(t-8) - 4*(t-14) .*u(t-14));

»f3 = g1+g2+g3+g4;

»plot(t,f3)

»grid


0 2 4 6 8 10 12 14 16 18 20-40

-30

-20

-10

0

10

20

30

40

(b) Using the code of problem 31, we have

»f1 = [4 4 4];

»f2 = [4 -4 -4 4];

»T = 2;

»tstep = T;

»f3 = [0 conv(f1,f2)*tstep 0];

»t = 0: tstep : tstep* (length(f1) + length(f2));

»plot(t,f3)

»grid


0 2 4 6 8 10 12 14-40

-30

-20

-10

0

10

20

30

40


SOLUTION 16.35.(a) In order to compute the area beneath f1(t − ) ⋅ f1( ) four regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, and 2 ≤ t.Step 1: t < 0. Here f1(t − ) ⋅ f1( ) = 0 for all . Hence

f3(t) = f1(t) ∗ f1( t) = 0 for t < 0.

Step 2: 0 ≤ t <1. In this case f1(t − ) ⋅ f1( ) = 1 for 0 ≤ ≤ t and is zero otherwise. Therefore the areabeneath f1(t − ) ⋅ f1( ) equals

f3(t) = t for 0 ≤ t <1.

Step 3: 1 ≤ t < 2. In this case f1(t − ) ⋅ f1( ) = 1 for t −1 < <1 and is zero otherwise. Therefore thearea beneath f1(t − ) ⋅ f1( ) equals

f3(t) = 1− (t −1)= 2 − t for 1 ≤ t < 2.

Step 4: 2 ≤ t. Here f1(t − ) ⋅ f1( ) = 0 for all . Hence

f3(t) = f1(t) ∗ f1( t) = 0 for2 ≤ t.In sum,

f3(t) =t, 0 ≤ t <1

2 − t, 1 ≤ t < 2

0, otherwise


(b) In order to compute the area beneath f1(t − ) ⋅ f2( ) four regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, and 2 ≤ t.Step 1: t < 0. Here f1(t − ) ⋅ f2( ) = 0 for all . Hence

f4(t) = f1(t) ∗ f2(t) = 0 for t < 0.

Step 2: 0 ≤ t <1. In this case f1(t − ) ⋅ f2( ) = for 0 ≤ ≤ t and is zero otherwise. Therefore the areabeneath f1(t − ) ⋅ f2( ) equals

f4(t) = 0.5t2 for 0 ≤ t <1.

Step 3: 1 ≤ t < 2. For t in this region f1(t − ) ⋅ f2( ) = for t −1 < <1 and is zero otherwise.Therefore the area beneath f1(t − ) ⋅ f2( ) equals

f4(t) = 0.5 − 0.5t2 for 1 ≤ t < 2.

Step 4: 2 ≤ t. Here f1(t − ) ⋅ f2( ) = 0 for all . Hence

f4(t) = f1(t) ∗ f1( t) = 0 for2 ≤ t.

In sum,

f4(t) =0.5t2, 0 ≤ t < 1

0.5 − 0.5t2, 1≤ t < 2

0, otherwise

(c) In order to compute the area beneath f2( t − ) ⋅ f2( ) four regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, and 2 ≤ t.

Step 1: t < 0. Here f2( t − ) ⋅ f2( ) = 0 for all . Hence

f5(t) = f2( t) ∗ f2( t) = 0 for t < 0.

Step 2: 0 ≤ t <1. In this case f2( t − ) ⋅ f2( ) = (t − ) for 0 ≤ ≤ t and is zero otherwise. Thereforethe area beneath f2( t − ) ⋅ f2( ) equals

f5(t) = ( t − ) d = −0.3333 3 + 0.5t 2[ ]0

t=

0

t

∫ 0.1667t3 for 0 ≤ t <1.

Step 3: 1 ≤ t < 2. For t in this region f2( t − ) ⋅ f2( ) = (t − ) for t −1 < <1 and is zero otherwise.Therefore the area beneath f2( t − ) ⋅ f2( ) equals

f5(t) = (t − ) d =t−1

1

∫ −0.3333 3 + 0.5t 2[ ]t−1

1

= −0.1667t3 + t − 0.6667 for 1 ≤ t < 2.

Step 4: 2 ≤ t. Here f2( t − ) ⋅ f2( ) = 0 for all . Hence


f5(t) = f2( t) ∗ f2( t) = 0 for2 ≤ t.In sum,

f5(t) =0.1667t3, 0 ≤ t <1

−0.1667t3 + t − 0.6667, 1 ≤ t < 2

0, otherwise

SOLUTION 16.36.In order to compute the area beneath f1(t − ) ⋅ f2( ) five regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, 2 ≤ t < 3and 3 ≤ t.

Step 1: t < 0. Here f1(t − ) ⋅ f2( ) = 0 for all . Hencef3(t) = f1(t) ∗ f2(t) = 0 for t < 0.

Step 2: 0 ≤ t <1. In this case f1(t − ) ⋅ f2( ) = for 0 ≤ ≤ t and is zero otherwise. Therefore the areabeneath f1(t − ) ⋅ f2( ) equals

f3(t) = 0.5t2 for 0 ≤ t <1.

Step 3: 1 ≤ t < 2. For t in this region

f1(t − ) ⋅ f2( ) =, t −1< < 1

2 − , 1≤ ≤ t

0, otherwise

Therefore the area beneath f1(t − ) ⋅ f2( ) equals

f3(t) = 0.5 − 0.5( t −1)2[ ] + 0.5 − 0.5(2 − t)2[ ]= −t2 + 3t −1.5 for 1 ≤ t < 2.

Step 4: 2 ≤ t < 3. For t in this region f1(t − ) ⋅ f2( ) = 2 − for all t −1 < < 2. Hence

f3(t) = f1(t) ∗ f2(t) = 0.5 − 0.5(2 − t)2 = −0.5t2 + 2t −1.5 for 2 ≤ t < 3.

Step 5: 3 ≤ t. Here f1(t − ) ⋅ f2( ) = 0 for all . Hencef3(t) = f1(t) ∗ f2(t) = 0 for 3 ≤ t.

In sum,

f3(t) =

0.5t2, 0 ≤ t < 1

− t2 + 3t −1.5, 1≤ t < 2

−0.5t2 + 2t −1.5, 2 ≤ t < 3

0, otherwise

SOLUTION 16.37.


(a) In order to compute the area beneath f1( ) ⋅ f2( t − ) three regions will be considered: t < 0,0 ≤ t < 2, and 2 ≤ t.Step 1: t < 0. Here f1( ) ⋅ f2( t − ) = 0 for all . Hence

f3(t) = f1(t) ∗ f2(t) = 0 for t < 0.

Step 2: 0 ≤ t < 2. In this case f1( ) ⋅ f2( t − ) = 8 for 0 ≤ ≤ t and is zero otherwise. Therefore thearea beneath f1( ) ⋅ f2( t − ) equals

f3(t) = 0.5(t ⋅ 8t) = 4t2 for 0 ≤ t < 2.

Step 3: 2 ≤ t. For t in this region

f1( ) ⋅ f2( t − ) =8 , 0 ≤ < 2

16, 2 ≤ < t

0, otherwise

Therefore the area beneath f1( ) ⋅ f2( t − ) equals

f3(t) = 16 +16( t − 2) =16( t −1) for 2 ≤ t.In sum,

f3(t) =4t2, 0 ≤ t < 2

16( t −1), 2 ≤ t

0, otherwise


(b) From figure P16.37 we observe that

f1(t) = 2tu(t) − 2( t − 2)u(t − 2)

From table 13.1 and the time shift property of the Laplace transform it follows that

F1(s) =2

s2 1− e−2s( )


F2(s) =4s

By the convolution theoremF3(s) = F1(s)F2(s)

Therefore

F3(s) =8

s3 1− e−2s( )Taking the inverse Laplace transform yields

f3(t) = 4t2u(t) − 4( t − 2)2u( t − 2)


SOLUTION 16.38.(a) In order to compute the area beneath f1(t − ) ⋅ f2( ) six regions will be considered: t < 0, 0 ≤ t < 2,2 ≤ t < 6, 6 ≤ t < 8, 8 ≤ t <10, and 10 ≤ t.Step 1: t < 0. Here f1(t − ) ⋅ f2( ) = 0 for all . Hence

f3(t) = f1(t) ∗ f2(t) = 0 for t < 0.

Step 2: 0 ≤ t < 2. In this case f1(t − ) ⋅ f2( ) = 8( t − ) for 0 ≤ ≤ t and is zero otherwise. Thereforethe area beneath f1( ) ⋅ f2( t − ) equals

f3(t) = 0.5(t ⋅ 8t) = 4t2 for 0 ≤ t < 2.

Step 3:2 ≤ t < 6. For t in this region

f1(t − ) ⋅ f2( ) =16, 0 ≤ < t − 2

8(t − ), t − 2 ≤ < t

0, otherwise

Therefore the area beneath f1(t − ) ⋅ f2( ) equals

f3(t) = 16 +16( t − 2) =16( t −1) for 2 ≤ t < 6.

Step 4: 6 ≤ t < 8. For t in this region

f1(t − ) ⋅ f2( ) =

16, 0 ≤ < t − 2

8(t − ), t − 2 ≤ < 6

−8( t − ), 6 ≤ < t

0, otherwise

Therefore, for 6 ≤ t < 8, the area beneath f1(t − ) ⋅ f2( ) equals

f3(t) = 16(t − 2) + 16 −8(t − 6)2

2

−

8( t − 6)2

2


= −8(t − 6)2 +16t −16 = −8t2 +112t − 304

Step 5: 8 ≤ t <10. For t in this region

f1(t − ) ⋅ f2( ) =

16, 0 ≤ < 6

−16, 6 ≤ < t − 2

−8( t − ), t − 2 ≤ < 8

0, otherwise

Therefore, for 8 ≤ t <10, the area beneath f1(t − ) ⋅ f2( ) equals

f3(t) = 96 −16(t − 8) − 16 −8(t − 8)2

2

= 4t2 − 80t + 464

Step 6: 10 ≤ t. For t in this region

f1(t − ) f2( ) =16, 0 ≤ < 6

−16, 6 ≤ < 8

0, otherwise

Therefore,f3(t) = 96 − 32 = 64 for 10 ≤ t.

In sum,

f3(t) =

0, t < 0

4t2, 0 ≤ t < 2

16t −16, 2 ≤ t < 6

−8t2 +112t − 304, 6 ≤ t < 8

4t2 − 80t + 464, 8 ≤ t < 10

64, 10 ≤ t


(b) From figure P16.38 we observe that


f2( t) = 4 u( t) − u( t − 6)[ ] − 4 u(t − 6)u − (t − 8)[ ]= 4u(t) − 8u( t − 6) + 4u(t − 8)

From table 13.1 and the time shift property of the Laplace transform it follows that

F2(s) =4s

−8s

e−6s +4s

e−8s

From problem 16.37 we have that

F1(s) =2

s2 1− e−2s( )By the convolution theorem

F3(s) = F1(s)F2(s)Therefore

F3(s) =8

s3 −8

s3 e−2s −16

s3 e−6s +24

s3 e−8s −8

s3 e−10s


f3(t) = 4t2u(t) − 4( t − 2)2u( t − 2) − 8( t − 6)2 u(t − 6)

+12(t − 8)2 u(t − 8) − 4( t −10)2u( t −10)


SOLUTION 16.39.p( t) ∗ q(t) will be computed using the techniques of convolution algebra. Therefore we can write

p( t) ∗ q(t) = p (−1)( t) ∗ q(1) (t)

where the superscript (−1)means integration and the superscript (1)means differentiation. By inspection

p( t) = ( t + 4) u(t + 4) − u( t)[ ] + (− t + 4) u( t) − u(t − 4)[ ] + 4 u( t − 4) − u(t − 8)[ ] == (t + 4)u( t + 4) + (−2t)u(t) + tu(t − 4) − 4u( t − 8)

Therefore,

p(−1)(t) = 0.5(t + 4)2 u(t + 4) − t2u( t) + (0.5t2 − 8)u( t − 4) − (4t − 32)u(t − 8)

By inspection we also have

q (1)(t) = 4 ( t)

By the sifting property of the delta function it follows that

p( t) ∗ q(t) = 4 p (−1)( t) == 2(t + 4)2 u(t + 4) − 4t2u(t) + 2( t2 −16)u(t − 4) −16(t − 8)u( t − 8).

SOLUTION 16.40.(a) First observe that


h(t) = 0.1u( t − 0.1) + 0.2u(t − 0.2) + 0.2u(t − 0.3) + 0.2u(t − 0.4) + 0.2u( t − 0.5) ++0.1u( t − 0.6) − 0.1u(t −1)− 0.2u(t −1.3)− 0.2u(t −1.5) −

−0.2u(t −1.7) − 0.2u( t − 2) − 0.1u( t − 2.2)Due to the fact that h(t) is a linear combination of terms of the type Ku(t − T ), the convolution of h(t)and vin (t) reduces to a linear combination of terms of the following type: K1u(t)[ ]∗ K2u( t − T )[ ] . Usingthe definition of the convolution, the previous convolution product is computed below

K1u(t)[ ]∗ K2u( t − T )[ ] = K1K2 u( )u(t − − T )d−∞

∞

∫ =

= K1K2 u(t − − T )d0

∞

∫ =

0, t < T

K1K2 d0

t−T

∫ , T ≤ t

=0, t < T

K1K2( t − T ), T ≤ t

= K1K2( t − T )u( t − T )

Therefore vout (t) is a linear combination of functions of type K1K2( t − T )u(t − T ),

vout (t) = h(t) ∗vin (t) == 10(t − 0.1)u(t − 0.1)+ 20(t − 0.2)u( t − 0.2) + 20(t − 0.3)u( t − 0.3) + 20( t − 0.4)u(t − 0.4) +

+20(t − 0.5)u( t − 0.5) +10(t − 0.6)u(t − 0.6) −10(t −1)u(t −1)− 20(t −1.3)u( t −1.3) − −20(t −1.5)u(t −1.5) − 20(t −1.7)u(t −1.7) − 20(t − 2)u(t − 2) −10(t − 2.2)u(t − 2.2)

At t = 0svout (0) = 0 V.

At t = 0.5svout (0.5) = 16 V.

At t =1svout (1)= 65 V.

At t =1.5svout (1.5)= 106 V.

(b) In this case vout (t) will be computed using the techniques of the convolution algebra. Hence we have

vout (t) = vin ( t) ∗ h(t) = vin(−1)(t) ∗ h (1)(t)

= 50t2u(t)[ ]∗ 0.1 ( t − 0.1) + 0.2 ( t − 0.2) + 0.2 (t − 0.3) + 0.2 ( t − 0.4) +[+0.2 ( t − 0.5) + 0.1 ( t − 0.6) − 0.1 ( t −1)− 0.2 (t −1.3)−

−0.2 ( t −1.5) − 0.2 (t −1.7)− 0.2 (t − 2) − 0.1 ( t − 2.2)]

Using the sifting property of delta function it follows that

+−−+−−+−−= )3.0()3.0(10)2.0()2.0(10)1.0()1.0(5)( 222 tuttuttuttvout

+10(t − 0.4)2u( t − 0.4) +10( t − 0.5)2 u(t − 0.5) + 5( t − 0.6)2u(t − 0.6) −−5(t −1)2u( t −1)−10( t −1.3)2u( t −1.3) −10(t −1.5)2u( t −1.5) −

−10(t −1.7)2u( t −1.7) −10( t − 2)2 u(t − 2) − 5(t − 2.2)2u( t − 2.2) V.


At t =1svout (1)= 22.25 V.

(c) From the expression of vout (t)obtained in part (a) we observe that vout (t)does not change aftert = 2.2s . Therefore it is sufficient to compute vout (t)for t ≤ 2.2s . Hence vin (t) can be considered to beequal to

vin (t) =100 u( t) − u(t − 2.2)[ ] V.Using the code of problem 16.31 we have

>> vin = 100*ones(1,22);>> h = [0, 0.1, 0.3, 0.5, 0.7, 0.9, 1, 1, 1, 1, 0.9, 0.9, 0.9, 0.7, 0.7, 0.5, 0.5, 0.3, 0.3, … …0.3, 0.1, 0.1];>> T = 0.1;>> tstep = T;>> y = tstep*conv(vin,h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(h)+length(vin));% After t = 2.2s vout(t) does not change>> t = t(1:length(h)+1);>> y = y(1:length(h)+1);>> plot(t,y)>> grid


Using the previous MATLAB code we the values of y(t) at the specified instants of time are:At t = 0s

y(0) = 0 V


At t = 0.5sy(0.5) = 16 V

At t =1sy(1)= 65 V

At t =1.5sy(1.5)= 106 V

The results of parts (a) and (c) coincide.

SOLUTION 16.41.Using the MATLAB code of problem 16.31 we have:

>> vin = [1];>> h = [9, -6, 3, -2];>> T = 1;>> tstep = T;>> y = tstep*conv(vin,h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(h)+length(vin));>> plot(t,y)>> grid

The breakpoints in y(t) of the above figure are [9, -6, 3 –2] as expected because the polynomialassociated with vin (t) is the constant 1 and the polynomial associated with h(t) is the polynomial

9x3 − 6x2 + 3x − 2, as it can be observed from figure P16.41.

SOLUTION 16.42.(a) Let vout ,40( t) denote the response that has been obtained in problem 16.40, part (a), to the input100u( t) .The expression of vout ,40( t) is (see problem 16.40, part (a)):

vout ,40( t) = h( t) ∗ 100u( t)[ ] == 10(t − 0.1)u(t − 0.1)+ 20(t − 0.2)u( t − 0.2) + 20(t − 0.3)u( t − 0.3) + 20( t − 0.4)u(t − 0.4) +


+20(t − 0.5)u( t − 0.5) +10(t − 0.6)u(t − 0.6) −10(t −1)u(t −1)− 20(t −1.3)u( t −1.3) −−20(t −1.5)u(t −1.5) − 20(t −1.7)u(t −1.7) − 20(t − 2)u(t − 2) −10(t − 2.2)u(t − 2.2)

Using the distributive property of the convolution product and the time invariance property it follows thatvout (t) = h(t) ∗vin (t) = h( t) ∗ 100u(t) −100u(t − 0.2)[ ] =

= h( t) ∗ 100u(t)[ ] − h(t) ∗ 100u( t − 0.2)[ ] == vout,40 (t) − vout,40 (t − 0.2)

Using the above expression of vout ,40( t) we have:vout ,40(0) = 0 V and vout ,40(−0.2) = 0V,

vout ,40(0.5) =16 V and vout ,40(0.3) = 4 V,vout ,40(1) = 65 V and vout ,40(0.8) = 45V,

vout ,40(1.5) =106 V and vout ,40(1.3)= 92V.

At t = 0svout (0) = vout ,40(0) − vout ,40(−0.2) = 0 V.

At t = 0.5svout (0.5) = vout ,40(0.5) − vout ,40(0.3) =12 V.

At t =1svout (1)= vout ,40(1)− vout ,40(0.8) = 20 V.

At t =1.5svout (1.5)= vout ,40(1.5)− vout ,40(1.3) =14 V.

(b) In this case vout (t) will be computed using the techniques of convolution algebra.We have

vout (t) = vin ( t) ∗ h(t) = vin(−1)(t) ∗ h (1)(t)

where the superscript (-1) means integration and the superscript (1) means differentiation.From figure P16.42 observe that

vin (t) =100t u( t) − u(t − 0.5)[ ] +100(1− t) u(t − 0.5) − u( t −1)[ ] == 100tu( t) +100(1− 2t)u(t − 0.5) +100(t −1)u( t −1)

Therefore

vin(−1)( t) = 50t2u(t) + (−100t2 +100t − 25)u( t − 0.5) + (50 t2 −100t + 50)u(t −1) = g(t)

By the sifting property of the delta function we have

vout (t) = vin(−1)(t) ∗ h (1)( t) = g( t) ∗ h (1) (t)

= g(t) ∗ 0.1 ( t − 0.1) + 0.2 (t − 0.2) + 0.2 (t − 0.3) + 0.2 (t − 0.4) +[+0.2 ( t − 0.5) + 0.1 ( t − 0.6) − 0.1 ( t −1)− 0.2 (t −1.3)−

−0.2 ( t −1.5) − 0.2 (t −1.7)− 0.2 (t − 2) − 0.1 ( t − 2.2)]=

= 0.1g(t − 0.1)+ 0.2g(t − 0.2) + 0.2g( t − 0.3) + 0.2g(t − 0.4) +

+0.2g( t − 0.5) + 0.1g( t − 0.6) − 0.1g( t −1)− 0.2g(t −1.3)−


−0.2g(t −1.5) − 0.2g( t −1.7) − 0.2g( t − 2) − 0.1g(t − 2.2)

The values ofvout (t) at the specified instants of time can be computed using MATLAB. The results are:vout (0) = 0V

vout (0.5) = 2.2Vvout (1)= 17.85V

vout (1.5)= 23.3V.

SOLUTION 16.43.

Using the techniques of convolution algebra we have

vout (t) = h(t) ∗vin (t) = h (−1)(t) ∗vin(1)(t)

where the superscript (-1) means integration and the superscript (1) means differentiation.We have

h (−1)( t) = 2(1− e−2t )u(t)and

vin(1)(t) = ( t) − ( t −1)

Using the sifting property of the delta function it follows that

vout (t) = 2(1− e−2 t )u(t)[ ]∗ ( t) − (t −1)[ ] =

= 2(1− e−2t )u( t) − 2(1− e−2(t−1))u( t −1)A picture of vout (t)is sketched in the next figure.

SOLUTION 16.44.(a) From table 13.1 it follows that the Laplace transform of vin (t) is

Vin (s) =1s

−1

s +1−

1s + 2

=−s2 + 2

s(s +1)(s + 2)


And the Laplace transform of vout (t) is

Vout (s) =1s

−2

s +1−

1

(s +1)2 +1

s + 2=

−s2 + 2

s(s +1)2(s + 2)

Therefore the transfer function of the circuit is


=1

s +1

A simple RC circuit that represents this transfer function is a series RC circuit with R =1Ω and C = 1F .vout (t)is represented by the capacitor voltage and vin (t) is the source voltage.

(b) The impulse response of the circuit is

h(t) = L−1 H (s)[ ] = L−1 1s +1

= e−tu( t).

(c) Assuming zero initial conditions we have

vout (t) = vin ( t) ∗ h(t) = vin(1)( t) ∗ h (−1)(t) = (t) ∗ (1− e−t )u( t)[ ] = (1− e−t )u( t) V.

(d) Using the techniques of convolution algebra the zero-state response can be computed as

vout (t) = vin ( t) ∗ h(t) = vin(2) (t) ∗ h (−2)(t)

where the superscript (2) means double differentiation and the superscript (-2) means double integration.First, from figure P16.44, observe that

vin(1)(t) = u(t −1)− u(t − 2)[ ] + u(t − 3)− u(t − 4)[ ]

Therefore

vin(2)( t) = (t −1)− (t − 2) + (t − 3) − ( t − 4)

h (−2) (t) is computed as the integral of h (−1)( t) .

h (−2) (t) = h (−1)( )d−∞

t

∫ = (1− e− )u( )d−∞

t

∫ = (1− e− )d

0−

t

∫ = (t + e−t −1)u(t)

The zero-state response can now be computed

vout (t) = vin(2) (t) ∗ h (−2)( t) =

= ( t −1)− (t − 2) + (t − 3)− (t − 4)[ ]∗ ( t + e−t −1)u( t)[ ]By the sifting property of the delta function it follows that

vout (t) = t − 2 + e−(t−1)( )u(t −1)− t − 3 + e−(t−2)( )u( t − 2)

+ t − 4 + e−(t−3)( )u(t − 3)− t − 5 + e−(t−4)( )u( t − 4) V.


SOLUTION 16.45.Using the techniques of the convolution algebra we have

y(t) = f ( t) ∗ g( t) = f (2) (t) ∗ g(−2)( t)where

g(−1)(t) = 2 cos( )d = sin( t)u( t)

0−

t

∫and

g(−2)( t) = 2 sin( )d = 1− cos( t)[ ]u( t)

0−

t

∫Differentiating f ( t) twice leads to

f (2) (t) = − ( t) + 2 (t −1)− 2 ( t − 3)+ (t − 4)Therefore

y(t) = f (2) (t) ∗ g(−2)( t) == − ( t) + 2 ( t −1)− 2 ( t − 3) + ( t − 4)[ ]∗ 1− cos( t)[ ]u(t)


y(t) = − 1− cos( t)[ ]u( t) + 2 1− cos[ (t −1)] u(t −1)−2 1− cos[ (t − 3)] u( t − 3) + 1− cos[ ( t − 4)] u(t − 4)

Simplifying the expression of y(t) yields

y(t) = − 1− cos( t)[ ]⋅ u(t) − u( t − 4)[ ] + 2 ⋅ 1 + cos( t)[ ]⋅ u(t −1)− u(t − 3)[ ] .



SOLUTION 16.46.(a) By the current division formula

IC (s) =Cs

Cs +1

Ls

⋅ Iin (s) =s2

s2 +1⋅ Iin (s)

The transfer function of the circuit can be computed as below

H(s) =VC (s)Iin (s)

=IC (s) ⋅ ZC (s)

Iin (s)=

s2

s2 +1⋅1s

=s

s2 +1

(b) The impulse response is computed as the inverse Laplace transform of the transfer function

h(t) = L−1 H (s)[ ] = cos(t)u( t).

(c) Assuming zero initial conditions it follows, by the impulse response theorem, that

vout (t) = iin (t) ∗ h(t)

Using the techniques of the convolution algebra we have

vout (t) = iin(2) (t) ∗ h (−2) (t)

By inspection

iin(1)( t) = u( t) − u( t − 2 )[ ] + u(t − 4 ) − u( t − 6 )[ ]

Therefore

iin(2) (t) = ( t) − ( t − 2 ) + (t − 4 ) − ( t − 6 )

And

h (−1)( t) = cos( )d = sin(t)u( t)

0−

t

∫Hence

h (−2) (t) = sin( )d = 1− cos(t)[ ]u(t)

0−

t

∫

Using the sifting property of the delta function we have

vout (t) = (t) − (t − 2 ) + ( t − 4 ) − (t − 6 )[ ] ∗ 1− cos(t)[ ]u( t) = 1− cos(t)[ ]u(t) − 1− cos(t − 2 )[ ]u(t − 2 )

+ 1− cos(t − 4 )[ ]u(t − 4 ) − 1− cos(t − 6 )[ ]u(t − 6 )= 1− cos(t)[ ]⋅ u( t) − u( t − 2 ) + u(t − 4 ) − u( t − 6 )[ ]


V.(d) A picture of vout (t) is sketched in the next figure.

SOLUTION 16.47.(a) The step response, vout (t), is computed using the convolution algebra techniques.We have

vout (t) = h(t) ∗vin (t) = h (1)( t) ∗ vin(−1)(t)

From figure P16.47 observe that

h(t) = 2u( t) − u(t −1)− 2u(t − 2) − u( t − 3)+ u(t − 5) + 2u(t − 6) + u( t − 7) − 2u( t − 8)

Differentiating we have

h (1)(t) = 2 ( t) − ( t −1)− 2 (t − 2) − (t − 3) + (t − 5) + 2 (t − 6) + ( t − 7) − 2 ( t − 8)Since

vin (t) = u(t),by integration it follows that

vin(−1)( t) = tu(t).


vout (t) = h (1) (t) ∗ vin(−1)( t)

= 2 ( t) − (t −1)− 2 (t − 2) − ( t − 3) + (t − 5) + 2 (t − 6) + (t − 7) − 2 ( t − 8)[ ] ∗ tu( t)[ ]= 2tu( t) − ( t −1)u( t −1)− 2( t − 2)u( t − 2) − ( t − 3)u(t − 3) +

+( t − 5)u(t − 5) + 2(t − 6)u( t − 6) + (t − 7)u( t − 7) − 2(t − 8)u(t − 8)V.

A picture of the step response is sketched in the next figure.


(b) Using the convolution algebra techniques we have

vout (t) = h(t) ∗vin (t) = h (1)( t) ∗ vin(−1)(t)

For computing vin(−1)( t) we have

vin(−1)( t) = e u(− )d =

e d−∞

t

∫ , t < 0

e d−∞

0

∫ , 0 ≤ t

−∞

t

∫ =et , t < 0

1, 0 ≤ t


vout (t) = h (1) (t) ∗ vin(−1)( t) =

= 2 ( t) − (t −1)− 2 (t − 2) − ( t − 3) + (t − 5) + 2 (t − 6) + (t − 7) − 2 ( t − 8)[ ] ∗ vin(−1)(t)[ ] =

= 2vin(−1)(t) − vin

(−1)(t −1)− 2vin(−1)(t − 2) − vin

(−1)( t − 3) + +vin

(−1)(t − 5) + 2vin(−1)( t − 6) + vin

(−1)(t − 7) − 2vin(−1)(t − 8) V.

Using the expression of vin(−1)( t) computed above it follows that

vout (7.5) = 0.7869 V,vout (6.5) = 1.1603V,vout (5.5) = 0.2720V,vout (0.5) = 0.8848V.

SOLUTION 16.48.First observe, from figure P16.48(a), that

h(t) = (1− t) ⋅ u(t) − u(t −1)[ ] =


= (1− t)u(t) − (1− t)u(t −1).Using the convolution algebra techniques it follows that the step response y(t) is

y(t) = v(t) ∗ h( t) = v(1) (t) ∗ h (−1)( t) == (t) ∗ h (−1)(t) = h (−1)(t)

where

h (−1)( t) = (1− )u( )d − (1− )u( −1)d =−∞

t

∫−∞

t

∫

= − 0.5 2[ ]0

t⋅ u(t) − − 0.5 2[ ]1

t⋅ u( t −1) =

= (t − 0.5t2)u( t) − ( t − 0.5t2 − 0.5)u(t −1).Hence the step response is

y(t) = (t − 0.5t2)u(t) − ( t − 0.5t2 − 0.5)u( t −1).

Let y v( t) denote the zero-state response to the input v(t).From figure P16.48(b) we observe that

v(t) = u( t) + u( t −1)− 2u( t − 2) .Using the distributive property of the convolution it follows that

y v( t) = h( t) ∗ v(t) == h( t) ∗ u(t) + u(t −1)− 2u(t − 2)[ ] =

= h( t) ∗ u( t) + h(t) ∗ u( t −1)− 2 ⋅ h( t) ∗ u(t − 2).Due to the fact that

y(t) = h(t) ∗ u(t)by the linearity and time invariance properties it follows that

y v( t) = y( t) + y(t −1)− 2y( t − 2) = = (t − 0.5t2)u( t) + (t −1)u(t −1)+

+(1.5t2 − 8t +12)u(t − 2) + (0.5t2 − 3t + 4)u( t − 3).

SOLUTION 16.49.(a) By the voltage division formula it follows that

Vout (s) =

1

Cs1

Cs+ Ls

⋅Vin (s) =1

LCs2 +1⋅Vin (s)

Therefore the transfer function is


=1

LCs2 +1Taking the inverse Laplace transform yields

h(t) = L−1 1

LCs2 +1

=

1

LCL−1

1

LC

s2 +1

LC

=1

LCsin

1

LCt

u( t) .

(b) The step response is computed as the convolution of the impulse response and the step function.vout (t) = h(t) ∗ u( t)

Using the techniques of the convolution algebra it follows that


vout (t) = h (−1)(t) ∗ u(1) (t)where the superscript (-1) means integration and the superscript (1) means differentiation.Taking the integral of h(t) we have

h (−1)( t) =1

LCsin

1

LC

u( )d

−∞

t

∫ =

= u(t) ⋅ −cos1

LC

0

t

= 1− cos1

LCt

⋅ u( t) .

Therefore

vout (t) = h (−1)(t) ∗ u(1) (t) =

= 1− cos1

LCt

⋅ u(t)

∗ (t) =

= 1− cos1

LCt

⋅ u( t) V.

(c) We denote by voutT (t) the output to the rectangular pulse in figure P16.49(b).

Observe, from figure P16.49(b), that

vin (t) =1T

⋅ u( t) − u(t − T )[ ] V.

By linearity and time invariance it follows that

voutT (t) =

1T

⋅ vout (t) − vout (t − T )[ ]where vout (t)is the step response obtained in part (b). Therefore

voutT (t) =

1

2 LC1− cos

1

LCt

⋅ u(t) −

1

2 LC1− cos

1

LCt − 2

⋅ u( t − 2 LC )

=1

2 LC1− cos

1

LCt

⋅ u(t) − u( t − 2 LC[ ]

A picture of voutT (t), for L = 1H and C = 1F , is sketched in the next figure.


SOLUTION 16.50.The impulse response of the configuration in the figure P16.50 is

h(t) = h1(t) ∗ h2(t) + h3( t)[ ] ∗ h4( t)Due to the fact that h4(t) = 2 ( t) , the sifting property of the delta function can be applied and it followsthat

h(t) = 2 ⋅ h1(t) ∗ h2(t) + h3( t)[ ]Using the techniques of the convolution algebra we can further write

h(t) = 2 ⋅ h1(1) (t) ∗ h2(t) + h3( t)[ ](−1) =

h(t) = 2 ⋅ h1(1) (t) ∗ h2

(−1)(t) + h3(−1)( t)[ ]

We have

h1(1)(t) = u (1)(t) = ( t)

h2(−1)( t) = 2e−2 u( )d = u( t) ⋅ −e−2[ ]0

t= 1− e−2t( )u(t)

−∞

t

∫

h3(−1)( t) = 8e−4 u( )d = u(t) ⋅ −2e−4[ ]0

t= 2 1− e−4 t( )u(t)

−∞

t

∫

Substituting the expressions of h1(−1)( t) , h2

(−1)( t) and h3(−1)( t) in the expression of h(t), and using the

sifting property of the delta function we have

h(t) = 2 1− e−2 t( )u( t) + 4 1− e−4 t( )u(t).

A picture of h(t) is sketched in the next figure.


SOLUTION 16.51.Observe first that h2( t) , h3(t) and h4(t) have the same expressions as in the problem 16.50. In problem16.50 the following convolution has been computed

u( t) ∗ h2(t) + h3(t)[ ]∗ h4 (t) = 2 1− e−2t( )u( t) + 4 1− e−4 t( )u(t)

Using the time shift property of the convolution (see problem 16.22, part (a)) it follows that

u( t − 2) ∗ h2(t) + h3(t)[ ]∗ h4 (t) = 2 1− e−2(t−2)( )u(t − 2) + 4 1− e−4( t−2)( )u( t − 2)

Using the above expressions and the distributive property of the convolution it follows that the overallimpulse response can be computed as below

h(t) = h1(t) ∗ h2(t) + h3( t)[ ] ∗ h4( t) = u( t) − u(t − 2)[ ]∗ h2( t) + h3(t)[ ]∗ h4( t)

= u(t) ∗ h2( t) + h3(t)[ ] ∗ h4( t) − u(t − 2)∗ h2(t) + h3( t)[ ]∗ h4 (t) =

= 21 − e−2t( )u( t) + 4 1− e−4 t( )u(t) − 2 1− e−2(t−2)( )u(t − 2) + 4 1− e−4(t−2)( )u(t − 2).



SOLUTION 16.52.The overall impulse response of the configuration is

h(t) = h1(t) ∗ h2(t) + h3( t)[ ] ∗ h4( t)

Using the distributive property of the convolution we have

h(t) = h1(t) ∗ h2(t) ∗ h4 (t) + h1( t) ∗ h3(t) ∗ h4 (t)

Replacing the expressions for h2( t) and h3(t), and using the sifting property of the delta function itfollows that

h(t) = 2 ⋅ h1(t) ∗ (t) ∗ h4 (t) − 2 ⋅ h1(t) ∗ (t − 2) ∗ h4 (t)= 2 ⋅ h1( t) ∗ h4 (t) − 2 ⋅ h1(t) ∗ h4( t)[ ]t=t−2

Using the techniques of the convolution algebra it follows that

h1(t) ∗ h4( t) = h1(1)(t) ∗ h4

(−1)(t)

where the superscript (1) means differentiation and the superscript (-1) means integration.We have

h1(1)(t) = u (1)(t) = ( t)

and

h4(−1)( t) = 2e− u( )d

−∞

t

∫ = 2u( t) e− d0

t

∫ = 2u(t) −e−[ ]0

t= 2(1− e−t )u(t)

Therefore

h1(t) ∗ h4( t) = (t) ∗ 2(1− e−t )u(t)[ ] = 2(1− e−t )u( t)


Replacing the above expression into the expression of h(t) it follows that

h(t) = 4(1− e−t )u( t) − 4 (1− e−t )u( t)[ ]t=t−2= 4(1− e−t )u( t) − 41 − e−(t−2)[ ]u(t − 2).

A picture of h(t) is sketched in the next picture.

SOLUTION 16.53.(a) The overall impulse response of the configuration in figure P16.53 is

h(t) = h1( t) + h2(t)[ ]∗ h3(t)

By the distributive property of convolution it follows that

h(t) = h1(t) ∗ h3( t) + h2(t) ∗ h3( t) = (t) ∗cos( t)u( t) + (t −1)∗cos( t)u( t)


h(t) = cos( t)u(t) + cos (t −1)[ ]u( t −1).



(b) The response y(t) is computed as

y(t) = h(t) ∗ u(t) = cos( t)u( t) + cos ( t −1)[ ]u(t −1) ∗ u(t)

Using the distributive property and the time shift property of convolution(see problem 16.22, part (a) ) wehave

y(t) = cos( t)u(t)[ ]∗ u( t) + cos ( t −1)[ ]u(t −1) ∗ u(t) == cos( t)u(t)[ ]∗ u(t) + cos( t)u(t)[ ] ∗ u(t) t=t−1

Using the techniques of convolution algebra, the convolution cos( t)u(t) ∗ u(t) is computed as

cos( t)u( t)[ ] ∗ u( t) = cos( t)u(t)[ ](−1) ∗ u (1)(t)

where the superscript (-1) means integration and the superscript (1) means differentiation.We have

cos( t)u( t)[ ](−1) = cos( )u( )d−∞

t

∫ = u(t) cos( )d =0

t

∫ u(t)1

sin( t)[ ]0t =

sin( t)u( t) .

Therefore

cos( t)u( t)[ ] ∗ u( t) =sin( t)

u( t)

∗ (t) =

sin( t)u( t)

Hence the step response is

y(t) =sin( t)

u(t) +sin (t −1)[ ]

u(t −1) =sin( t)

u(t) − u( t −1)[ ] .



h(t) = h1( t) + h2(t)[ ]∗ h3(t)


h(t) = h1(t) ∗ h3( t) + h2(t) ∗ h3( t) = (t) ∗cos( t)u( t) + (t − 3) ∗cos( t)u( t)


h(t) = cos( t)u(t) + cos (t − 3)[ ]u( t − 3).



y(t) = h(t) ∗ u(t) = cos( t)u( t) + cos ( t − 3)[ ]u(t − 3) ∗ u(t)

Using the distributive property and the time shift property of convolution(see problem 16.22, part (a) wehave

y(t) = cos( t)u(t)[ ]∗ u( t) + cos ( t − 3)[ ]u(t − 3) ∗ u(t)

= cos( t)u(t)[ ]∗ u(t) + cos( t)u(t)[ ] ∗ u(t) t=t−3



where the superscript (-1) means integration and the superscript (1) means differentiation. We have

cos( t)u( t)[ ](−1) = cos( )u( )d−∞

t

∫ = u(t) cos( )d =0

t

∫ u(t)1

sin( t)[ ]0t =

sin( t)u( t) .

Therefore


cos( t)u( t)[ ] ∗ u( t) =sin( t)

u( t)

∗ (t) =

sin( t)u( t)


y(t) =sin( t)

u(t) +sin (t − 3)[ ]

u(t − 3) =sin( t)

u(t) − u( t − 3)[ ] .


h(t) = h1( t) + h2(t) + h3( t) + h4 (t)[ ] ∗ h5( t)


h(t) = h1(t) ∗ h5( t) + h2(t) ∗ h5( t) + h3(t) ∗ h5(t) + h4( t) ∗ h5( t) = = (t) ∗cos( t)u( t) + (t −1)∗cos( t)u( t) −

− ( t − 3)∗ cos( t)u(t) − (t − 4) ∗cos( t)u(t)

By the sifting property of the delta function it follows thath(t) = cos( t)u(t) + cos (t −1)[ ]u( t −1)−

−cos (t − 3)[ ]u( t − 3)− cos (t − 4)[ ]u(t − 4) .



y(t) = h(t) ∗ u(t) = cos( t)u( t) + cos ( t −1)[ ]u(t −1)−cos (t − 3)[ ]u( t − 3)− cos (t − 4)[ ]u(t − 4)∗ u( t)


Using the distributive property and the time shift property of convolution (see problem 16.22, part (a)) wehave

y(t) = cos( t)u(t)[ ]∗ u( t) + cos ( t −1)[ ]u(t −1) ∗ u(t) − − cos (t − 3)[ ]u( t − 3) ∗ u(t) − cos ( t − 4)[ ]u( t − 4) ∗ u(t) =

= cos( t)u(t)[ ]∗ u(t) + cos( t)u(t)[ ] ∗ u(t) t=t−1 − − cos( t)u( t)[ ]∗ u( t) t=t−3 − cos( t)u( t)[ ] ∗ u(t) t=t−4



where the superscript (-1) means integration and the superscript (1) means differentiation. We have

cos( t)u( t)[ ](−1) = cos( )u( )d−∞

t

∫ = u(t) cos( )d =0

t

∫ u(t)1

sin( t)[ ]0t =

sin( t)u( t) .

Therefore

cos( t)u( t)[ ] ∗ u( t) =sin( t)

u( t)

∗ (t) =

sin( t)u( t)


y(t) =sin( t)

u(t) +sin (t −1)[ ]

u(t −1) −sin (t − 3)[ ]

u(t − 3)−sin ( t − 4)[ ]

u(t − 4)

=sin( t)

u(t) − u( t −1)+ u(t − 3) − u( t − 4)[ ] .


h(t) ∗ f (t) = h(t − ) f ( )d−∞

∞

∫By making a change of variable

1 = t −we have

h(t) ∗ f (t) = h( 1) f (t − 1)d 1−∞

∞

∫ = f ( t − 1)h( 1)d 1−∞

∞

∫By definition

f ( t) ∗ h(t) = f (t − )h( )d−∞

∞

∫From the above expressions we observe that

h(t) ∗ f (t) = f ( t) ∗ h( t)because and 1 are only variables of integration.


(b) Using the definition of convolution we have

h( t) ∗ f ( t)[ ] ∗ g( t) = h( ) f (t − )d−∞

∞

∫

∗ g(t) = h( ) f (t − 1 − )d−∞

∞

∫

g( 1)

d 1

−∞

∞

∫ =

= h( ) f (t − 1 − )g( 1)[ ]d−∞

∞

∫

d 1

−∞

∞

∫

Changing the order of integration we have

h( t) ∗ f ( t)[ ] ∗ g( t) = h( ) f (t − 1 − )g( 1)d 1−∞

∞

∫

d

−∞

∞

∫

By the definition of the convolution we have

f ( t) ∗ g( t) = f (t − )g( )d−∞

∞

∫Therefore

f (t) ∗ g(t)[ ]t=t− = f ( t − − )g( )d−∞

∞

∫Hence

h( t) ∗ f ( t)[ ] ∗ g( t) = h( ) ⋅ f ( t) ∗ g( t)[ ]t=t− d−∞

∞

∫

By the definition of the convolution product we have

h(t) ∗ f ( t) ∗ g(t)[ ] = h( ) ⋅ f ( t) ∗ g( t)[ ]t=t− d−∞

∞

∫Therefore

h( t) ∗ f ( t)[ ] ∗ g( t) = h( t) ∗ f (t) ∗ g( t)[ ] .

Thus the associative property of convolution is proved.

SOLUTION 16.57.We have

f ( t) ⋅ h( t) = f (t) (t − kT )k=0

∞∑

For some nonnegative k , the Laplace transform of f ( t) (t − kT) is

L f (t) ( t − kT)[ ] = f (kT)e−skT


by the sifting property of the delta function. Therefore we have

L f (t)h(t)[ ] = L f ( t) ( t − kT)k =0

∞∑

= = f (kT )e−skT

k =0

∞∑

Using the notation z = esT we have

L f (t)h(t)[ ] = f (kT)z−k

k =0

∞∑ .

SOLUTION 16.58.(a) By the voltage division formula we have

Vout (s) =

1

Cs

R +1

Cs

⋅Vin (s)

Therefore the transfer function of the circuit is


=

1

Cs

R +1

Cs

=1

CRs +1=

12s +1

.


h(t) = L-1 12s +1

= = L-1

1

2

s +1

2

= 0.5e−0.5tu(t).

(b) By the impulse response theorem it follows thatvout (t) = h(t) ∗vin (t) =

= 0.5e−0.5tu(t)[ ]∗ ( t) + ( t −1) + (t − 2) +K[ ]


vout (t) = 0.5e−0.5tu( t) + 0.5e−0.5(t−1)u(t −1) + 0.5e−0.5(t−2)u(t − 2) +K V.

Therefore for 0 < t <1

vout (t) = 0.5e−0.5t V

because only u( t) is nonzero for 0 < t <1.

(c) From the expression of vout (t)obtained in the part (b) it follows that


vout (t) = 0.5e−0.5t + 0.5e−0.5(t−1) V, for 1 < t < 2

because only u( t) and u( t −1) are nonzero for 1 < t < 2.

(d) For t in the interval (4,5) , only u( t) , u( t −1), u( t − 2) , u( t − 3) and u( t − 4) are nonzero. Therefore

vout (t) = 0.5e−0.5t + 0.5e−0.5(t−1) + 0.5e−0.5(t−2) + 0.5e−0.5(t−3) + 0.5e−0.5(t−4) V.

The above expression can be written as

vout (t) = 0.5e−0.5(t−4) 1+ e−0.5( ) + e−0.5( )2+ e−0.5( )3

+ e−0.5( )4

V.

(e) Using the sum formula for geometric series with = e−0.5 we have

1 + e−0.5( ) + e−0.5( )2+ e−0.5( )3

+ e−0.5( )4=

1− e−0.5( )5

1− e−0.5

vout (t) = 0.5e−0.5(t−4) ⋅1− e−0.5( )5

1− e−0.5 V.

Therefore

vout (t) = 8.6189 ⋅ e−0.5t V.

(f) Using the expression of vout (t)obtained in part (b) it follows that, for n < t < n +1,

vout (t) = 0.5e−0.5t + 0.5e−0.5(t−1) + 0.5e−0.5(t−2) +K + 0.5e−0.5(t−n) =

= 0.5e−0.5(t−n) 1+ e−0.5( ) + e−0.5( )2

+K + e−0.5( )n

V.

Using the sum formula for geometric series we have

vout (t) = 0.5e−0.5(t−n) ⋅1− e−0.5( )n+1

1− e−0.5 V.

(g) For large n , e−0.5( )n+1≅ 0. Therefore, for large n we have

vout (t) ≅ e−0.5(t−n) ⋅0.5

1− e−0.5 V for n < t < n +1.

A picture of vout (t), for large n , is sketched in the next figure.


SOLUTION 16.59.By the impulse response theorem we have

vout (t) = h(t) ∗vin (t) = 0.5e−0.5tu(t)[ ]∗ ( t) + ( t +1)+ (t + 2) +K[ ] V.


vout (t) = 0.5e−0.5tu( t) + 0.5e−0.5(t+1)u(t +1) + 0.5e−0.5(t+ 2)u(t + 2) +K V.

For 0 < t we have

vout (t) = 0.5e−0.5t + 0.5e−0.5(t+1) + 0.5e−0.5(t+ 2) +K =

= 0.5e−0.5t 1+ e−0.5( ) + e−0.5( )2

+K

V.

Using the sum formula for geometric series( for n = ∞ ) we have

vout (t) = 0.5e−0.5t ⋅1

1− e−0.5 V for 0 < t.

Simplifying the expression of vout (t) it follows that

vout (t) = 1.2707 ⋅ e−0.5t V for 0 < t.

SOLUTION 16.60. First we plot for reader convenience vin(t) and its staircase approximation.


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8vin and its Staircase Approximation

The transfer function of the circuit of figure 16.58 is H(s) =1

2s + 1 in which case

h(t) = 0.5e−0.5tu(t). Because we only want the output for 0 ≤ t ≤ 2, we only need h(t) for 0 ≤ t ≤ 2 s.

Hence we need to generate staircase approximations to both vin(t) and h(t) as follows:

t = 0:0.05:2;

vin = exp(t .^2) .* (u(t) - u(t - 1));

h = 0.5*exp( - 0.5*t) .* (u(t) - u(t - 2));

T = 0.05;

tstep = T;

y = [0 conv(vin,h)*tstep 0];

t = 0:tstep:tstep*(length(vin)+length(h));

% For plotting through time 2 s we set

t=t(1:41);

y = y(1:41);

plot(t,y)

grid


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7Convolution of vin and h

5/15/01 P17-1 R.A. DeCarlo & P.M. Lin


SOLUTION PROBLEM 17.11. Case 1: suppose R1 > R2. From example 17.3, page 696, if L

and C are connected as indicated in part (a), then Z1 can be made real and larger than R2. This

means we can solve the problem at least for Z1. Specifically, consider the figure below

From example 17.3, at a specified frequency, ωr, for which Z1 is real, then L, C, and R2 must

satisfy,

ωr =1

LC−

R22

L2 (1)

Further from example 17,3, at ωr,

Z1( jωr ) =L

R2C

We require that Z1( jωr ) = R1 in which case

R1R2 =L

C(2)

It is necessary to solve equations (1) and (2) simultaneously for L and C. From (2), L = R1R2C .

Substituting into the square of (1) yields

ωr2 =

1

R1R2C2 −R2

2

R12R2

2C2

Hence

C =1

ωrR1

R1 − R2

R2

It follows that


L = R1R21

ωr R1

R1 − R2

R2

=1

ωrR2 R1 − R2( )

Observe that since R1 > R2, both C and L are real, i.e., exist. Please note that this connection would

not result in real values of C and L had R1 < R2. If we can now show that Z2 = R2, then parts (a)

and (b) are valid for this case.

By direct computation

Z2 ( jωr) = jωr L +1

jωrC + 1

R1

= jωrL +R1

jωrCR1 +1= j R2 R1 − R2( ) +

R1

jR1 − R2

R2+1

= j R2 R1 − R2( ) +R1 − jR1

R1 − R2

R2

1+ R1 − R2

R2

= j R2 R1 − R2( ) + R2 − jR2R1 − R2

R2= R2

Thus, (a) and (b) are true for the case R1 > R2.

We can also arrive at the conclusion that Z2 = R2 using maximum power transfer concepts.

Since Z1 is constructed so that Z1 = R1, we have set up the conditions for maximum power transfer

of a V-source in series with R1 to the "load" Z1. Since the LC coupling network is lossless,

whatever average power is received by the network to the right of R1, will be dissipated by R2.

Therefore maximum power is transferred to the load R2. Looking back from R2, it must be that R2

sees a Thevenin resistance Z2 = R2 since it is known that there is a non-zero R1.

Case 2, R1 < R2 . Now consider the configuration


Interchanging the subscripts of 1 and 2 in case 1 produces the derivation for this case.

Combining cases 1 and 2 using the text notation produces

C =1

ωRlarge

Rlarge − Rsmall

RsmallL =

1

ωRsmall Rlarge − Rsmall( )

SOLUTION PROBLEM 17.16.

(a)

»C = 0.1e-6;

»R = 510;

»wr = 2*pi*1.56e3;

»%From equation 17.4 in example 17.3 which analyzes

»%this particular circuit,

»%L must satisfy

»% wr^2*L^2 - L/C + R^2 = 0.

»% Therefore

»

»L = roots([wr^2 -1/C R^2])

L =

5.3133e-02

5.0952e-02

(b) From HW problem 17.9,

ωr( )2 =1

LC−

1

R2C2

Hence, in MATLAB,

»Linv = C*(wr^2 + 1/(R^2*C^2))

Linv =

4.8054e+01

»L = 1/Linv

L = 2.0810e-02


SOLUTION PROBLEM 17.22. (Correction: change RL in figure P17.22 to R.) We solve this

problem using phasors rather than the Laplace transform approach as it is simpler. Consider

As such, since by definition ω0 = 1 LC , the resonant frequency,

VC = VL = Zin ( jω0 )Iin = RIin = RIm

Hence

vC (t) = vL (t) = RIm cos(ω0t) and iL (t) =RImω0 L

sin(ω0t).

(a) By direct computation

wC (t) =1

2CvC

2 (t) =1

2CR2Im

2 cos2(ω0t)

(b) By direct computation

wL( t) =1

2LiL

2(t) =R2Im

2

2ω02 L

sin2(ω0t)

(c) Also by direct computation, since ω02 = 1 LC ,

wC (t) + wL (t) =1

2CR2Im

2 cos2(ω0t) +R2Im

2

2ω02L

sin2(ω0t)

=1

2R2Im

2C cos2(ω0t) +1

ω02 LC

sin2(ω0t)

=1

2R2Im

2 C

(d) The energy dissipated in the resistance in one period is

wR(0,T ) = R iR2(t) dt

0

T

∫ = R iin2 (t) dt

0

T

∫ = RIm2 cos2(ω0t)dt

0

T

∫


=RIm

2

2dt

0

T

∫ +RIm

2

2cos(2ω0t) dt

0

T

∫ =RIm

2

2T =

RIm2 π

ω0

(e) Finally

2πmaximum energy stored

total energy lost per period= 2π

0.5R2Im2 C

RIm2 π

ω0

= ω0RC = Q

by equation 17.13.


(a) H1(s) =IL

Vs=

1

Zin (s)=

1

R + Ls + 1

Cs

=

1

Ls

s2 + R

Ls + 1

LC

H2(s) = H1(s) ×1

Cs=

1

VsIL ×

1

Cs

=

VC

Vs=

s

L×

1

Cs

s2 + R

Ls + 1

LC

=

1

LC

s2 + R

Ls + 1

LC

(b) H1(s) is precisely of the form of equation 17.18 with a single zero at the origin. It follows

that ω p2 =

1

LC and 2σ p =

R

L. Hence, from equation 17.19, ωm = ω p =

1

LC, H1( jωm ) =

1

R,

Bω = 2σ p =R

L, and Qcir = Qp =

ω p

2σ p=

1

LC×

L

R=

1

R

L

C.

(c) With s = jω,

H2( jω)2 =

1

LC

2

1

LC−ω 2

2+ R

Lω

2 =1

1− LCω2( )2+ RCω( )2

Instead of maximizing H2( jω)2

we minimize its reciprocal, through differentiation. Let


f (C) =1

H2( jω)2 = 1 − LCω2( )2

+ RCω( )2

Then

f '(C) = 2C(ωR)2 + 2 ω2LC −1( )ω2L = 0

implies that

C =L

R2(ωL)2 =1

Lω21

R2

L2ω2 +1

=1

Lω21

1

Qcoil2 +1

If the coil has high Q, then

C ≅1

Lω2

in which case

H2( jω) =VC

Vs≅

LωR

= Qcoil

Therefore

VC max ≅ Qcoil Vs

SOLUTION PROBLEM 17.31. Here

H( jω) 2 = Ka + jω

(ωp2 −ω 2) + j2σ pω

2

= K2 a2 +ω 2

(ω p2 −ω 2)2 + 4σ p

2ω2 ≡ K2 f ω2( )

(a) Differentiating with respect to ω2 and setting the derivative equal to zero yields

0 =d f ω2( )

dω2 =d

dω2a2 +ω2

(ω p2 −ω 2 )2 + 4σ p

2ω2


= 1

(ω p2 −ω 2)2 + 4σ p

2ω2 −a2 +ω 2( ) −2(ω p

2 −ω 2) + 4σ p2( )

(ω p2 −ω 2)2 + 4σp

2ω2[ ]2

Given that the denominator of the first term is non-zero, this is equivalent to

0 = (ω p2 −ω 2)2 + 4σp

2 ω2 − a2 +ω 2( ) −2(ω p2 −ω2 ) + 4σp

2( )

= ω2( )2+ω p

4 − 2ω p2ω2 + 4σ p

2ω2 + 2 a2 +ω 2( )(ω p2 −ω 2) − 4 a2 +ω 2( )σp

2

= ω2( )2+ω p

4 − 2ω p2ω2 + 4σ p

2ω2 + 2a2ω p2 − 2a2ω2 + 2ωp

2 ω2 − 2 ω2( )2− 4a2σp

2 − 4σ p2ω2

= − ω2( )2− 2a2ω2 +ω p

4 + 2a2ω p2 − 4a2σ p

2

Hence

ω2( )2+ 2a2ω2 + a4 =ω p

4 + 2a2ω p2 − 4a2σp

2 + a4

where we have added a4 to produce perfect squares, i.e.,

ω2 + a2( )2= ω p

2 + a2( )2− 2aσp( )2

This implies that

ω2 = −a2 ± ω p2 + a2( )2

− 2aσp( )2

Thus, to achieve a real positive solution we obtain,

ωm = −a2 + ω p2 + a2( )2

− 2aσ p( )2

(b) Here, as in part (a), the arithmetic is simpler if we deal with 1/H(s) rather than H(s).

Specifically

1

H(s)=

s2 + 2σ ps +ω p2

K(s + a)=

s2 + 2σ p(s + a) − 2σ pa +ω p2

K(s + a)=

2σp

K+

s2 − 2σ pa +ω p2

K(s + a)


The problem asks for a non-zero value of ω. For zero phase shift of H(jω), Im[H(jω)] = 0

Im[H(jω)] > 0. This necessarily requires that the imaginary part of the above expression at s = jω

must be zero. Thus

Im1

H( jω)

= Im

−ω2 − 2σ pa +ω p2

K( jω + a)

= Im−ω2 − 2σ pa +ω p

2( )(a − jω)

K(ω2 + a2)

=−ω2 − 2σ pa +ω p

2( )(−ω)

K(ω2 + a2)= 0

The solutions to this are dc, i.e., ω = 0, and

ω (for zero phase shift) = ω p2 − 2σ pa

Note that such a frequency may not exist if the quantity under the radical is negative.

SOLUTION PROBLEM 17.40. (a) Compute the transfer function:

H(s) =Vout

Iin=

1

Yin(s)=

11

Ls + Rs+ Cs + 1

Rp

=1

C×

s +Rs

L

s2 + RsL

+ 1RpC

s + RsRp

+1

1

LC

= 8 ×105 s +1.333 ×103

s2 + 2000s +1.0006ω02

where ω02 = 1.6 ×109 and ω p

2 = 1.0006ω02 ≅1.60089 ×109 and ω p = 4.001×104 rad/sec.

(b) Qp =ω p

2σ p= 20 which is high Qp. Further a =

Rs

L= 1.333 ×103 <<ω p = 40.011109 ×103 .

(c) Equation 17.28,


ωm = −a2 + ω p2 + a2( )2

− 2aσ p( )2

gives the exact value of ωm. In MATLAB,

»wmsqrd = -a^2 + sqrt((wp^2+a^2)^2 - (twosig*a)^2)

wmsqrd = 1.600886670362673e+09

»wm = sqrt(wmsqrd)

wm = 4.001108184444246e+04

Clearly, ωm approximates ωp. Finally,

H( jωm) ≅ H( jω p) = 8 ×105 jω p +1.333 ×103

−ω p2 + j2000ω p +ω p

2 =8 ×105

20001− j

1.333 ×103

ω p

≅ 400 Ω

(d) From equation 17.30,

Bω ≅ 2σ p = 2000 rad/sec, ω1 ≅ ωm − 0.5Bω = 39.011 ×103 rad/sec, and

ω2 ≅ω m + 0.5Bω = 41.011 ×103 rad/sec.

(e) From equation 17.30,

Qcir ≅ Qp =ω p

2σ p=

Rs

Rp+ 1

1

LC

RsL

+ 1RpC

≅ω0

RsL

+ 1RpC

1

Qcir≅

Rs

L+

1

RpC

ω0=

Rs

Lω0+

1

RpCω0=

1

Qcoil+

1

Qcap=

Qcoil + Qcap

QcoilQcap

Hence,


Qcir ≅QcoilQcap

Qcoil + Qcap

SOLUTION PROBLEM 17.65. Here we consider the equivalent circuit valid for t > 0. Note that

iL(0-) = iL(0

+) = E/Rs and vC(0

-) = vC(0

+) = 0. Hence

It follows that

VC (s) =1

1

R+ 1

Ls+ Cs

×E

Rss=

1

s2 + 1

RCs + 1

LC

×E

CRs

Complex roots occur when

1

RC

2−

4

LC=

1

RC

2− 4ω0

2 < 0

Equivalently,

ω0RC >1

2

as was to be shown. Further, since vC(0+) = 0, the general form of the capacitor voltage for constant

excitation is (as per chapter 10)

vC (t) = e−σt Acos(ωdt) + Bsin(ωdt)( ) = Be−σt sin(ωdt)

From the characteristic equation, the complex roots are

−σ ± jωd =−1

2RC± j

1

LC−

1

2RC

2=

−ω0

2Q± jω0 1−

1

4Q2

Here


a = σ =ω0

2Q ωd =ω 0 1−

1

4Q2

Letting B = Vm, for the appropriate value of Vm, we obtain the desired result.

(b) If Q is large, ωd ≅ω 0 . The Vme−at will drop to 1/e of its peak value in t =1

a=

2Q

ω0≅

2Q

ωd

seconds. The period of oscillation of the damped sinusoid is 2πωd

. Therefore the number of cycles

contained in this interval is

2Q

ωd2πωd

=Q

π

5/31/01 Mag Crt Probs P18-1 © R. A. DeCarlo & P.M. Lin


SOLUTION PROBLEM 18.34. There is a correction to this problem: set M = 3 H.

(a) The stored energy at t = 0 is:

W(0) = 0.5L1i12(0) + 0.5L2i2

2(0) + Mi1 (0)i2(0) = 8 J

»L1 = 10; L2 = 2; M = 3; i10 = 1; i20 = -3;»W0 = 0.5*L1*i10^2 + 0.5*L2*i20^2 + M*i10*i20W0 = 5

(b) Writing two differential mesh equations we obtain

L1di1dt

+ Mdi2dt

+ R1i1 = 10di1dt

+ 3di2dt

+ i1 = 0

and

L2di2dt

+ Mdi1dt

+ R2i2 = 2di2dt

+ 3di1dt

+ i2 = 0

Taking the Laplace transform of these equations yields

10sI1 −10i1(0) + 3sI2 − 3i2(0) + I1 = (10s +1)I1 + 3sI2 −10i1(0) − 3i2 (0) = 0and

2sI2 − 2i2(0) + 3sI1 − 3i1(0) + I2 = (2s +1)I2 + 3sI1 − 3i1(0) − 2i2(0) = 0

Putting these equations in matrix form yields

(10s +1) 3s

3s (2s +1)

I1I2

=

10i1(0) + 3i2(0)

3i1(0) + 2i2(0)

=

1

−3

Solving yields

I1I2

=

(10s +1) 3s

3s (2s + 1)

−1 1

−3

=

1

11s2 +12s +1

2s +1 −3s

−3s 10s +1

1

−3

=1

(11s +1)(s +1)

11s +1

−33s − 3

=

1

s +1

1

−3

Therefore, by inspection,

i1(t) = e−tu(t) A and i2(t) = −3e−tu(t) A

Remark: normally, i1(t) and i2(t) would have two exponential terms present. Because of the specialchoice of initial conditions, a pole cancelled out.

(c) From equation 18.24 with the lower limit changed to zero and the upper limit changed to ∞, we have


W(0,∞) = v1i1 + v2i2( )0

∞∫ dt = 0.5L1i1

2(∞) + 0.5L2i22(∞) + Mi1(∞)i2(∞)

−0.5L1i12(0) + 0.5L2i2

2(0) + Mi1(0)i2(0)

From part (b) all currents at t = ∞ are zero, hence

W(0,∞) = −0.5L1i12 (0) + 0.5L2i2

2(0) + Mi1(0)i2(0) = −5 J

The result of part (a) indicates that the initial store energy is 5 J. The result of part (c) indicates that theenergy returned to the circuit is also 5 J, i.e., the total energy accumulated in the inductors over [0,∞) is–5 J. Hence 5 J is dissipated in the resistors.

Remark: the interested student might computer the integral R1i12( t) + R2i2

2( t)( )0

∞

∫ dt , the actual energy

dissipated in the resistors over [0,∞), and show that this is 5 J.


(a)»L1 = 4; L2 = 9; M = 3;»I1 = 2; I2 = -3;»W = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2W = 6.6500e+01

(b)»K = 0.5*L1*I1^2K = 8»% Minimize (over I2) K + 0.5*9*I2^2 - 3*2*I2»% Take Derivative and set to zero; then solve for I2.»% Derivative is: 9*I2 – 6 = 0»% The result is I2 = 2/3 A.»I2 = 2/3;»Wmin = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2Wmin = 6

(c)


(d)»L1 = 4; L2 = 9; M = 3;»k = M/sqrt(L1*L2)k = 5.0e-01


»k = 0.5;»L1 = 9; L2 = 4;L3 = 1;»M = 0.5*sqrt(L1*L2)M = 3»Lcpld = L1 + L2 + 2*MLcpld = 19»Leq = Lcpld + L3Leq = 20»Imax = 2;»Wmax = 0.5*Leq*Imax^2Wmax = 40 J


»RL = 100; Rs = 300e3; R = 10e3;»m = 20; n = 5;

(a)»Z2 = RL*m^2Z2 = 40000»Req1 = Z2*R/(Z2+R)Req1 = 8000»Z1 = Req1*n^2Z1 = 200000


(b)»%Gv1 = v1/vin»%v1 = [Z1/(Rs + Z1)]vin»Gv1 = Z1/(Rs+Z1)Gv1 = 4.0000e-01

»% Gv2 = v2/vin»% Gv2 = v2/v1 * v1/vin = (1/n) * G1»Gv2 = G1/nGv2 = 8.0000e-02

»% Gv3 = v3/vin»Gv3 = -Gv2/mGv3 = -4.0000e-03

(c)»% Gi2 = i2/iin»Gi2 = n*R/(R+Z2)Gi2 = 1

»% Gi3 = i3/iin = i3/i2 * i2/iin = –m*Gi2»Gi3 = -m*Gi2Gi3 = -20

SOLUTION PROBLEM 18.55. (a) The parameters in the circuit of figure P18.55b are given byequations in figure 18.22b. Specifically, since k = M L1L2 = 0.16 3.5 × 0.008 = 0.95618

»M = 0.16; L1 = 3.5; L2 = 0.008;»k=M/sqrt(L1*L2)k = 9.5618e-01»La = (1 - k^2)*L1La = 3.0000e-01»Lb = k^2 * L1Lb = 3.2000e+00»N =M/L2N = 20

(b)»R = 500;»w = 2*pi*60;»Vseff = 110;»Zin = R + j*La*w + j*Lb*wZin = 5.0000e+02 + 1.3195e+03i»Iseff = Vseff/ZinIseff = 2.7624e-02 - 7.2899e-02i»Pave = R*abs(Iseff)^2


Pave = 3.0387e+00

(c)»Zin2 = R + j*La*wZin2 = 5.0000e+02 + 1.1310e+02i»Iseff2 = Vseff/Zin2Iseff2 = 2.0929e-01 - 4.7341e-02i»Is2mag = abs(Iseff2)Is2mag = 2.1458e-01

» % The current in the secondary is (in A):

»Isecmag = Ismag*NIsecmag = 4.2916e+00

(d)»% Our first step is to compute the reflected impedance:»Zrefl = 100*N^2Zrefl = 40000

»% We now compute the impedance of the parallel combination»% of Lb and Zrefl denoted Zpar»Zpar = 1/(1/Zrefl + 1/(j*w*Lb))Zpar = 3.6350e+01 + 1.2053e+03i

»% We now compute the input impedance:

»Zin = R+j*w*La + ZparZin = 5.3635e+02 + 1.3184e+03i

»% Now we compute the voltage across the primary of the»% ideal transformer, by voltage division:

»Vpar = Vseff*Zpar/ZinVpar = 8.7342e+01 + 3.2500e+01i

»% Now we compute the voltage across the load:»Vload = Vpar/NVload = 4.3671e+00 + 1.6250e+00i»Vloadmag = abs(Vload)Vloadmag = 4.6596e+00»Iloadmag = Vloadmag/100Iloadmag = 4.6596e-02


SOLUTION PROBLEM 18.65. (a) The equivalent circuit accounting for initial conditions is givenbelow:

(b) From the definition of coupled inductors

V1

V2

=

0.6s 0.1472s

0.1472s 0.1472s

I1'

I2'

(c) Hence

I1'

I2'

=0.6s 0.1472s

0.1472s 0.1472s

−1 V1

V2

=

15

s

0.1472 −0.1472

−0.1472 0.6

V1

V2

(d) Writing nodal equations we obtain,

Vs

0

=

2 −1

−1 2

V1

V2

+

I1

I2

=

2 −1

−1 2

V1

V2

+

1

s

i1(0− )

i2(0− )

+I1'

I2'

(e) Now we substitute our result of part (c):

Vs

0

=

2 −1

−1 2

V1

V2

+

1

s

−2

−2

+

1

s

2.208 −2.208

−2.208 9

V1

V2

which simplifies toVs

0

−

1

s

−2

−2

=

1

s

2s + 2.208 −(s + 2.208)

−(s + 2.208) 2s + 9

V1

V2

or equivalently

2s + 2.208 −(s + 2.208)

−(s + 2.208) 2s + 9

V1

V2

=

sVs + 2

2

(f) Solving these equations we obtain:


V1

V2

=

2s + 2.208 −(s + 2.208)

−(s + 2.208) 2s + 9

−1 sVs + 2

2

=1

3s2 +18s + 15

2s + 9 (s + 2.208)

(s + 2.208) 2s + 2.208

sVs + 2

2

(g) If vs(t) = 10u(t) V, then

V1

V2

=

1

3s2 +18s +15

2s + 9 (s + 2.208)

(s + 2.208) 2s + 2.208

12

2

=

1

3s2 +18s + 15

26s +112.42

16s + 30.912

From MATLAB»[r,p,k]=residue([16 30.912], [3 18 15])r = 4.0907e+00 1.2427e+00p = -5 -1k = []

Therefore,

v2( t) = 4.0907e−5t +1.2427e−t( )u(t) V

SOLUTION PROBLEM 18.67. The solution to this problem is based upon the following: (i) Leq = L1 +L2 + 2M for series aiding connection (see example 18.4) and (ii) k = M L1L2 (a definition), and (iii) k= 1 (an assumption).

(a) Given L1 = L2 = L and k = 1, Leq = L1 + L2 + 2M = L + L + 2k L2 = 4L . Hence, when thenumber of turns is doubled, the inductance is quadrupled.

(b) For this part, let us first consider L2 which has 2N turns. We can view L2 as two coils of N turnseach connected in series aiding with coupling coefficient k = 1. Hence, according to part (a), theinductance of L2 is four times that of L1 which only has N turns. Hence,

Leq = L1 + L2 + 2M = L + 4L + 2k 4L2 = 9L

Observe that the coil has 3N turns yielding an inductance of 9L = 32L.

(c) Suppose coil 1 and coil 2 consist of one turn each. Here the total number of turns is 2N where N =

1 turn. Suppose further that L1 = L. From part (a), Leq = 4L = (2)2 L . Now suppose coil one consistsof one turn and coil 2 consists of M turns. We assume here as an induction hypothesis that


Leq = M +1( )2 L

We must show that if coil 2 has (M+1)N turns then,

Leq = M + 2( )2 L

Our first step is to compute the equivalent inductance of coil 2. However, coil 2 consists of a single turncoupled to an M-turn coil, which by the induction hypothesis means that

L2 = M +1( )2 L

Thus coil 1 in a series aiding connection with L2 leads to

Leq = L1 + L2 + 2M = L + M + 1( )2 L + 2 M + 1( )2 L2 = L + M +1( )2 L + 2 M +1( )L

= L M +1( )2 + 2 M +1( ) +1[ ] = L M +1( ) +1[ ]2 = M + 2( )2 L

Given this relationship, if coil 1 consists of N1 turns, and one turn has an inductance L, then

L1 = N1( )2L . Similarly, L2 = N2( )2

L , and M = k L1L2 = L1L2 = N1( )2N2( )2

L2 = N1N2L . Itimmediately follows that

L1 : L2 : M = N12 : N2

2 : (N1N2)

SOLUTION PROBLEM 18.70. (a) k = M / L1L2»M = 1.5; L1 = 1.5; L2 = 6;»k = M/sqrt(L1*L2)k = 5.0000e-01

(b) For this part and the remaining parts consider the following equivalent circuit where the coupledcoils have been replaced by the model of figure 18.22(b).

»La = (1 - k^2)*L1La = 1.1250e+00»Lb = k^2 * L1Lb = 3.7500e-01


»% The turns ratio is M:L2, i.e.,»1.5/6ans = 2.5000e-01»% Therefore the turns ratio is 1:4. It follows that

»Rb = 200*(1/4)^2Rb = 1.2500e+01

To compute Z(s) we have,

Z(s) = 20 +1

Cs+ Las +

RbLbs

Lbs + Rb= 1.125s + 20 +

12.5s

s + 100

3

+1

Cs

Hence,

Z( jω) = j1.125ω + 20 +j12.5ω

jω + 100

3

−j

Cω= 20 +

12.5ω2

ω2 + 104

9

+ j 1.125ω −1

Cω+

416.67ω

ω2 + 104

9

(c) For this part we need to make the imaginary part of Z(jw) real. To this end:

»K1 = 12.5*100/3K1 = 4.1667e+02»w = 1333;»K2 = 1.125*w + K1*w/(w^2 + 1e4/9)K2 = 1.4999e+03»C = 1/(K2*w)C = 5.0015e-07

Hence, we take C = 5 µF.

(d) At resonance, we have

Z( jωr ) = Z( j1333) = 20 +12.5ωr

2

ωr2 + 104

9

= 32.5 Ω

and

Zb ( jωr) =j12.5ωr

jωr + 100

3

=j16,662

33.33 + j1333= 12.492 + j0.31238

By voltage division

Vout ( jωr)

Vs( jωr)=

Vout

Vb×

Vb

Vs=

4

1×

Zb( jωr )

Z( jωr )= 1.5379 + j0.038456

»Zb = j*12.5*w/(j*w + 100/3)Zb = 1.2492e+01 + 3.1238e-01i

»Zwr=20 + 12.5*w^2/(w^2 + 1e4/9)Zwr =


3.2492e+01

»Gv = 4*Zb/ZwrGv = 1.5379e+00 + 3.8456e-02i

»MagGv = abs(Gv)MagGv = 1.5384e+00

»AngGv = angle(Gv)*180/piAngGv = 1.4325e+00


(a) Following the hint we apply a source transformation to obtain

Writing two mesh equations we obtain the following matrix form of the mesh equations:

s +β+ 1 s ks

ks s +β+ 1 s

I1

I2

=

Iin s

0

Solving for I2 yields

I2 =det

s + β + 1 s Iin s

ks 0

dets + β + 1 s ks

ks s + β + 1 s

=−kIin

s + β + 1 s( )2 − ks( )2

To find H(s) we have

H(s) =Vout

Iin=

− I2 s

Iin=

k s

s + β + 1 s( )2 − ks( )2 =k s

s + ks + β + 1 s( ) s − ks + β + 1 s( )

=ks

(1+ k)s2 +β s +1[ ] (1− k)s2 +βs +1[ ]


(b) R = 0.02 Ω, b = R = 0.02 Ω, Q = 50, and k = 0.01, 0.02, and 0.04.

»beta = 0.02; k1 = 0.01; k2 = 0.02; k3 = 0.04;»p11 = roots([(1+k1) beta 1]);»p12 = roots([(1-k1) beta 1]);»p1 = [p11;p12]p1 = -9.9010e-03 + 9.9499e-01i -9.9010e-03 - 9.9499e-01i -1.0101e-02 + 1.0050e+00i -1.0101e-02 - 1.0050e+00i

»p21 = roots([(1+k2) beta 1]);»p22 = roots([(1-k2) beta 1]);»p2 = [p21;p22]p2 = -9.8039e-03 + 9.9010e-01i -9.8039e-03 - 9.9010e-01i -1.0204e-02 + 1.0101e+00i -1.0204e-02 - 1.0101e+00i

»p31 = roots([(1+k3) beta 1]);»p32 = roots([(1-k3) beta 1]);»p3 = [p31;p32]p3 = -9.6154e-03 + 9.8053e-01i -9.6154e-03 - 9.8053e-01i -1.0417e-02 + 1.0206e+00i -1.0417e-02 - 1.0206e+00i

(c)

»f = 0.14:.0001:.18;»n1 = [k1/(1-k1)^2 0];n2 = [k2/(1-k2)^2 0];n3 = [k3/(1-k3)^2 0];»h1 = freqs(n1, poly(p1), 2*pi*f);»h2 = freqs(n2, poly(p2), 2*pi*f);»h3 = freqs(n3, poly(p3), 2*pi*f);»plot(f, abs(h1), f, abs(h2), f, abs(h3))»grid


(d) An inspection of [f, abs(h2)] (i.e., a tabulation of the values) in part (c) indicates that fpeak = 0.15865Hz and Hpeak = 26.148 Ω. The frequency scale factor Kf is defined according to:»Kf = 455e3/fpeakKf = 2.8679e+06

Further,

»Km = Kf*2.35e-3Km = 6.7397e+03

»Lnew = Km*1/KfLnew = 2.3500e-03»Cnew = 1/(Km*Kf)Cnew = 5.1736e-11»Rnew = Km*0.02Rnew = 1.3479e+02

The 3 dB down value of h2 is Hpeak/sqrt(2). Hence


»Hmax = max(abs(h2))Hmax = 2.6148e+01»H3db = Hmax/sqrt(2)H3db = 1.8490e+01

Again, inspecting the tabulated values indicates that the 3 dB frequencies are: f1 = 0.1569 Hz andf2 = 0.1614 Hz. Finally»Bf = f2 - f1Bf = 4.5000e-03 Hz»Bfnew = Kf*BfBfnew = 1.2906e+04 Hz

(e) For this part we redo part (a) with R, L, and C as literals.

Writing two mesh equations we obtain the following matrix form of the mesh equations:

Ls + R +1 Cs kLs

kLs Ls + R +1 Cs

I1I2

=

Iin Cs

0

Solving for I2 yields

I2 =det

Ls + R + 1 Cs Iin Cs

kLs 0

detLs + R +1 Cs kLs

kLs Ls + R +1 Cs

=−kL C

Ls + R +1 Cs( )2 − kLs( )2 Iin

To find H(s) we have

H(s) =Vout

Iin=

− I2 Cs

Iin=

kL C2s

Ls + R + 1 Cs( )2 − kLs( )2 =kL C2s

Ls + kLs + R +1 Cs( ) Ls − kLs + R +1 Cs( )

=kLs

LC(1+ k)s2 + CRs +1( ) LC(1− k)s2 + CRs +1( ) =ks

LC2

(1+ k)s2 + R

Ls + 1

LC

(1− k)s2 + R

Ls + 1

LC

=kω0

2sC

(1+ k)s2 + ω0Q

s +ω 02

(1− k)s2 + ω0

Qs +ω 0

2

Evaluating this expression at s = jω0, yields


H( jω0 ) =jkω0

3

C

−(1+ k)ω02 + jω0

2

Q+ω 0

2

−(1− k)ω02 + jω0

2

Q+ω 0

2

=− jk

Cω0

k − jQ

k + j1

Q

=− jk

Cω0

k2 + 1

Q2

Therefore

H( jω0) =1

Cω0×

k

k 2 + 1

Q2

(f) To solve this part we differentiate and set equal to zero as follows:

d H ( jω0)

dk=

1

Cω0×

d

dk

k

k2 + 1

Q2

=1

Cω0×

1

k 2 + 1

Q2

−2k2

k2 + 1

Q2

2

= 0

It follows that2k2

k 2 + 1

Q2

= 1

Hence

k 2 =1

Q2

or k = 1/Q. With this value of k,

H( jω0) max =1

Cω0×

1/ Q2

Q2

=Q

2Cω0

At ω0, the magnitude of the transfer function increases with increasing k, reaching a peak at k = 1/Q andthen decreases with a further increase in k as born out in the plots of part (c).

Prbs Chap 19, 1/7/02 P19-1 © R. A. DeCarlo, P. M. Lin


SOLUTION PROBLEM 19.1. Refer to figure 19.3.

»Vs = 100; ZL = 20; Rs = 1e3;

»beta = 149;

»Zinbox = (beta + 1)*ZL

Zinbox =

3000

»% By voltage division

»V1 = Vs*Zinbox/(Zinbox + Rs)

V1 = 75

»% To obtain the power delivered by the source

»I1 = Vs/(Zinbox +Rs)

I1 = 2.5000e-02

»Psource = I1*Vs

Psource = 2.5000e+00


»Vs = 100; Z1 = 30e3; Rs = 50; beta = 149;

»Zboxin = Z1/(beta+1)

Zboxin = 200

»V1 = Vs*Zboxin/(Zboxin + Rs)

V1 = 80

»Psource = Vs^2/(Rs + Zboxin)

Psource = 40


(a) »C = 0.1e-3; vc0 = 10; Z1 = 300; Z2 = 1e3;

»Z3 = 1e3; gm = 9e-3;

»

»Zin = Z1 + (1 + gm*Z1)*Z2

Zin = 4.0000e+03

(b)

»tau = Zin*C


tau = 4.0000e-01

Hence, vC (t) = vC (0)e−t / =10e−2.5t V.

SOLUTION PROBLEM 19.4. (a) First observe that since no current can flow into the secondary

we have

Voc = aVpri = aRIin = 800 Vrms

Now

Zth =1

j C+ a2R = 640 − j 360 Ω

(b) ZL = Zth( )* = 640 + j360 Ω.

»Voc = 800; Rth = 640;

»Pmax = Voc^2/(4*Rth)

Pmax = 250

(c) By inspection the circuit is a 640 Ω resistor in series with a 3.6 H inductor.

SOLUTION PROBLEM 19.5. Because the output is open circuited, no current flows into the

secondary of the transformer, hence

voc = vsec + sin(3t)u( t) = 2v pri + sin(3t)u(t) = cos(3t) + sin(3t)[ ]u( t).

Additionally

Zth (s) =10s

+ s + 4 + 42.5s

+ 0.25s + 9

=

20s

+ 2s + 40

SOLUTION PROBLEM 19.6. Using Cramer's rule,


I1 =

det

V1 1 −a

0 0.5 0

0 0 0.5

1.5a=

0.25V11.5a

=16a

V1

Therefore Rin = 6a Ω.

To compute the average power, V1,eff =10 V. Hence Pave =V1,eff

2

Rin=

1006a

watts.

SOLUTION PROBLEM 19.7. As per the hint, we write loop equations as follows:

−Vout

10 s

−40 s

=s +1 −1 0

−1 6 −2

0 −2 4

Iout

I2

I3


−Vout = s +1− −1 0[ ] 0.2 0.1

0.1 0.3

−1

0

Iout + −1 0[ ] 0.2 0.1

0.1 0.3

10 s

−40 s

= (s + 0.8)Iout +2

s

Therefore

Vout = (s + 0.8) −Iout( ) −2s

= Zth − Iout( ) +Voc

i.e., Zth = s + 0.8, and Voc = −2s

.

SOLUTION PROBLEM 19.8. (a) Let the node voltages from left to right be V1, V2, and Vout.

Also inject a current I3 into node 3. Writing nodal equations by inspection we have:


Iin

0

I3

=1.5 −1 −0.25

−1 2 −0.5

−0.25 −0.5 0.0625s + 0.75

V1

V2

Vout

Using equation 19.11, we have

I3 = W22 − W21W11−1W12( )Vout + W21W11

−1 Iin

0

= 0.0625s + 0.75 − 0.25 0.5[ ] 1 0.5

0.5 0.75

0.25

0.5

Vout − 0.25 0.5[ ] 1 0.5

0.5 0.75

Iin

0

Thus

I3 = 0.0625s + 0.375( )Vout − 0.5Iin

Therefore Isc = −I3]Vout =0 = 0.5Iin . Further Zth (s) =1

0.0625s + 0.375=

16s + 6

Ω.

(b) Vout (s) = Zth (s)Isc =16Iscs + 6

=8Iins + 6

. By inspection, the impulse response is

vout ,imp(t) = 8e−6tu( t) V. Further, from MATLAB

»n = 8; d = [1 6 0];

»[r,p,k] = residue(n,d)

r =

-1.3334e+00

1.3334e+00

p =

-6

0

k =

[]

Hence the step response is:

vout ,step( t) =43

1− e−6t( )u(t) V

SOLUTION PROBLEM 19.9. (a) Consider the following figure:


Let Iout enter the output terminal and I1 and I2 be the currents entering the primary and secondary

of the transformer respectively. It follows that

Iin =bVout

R+ I1 +

(b −1)VoutR

=(2b −1)Vout

R+ I1

which implies that

I1 = Iin −(2b −1)Vout

R

Further,

Iout =(1− b)Vout

R+ I2 =

(1− b)VoutR

− bI1

Therefore

Iout =(1− b)Vout

R+ I2 =

(1− b)VoutR

− bI1 =(1− b)Vout

R− b Iin −

(2b −1)VoutR

=2b2 − 2b +1

RVout − bIin

Equivalently

Vout =R

2b2 − 2b +1Iout +

bR

2b2 − 2b +1Iin

Therefore

Zth =R

2b2 − 2b +1, Voc =

bR

2b2 − 2b +1Iin , Isc = bIin

(b) Isc = b∠45o , Zth =R

2b2 − 2b +1.


(c) B =10 =1

ZthC=

1R

2b2 − 2b +1C

=2b2 − 2b +1

RC=

525C

. Hence C = 0.02 F. Further,

02 = 25 =

1LC

=1

0.02L. Therefore L = 2 H.

(d) H(s) =VoutIin

=Vout

0.5Isc= 2

1

2b2 − 2b +1

R+

1

Ls+ Cs

=2

0.2 +1

2s+ 0.02s

=100s

s2 +10s + 25

and Iin (s) =2

2×

s−10

s2 +100. Hence,

Vout (s) =50 2 s(s−10)

(s2 +10s + 25)(s2 +100)

To compute vout(t) in MATLAB,

%partial fraction expansion of Vout/√2 = = n(s)/d(s)

n =50*[ 1 -10 0];d = conv([ 1 0 100], [ 1 10 25]);[r ,p, k] = residue(n,d)

%numerator polynomial of combined complex pole terms

num = [1 -p(1)]*r(2) +[1 -p(2)]*r(1)

Output from MATLAB

r = 2.8000 + 0.4000i 2.8000 - 0.4000i -5.6000 30.0000p = -0.0000 +10.0000i -0.0000 -10.0000i -5.0000 -5.0000num = 5.6000 -8.0000

Laplace transform of ouput

Vout (s) =−5.6 2

s + 5+

30 2

(s + 5)2 +(5.6s − 8) 2

s2 +100


Taking the inverse Laplace transform using table 13.1 on page 515:

vout (t) = 30 2te−5t − 5.6 2e−5t + 5.6 2 cos(10t) − 0.8 2 sin(10t)[ ]u( t) V

The steady state part consists of the cosine and sine terms only. Since the parallel RLC acts like

a band pass circuit and the peak value occurs at 5 rad/s, one expects the magnitude of the steady

state output to be much smaller at 100 rad/s.

SOLUTION PROBLEM 19.10. (a) Write two nodal equations by inspection:

I1

I2

=

Y1 + Y3 −Y3 + gm

−Y3 Y2 + Y3

V1

V2

=

y11 y12

y21 y22

V1

V2

(b) When port-2 is shorted, y11 is the input admittance. Therefore

Zin =1

y11=

1Y1 + Y3

and since V2 = 0,

I2 = y21V1 =−KY3

s

SOLUTION PROBLEM 19.11. (a) By inspection

I1

I2

=

Y1 + Y3 −Y3

−Y3 + gm Y2 + Y3

V1

V2

=

y11 y12

y21 y22

V1

V2

Clearly, Y3 = −y12 . Then, Y1 = y11 − Y3 = y11 + y12 and Y2 = y22 − Y3 = y22 + y12 . Finally,

gm = y21 +Y3 = y21 − y12 .

(b) Recall that


I1

I2

=

Y1 + Y3 −Y3 + gm

−Y3 Y2 + Y3

V1

V2

=

y11 y12

y21 y22

V1

V2

Clearly, Y3 = −y21 . Then, Y1 = y11 − Y3 = y11 + y21 and Y2 = y22 − Y3 = y22 + y21 . Finally,

gm = y12 + Y3 = y12 − y21 .

SOLUTION PROBLEM 19.12. (a) By definition of coupled inductors

V1

V2

=

L1s Ms

Ms L2s

I1I2

⇒

I1I2

=

1

L1L2 − M2( )s

L2 −M

−M L1

V1

V2

Hence, the y-parameters are:

1

L1L2 − M 2( )sL2 −M

−M L1

(b) By definition of coupled inductors

V1

V2

=

L1s −Ms

−Ms L2s

I1

I2

⇒

I1

I2

=

1

L1L2 − M 2( )sL2 M

M L1

V1

V2


1

L1L2 − M 2( )sL2 M

M L1

If the coupling coefficient is 1, L1L2 = M2 and the y-parameters do not exist since the

determinant of the z-parameter matrix is zero.

SOLUTION PROBLEM 19.13. Let I2' denote the current entering the dotted terminal of the

secondary of the coupled inductors. Then using the result of problem 12a,


I1

I2'

=

1

L1L2 − M 2( )sL2 −M

−M L1

V1

V2

=

16s

4 −1

−1 1

V1

V2

From the given circuit I2 = I2' + 2I1 =

−16s

V1 +16s

V2 +86s

V1 −26s

V2 =76s

V1 −16s

V2 . Therefore

the y-parameter matrix is:

16s

4 −1

7 −1


(a) By definition and the properties of the ideal transformer y11 =I1V1

V2 =0

= G1 and

y22 =I2

V2

V1=0

=G1 + G2

a2 . Additionally, since the circuit is obviously reciprocal,

y12(= y21) =I1V2

V1=0

=G1

a.

(b) V2 reflected to the primary side, denoted by ˆ V 1, is

ˆ V 1 =−2K a

s2 + 4

Hence

I1 =2G1K a

s2 + 4

To compute I2, we reflect the parallel of G1 and G2 to the secondary of the ideal transformer.

Hence the impedance in parallel with V2, denoted Zsec, is

Zsec =a2

G1 + G2

Therefore,

Prbs Chap 19, 1/7/02 P19-1 0 © R. A. DeCarlo, P. M. Lin

I2 =V2

Zsec=

2K G1 + G2( ) a2

s2 + 4

SOLUTION PROBLEM 19.15. (a) By definition

y11 =I1V1

V2 =0

=1

Zin

V2 =0

=19

S

where

Zin = 6 +12/ /1

16× (320/ /80)

= 9 Ω

Similarly, by definition

y22 =I2V2

V1=0

=1

Zout

V1=0

=3

400 S

where

Zout = 80 + 320/ /16 × (6 //12)( ) =4003

Ω

SOLUTION PROBLEM 19.16. Write nodal equations:

I1

I2

0

=2s + 2 −1 −2s

−1 s + 2 −s

−2s −s 5s

V1

V2

V3

Using the matrix partitioning method, we obtain the 2-port y parameters

I1

I2

=

2s + 2 −1

−1 s + 2

−

15s

−2s

−s

−2s − s[ ]

V1

V2

=1.2s + 2 −0.4s−1

−0.4s−1 0.8s + 2

V1

V2

=

y11 y12

y21 y22

V1

V2


SOLUTION PROBLEM 19.17. (a) The ideal transformer yields the constraints

I1 = −aˆ I 2 and ˆ V 2 = aV1

The three resistors have nodal equations

− ˆ I 2I2

=2 −1

−1 2

ˆ V 2V2

Substituting the first two equations into the last one, we obtain

I1

I2

=

2a2 −a

−a 2

V1

V2

=

y11 y12

y21 y22

V1

V2

(b)

Yin = y11 −y12y21

y22 + YL= 2a2 −

a2

4=1.75a2 S

Zin =1

Yin=

4

7a2 Ω

and

Gv = V2

V1 = -y21

y22 + yL1 = a

4

SOLUTION PROBLEM 19.18. This problem is solved in MATLAB.

Part (a)

»% Parameter specification

»Ys = 1e-3;YL = 1e-3;»y11=4e-3; y12 = -0.1e-3;»y21 = 50e-3; y22 = 1e-3;

»% Calculation of input admittance and impedance

»Yin = y11 - y12*y21/(y22 + YL)Yin = 6.5000e-03»Zin = 1/YinZin = 1.5385e+02

»% Calculation of output admittance


»Yout = y22 - y12*y21/(y11 + Ys)Yout = 2.0000e-03

Part (b)»% Calculation of voltage gain

»Gv = (Ys/(Ys + Yin))*(-y21/(y22 + YL))Gv = -3.3333e+00

Part (c)»V2 = Gv*10V2 = -3.3333e+01

Therefore, v2(t) = −33.333u(t) V. Finally,

»PL = V2^2/1e3PL = 1.1111e+00


(a) V1

Vs = Z in

Z in + Zs = Z in

Z in + 10 = 0.5 ⇒ Z in = 10 Ω or 0.1 S

Now

Yin = y11 - y12y21

y22 +YL = y11 -

0.02×20.2 + 0.1

= 0.1

Solving for y11 yields y11 = 0.2333 S.

(b) v2v1

= - y21

y22 +YL = - 2

0.2 +0.1 = - 6.667

Hence v2(t) = - 6.667 v1(t) = - 3.333 vs(t) = - 33.33 u(t) V

and

PL = V22

RL = 33.332

10 = 111.11 W

SOLUTION PROBLEM 19.20.(a) Writing a node equation at port 1 and mesh equation at port 2, we obtain by inspection

I1 = 2V1 + 3I 2

V2 = 2V1 + 2I2

Rearranging in matrix form , we have


1 -3

0 2

I1

I2

= 2 0

-2 1

V1

V2Therefore

I1

I2

= -1 1.5

-1 0.5

V1

V2

= y11 y12

y21 y22

V1

V2

(b) Yin = y11 - y12y21

y22 +YL = -1 + 1.5×1

0.5 + 0.25 = 1 S

(c) Here we compute Yout seen looking into port-2, i.e.,

Yout = y22 - y12y21

y11 +Ys = 0.5 + 1.5×1

-1 + 0 = -1 S

From current division,

i2(t) = Yout

Yout+YLis(t) = -1

-1 + 0.25×5u(t) = 6.667u(t) A

Finally from Ohm's law

v1(t) = - 0.5 ×3i2(t) = - 0.5 ×3×6.667u(t) = - 10u(t) V

REMARK: Because the current source sees a negative resistance, the circuit is unstable as itstands.

SOLUTION PROBLEM 19.21. (a) With port 2 shorted, the Laplace transform of the given data

are:

I1(s) =1s, V1(s) =

1s

−1

s + 4=

4s(s + 4)

, I2(s) =−1

s + 3Hence

y11 =I1V1

=1 s4

s(s + 4)

=s + 4

4

and

y21 =I2V1

=−1 (s + 3)

4

s(s + 4)

=− s(s + 4)4(s + 3)

Next, with port-2 terminated in a 1-Ω reistor, the Laplace transform of the given data are:


ZL=YL = 1

I1(s) = 1/s,

V1(s) = 1s

- 1s+4

+ 1(s+4)2

= 5s + 16s(s + 4)2

I2 = - 1s +7

Now

I2 = - V2

ZL = - V 2 = y21V1

y22 +YL

Solving for y22, and using y21 expression found earlier, we obtain

y22 = y21V1/I2 - YL= - s(s + 4)4 (s + 3)

× 5s + 16s(s + 4)2

×s + 7 -1

- - 1

= s2 + 23s + 644(s + 3)(s+ 4)

Finally, to compute y12, we use the defining equation for y-parameters, i.e.,

I1 = y11V1 + y12V2 from which we obtain

y12 = I1 - y11V1

V2 =

1s - s + 4

4× 5s + 16

s(s + 4 )2

1s + 7

= - (s + 7)4 ( s + 4)

(b) Given YL =1 S, the input admittance is

Yin = y11 - y12y21

y22 +YL = s + 4

4 -

- (s + 7)4 ( s + 4)

×- s(s + 4)4(s + 3)

s2 + 23s + 644(s + 3)(s+ 4)

+1 = (s + 4)2

5(s+3.2)

and

Zin = 1Yin

= 5(s+3.2)

(s + 4)2


(c) For this part, we use phasors to do the sinusoidal steady state analysis: ω = 10 rad/s and I1 =1. Also,

V1

I1 = Zin = 5(j10+3.2)

(j10 + 4)2 = 0.1973 - j0.4072

and

V2

V1 = -y21

y22 +YL =

- j10(j10 + 4)4(j10 + 3)

-100 + 230j + 644(j + 3)(j+ 4)

+1 = -0.2873+ j0.1787

Thus

V2

I1 = V2

V1 ×V1

I1 = (-0.2873+ j0.1787)(0.1973 - j0.4072)

from which we obtain in Ohms,

V2

I1 = -0.2873+ j0.1787 ×0.1973 - j0.4072 = 0.8192

SOLUTION PROBLEM 19.22. Looking into port-1, the admittance is:

Yport 1 (s) = y11 - y12y21

y22 +YL = 0 + 2×2

0 + 0.1s = 40

s

andZport 1(s) = s

40

Therefore

Zin(s) = 1000s

+ s40

= s2 + 2002

40sHence

Zin ( j ) =4 ×104 − 2

j40= − j

4 ×104 − 2

40

The imaginary part is zero when = 200 rad/s. Hence, the resonant frequency is 200 rad/s.

SOLUTION PROBLEM 19.23. (a) By writing nodal equations for the boxed 2-port, we have by

inspection (note passive circuit in which y21 = y12):


I1

I2

= 3s + 2 -2s -2

-2s -2 3s + 3

V1

V2

= y11 y12

y21 y22

V1

V2

(b) In Ohms, ZL = s+1; Zs = 2. Therefore,

Yin = y11 - y12y21

y22 +YL = 3s+2 - (2s +2)2

3s + 3 + s + 1 = 2s2 + 3s + 1

s + 1 = 2s + 1

V1

Vs = Z in

Z in + Zs = 1

1 + Z s Yin = 1

1 + 2×(2s + 1) = 1

4s + 3and

V2

V1 = -y21

y22 + yL = 2s +2

3s + 3 + s + 1 = 0.5

Thus,

V2

Vs = V1

Vs ×V2

V1 = 1

4s + 3× 0.5 = 1

8s2 + 14s + 6 = 1

8s+ 6

(c) The impulse reponse is

h(t) = L−1 H (s) = L−1 18s + 6

= L−1 0.125s + 0.75

= 0.125e−0.75tu(t)

For the step response,

v2(t) = L−1 H (s)s

= L−1 0.125s(s + 0.75)

=16

1− e−0.75t( )u(t) V

(d) We must compute the complete Laplace transform and invert. Here

Vs(s) = 12.75 × 2s2 + 4

and

V2(s) = H(s)V2(s) = 0.125s+ 0.75

×12.75× 2s2 + 4

= 3.1875(s + 0.75) (s2 + 4)

We use MATLAB to compute the partial fraction expansion

n=3.1875;d= conv([1 0.75], [ 1 0 4]);[r p k ] = residue (n,d)r = -0.3493 - 0.1310i -0.3493 + 0.1310i 0.6986p = -0.0000 + 2.0000i


-0.0000 - 2.0000i -0.7500

Hence, after combining the two complex terms, we obtain

V2(s) = 0.6986s + 0.75

+ - 0.6986s + 0.524s2 + 4

From Table 13.1, the steady state response is

v2,ss(t) = [-0.6986cos(2t) + 0.262sin(2t)]u(t) V

and the transient response is

v2,tran(t) = 0.6986e-0.75tu(t) V

SOLUTION PROBLEM 19.24. (a) Using the y-parameters of stage 2,

Yin2 = y11 - y12y21

y22 +YL = 500 + 0.1×75000

185 +4000 = 501.8 µS

The load for stage 1 is the parallel combination of Zin2 and the 2 kΩ resistor. Hence, using the y-parameters of stage 1, we obtain

Yin1 = y11 - y12y21

y22 +YL = 2000 + 0.5×24×104

100 +501.8 + 500 = 2108.9 µS

(b) We compute the following voltage gains:

V1

Vs = Z in1

Z in1 + Zs = 1

1 + Z s Yin1 = 1

1 + 25×2108.9×10-6 = 0.9499

V2V1

stage1

=−y21

y22 + YL,stage1=

−0.24

(100 + 500 + 501.8) ×10−6 = −217.8

andV2V1

stage2

=−y21

y22 +YL,stage2=

−0.075

(185 + 4000) ×10−6 = −17.92

Finally Gv is the product of the three gains calculated above

Gv = 0.9499×(- 217.8) ×(-17.92) = 3708.2


SOLUTION PROBLEM 19.25. (a) With the switch in position A, the load to the 2-port is YL =Cs = 0.25s. Hence,

V2V1

=−y21

y22 +YL=

−y21y22 + 0.25s

(b) From the given data, V1(s) = –2/s and hence,

V2(s) =−y21

y22 + 0.25sV1(s) =

−1 s−1 s + 0.25s

×−2s

=8

s(s2 − 4)=

−2s

+1

s + 2+

1s − 2

Hence, for t ≥ 0,

v2(t) = ( -2 + e-2t + e2t) u(t) V

(c) The circuit is not stable in the time interval 0 to 1 s, because the transfer function has a polein the right half plane.

(d) v2(1-) = -2 + e-2 + e2 = 5.524 V

(e) Replace the charged capacitor by the parallel combination of an admittance of 0.25s and acurrent source of value 0.25×8.524 (in accordance with figure 14.16

(f) Z1(s) = b2 × 1 = 4 Ω.

(g) For t ≥ 1s, the capacitor is discharging through a 4-Ω equivalent resistance, with a timeconstant 0.25×4 = 1 s, and an initial voltage v2(1-) = 5.542 V.Hence

v2(t) = 5.524e−(t−1)u(t −1)Vand by the ideal transformer voltage ratio property,

v3( t) = 2.762e−(t−1)u(t −1)V

SOLUTION PROBLEM 19.26. (a) By definition of coupled inductors

V1

V2

=

L1s Ms

Ms L2s

I1I2

Hence, the z-parameters are:


L1s Ms

Ms L2s

(b) By definition of coupled inductors

V1

V2

=

L1s −Ms

−Ms L2s

I1

I2


L1s −Ms

−Ms L2s

The z-parameters exist independent of the values of M, L1, and L2.

SOLUTION PROBLEM 19.27. (a) By definition and the properties of the ideal transformer

z11 =V1I1

I 2=0

= R1 + R2 and z22 =V2I2

I1=0

= a2R2 . Additionally, since the circuit is obviously

reciprocal, z21(= z12) =V2I1

I2=0

= aR2 .

(b) The input impedance is given by the formula

Zin = z11 −z12z21

z22 + ZL= R1 + R2 −

a2R22

a2R2 + R2= R1 +

1

a2 +1R2

(c) If port-1 is open circuited, I1 = 0. Hence,

V1 = z12I2 =2aKR2

s2 + 4 and V2 = z22I2 =

2a2KR2

s2 + 4

SOLUTION PROBLEM 19.28. (a) For this part consider the figure below:


With port 2 open, it follows that Zpr = 320/16 = 20 Ω and

z11 =V1

I1

I2 =0

= 6 +20 ×12

20 + 12=13.5 Ω

With port 1 open and I2 injected into port 2, we have Zsec = 12*16 = 192 Ω and

z12 =V1

I2

I1= 0

=Vpr

I2

I1=0

=−Vsec

4I2

I1=0

=−I2 320 / /192( )

4I2

I1=0

= −30 Ω

With port 2 open and I1 injected into port 1, we have Zpr = 20 Ω and

z21 =V2

I1

I2 =0

=Vsec

I1

I2 =0

=−4Vpr

I1

I2 =0

=−4I1 12 //20( )

I1

I2 =0

= −30 Ω

With port 1 open, it follows that Zsec = 12*16 = 192 Ω and

z22 =V2

I2

I1=0

= 80 +320 ×192

320 + 192= 200 Ω

(b) V1(s) = z12I2 (s) =−300

s2 + 4 and V2(s) = z22I2(s) =

2000

s2 + 4.


SOLUTION PROBLEM 19.29. (a) Writing two mesh equations we have by inspection,

V1

V2

=

Z1 + Z3 Z3

Z3 + rm Z2 + Z3

I1

I2

=

z11 z12

z21 z22

I1

I2

(b) V1 = z11I1 =K Z1 + Z3( )

s and V2 = z21I1 =

K Z3 + rm( )s

SOLUTION PROBLEM 19.30. From the result of problem 29,

Z1 + Z3 Z3

Z3 + rm Z2 + Z3

=

z11 z12

z21 z22

Therefore, Z3 = z12 , Z1 = z11 − Z3 = z11 − z12 , Z2 = z22 − Z3 = z22 − z12 , and

rm = z21 − Z3 = z21 − z12 .

SOLUTION PROBLEM 19.31. (a) The z-parameters can be computed by inspection (first writethe z-parameters of the passive part of the network, i.e., with the dependent source ignored; thenadd the effect of the dependent source to the resulting equations.) As such, using loop equations,

Z =5 +

10

s10 +

10

s10

s10 +

10

s

=10s

0.5s +1 s +1

1 s +1

Ω

(b) Zin =5s +10

s−

10

s×

10

s(s +1)

10(s +1)

s+10

=5s +10

s−

10(s +1)s(2s +1)

=(5s +10)(2s +1)−10(s +1)

2s(s + 0.5)

=5s2 + 7.5ss(s + 0.5)

=5s + 7.5s + 0.5

(c) I1(s) =V1(s)Zin (s)

=10s

×s + 0.5

5s + 7.5=

0.6667s

+1.3334s +1.5

. Hence


i1(t) =23

+43

e−1.5t

u(t) A

SOLUTION PROBLEM 19.32. (a) To find the resonant frequency we first find

Zin (s) = z11 −z12z21

z22 + ZL= z11 −

z11z11

z11 + 10=

10z11

z11 +10=

10

1 + 10

z11

=10

1 + s2 + 25

s

=10s

s2 + s + 25

This is of the form of equation 17.18 with K = 10. Here according to equation 17.19f,

ωr = ω p = 25 = 5 rad/s

(b) To find Q we use equation 17.19e, i.e.,

Q = Qp =ω p

2σ p=

5

1= 5

(c) Using MATLAB we obtain the frequency response plots below:


0 2 4 6 8 10 12 14 16 18 200

1

2

3

4

5

6

7

8

9

10

Frequency in rad/s

Mag

nitu

de o

f Zin

Ω

TextEnd


0 2 4 6 8 10 12 14 16 18 20-100

-80

-60

-40

-20

0

20

40

60

80

100

Frequency in rad/s

Pha

se o

f Zin

deg

rees

TextEnd

SOLUTION PROBLEM 19.33. (a) Since Z =z11 z12

z21 z22

=

2 −3

30 4

Ω. Assuming that Zin does

not include Zs, it follows that

Zin = z11 −z12z21

z22 + ZL= 2 +

9024

= 5.75 Ω

Assuming that Zout does not include the parallel connection of ZL, then

Zout = z22 −z12z21

z11 + Zs= 4 +

903

= 34 Ω

(b) From equation 19.27

GV = Gv2Gv1 =ZL

z22 + ZL

z21

Zin + ZS

.


Thus in MATLAB»Gv = (20/(4 + 20))*(30/(5.75+1))Gv = 3.7037e+00»% Therefore»v2 = Gv*30v2 = 1.1111e+02

Hence, v2(t) = 111.11u( t) V. The power absorbed by ZL is therefore (in watts):

»PZL = v2^2/20PZL = 6.1728e+02

SOLUTION PROBLEM 19.34. (a) Before determining b, it is necessary to compute the

impedance seen at the secondary of the 2-port. Here,

Zout = z22 −z12z21

z11 + Zs= 33 2R0 −

2 3R02

R0 + R0= 33 2R0 − 2R0 = 32 2R0

Thus b must be chosen so that the impedance reflected to the secondary of the transformer is 2

Ω, i.e.,

2 =32 2R0

b2 ⇒ b = 4 R0

(b) First do a source transformation on the front end of the two port to obtain

vs(t) = R0 2 cos(2t) = 32 2 cos(2 t) V

Therefore, Vs,eff = 32 V. Also note that the impedance looking into the primary of the two port

is

Zin = z11 −z12z21

z22 + ZL= R0 −

2 3R02

33 2R0 + 32 2R0= 31.015 Ω

Hence

GV = Gv2Gv1 =ZL

z22 + ZL

z21Zin + ZS

= 0.5

Consider the following MATLAB calculations:

»a=2; R0 = 16;

»Zin = a*R0-2*a^3 * R0^2/(65*a^2 *R0)


Zin = 3.1015e+01

»ZL = 32*a^2 * R0

ZL = 2048

»Zs = a*R0

Zs = 32

»z22 = 33*a^2*R0

z22 = 2112

»z21 = 2*a*R0

z21 = 64

»Gv = (ZL/(z22+ZL))*(z21/(Zin+Zs))

Gv = 5.0000e-01

»V1eff = 32;

»V2eff = Gv*V1eff

V2eff = 16

»b = 4*a*sqrt(R0)

b = 32

»VLoadeff=V2eff/b

VLoadeff = 5.0000e-01

»Pmax = VLoadeff^2/2

Pmax = 1.2500e-01

Hence, max power transferred to the load is 125 mW.

SOLUTION PROBLEM 19.35. (a) Using the usual formula

Z(s) = z11 −z12z21

z22 + ZL= 0 −

1000 × (−1000)

0 +108 s= 0.01s

Yes, the input to the 2-port looks like a 0.01 H inductor.

(b) Under the given conditions, the circuit reduces to parallel RLC with Req = 50 kΩ, L = 0.01

H, and C = 100 pF. Therefore, m =1

LC= 106 rad/s and B =

1ReqC

= 2 ×105 rad/s.

(c) For this circuit


V1(s) =1

1

Req+

1

Ls+ Cs

×VinR1

=s C

s2 +1

ReqCs +

1

LC

×VinR1

=106s

s2 + 2 ×106s +1012 Vin

For the impulse response, Vin = 1 and from MATLAB

»n = [1e6 0];d = [1 2e6 1e12];

»[r,p,k] = residue(n,d)

r =

1.0000e+06

-1.0000e+12

p =

-1000000

-1000000

k =

[]

Therefore, the impulse response is:

h(t) = 106 −1012t( )e−106 tu(t) V

(d) y =0 1000

−1000 0

−1

=0 −0.001

0.001 0

S

(e) V2(s) =−y21

y22 +YLV1(s) =

−0.001

0 +10−8 s×

106 s

s2 + 2 ×106s +1012 =−1011

(s +106)2

Hence

v2(t) = −1011te−106 tu( t) V.



(a) Zin2 = z11 −z12z21

z22 + ZL= 62.582 −

1.2075 × 63.751.25 + 0.016

= 1.7778 kΩ

Because z12 for stage 1 is zero,

Zin = z11 −z12z21

z22 + Zin2 / /2= z11 = 2 kΩ

(b) Gv1 =V1Vs

=Zin

Zin + 75= 0.96386 . Let ZL1 = Zin2 //2 = 0.94118 kΩ. Then

Gv2 =V2V1

=ZL1

ZL1 + z22×

z21Zin

= −22.472. Finally Gv3 =VoutV2

=ZL

ZL + z22×

z21Zin2

= 0.45319.

Thus

Gv = Gv1Gv 2Gv 3 = −9.816

SOLUTION PROBLEM 19.37. (a) By inspection via mesh standard equations

z =L1s Ms

Ms L2s

(b) Utilizing the properties of an ideal transformer,

z11 =V1I1

I 2=0

= L1s z22 =V2I2

I1=0

=k2L2L1

L1s + (1− k2)L2s = L2s

z12 =V1I2

I1=0

=k L2

L1L1s = k L1L2s = Ms

Finally,

z21 =V2I1

I2 =0

=k L2


(c) Utilizing the properties of an ideal transformer,


z22 =V2I2

I1=0

= L2s z11 =V1I1

I 2=0

=k2L1L2

L2s + (1− k2)L1s = L1s

z12 =V1I2

I1=0

=k L1


Finally,

z21 =V2I1

I2 =0

=k L1


(d) For this circuit k = 1 and the turns ratio a =10−2

1= 0.1. Under this condition the given

circuit reduces to a current source of value Is = Vin 5000 driving a parallel RLC with R =

5000 Ω, L = L1 = 0.01 H, the capacitance reflected to the primary of value C = 10−8 F.

Therefore

m =1

LC= 105 rad/s and B =

1ReqC

= 2 ×104 rad/s.

Finally, at w = wm, the circuit is resonant and Vin appears across the primary of the transformer.

This voltage is then stepped up by a factor of 10. Therefore VoutVin max

= 10.

SOLUTION PROBLEM 19.38. (a) By inspection

V1

I2

=

Z1 1

−1 Y2

I1

V2

=

h11 h12

h21 h22

I1V2

(b) By inspection

V1 = V2 − Z1I2 ⇒ V1 + Z1I2 = V2

and

I1 = Y2V1 − I2 ⇒ Y2V1 − I2 = I1


In matrix form,

V1

I2

=

0 1

1 0

I1V2

⇒

V1

I2

=

1 Z1

Y2 −1

−1 0 1

1 0

I1

V2

Thus

V1

I2

=

1Z1Y2 +1

Z1 1

−1 Y2

I1

V2

SOLUTION PROBLEM 19.39. (a) From problem 19.38b we have

V1

I2

=

1Z1Y2 +1

Z1 1

−1 Y2

I1

V2

=

11

RCs+1

1

Cs1

−11

R

I1V2

=

1RCs +1

R RCs

−RCs Cs

I1

V2

(b) This part is a cascade of part (a) and an ideal transformer. Label the voltage and current at

the port 2 of N1 as ˆ V 2 and ˆ I 2 . From the properties of the ideal transformer, V2 = n ˆ V 2 and

I2 = ˆ I 2 n . Hence

V1ˆ I 2

=

1RCs +1

R RCs

−RCs Cs

I1ˆ V 2

⇒

V1

nI2

=

1RCs +1

R RCs

−RCs Cs

I1

V2 n

Therefore

V1

I2

=

1RCs +1

R RCs n

−RCs n Cs n2

I1V2

From table 19.1, if h22 = Cs

n2(RCs +1) ≠ 0, then the z-parameters exist and if h11 =

RRCs +1

≠ 0,

the y-parameters exist, i.e., if C ≠ 0 and R ≠ 0 respectively.



SOLUTION PROBLEM 19.40. (a) Let Z1 = R and Z2 = 1/Cs or Y2 = Cs. From problem 38,

h =Z1 1

−1 Y2

=

R 1

−1 Cs

(b) This part is a cascade of an ideal transformer and part (a). Label the voltage and current at

the port 1 of N1 as ˆ V 1 and ˆ I 1 . From the properties of the ideal transformer, V1 = −b ˆ V 1 and

I1 = − ˆ I 1 b . Hence

ˆ V 1I2

=R 1

−1 Cs

ˆ I 1V2

⇒

−V1 b

I2

=

R 1

−1 Cs

−bI1V2

Therefore

V1

I2

=

b2R −b

b Cs

I1

V2

From table 19.1, if h22 = Cs ≠ 0, then the z-parameters exist and if h11 = b2R ≠ 0, the y-

parameters exist, i.e., if C ≠ 0 and R ≠ 0 (assuming reasonably that b ≠ 0) respectively.

SOLUTION PROBLEM 19.41. For this solution we apply the definition of h-parameters: by

inspection

h11 =V1I1 V2 =0

=1

2 + 2s

h21 =I2I1 V2 =0

=2V1 − 2sV1

I1 V2=0

=(2 − 2s)V1(2 + 2s)V1 V2 =0

=1− ss +1

When I1 = 0, then I2 = 2V1 + 2V1 = 4V1 and V2 = 0.5(2V1) +12s

(2V1) =s +1

sV1.


Therefore, h12 =V1V2 I1=0

=s

s +1 and h22 =

I2V2 I1=0

=4V1V2 I1=0

=4s

s +1.

SOLUTION PROBLEM 19.42. (a) In MATLAB

»h11 = 250; h12 = 0.025; h21 = 12.5; h22 = 2.25e-3;

»Zs = 1e3; ZL = 500;

»YL = 1/ZL

YL = 2.0000e-03

»Zin = h11 - h12*h21/(h22 + (1/ZL))

Zin = 1.7647e+02

»Yout = h22 - h12*h21/(h11 + Zs)

Yout = 2.0000e-03

»Zout = 1/Yout

Zout = 500

(b)

»% Gv1 = V1/Vs

»Gv1 = Zin/(Zin + Zs)

Gv1 = 1.5000e-01

»% Gv2 = V2/V1

»Gv2 = -h21/(Zin*(h22 + YL))

Gv2 = -1.6667e+01

»Gv = Gv1*Gv2

Gv = -2.5000e+00

(c) Given the above, the Thevenin equivalent seen by the capacitor is Voc = −2.5Vin and

Rth = 500 Ω.

In MATLAB

»Zth = ZL*Zout/(ZL + Zout)

Zth = 250

»Vin = 10;

»Voc = -2.5*Vin;

»w = 400;


»Zc = 1/(j*w*10e-6)

Zc = 0 - 2.5000e+02i

»Vc = Voc*Zc/(Zth + Zc)

Vc = -1.2500e+01 + 1.2500e+01i»V2mag = abs(Vc)

»V2mag = abs(Vc)

V2mag =

1.7678e+01

»V2ang = angle(Vc)*180/pi

V2ang =

135

From above,

v2(t) = 17.678 2 cos(400 t +135o) V

Therefore

»Pave = V2mag^2/500

Pave =

6.2500e-01


(a) Using the h-parameters of stage 2

Zin2 = h11 - h12h21

h22 +YL = 1000+ 0.966×51

0.0008 + 1/64 = 4000 Ω

The load for stage 1 is the parallel combination of Zin2 and the 3 kΩ resistance. However,because h12 = 0, the input impedance is unaffected by the load, and hence for stage 1,

Zin1 = h11 = 2000 Ω

(b) For stage 1, because h12 = 0, the output impedance is unaffected by the source impedance.Thus,

Yout1 = h22 = 0.05 ×10-3 S,

Zout1 =1/Yout1 = 20 ×103 Ω


For stage 2, the source impedance is the parallel combination of Zout1 and the 3 kΩ resistance.Thus

Zs2 = 20000×300020000 + 3000

=2608.7 Ω

and

Yout2 = h22 - h12h21

h11 +Zs2 = 0.0008+ 0.966×51

1000 + 2608.7 = 0.0145 S

Zout2 =1/Yout2 = 69.19 Ω

(c)

V1

Vs = Z in1

Z in1 + Zs = 2000

2000 + 2000 = 0.5

The load of stage 1 is the parallel combination of Zin2 and Zm. Thus

YL1 = Y in2 + 1/3000 = 5.834 ×10-4 S

HenceZL1 = 1/ YL1 = 1714 Ω

and

V2

V1 stage 1 = 1

Z in1× -h21

h22 + YL1 = 1

2000× -50

(0.05 + 0.5834) ×10-3 = - 39.46

For stage 2, the load is 64 Ω. Hence

V2

V1 stage 2 = 1

Z in2× -h21

h22 + YL2 = 1

4000× 51

(0.8 + 1000/64) ×10-3 = 0.7762

Finally, the overall voltage gain is the product of the three gains calculated above

Vout

Vs = 0.5 ×(- 39.46) ×(0.7762) = - 15.32

(d) The input circuit consists of a series connection of Vs, Zs, C and Zin1. The remainder of thecircuit is resistive and has no effect on the frequency response. The magnitude response is of thehigh pass type with

f3dB =1

2π(Rs + Rin1)C=

1

2π(2000 + 2000)10−6 = 39.789 Hz

SOLUTION PROBLEM 19.44. To meet the required matching, we must have


Zout = Zout2 = ZL = 64 = 1Yout2

This requires that

Yout2 = h22 - h12h21

h11 +Zs2 = 0.0008+ 0.966×51

1000 +Zs2 = 1

64

Solving for Zs2, we obtain Zs2 = 2323.2 Ω. Now for stage 2, Zs2 is the parallel combination ofZout1 = 20 kΩ and Zm:

Zs2 = 20000×Zm

20000 +Zm = 2323.2 Ω

from which Zm = 2628.5 Ω. With this new value of Zm, we repeat the calculations of problem19.43 to obtain Zin = 2000 Ω and Vout/Vs = -14 .26. Details follow.

Using the h parameters of stage 2

Zin2 = h11 - h12h21

h22 +YL = 1000+ 0.966×51

0.0008 + 1/64 = 4000 Ω

For stage 1, because h12 = 0, the input impedance is not affected by the load and

Zin1 = h11 = 2000 Ω. The voltage gains of the various stages are:

V1

Vs = Z in1

Z in1 + Zs = 2000

2000 + 2000 = 0.5


YL1 = Y in2 + 1/2628.5 = 6.3044 ×10-4 SHence

V2

V1 stage 1 = 1

Z in1× -h21

h22 + YL1 = 1

2000× -50

(0.05 + 0.63044) ×10-3 = - 36.74


V2

V1 stage 2 = 1

Z in2× -h21

h22 + YL2 = 1

4000× 51

(0.8 + 1000/64) ×10-3 = 0.7762


Vout

Vs = 0.5 ×(- 36.74) ×(0.7762) = - 14.26



(a) Using the h-parameters of stage 2

Zin2 = h11 - h12h21

h22 +YL = 500 + 0.966×51

0.0016 + 1/32 = 2000 Ω

The load for stage 1 is the parallel combination of Zin2 and the 1.5 kΩ resistance. However,

because h12 = 0, the input impedance is unaffected by the load, and Zin1 = h11 = 1000 Ω.

(b) For stage 1, because h12 =0, the output impedance is unaffected by the sourceimpedance, and

Yout1 = h22 = 0.1 ×10-3 S,

Zout1 =1/Yout1 =104 Ω

For stage 2, the source impedance is the parallel combination of Zout1 and the 1.5 kΩ resistor

Zs2 =104 ×1.5 ×103

104 +1.5 ×103 =1304.3 Ω

and

Yout2 = h22 - h12h21

h11 +Zs2 = 0.0016+ 0.966×51

500 + 1304.3 = 0.0289 S

Zout2 =1/Yout2 = 34.6 Ω

(c)

V1

Vs = Z in1

Z in1 + Zs = 1000

1000 + 1000 = 0.5


YL1 = Y in2 + 1/1500 = 1.1667 ×10-3 SHence

ZL1 = 1/ YL1 = 857.1 Ω

V2

V1 stage 1 = 1

Z in1× -h21

h22 + YL1 = 1

1000× -50

(0.1 + 1.1667) ×10-3 = - 39.47



V2

V1 stage 2 = 1

Z in2× -h21

h22 + YL2 = 1

2000× 51

0.0016 + 1/32 = 0.7762


Vout

Vs = 0.5 ×(- 39.47) ×(0.7762) = - 15.32

(d) The input circuit consists of a series connection of Vs, Zs, C and Zin1. The remainder of thecircuit is resistive and has no effect on the frequency response. The magnitude response is of thehighpass type with

f3dB =1

2π(Rs + Rin1)C=

1

2π(1000 +1000)10−6 = 79.58 Hz

SOLUTION PROBLEM 19.46. (a) Since the currents through YL and h22 are the same, h22 = YL.

(b) From current division, I2 =YL

YL + h22h21I1 ⇒

I2I1

=YL h21

YL + h22.

(c) 150 =I2I1

=YL h21

YL + h22= 0.5h21 ⇒ h21 = 300.

(d) h12 =V1V2

I1=0

=−12

= −0.5 .

(e) I1Is

=Zs

Zs + Zin=

9 ×103

9 ×103 + Zin= 0.9 ⇒ Zin =1000 Ω. Given this quantity,

h11 = Zin +h12h21

h22 +YL=1000 −

1500.25

= 400 Ω.

SOLUTION PROBLEM 19.47. Recall that

V1

I2

=

h11 h12

h21 h22

I1

V2


(a) From this expression and specification 1, h12 =V1

V2

I1=0

= 0 .

(b) From the formula for Yout (equation 19.50), specification 2, and the result of part (a), we

have

Yout =1

Zout=

1

800= h22 −

h12h21

h11 + Zs= h22

Hence, h22 =1

800= 1.25 mS.

For maximum power transfer from amplifier to the load,

Zout = 800 = b2ZL = 8b2

Therefore, b = 10.

(c) and (d) Observe that

Zin = h11 −h12h21

h22 + YL b2 = h11

From specification 3 and voltage division,

V1

Vs=

24

25=

Z in

Zin + 40

Equivalently,

Vs

V1=

25

24= 1+

40

Zin

Hence h11 = Z in = 40 × 24 = 960 Ω.

(e) From equation 19.51,

Gv2 =V2

V1= −100 =

− h21

Zin h22 + YL b2( ) =−h21

960 1.25 ×10−3 +1.25 × 10−3( ) =−h21

2.4

Hence, h21 = 240 .


(f) The power delivered to the load is PL =V2

2

800 and the power delivered to the amplifier is

Pamp =V1

2

Zin=

V12

960. Therefore the power gain is

PL

Pamp=

96

80

V2

V1

2

= 1.2 ×104

SOLUTION PROBLEM 19.48. Recall that

V1

I2

=

h11 h12

h21 h22

I1

V2

(a) From this expression and specification 1, h12 =V1V2

I1=0

= 0.01.

(b) For maximum power transfer from amplifier to the load,

Zout = 800 = b2ZL = 8b2

Therefore, b = 10.

Now we find Zin. From specification 3 and voltage division,

V1

Vs=

24

25=

Z in

Zin + 40 (1)

Equivalently,

VsV1

=2524

=1 +40Zin

⇒ Zin = 40 × 24 = 960 Ω

Using the formula for Zin we have the following equation

Zin = 960 = h11 −0.01h21

h22 +1.25 ×10−3 ⇒ h11 = 960 + 0.01h21

h22 +1.25 ×10−3 (2)


But from the given specs,

V2V1

= −100 =−h21

Zin h22 + YL b2( ) =−h21

960 h22 +1.25 ×10−3( )which implies that

h21

h22 +1.25 ×10−3( ) = 960 ×102 (3)

Substituting (3) into (2) allows us to solve for h11:

h11 = 960 + 0.01× 960 ×102 =1920 Ω

Now let us rewrite equation (3) as:

h21 − 960 ×102h22 = 960 ×102 ×1.25 ×10−3 = 120 (4)

Also

Yout = 1.25 ×10−3 = h22 −h12h21

h11 + Zs= h22 −

0.01h211960

Equivalently,

1960 ×1.25 ×10−3 = 2.45 =1960h22 − 0.01h21 (5)

Solving equations (4) and (5) simultaneously in MATLAB yields

»A = [1 -960e2;-0.01 1960]

A =

1.0000e+00 -9.6000e+04

-1.0000e-02 1.9600e+03

»b = [120; 2.45]

b =

1.2000e+02

2.4500e+00

»x = A\b

x =


4.7040e+02

3.6500e-03

»h21 = x(1); h22 = x(2);

h = [1920 0.01;h21 h22]

h =

1.9200e+03 1.0000e-02

4.7040e+02 3.6500e-03

We can verify these results as well as compute the overall amplifier gain using the following m-

file:

% two-port analysis in terms of h-parameters

function [zin, zout] =twoport(h, zL, zs)

['twoport analysis using h-parameters']

h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2);

zin = h11 - h12*h21/(h22+ 1/zL)

yout= h22 - h12*h21/(h11+zs);

zout= 1/yout

v1tovs= zin/(zin+zs)

v2tov1= -h21/(zin*(h22+1/zL))

v2tovs= v1tovs*v2tov1

»twoporth(h,ZL,Zs)

ans =

twoport analysis using h-parameters

zin =

960

zout =

8.0000e+02

v1tovs =

9.6000e-01

v2tov1 =

-100

v2tovs =

-96


Hence the overall voltage gain is VL/Vs = –96/10 = –96 because of the transformer.

Finally to compute power gains,

»Vs = 1; Vin = 24/25;

»VL = -9.6;

»Pin = Vin^2/960

Pin =

9.6000e-04

»Pload = VL^2/8

Pload =

1.1520e+01

»Pgain = Pload/Pin

Pgain =

12000


(a) h21 =I2I1 V2 =0

=−C sVπ + gmVπ

1

Rπ+ Cπ + C( )s

Vπ

=−C s + gm

Cπ + C( )s +1

Rπ

(b) h11 =V1I1 V2 =0

=

Rx +1

1

Rπ+ Cπ + C( )s

I1

I1= Rx +

11

Rπ+ Cπ + C( )s

(c) Under the condition that I1 = 0, V1 = Vπ. Using voltage division from V2 to Vπ:


h12 =V1V2 I1=0

=VπV2

=

11

Rπ+ Cπs

1

C s+

11

Rπ+ Cπs

=C s

1

Rπ+ Cπs + C s

=C s

Cπ + C( )s +1

Rπ

(d) h22 =I2V2 I1=0

=gmVπ +

1

Rπ+ Cπs

Vπ

Cπ + C( )s +1

RπC s

Vπ

=C s Cπs + gm +

1

Rπ

Cπ + C( )s +1

Rπ

SOLUTION PROBLEM 19.50: (a) Recall t-parameter relationship:

V1

I1

=

t11 t12

t21 t22

V2

−I2

For the given network,V2 = ZL (−I2) and

V1 = t11V2 + t12(−I2) = t11ZL (− I2) + t12(− I2) = (t11ZL + t12)(−I2)

Further,

I1 = t21V2 + t22(− I2) = ( t21ZL + t22)(−I2)

Hence,

Zin =V1I1

=( t11ZL + t12)(−I2)( t21ZL + t22)(−I2)

=t11ZL + t12t21ZL + t22

(b) For the output impedance relationship, from the t-parameter relationships

V1 = Zs(− I1) = t11V2 + t12(−I2) = −Zs(t21V2 + t22(− I2))

Grouping V2 and I2 terms together on separate sides of the equation implies

(t21Zs + t11)V2 = (t22Zs + t12)I2Thus


Zout =V2I2

=t22Zs + t12t21Zs + t11

SOLUTION PROBLEM 19.51:

45. From the z-parameter relationships

V1 = z11I1 + z12I2 ⇒ V1 − z11I1 = z12I2 = −z12(− I2)and

V2 = z21I1 + z22I2 ⇒ z21I1 = V2 + z22(− I2)

These two equations in matrix form are:

1 −z11

0 z21

V1

I1

=

0 − z12

1 z22

V2

−I2

Solving for V1 I1[ ]T yields

V1

I1

=

1z21

z21 z11

0 1

0 − z12

1 z22

V2

− I2

=

1z21

z11 ∆z

1 z22

V2

− I2

SOLUTION PROBLEM 19.52: For figure 19.52a, by inspection

V1

I1

=

1 Z1

0 1

V2

−I2

Therefore T is as indicated in the problem.For figure 19.52b, by inspection

V1

I1

=

1 0

Y2 1

V2

− I2

Therefore T is as indicated in the problem.

SOLUTION PROBLEM 19.53: Here we use the results of problem 19.52:

(a) For figure (a)

Tnew =1 Z1

0 1

1 0

Y2 1

=

1+ Z1Y2 Z1

Y2 1

(b) For figure (b)


Tnew =1 0

Y2 1

1 Z1

0 1

=

1 Z1

Y2 1 + Z1Y2

SOLUTION PROBLEM 19.54: By the properties of an ideal transformer,

V1 = nV2 I1 =−1n

I2 =1n

(− I2)

ThereforeV1

I1

=

n 0

0 1 n

V2

−I2

with the t-parameters given by the 2x2 matrix.

SOLUTION PROBLEM 19.55: This problem uses the results of the previous two problems.(a)

T =1+ Z1Y2 Z1

Y2 1

n 0

0 1 n

=

n 1+ Z1Y2( ) Z1 n

nY2 1 n

(b)

T =n 0

0 1 n

1 + Z1Y2 Z1

Y2 1

=

n 1+ Z1Y2( ) nZ1

Y2 n 1 n

SOLUTION PROBLEM 19.56:(a)

T =0.25 0

0 4

8 4

2 5

=

2 1

8 20

(b)

T =1 0

1 R 1

8 4

2 5

=

8 4

2 + 8 R 5 + 4 R

(c)

T =1+ Z1Y2 Z1

Y2 1

8 4

2 5

=

1+ 0.25s 0.5s

0.5 1

8 4

2 5

=

8 + 3s 4 + 3.5s

6 7

SOLUTION PROBLEM 19.57: For each 2-port of the form of figure P19.53a, we have that the t-

parameters are given by


T =1+ Z1Y2 Z1

Y2 1

The given network consists of three such sections in cascade whose t-parameters are respectiely,

T1 =1 + s2 s

s 1

, T2 =

1+1× 0.5s 1

0.5s 1

=

1 + 0.5s 1

0.5s 1

, and

T3 =1 + 2s ×

1

4s 2s

1

4s 1

=1+

1

2s2 2s

1

4s 1

Observe that

T1T2 =

1

2s3 +

3

2s2 +

1

2s +1 s2 + s +1

1

2s2 +

3

2s s +1

and the overall t-parameters are

T = T1T2T3 =

1

4s5 +

3

4s4 + s3 +

9

4s2 +

3

4s +1 s4 + 3s3 + 2s2 + 3s +1

1

4s4 +

3

4s3 +

3

4s2 +

7

4s s3 + 3s2 + s +1

SOLUTION PROBLEM 19.58: This poblem is done primarily in MATLAB.

Part (a)% The following code solves part (a) of the problem.% Parameter Specificationt11= 0.895+j*0.022;t22= t11;t12= 40 + j*180;

% t21= (t11*t22 -1)/t12;% The above formula follows because it is a reciprocal network.% The actual value is specified.t21= -2.6175e-05+j*1.1023e-03;t=[t11 t12; t21 t22]

% Part (a) calculationsvr= 115200ir = 361

vs= t11*vr +t12*irmagvs=abs(vs)angvs= angle(vs)*180/pi


is=t21*vr+t22*irmagis= abs(is)angis= angle(is)*180/pipscomp=vs*conj(is)ps=real(pscomp)pr=real(vr*conj(ir))eff= pr/pspf= ps/abs(pscomp)ploss= ps- pr

The MATLAB output is as follows:

T = 8.9500e-01 + 2.2000e-02i 4.0000e+01 + 1.8000e+02i -2.6175e-05 + 1.1023e-03i 8.9500e-01 + 2.2000e-02ivr = 115200ir = 361

vs = 1.1754e+05 + 6.7514e+04imagvs = 1.3555e+05angvs = 2.9872e+01

is = 3.2008e+02 + 1.3493e+02imagis = 3.4736e+02angis = 2.2857e+01

pscomp = 4.6733e+07 + 5.7501e+06ips = 4.6733e+07pr = 41587200

eff = 8.8989e-01pf = 9.9252e-01ploss = 5.1458e+06

Part (b)% The following code solves part (b) of the problem.

zL=500;zin= (t11*zL + t12)/( t21*zL + t22)yin = 1/zinvsnew=134000;iin= yin*vsnewpsnew= vsnew^2*real(yin)m= inv(t)v2=m(1,1)*vsnew +m(1,2) *iinmagv2=abs(v2)iload= m(2,1) *vsnew + m(2,2)* iinmagild = abs(iload)% Check value of rloadrload=abs(v2)/abs(iload)

The MATLAB output for part (b) is:


zin = 4.8759e+02 - 1.0031e+02iyin = 1.9676e-03 + 4.0478e-04iiin = 2.6366e+02 + 5.4241e+01ipsnew = 3.5331e+07

% m is the inverse of T-matrix.

m = 8.9500e-01 + 2.2000e-02i -4.0000e+01 - 1.8000e+02i 2.6175e-05 - 1.1023e-03i 8.9500e-01 + 2.2000e-02i

v2 = 1.1915e+05 − 4.6681e+04imagv2 = 1.2796e+05iload = 2.3829e+02 - 9.3362e+01imagild = 2.5593e+02rload = 5.0000e+02

SOLUTION PROBLEM 19.59. From the given information, the circuit is linear and reciprocal.(a) Here i2( t) is the integral of i1(t). Therefore, the new v1( t) is the integral of the old v2(t).The result for t ≥ 0 is:

v1( t) = 3.005 − 3e−t + e−t[0.00865sin(500 t) − 0.005cos(500 t)]

= 3.005 − 3e−t + e−t −2 ⋅10−5 cos(500t − π /6) + 0.01sin(500t − π /6)[ ](b) From the problem statement

z21(s) =3

s +1+ 5

cosπ6

s +1( ) + 250

s +1( )2 + 500( )2

From reciprocity, z12(s) = z21(s). For steady state analysis, we use phasors to obtain

V1 = z12( j500)I2 =3

1+ j500+ 5

cosπ6

(1+ j500) + 250

(1+ j500)2 + (500)2

v1ss (t) = 2.505cos(500 t − 30.15o) V

SOLUTION PROBLEM 19.60. Writing loop equations we have:

(i) For the left loop,

V1 − aV2 − 3(I1 + I3) = 0 ⇒ V1 − aV2 = 3I1 + 3I3

(ii) For the right loop,


V2 + bI1 − (I2 − I3) = 0 ⇒ V2 = −bI1 + I2 − I3

(iii) For the middle loop,

bI1 + (I3 − I2) + I3 + 3(I1 + I3) = 0 ⇒ 0 = (b + 3)I1 − I2 + 5I3

Writing the first two equations in matrix form yields

1 −a

0 1

V1

V2

=

3 0 3

−b 1 −1

I1

I2

I3

whose solution is

V1

V2

=

1 a

0 1

3 0 3

−b 1 −1

I1

I2

I3

=3 − ab a 3 − a

−b 1 −1

I1

I2

I3

HenceV1

V2

=

3− ab a

−b 1

I1

I2

+

3 − a

−1

I3

From the third equation

I3 = −0.2b − 0.6 0.2[ ] I1

I2

ThusV1

V2

=

3 − ab a

−b 1

+

3− a

−1

−0.2b − 0.6 0.2[ ]

I1

I2

V1

V2

=

−0.6b − 0.8ab + 0.6a +1.2 0.6 + 0.8a

−0.8b + 0.6 0.8

I1I2

For reciprocity,z12 = z21 ⇒ a = −b

SOLUTION PROBLEM 19.61. (a) and (b) together: Observe that

V1 − 0.5V1 − I1 = V2 ⇒ I1 = 0.5V1 − V2 ⇒ V1 = 2I1 + 2V2and

I1 + I2 = 0.5(V2 − aI1) ⇒ I2 = −(0.5a +1)I1 + 0.5V2The h-parameters are

H =2 2

−(0.5a +1) 0.5


Reciprocity requires thath12 = −h21 ⇒ 2 = 0.5a +1 ⇒ a = 2

Thus

H =2 2

−2 0.5

SOLUTION PROBLEM 19.62. There are 2 corrections in the problem statement concerning thesecond set of expressions:

(1) v1(t) = 2e-t - 1.5e-1.5t V(2) i2(t) = 0.5 e-1.5t A

For both parts, recall, the y-parameters:

I1

I2

=

y11

y21

y12

y22

V1

V2

(a) Part-1: From the first set of given data (V2 = 0)

I1 =1s, I2 =

0.5s +1

, V1 =0.5s +1

+0.5

(s +1)2 =0.5(s + 2)

(s +1)2

Hence,

y11 =I1V1

V2 =0

=2(s +1)2

s(s + 2), y21 =

I2V1

V2=0

=s +1s + 2

(a) Part-2: From the second set of given data (ZL = 1 Ω, I1 = 1 s , etc.), we have

I2 =0.5

s +1.5, V1 =

2s +1

−1.5

s +1.5=

0.5(s + 3)(s +1)(s +1.5)

For a terminated 2-port,

I2 =−V2ZL

= −V2 =y21

y22 + YLV1 =

y21y22 +1

V1

Therefore

y22 = y21V1I2

−1 =s +1s + 2

×0.5(s + 3)

(s +1)(s +1.5)×

s +1.50.5

−1=(s + 3)(s + 2)

−1 =1

s + 2

Also, I1 = y11V1 + y12V2 ⇒ y12 =I1 − y11V1

V2=

I1 − y11V1−I2

. Hence

y12 = −s +1.5

0.5×

1s

−2(s +1)2

s(s + 2)0.5(s + 3)

(s +1)(s +1.5)

= −(2s + 3)(s + 2)

s(s + 2)+

2(s +1)(s + 3)s(s + 2)


=1

(s + 2)In conclusion

y11

y21

y12

y22

=

2(s +1)2

s(s + 2)

1

(s + 2)s +1

s + 2

1

(s + 2)

=1

(s + 2)

2(s +1)2

s1

s +1 1

(b) This is a straightforward application of the conversion table 19.1.

SOLUTION PROBLEM 19.63. (a) Consider figure (a). Write two mesh equations:

V1 =16I1 +2s

(I1 + I2) + 4I1 = 20 +2s

I1 +

2s

I2

V2 = sI2 +2s

(I1 + I2) + 4I1 = 4 +2s

I1 + s +

2s

I2

Therefore

Z =1s

20s + 2

4s + 2

2

s2 + 2

Taking the inverse yields the y-parameters

Y =1

∆Z

z22 −z12

−z21 z11

=

1

20s2 + 2s + 32

s2 + 2 −2

−(4s + 2) 20s + 2

where

∆Z =(20s + 2)(s2 + 2)− 2(4s + 2)

s=

20s3 + 2s2 + 32ss

= 20s2 + 2s + 32

Finally, the h-parameters are given as

H =1 y11 z12 z22

− z21 z22 1 z22

=

1

s2 + 2

20s2 + 2s + 32 2

−(4 s + 2) s

(b) Now consider figure (b). z11 is V1 when I2 = 0. But if I2 = 0, then because of the idealtransformer I1 = 0, meaning that the ratio is not defined. Hence the z-parameters do not exist.

To find the h-parameters, observe that because of the ideal transformer, I2 = −0.5I1,I1 = −2I2 , and Vpri = 0.5Vsec . Writing a mesh equation at the right mesh first we obtain

V2 = RI2 + Vsec + (I1 + I2)R = Vsec


Now writing a mesh equation on the left we have

V1 = RI1 + Vpri + (I1 + I2)R = 1.5RI1 +Vpri = 1.5RI1 + 0.5V2

Therefore

H =1.5R 0.5

−0.5 0

To obtain the y-parameters we use table 19.1:

Y =1

h11

1 −h12

h21 ∆h

=

16R

4 −2

−2 1

Note: the det[Y] = 0 implying again that the z-parameters do not exist.

(c) For this network we consider it as a cascade (left to right) of an ideal transformer, the middlenetwork of a transformer and an inductor across the top, and finally another ideal transformer.The t-parameters of these two ports are respectively:

T1 =a 0

0 1 a

, T2 =

A B

C D

, T3 =

1 b 0

0 b

To find T2, we replace the mutually coupled inductors by the pi-equivalent circuit of figure

18.25c where L1 = 4 H, L2 = 9 H, M = k L1L2 = 3 H, and ∆ = 27. Thus Lleft =∆

L2 − M= 4.5

H, Lright =∆

L1 − M= 27 H, and Ltop =

∆M

= 9 H. Notice that the external 9 H inductor is in

parallel with Ltop leading to Lpar = 4.5 H. The y-matrix of this new pi-network is by inspection:

Ymid =1s

2 4.5 −1 4.5

−1 4.5 1 27 +1 4.5

=

127s

12 −6

−6 7

To compute the t-parameters we have

T2 =

−y22

y21

−1

y21−∆y

y21

−y11

y21

=7 6 4.5s

8 27s 2

Therefore

T = T1T2T3 =a 0

0 1 a

7 6 4.5s

8 27s 2

1 b 0

0 b

=

7a 6b 4.5abs

8 27abs 2b a

From table 19.1, we obtain


H =a22.25s 0.5a / b

−0.5a / b 4 27b2

, Z = s

63a2

16

27ab

827ab

8

27b2

4

, Y =1

27a2b2s

12b2 −6ab

−6ab 7a2


(a) The defining equation for the g-parameters is:

I1V2

=

g11 g12

g21 g12

V1

I2

Because of the load impedance ZL, we have V2 = −ZLI2 . Hence substituting for V2 in thesecond equation yields

V2 = g21V1 + g22I2 = −ZLI2 ⇒ I2 =−g21V1

g22 + ZL

Substituting this equation into I1 = g11V1 + g12I2 we obtain

I1 = g11V1 − g12g21V1

g22 + ZL

= g11 −

g12g21g22 + ZL

V1

Therefore,

Yin = g11 −g12g21

g22 + ZL

(b) Because of the source impedance Zs, we have V1 = −ZsI1 or I1 = −YsV1 . Hence substitutingfor V1 in the first equation yields

I1 = g11V1 + g12I2 = −YsV1 ⇒ V1 =−g12I2g11 + Ys

Substituting this equation into V2 = g21V1 + g22I2 we obtain

V2 = g21−g12I2g11 + Ys

+ g22I2 = g22 −

g12g21g11 + Ys

I2 ⇒ Zout = g22 −

g12g21g11 + Ys

(c)

G1 =V1Vs

=Zin

Zs + Zin=

YsYs +Yin

=Ys

Ys + g11 −g12g21

g22 + ZL

=Ys g22 + ZL( )

g22 + ZL( ) g11 +Ys( ) − g12g21

(d) Refer to figure P19.64, where V2 =ZL

g22 + ZLg21V1 ⇒ G2 =

V2V1

=g21ZL

g22 + ZL.


(e) Gv = G1G2 =Ys g22 + ZL( )

g22 + ZL( ) g11 +Ys( ) − g12g21×

g21ZLg22 + ZL

=g21YsZL

g22 + ZL( ) g11 + Ys( ) − g12g21

SOLUTION PROBLEM 19.65. Recall

I1V2

=

g11 g12

g21 g22

V1

I2

For the given circuit

V2 =14

(I2 − 4V2) + V1 ⇒ V2 =12

V1 +18

I2

Also

V2 = V1 −14

I1 − sV1( )This implies that

I1 = (4 + s)V1 − 4V2 = (4 + s)V1 − 412

V1 +18

I2

= (2 + s)V1 −

12

I2

Thus the g-parameter matrix is:

G =2 + s −

1

21

2

1

8

SOLUTION PROBLEM 19.66. We first convert the y-parameters to t-parameters using table 19.1:

YN1 → TN 1 =

−y22

y21

−1

y21−∆y

y21

−y11

y21

=−0.1 −0.5

−3.82 −25.1

where ∆y = 50.2 × 0.2 − 2 ×1.2 = 7.64 . Now we convert the h-parameters to t-parameters usingtable 19.1:

HN 2 → TN 2 =

−∆h

h21

−h11

h21−h21

h21

−1

h21

=15

56 13

2 1

=

11.2 2.6

0.4 0.2

where ∆h =13 × 2 + 6 × 5 = 56 . To obtain the cascaded t-parameters we compute


Tcas = TN1TN 2 =15

−6.6 −1.8

−264.12 −74.76

=

−1.32 −0.36

−52.824 −14.952

Thus,

Yin =1

Zin=

t21ZL + t22t11ZL + t12

=

603

515

5

= 40.2 S

To obtain the voltage gain, we first convert the t-parameters back to y-parameters (table19.1) and then use the derived voltage gain formula:

Tcas =−1.32 −0.36

−52.824 −14.952

→ Ycas =

41.533 2

2.7778 3.6667

S

Hence,

Gv =VLV1

=−y21

y22 + YL=

−2.77783.6667 + 0.5

= −0.66667

Alternately, one could consider the load as a 2-port, compute its t-parameters, construct

the overall t-parameters as a cascade of three networks, and then use Gv =1t11

.


(a) Since only one C or L , weh have in general: H(s) =as + bcs + d

. Since H(∞) = 0 =ac

, we

have that a = 0. Therefore,

H(s) =

b

c

s +d

c

=K

s + c

(b) To prove that c =1

RthC or

RthL

where Rth is the Thevenin resistance seen by the energy

storage element, we refer the reader to problem 19.69 which provides a general derivation withH(∞) arbitrary; hence this problem is the special case of H(∞) = 0.

(c) HereRth = 2000/ /(50 +1000/ /200)) = 195.49 Ω. Hence

c =1

RthC= 25.58 ×106 rad/s



(a) Because there is only one energy storage element, it follows that the most general form of

the transfer function is: H(s) =as + bcs + d

. Since H(0) = 0 =bd

, we have b = 0. Further, since the

transfer function is high pass, c ≠ 0, and

H(s) =as

cs + d=

a

c

s

s +d

c

=Ks

s + c

(b) To prove that c =1

RthC or

RthL

where Rth is the Thevenin resistance seen by the energy

storage element, we refer the reader to problem 19.69 which provides a general derivation withH(0) arbitrary; hence this problem is the special case of H(0) = 0.

(c) Let us apply the result of example 19.1 to that part of the circuit to the right of the 1 kΩresistor and call the associated resistance Zin. Here Z1 = ∞, Z2 = 2 kΩ, ZL = 100 Ω, and beta =50. Hence

Zin = beta +1( )ZL = 5.1 kΩThus

Rth = 200 +1000 //(2000 + 5100) = 1076.5 ΩHence,

c =1

RthC=

1

1076.54 × 2 ×10−6 = 464.45 rad/s

SOLUTION PROBLEM 19.69. In this problem we assume (i) a single input single output system

and that linear circuit seen by the energy storage element has a Thevenin equivalent or a

Norton equivalent. For simplicity we will presume the existence of a Thevenin equivalent.

(a) The Thevenin equivalent seen by the dynamic element L or C consists of Zth(s) in series with

Voc(s). Since the remainder network seen by L or C is non-dynamic (resistive, resistive

with dependent sources and ideal op amps, etc), we have Zth(s) = Rth and Voc(s) =

Ko×Input(s), Rth and Ko being real constants.


By voltage division,

VL =Ls

Ls + RthVoc =

K0s

s + Rth

L

× Input(s) (1)

and

VC =

1

Cs1

Cs+ Rth

Voc =

K0

RthC

s + 1

RthC

× Input(s) (2)

After VL(s) or VC(s) has been determined, we can find the Laplace transform of any other output

(voltage or current) using the voltage source substitution theorem (chapter 6) and linearity

(chapter 5):

Ouput(s) = K1 × Input(s) + K2 VL (s) or VC(s)( ) (3)

For the case of VL(s),

Ouput(s) = K1 + K2K0s

s + Rth

L

× Input(s) (4)

and for the case of VC(s),

Ouput(s) = K1 + K2

K0

RthC

s + 1

RthC

× Input(s) (5)

For either case, the transfer function H(s) has the form

H(s) =Ouput(s)

Input(s)=

K3s + K4

s +ω c(6)


where ωc =Rth

L or

1

RthC with K3 and K4 real constants.

It remains to give K3 and K4 some physical interpretations. In (6), let s = ∞, we have

H(∞) =K3s + K4

s + ωc

s=∞= K3

On the other hand, letting s = 0 in (6) produces

H(0) =K3s + K4

s +ω c

s=0

=K4

ωc

Substituting K3 and K4 into (6), we obtain the desired result

H(s) =Ouput(s)

Input(s)=

H (∞)s +ω cH(0)

s +ω c(7)

(b) When s = ∞, the impedance of C is

ZC (∞) =1

Cs

s=∞= 0

and the impedance of L isZL (∞) = Ls]s=∞ = ∞

Therefore in calculating H(∞), we may replace C by a short circuit and L by an open circuit. Onthe other hand, when s = 0, the impedance of C is

ZC (0) =1

Cs

s=0= ∞

and the iimpedance of L is

ZL (0) = Ls]s=0 = 0

Therefore in calculating H(0), we may replace C by an open circuit and L by a short circuit.


(c) For figure P19.69a, by inspection

Rth = 3//(2 + 4) = 3//6 = 2 Ωωc = 1/(Rth C) = 1/(2 × 0.5) = 1H(∞) = 4/(2 + 4) = 2/3H(0) = 4/(2 + 3 + 4) = 4/9

Therefore

H(s) =H(∞)s + ωcH(0)

s +ω c=

2

3s +

4

9s + 1

=

2

3s +

2

3

s +1

For figure P19.69b

Rth = 3//(2 + 4) = 3//6 = 2 Ωωc = Rth/ L = 2/2 =1H(∞) = 4/(2 + 4) = 2/3H(0) = 4/(2 + 3 + 4) = 4/9

Therefore

H(s) =H(∞)s + ωcH(0)

s +ω c=

4

9s +

2

3s + 1

=

4

9s + 1.5( )

s +1

SOLUTION PROBLEM 19.70. According to problem 19.68, the transfer function of the circuit is

H(s) =Ks

s + c

where c =1

RthC and Rth is the Thevenin resistance seen by the storage element C. To find Rth

we make use of figure 19.4 and the associated formula. The details are in the MATLAB code

below:

»R1 = 200*1e3/1200

R1 =

1.6667e+02

»Z1 = R1+2e3

Z1 =

2.1667e+03

»Z3 = 100;


»beta = -50;

»Zout = Z1/(1+beta)

Zout =

-4.4218e+01

»Rth = 2000 + Zout

Rth =

1.9558e+03

»wc = 1/(Rth*2e-6)

wc =

2.5565e+02

Hence

c =1

1955.8 × 2 ×10−6 = 255.65 rad/s

SOLUTION PROBLEM 19.71. (a) Except for the terminating resistor, let the other element

branches of the circuit be given by

Ζ1 =1

3s1, Y2 =

1

2s +1

2s

=2s

4s2 +1, and Z3 =1s

Consider the circuit

Here from problem 19.53a and from 19.52 a we obtain


V1

I1

=

1+

23

4s2 +1

1

3s2s

4s2 +11

V3

−I3

and

V3

−I3

=

11

s0 1

V2

− I2

which implies that

V1

I1

=

1

4s2 +14s2 + 5/ 3

16s2 + 6( )3s

2s 4s2 + 3

V2

−I2

(b)

GV (s) =−y21

y22 + yL

(c)

y21 = −1t12

= −3s(4s2 +1)

16s2 + 6, y22 =

t11t12

=3s 4s2 +

5

3

16s2 + 6=

s(12s2 + 5)

16s2 + 6, and YL = 1

(d) Hence,

GV (s) =−y21

y22 + yL=

3s(4s2 +1)

16s2 + 6s(12s2 + s)

16s2 + 6+1

=3s(4 s2 +1)

12s3 +16s2 + 5s + 6

SOLUTION PROBLEM 19.72. For part (a), treating each capacitor as a short circuit yields the

equivalent circuit below.


»h11T = 4.2e3; h12T = 0; h21T = 150; h22T = 0.1e-3;

»hT = [h11T, h12T;h21T h22T];

»yT = htoy(hT)

yT =

2.3810e-04 0

3.5714e-02 1.0000e-04

By inspection, the y11 parameter of the overall two-port consists of the sum of y11T plus the

conductances of the two front end resistors. Also, the y22 parameter of the overall two port

is y22T plus the conductance of the 4.7k Ω resistor. Hence,

»y = yT + [1/1e4+1/1e5 0;0 1/4700]

y =

3.4810e-04 0

3.5714e-02 3.1277e-04

Hence, the overall h-parameters are:

»h = ytoh(y)

h =

2.8728e+03 0

1.0260e+02 3.1277e-04

(b)

»yout = h(2,2) - (h(1,2)*h(2,1)/(h(1,1)+50))

yout =


3.1277e-04

»Zout = 1/yout

Zout =

3.1973e+03

»a = sqrt(8/Zout)

a =

5.0021e-02

(c) To compute the gain we first need Zin . From equation 19.49, since h12 = 0, Zin = h11.

» Zin = h(1,1)

Zin =

2.8728e+03

From equation 19.51,

»Gv2 = -h(2,1)/(Zin*(h(2,2)+h(2,2)))

Gv2 =

-5.7094e+01

From equation 19.52,

»Gv1 = Zin/(Zin + 50)

Gv1 =

9.8289e-01

To compute VL/Vs we use:

»Gv = Gv1*Gv2*a

Gv =

-2.8071e+00

To compute VL/V1 we use:

»Gvv = Gv2*a

Gvv =


-2.8559e+00

(d) For these calculations, we assume Vs is normalized to 1 V. Since we are computing gains,

we may do this without loss of generality.

»Vs = 1;

»Is = 1/(50 + Zin)

Is =

3.4214e-04

Now we compute the normalized power delivered by the voltage source.

»Psnorm = Vs*Is

Psnorm =

3.4214e-04

Next we compute the normalized power absorbed by the load.

»VL = Gv*1

VL =

-2.8071e+00

»PLnorm = VL^2/8

PLnorm =

9.8496e-01

Next, the power gain from source to load is:

»GpLs = PLnorm/Psnorm

GpLs =

2.8788e+03

Further, we compute the power gain from input to the two port to the load as follows:

»V1 = Vs*Zin/(Zin + 50)

V1 =


9.8289e-01

»P1 = V1*Is

P1 =

3.3629e-04

»GpL1 = PLnorm/P1

GpL1 =

2.9289e+03

(e) SPICE Simulation Because the frequency response is flat for freqency above 800 Hz, we

only plotted up to 1.6k Hz. The circuit diagram reflects the load back to the primary of the

ideal transformer. In general, this is not possible. Hence for a SPICE simulation, it is

necessary to use one of the models given in Figure 18.15 consisting of two controlled

sources. For this example, this is not necessary. Note however that the actual output

voltage is 0.05 times the values on the graph given below. This simulation assumes a 1 V

source voltage and the parameter of GVCCS is 0.035714. Notice that in this problem


MAG(V(IVM2))

Frequency (Hz)Prb 19.72-Small Signal AC-3

+0.000e+000

+10.000

+20.000

+30.000

+40.000

+50.000

+60.000

+200.000 +400.000 +600.000 +800.000 +1.000k +1.200k +1.400k +1.600k

(f) For this part, we change 100 µF to 10 µF. The resulting plot shows degradation of the low

end frequency response.

MAG(V(IVM2))

Frequency (Hz)Prb 19.72-Small Signal AC-4

+0.000e+000

+10.000

+20.000

+30.000

+40.000

+50.000

+60.000

+200.000 +400.000 +600.000 +800.000 +1.000k +1.200k +1.400k +1.600k

SOLUTIONS PROBLEMS CHAPTER 20

USEFUL MATLAB M-FILES FOR USE IN THE SOLUTION TO PROBLEMS IN THIS CHAPTER.

Program 1: converts y-parameters to t-parameters% convert y parameters to t parametersfunction [t,t11,t12,t21,t22] = ytot(y)y11=y(1,1);y12=y(1,2);y21=y(2,1);y22=y(2,2);deltay= y11*y22-y12*y21;t11=-y22/y21;t12 = -1/y21;t21= -deltay/y21;t22= -y11/y21;t= [ t11 t12; t21 t22];

Program 2: Computes Zin, Zout, and gains using two port h-parameters.»% two-port analysis in terms of h-parameters»function [zin, zout] =twoport(h, zL, zs)»['twoport analysis using h-parameters']»h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2);»zin = h11 - h12*h21/(h22+ 1/zL)»yout= h22 - h12*h21/(h11+zs);»zout= 1/yout»v1tovs= zin/(zin+zs)»v2tov1= -h21/(zin*(h22+1/zL))»v2tovs= v1tovs*v2tov1

Program 3: Computes Zin, Zout, and gains using two port y-parameters.% two-port analysis in terms of y-parametersfunction [zin, zout] =twoporty(y, zL, zs)['twoport analysis using y-parameters']y11= y(1,1); y12=y(1,2); y21=y(2,1); y22=y(2,2);yin = y11 - y12*y21/(y22+ 1/zL)zin= 1/yinyout= y22 - y12*y21/(y11+1/zs)zout= 1/youtv1tovs= zin/(zin+zs)v2tov1= -y21/(y22+1/zL)v2tovs= v1tovs*v2tov1

Program 4: Computes Zin, Zout, and gains using two port t-parameters.% two-port analysis in terms of t-parametersfunction [zin, zout] =twoportt(t, zL, zs)

['analysis of terminated twoport using t-parameters']t11= t(1,1); t12=t(1,2); t21=t(2,1); t22=t(2,2);zin= (t11*zL + t12)/(t21*zL + t22)zout= (t22*zs + t12)/(t21*zs + t11)v2tov1= zL/(t11*zL + t12)v1tovs= zin/(zin+zs)v2tovs= v2tov1*v1tovs

Program 5: converts z-parameters to t-parameters%converting z to t paramters (same sormulas as%converting t to z parameters)

function [t,t11,t12,t21,t22] = ztot(z)z11=z(1,1); z12=z(1,2); z21=z(2,1); z22=z(2,2);deltaz= z11*z22 - z12*z21;t11= z11/z21;t12= deltaz/z21;t21= 1/z21;t22= z22/z21;t= [ t11 t12; t21 t22];

SOLUTION 20.1.

For network a, the z-parameters are by inspection

Za =R1 + R3 R3

R3 R2 + R3

Similarly, for network b, the z-parameters are the same:

Zb =R1 + R3 R3

R3 R2 + R3

The interconnection of networks a and b conforms to figure 20.2b, which is a seriesinterconnection. Hence, the new overall z-parameters are

Znew = Za + Zb = 2R1 + R3 R3

R3 R2 + R3

SOLUTION 20.2. For networks a and b, the y-parameters are by inspection

Ya = Yb =3 −2

−2 4

S

Hence, their z-parameters are the inverse of the y-parameter matrix:

Za = Zb =18

4 2

2 3

Ω


Znew = Za + Zb =14

4 2

2 3

Ω

SOLUTION 20.3. For network a consisting of the single inductor,

Za =s s

s s

Ω

For network b, we have

Zb = Yb[ ]−1 =12

1 1

2 4

=

0.5 0.5

1 2

Ω


Znew = Za + Zb =s + 0.5 s + 0.5

s +1 s + 2

Ω

SOLUTION 20.4. For network a consisting of the single capacitor,

Za =1s

1 1

1 1

Ω


Znew = Za + Zb =1s

1 + 0.5s 1+ 0.5s

1+ s 1 + 2s

Ω

SOLUTION 20.5. For network a, the y-parameters are by inspection

Ya =3 −2

−2 3

S


Za =15

3 2

2 3

=

0.6 0.4

0.4 0.6

Ω


Znew = Za + Zb =1.6 0.9

0.4 −0.4

Ω

SOLUTION 20.6. For networks Na and Nb, the y-parameters are:

YNa = YNb =0.7 −0.2

−0.2 0.7

S


ZNa = ZNb =209

0.7 0.2

0.2 0.7

=

19

14 4

4 14

Ω

The network Na* has the same z-parameters as Na and continues to act as a two when seriesinterconnected to another 2-port because of the transformer. Hence, the interconnection ofnetworks Na* and Nb forms a valid series interconnection in which cas

Znew = ZNa * + ZNb =29

14 4

4 14

Ω

SOLUTION 20.7. For network Nb, the y-parameters are the same as in problem 20.6, i.e.,

YNb =0.7 −0.2

−0.2 0.7

S and ZNb =

19

14 4

4 14

Ω

For network Na, consider the purely resistive part without the transformer. The y-parameters ofthis part are half the y-parameters of Nb, i.e.,

YR =0.35 −0.1

−0.1 0.35

S

In MATLAB we use the m-file which converts y-parameters to t-parameters:

»y = [0.35 -0.1;-0.1 0.35];»t = ytot(y)t = 3.5000e+00 1.0000e+01 1.1250e+00 3.5000e+00

From figure 19.34, the transformer has to parameters with a = 2 given by

ttrans =1 a 0

0 a

=

0.5 0

0 2

»ttrans = [0.5 0;0 2];»tNa = ttrans*ttNa = 1.7500e+00 5.0000e+00 2.2500e+00 7.0000e+00»»zNa = ttoz(tNa)zNa = 7.7778e-01 4.4444e-01 4.4444e-01 3.1111e+00»»zNb = [14 4;4 14]/9;»znew = zNa + zNbznew = 2.3333e+00 8.8889e-01 8.8889e-01 4.6667e+00»

SOLUTION 20.8. This problem is identical numerically to problem 20.6. Here however theisolation transformer is on the right hand side which makes no difference to the interconnection.Therefore,

Znew = ZNa * + ZNb =29

14 4

4 14

Ω

SOLUTION 20.9. Figure 20.4 is

Using the values in figure 20.3, we obtain the following mesh equation matrix

V1

V2

0

=4 + R1 2 −R1

2 4 + R2 R2

−R1 R2 R1 + R2

I1I2

I3

For I3 to be zero, the third equation implies that R1I1 = R2I2. Therefore,

V1 = 4 + R1( )I1 + 2R1R2

I1 = 4 + R1 +2R1R2

I1

Similarly

V2 = 2I1 + 4 + R2( ) R1R2

I1+ = 2 +(4 + R2)R1

R2

I1

Therefore

V1V2

=4 + R1 +

2R1

R2

2 +( 4 + R2)R1

R2

=4R2 + R1R2 + 2R1( )2R2 + R1R2 + 4R1( ) =

4248

=78

as was to be shown.

Now suppose, V1V2

=78

or equivalently V1 =78

V2. With specific values

V1

V2

0

= V2

7 8

1

0

=10 2 −6

2 7 3

−6 3 9

I1

I2

I3

In MATLAB»Z = [10 2 -6;2 7 3;-6 3 9];»b = [7/8 1 0]';»I = inv(Z)*bI = 6.2500e-02 1.2500e-01

2.7756e-17Thus

I1

I2

I3

= V2

0.0625

0.124

0

SOLUTION 20.10(a) Write two loop equations assuming I1 and I2 are the usual port currents. Here

V1 = Z1I1 + V13 + Z3 I1 + I2( )and

V2 = Z2I2 +V23 + Z3 I1 + I2( )which in matrix form are

V1

V2

=

Z1 + Z3 Z3

Z3 Z2 + Z3

I1I2

+

V13

V23

However,V13

V23

= ZA

I1I2

ThereforeV1

V2

=

Z1 + Z3 Z3

Z3 Z2 + Z3

I1I2

+ ZA

I1I2

=

Z1 + Z3 Z3

Z3 Z2 + Z3

+ ZA

I1I2

(b) The procedure of part (a) is repeated to produce the same result.

REMARK: this problem states that the two networks are equivalent two ports. Thus theconfigurations can be used interchangeably.

SOLUTION 20.11. Because of the isolation transformers, the overall Z-parameters are the sum of

the three component Z-parameters. For Na,

Za =2 1

1 2

Ω

For Nb,

Zb =8 1

1 5

Ω

For Nc,

Zc = Za =2 1

1 2

Ω

Therefore

Z = Za + Zb + Zc =12 3

3 9

Ω

SOLUTION 20.12. (a) Given the Z-parameters of N, then

YN = ZN−1 =

7 2

10 3

−1

=3 −2

−10 7

S

The 1-F capacitor considered as a two port has y-parameters

YC =s − s

− s s

S

Therefore, the parallel connection of the two ports has y-parameters

Y = YN +YC =s −s

−s s

+

3 −2

−10 7

=

s + 3 −(s + 2)

−(s +10) s + 7

S

(b) For this part, the same reasoning applies with s replaced by 1/s.

SOLUTION 20.13. The isolation transformer allows for the valid parallel connection of Na* and

Nb in the sense that the overall y-parameters are the sum of the individual y-parameters. Further,

because the ideal transformer is 1:1 with the standard dot locations, the y-parameters of Na* are

those of Na. Further, the y-parameters of Nb are the same as those of Na as the circuits aresimple vertical flips of each other. Therefore

Ya* = Ya =

0.7 −0.2

−0.2 0.7

= Yb

Hence, the overall y-parameters are:

Y = 2 ×0.7 −0.2

−0.2 0.7

=

1.4 −0.4

−0.4 1.4

S

True-False: FALSE because the connection does not conform to figure 20.2a.

SOLUTION 20.14. From the previous example, the y-parameters of Nb are

Yb =0.7 −0.2

−0.2 0.7

S

The resistance values of the resistive part of Na are twice those of Nb. Hence, the y-parameters

of the resistive part are half those of Nb, i.e.,

I1'

I2

= Ya ,RV1

'

V2

=0.35 −0.1

−0.1 0.35

V1'

V2

S

We obtain the y-parameters of Na by considering the effect of the transformer on these y-parameters. Observe that

I1 = 2I1' and V1 = 0.5V1

'

HenceI1

I2

=

0.7 −0.2

−0.1 0.35

V1'

V2

=0.7 −0.2

−0.1 0.35

2V1

V2

=

1.4 −0.2

−0.2 0.35

V1

V2

Thus

Ya =1.4 −0.2

−0.2 0.35

S

It follows that

Y = Ya + Yb =2.1 −0.4

−0.4 1.05

S

SOLUTION 20.15. (a)

MAG(V(IVM))

Frequency (Hz)Prob 20.15-Small Signal AC-10

+0.000e+000

+200.000m

+400.000m

+600.000m

+800.000m

+1.000

+100.000m +200.000m +300.000m +400.000m +500.000m

(b)

C31.8n

R1k

R01k

C015.9n

C115.9n

R1500

IVm

V0

MAG(V(IVM))


+0.000e+000

+200.000m

+400.000m

+600.000m

+800.000m

+1.000

+10.000k +20.000k +30.000k

SOLUTION 20.16. (a) For network Na, the y-parameters by inspection are:

Ya =G1 + j C1 0

gm G0

=

2 + j10.21 0

95 0.07143

mS

For network Nb, the y-parameters by inspection are:

Yb = G f + j C2( ) 1 −1

−1 1

= 0.8333 + j1.021( ) 1 −1

−1 1

mS

Therefore

Y = Ya + Yb =2.8333 + j11.23 −0.8333− j1.021

94.167 − j1.021 0.9048 + j1.021

mS

(b), (c), and (d). Here we use the MATLAB m-file for two port analysis in terms of y-parameters:

»zs = 50; zL = 50;»twoporty(Y, zL, zs)ans =twoport analysis using y-parametersyin = 6.8501e-03 + 1.5594e-02izin = 2.3614e+01 - 5.3756e+01iyout = 5.3617e-03 + 3.0023e-03izout = 1.4199e+02 - 7.9506e+01iv1tovs = 5.5701e-01 - 3.2349e-01iv2tov1 = -4.4915e+00 + 2.6821e-01iv2tovs = -2.4150e+00 + 1.6023e+00i»

SOLUTION 20.17. The t-parameters of the LR circuit follow from problem 19.53 with Z1 = Ls =

s Ω and Z2 = 0.5 Ω:

TLR =1+ 2s s

2 1

Therefore

Tcascade = T * TLR =5 + 2s 2 + s

5 + 2s 2 + s

SOLUTION 20.18. This problem can be solved in many ways. Here we emphasize the cascadenature of the two ports.»% The y-parameters of Nb are:»Yb = [8 2;20 6];»% The z-parameters of Na are:»Za = [0.75 -0.25;-2.5 1];»% The t-parameters of Na are:»Ta = ztot(Za)Ta = -3.0000e-01 -5.0000e-02 -4.0000e-01 -4.0000e-01»% The t-parameters of Nb are:»Tb = ytot(Yb)Tb =

-3.0000e-01 -5.0000e-02 -4.0000e-01 -4.0000e-01»% The t-parameters of the cascaded two port are:»»Tab = Ta*TbTab = 1.1000e-01 3.5000e-02 2.8000e-01 1.8000e-01»% Doing a t-parameter analysis we obtain:»twoportt(Tab,0.25,0.5)ans =analysis of terminated twoport using t-parameterszin = 2.5000e-01zout = 5.0000e-01v2tov1 = 4v1tovs = 3.3333e-01v2tovs = 1.3333e+00ans = 2.5000e-01»

SOLUTION 20.19. Use the h-parameter analysis m-file.

»H1 = [1e3 0.001;50 6e-5]; H2 = [1e3 0.99;-5 8e-4];»zL = 100; zs = 2e3; zm = 10e3;»% For part (a), the common collector stage:»twoporth(H2,zL,0)ans =twoport analysis using h-parameterszin = 1.4583e+03zout = 1.7391e+02v1tovs = 1v2tov1 = 3.1746e-01v2tovs = 3.1746e-01

Remark: in the above, v2 is Vout and v1 is vm; zout has no signifigance.

»% We now compute the load on the first stage.»zin2 = 1.4583e+03;»zLm = zm*zin2/(zin2 + zm)zLm = 1.2727e+03»% zLm is the load impedance on stage 1.

»% For part (a), the common emitter stage:»twoporth(H1,zLm,zs)ans =twoport analysis using h-parameterszin = 9.4088e+02zout = 2.3077e+04v1tovs = 3.1993e-01v2tov1 = -6.2835e+01v2tovs = -2.0103e+01

Remark: in the above, v2 is vm and v1 is Vin; zout is the output impedance of stage 1.

Conclusion: the input impedance to stage 1 is 940.88 Ω and the input impedance to stage 2 is1.4583 kΩ.

(b) From the above output and remark, Vm/Vin = -62.835. Further Vout/Vm = 0.31746.

(c) From above Vm/Vs = -20.103. Therefore Vout/Vs = Vout/Vm * Vm/Vs = -6.38.

(d)»zout1 = 2.3077e+04;»zs2 = zm*zout1/(zm + zout1)zs2 = 6.9768e+03»twoporth(H2,zL,zs2)ans =twoport analysis using h-parameterszin = 1.4583e+03zout = 7.0395e+02v1tovs =

1.7289e-01v2tov1 = 3.1746e-01v2tovs = 5.4885e-02

Conclusion: Zout of amplifier is 704 Ω.

SOLUTION 20.20(a) Using MATLAB»na = 1.1514;»nb = 3.4012;»Zlprime = nb^2*75 + j*1042.94ZLprime = 8.6761e+02 + 1.0429e+03i

»Zsprime = 75/na^2 + j*30Zsprime = 5.6573e+01 + 3.0000e+01i

(b) Since the h-parameters of the transistors are given, we can again use MATLAB and the m-file twoporth defined earlier. Hence:

»h = [ 60-j*50 0.01; -j*2 0.0005+j*0.0004];»[Zin, Zout] = twoporth(h,ZLprime, Zsprime)Zin = 5.6569e+01 - 3.0000e+01iZout = 8.6763e+02 - 1.0429e+03i

(c) Observe that Zin = 5.6569e+01 - 3.0000e+01i and Zsprime = 5.6573e+01 + 3.0000e+01i,which are clearly conjugates of each other. Further, Zout = 8.6763e+02 - 1.0429e+03iand ZLprime = 8.6761e+02 + 1.0429e+03i, which are also conjugates of each other.Hence maximum power is transferred into and out of the transistor.

(d) For this part, we change all cascaded two ports to t-parameters. Specifically,

t0 = [1 75;0 1];t1 = [na 0; 0 1/na];t2 = [ 1 j*30; 0 1];t3 = htot(h);t4 = [ 1 j*1042.9; 0 1];t5 = [nb 0; 0 1/nb];t6 = [1 0;1/75 1];t = t0*t1*t2*t3*t4*t5*t6

t = 2.2176e-01 - 2.5959e-01i 8.3160e+00 - 9.7351e+00i 1.4785e-03 - 1.7307e-03i 6.6577e-02 - 7.4415e-02i»gain = 1/t(1,1)gain = 1.9024e+00 + 2.2270e+00i

»gainmag = abs(gain)gainmag = 2.9289e+00

»gainangle = angle(gain)*180/pigainangle = 4.9494e+01

In this case, Vout/Vs = gain = 1.9024e+00 + 2.2270e+00i = 2.9289∠49.494o

SOLUTION 20.21.(a)»Y2N = [1 0;20.1 0]*1e-3;»Y10k = [1 -1;-1 1]*1e-4;»Yshade = Y2N + Y10kYshade = 1.1000e-03 -1.0000e-04 2.0000e-02 1.0000e-04»(b) This is a series connection of two 2-ports. Hence we first convert the answer of part (a) to z-parameters.

»Y2N = [1 0;20.1 0]*1e-3;»Y10k = [1 -1;-1 1]*1e-4;»Yshade = Y2N + Y10kYshade = 1.1000e-03 -1.0000e-04 2.0000e-02 1.0000e-04»Zshade = inv(Yshade)Zshade = 4.7393e+01 4.7393e+01 -9.4787e+03 5.2133e+02»Z1k = [1 1;1 1]*1e3;»Zdashed = Zshade + Z1kZdashed = 1.0474e+03 1.0474e+03 -8.4787e+03 1.5213e+03»

(c)»twoportz(Zdashed,1e12,1e3)

ans =twoport analysis using z-parameterszin = 1.0474e+03zout = 5.8588e+03v1tovs = 5.1157e-01v2tov1 = -8.0950e+00v2tovs = -4.1412e+00

Conclusion: Vout/Vs = -4.1412.

SOLUTION 20.22.»Z = [3 1;5 2]*1e3;»Y = inv(Z)Y = 2.0000e-03 -1.0000e-03 -5.0000e-03 3.0000e-03»% Consider the parallel connection of Y with the 1 kΩ resistor»Y1 = Y + [1 -1;-1 1]*1e-3Y1 = 3.0000e-03 -2.0000e-03 -6.0000e-03 4.0000e-03»»% Now consider Y/Z in parallel with 1 kΩ connected between B and C»Y2 = Y + [0 0;0 1]*1e-3Y2 = 2.0000e-03 -1.0000e-03 -5.0000e-03 4.0000e-03»»% This combination is in series with another 1 kΩ resistor»% Hence we need to compute Z2 first»Z2 = inv(Y2)Z2 = 1.3333e+03 3.3333e+02 1.6667e+03 6.6667e+02»% Now we compute the series combo of Z2 and the 1 kΩ resistor»Z3 = Z2 + [1 1;1 1]*1e3Z3 = 2.3333e+03 1.3333e+03 2.6667e+03 1.6667e+03»

»»% We now convert Y1 and Z3 to t-parameters and then multiply»% together to obtain the overall t-parameters»»T1 = ytot(Y1)T1 = 6.6667e-01 1.6667e+02 0 5.0000e-01»T3 = ztot(Z3)T3 = 8.7500e-01 1.2500e+02 3.7500e-04 6.2500e-01»T = T1*T3T = 6.4583e-01 1.8750e+02 1.8750e-04 3.1250e-01»Y = ttoy(T)Y = 1.6667e-03 -8.8889e-04 -5.3333e-03 3.4444e-03»

SOLUTION 20.23. Refer to figure 20.13 for all problems.(a) By inspection,

I1

I2

I3

=6 −4 −2

−4 4 + 5s −5s

−2 −5s 2 + 5s

V1

V2

V3

The 3x3 coefficient matrix is the desired Yind(s).(b) Writing the nodal equation by inspection yields

I1

I2

I3

0

=

0.5 0 −0.25 −0.25

0 1 −0.5 −0.5

−0.25 −0.5 1.25 −0.5

−0.25 −0.5 −0.5 1.25

V1

V2

V3

V4

Using MATLAB,

»W11=[0.5, 0 -0.25;0 1 -0.5;-0.25 -0.5 1.25]W11 = 5.0000e-01 0 -2.5000e-01 0 1.0000e+00 -5.0000e-01 -2.5000e-01 -5.0000e-01 1.2500e+00»W12 = [-0.25 -0.5 -0.5]'

W12 = -2.5000e-01 -5.0000e-01 -5.0000e-01»W21 = [-0.25 -0.5 -0.5]W21 = -2.5000e-01 -5.0000e-01 -5.0000e-01»W22 = 1.25W22 = 1.2500e+00»Yind = W11 - W12*inv(W22)*W21Yind = 4.5000e-01 -1.0000e-01 -3.5000e-01 -1.0000e-01 8.0000e-01 -7.0000e-01 -3.5000e-01 -7.0000e-01 1.0500e+00»

(c) Again, writing the nodal equation by inspection yields

I1

I2

I3

0

=

5 0 0 −5

0 6 −7 1

0 −4 8 −4

−5 −2 −1 8

V1

V2

V3

V4

Again, using MATLAB,»W11 = [5 0 0;0 6 -7;0 -4 8]W11 = 5 0 0 0 6 -7 0 -4 8»W12 = [-5 1 -4]'W12 = -5 1 -4»W21 = [-5 -2 -1]W21 = -5 -2 -1»W22 = 8;»Yind = W11 - W12*inv(W22)*W21Yind = 1.8750e+00 -1.2500e+00 -6.2500e-01 6.2500e-01 6.2500e+00 -6.8750e+00 -2.5000e+00 -5.0000e+00 7.5000e+00»

(d) This part is similar to part (a) as it does not require the method of matrix partitioning. Byinspection,

IG

ID

IS

=(CGD + CGS )s −CGDs −CGS s

−CGDs + gm CGDs −gm

−CGSs − gm 0 CGSs + gm

VG

VD

VS

The 3x3 coefficient matrix is the desired Yind(s).

SOLUTION 20.24. (a) With regard to the given information, the associated indefinite admittancematrix is the coefficient matrix in the following nodal equation given reference to figure 20.13:

IG

ID

IS

=0 0 0

gm 0 −gm

−gm 0 gm

VG

VD

VS

We use property 5 to compute the remaining answers:

(a)-(a): YGD is as given.

(a)-(b): YSD =gm 0

−gm 0

S

(a)-(c): YGS =0 0

−gm gm

S

(a)-(d): YDG =0 gm

0 0

S

(a)-(e): YDS =0 −gm

0 gm

S

(a)-(e): YSG =gm −gm

0 0

S

(b) Transmission from port-1 to port-2 occurs when the 2-1 entry of the 2-port y-matrix isnonzero. Hence, the following all have the desired transmission: YGD , YSD, YGS .

SOLUTION 20.25. Using the zero-sum properties of the rows and columns, we have byinspection:

Yind =40 2 −42

? ? −50

? −22 92

=40 2 −42

30 20 −50

−70 −22 92

S

SOLUTION 20.26. (a) Yind =8 ? ?

−2 9 −7

−6 ? ?

=8 −2 −6

−2 9 −7

−6 −7 13

S, where the second equality arises

because a purely

resistive network has a symmetric indefinite admittance matrix.(b) Construct a common ground 2-port with input terminal C, common terminal A, and outputterminal B:

YCB =13 −7

−7 9

S

Therefore, since the output terminal B is shorted, YL = ∞, i.e.,

Yin = y11 −y12y21

y22 + YL= 13 S

Hence

Zin =1

13 Ω

(c) Construct a common ground 2-port with input terminal A, common terminal B, and outputterminal C:

YAC =8 −6

−6 13

S

Hence, the required voltage gain is

GV =V2V1

=−y21

y22 + YL=

−y21y22

=613

Therefore, V2 =613

V.

SOLUTION 20.27.(a) Using the zero-sum property of the rows and columns of an indefinite admittance matrix wecan write down by inspection (in mS)

IB

IE

IC

=1 −1 0

−100 100.1 −0.1

99 99.1 0.1

VB

VE

VC

(b) The y-parameters (also in mS) of the common emitter configuration are easily computed as

IB

IC

=

1 0

99 0.1

VBE

VCE

Hence in MATLAB»Y = [1 0;99 0.1];»Z = inv(Y)Z = 1.0000e+00 0 -9.9000e+02 1.0000e+01

where Z is in kΩ. It follows that VCE = −990IB + 10IC where IB and IC are in mA and VCE isin volts.

SOLUTION 20.28. Expanding the given y-parameter matrix into an indefinite admittance matrixyields

IG

ID

IS

= Yind

VG

VD

VS

=0.2 + j2.5 −0.01− j0.65 −0.19 − j1.85

3.1− j0.65 0.05 + j0.8 −3.15 − j0.15

−3.3 − j1.85 −0.04 − j0.15 3.34 + j2

VG

VD

VS

By inspection, with G as the common terminal, S as the input terminal, and D as the outputterminal, we obtain,

Ynew =3.34 + j2 −0.04 − j0.15

−3.15 − j0.15 0.05 + j0.8

mS

SOLUTION 20.29. Here we use the zero-sum properties of the columns and rows to complete theindefinite admittance matrix:

Yind =−s ? ?

s −1 2 ?

1 2s −1 −2s

=−s −2s −1 3s +1

s−1 2 −s−1

1 2s−1 −2s

S

In figure (b), the top 2-port has ytop =−s −2s−1

s −1 2

S and the bottom 2-port has y-parameters

ybot =−2s 2s −1

−s −1 2

S. Since these 2-ports are connected in parallel, the overall 2-port y-

parameters are

y = ytop + ybot =−s −2s−1

s −1 2

+

−2s 2s −1

−s −1 2

=

−3s −2

−2 4

S

SOLUTION 20.30. From problem 28,

IG

ID

IS

= Yind

VG

VD

VS

=0.2 + j2.5 −0.01− j0.65 −0.19 − j1.85

3.1− j0.65 0.05 + j0.8 −3.15 − j0.15

−3.3 − j1.85 −0.04 − j0.15 3.34 + j2

VG

VD

VS

Hence

yGS =0.2 + j2.5 −0.19 − j1.85

−3.3− j1.85 3.34 + j2

S

SOLUTION 20.31. (a) Here, by inspection we can compute the indefinite admittance matrix asthe coefficient matrix of the following nodal equations:

IA

IB

IC

= Yind

VA

VB

VC

=Y1 + Y2 + 2 −Y2 −Y1 − 2

Y2 − 2 −Y2 2

−Y1 − 2Y2 2Y2 Y1

VA

VB

VC

(b)

yAB =Y1 + Y2 + 2 −Y2

Y2 − 2 −Y2

(c)

yAC =Y1 + Y2 + 2 −Y1 − 2

−Y1 − 2Y2 Y1

SOLUTION 20.32. (a) Here, by inspection we can compute the indefinite admittance matrix asthe coefficient matrix of the following nodal equations:

IA

IB

IC

= Yind

VA

VB

VC

=Y1 + Y2 −Y2 −Y1

−Y2 + gm Y2 +Y3 −Y3 − gm

−Y1 − gm −Y3 Y1 + Y3 + gm

VA

VB

VC

(b)

yAB =Y1 + Y2 −Y2

−Y2 + gm Y2 +Y3

(c)

yAC =Y1 + Y2 −Y1

−Y1 − gm Y1 + Y3 + gm

SOLUTION 20.33. (a) Here the nodal equation matrix is:

IA

IB

IC

0

=W11 W12

W21 W22

VA

VB

VC

VD

=

8 0 0 −8

0 2 + 2s −2s −10 8

0 −2s 16 + 2s −16

−8 −2 −6 16

VA

VB

VC

VD

where VD is the internal node voltage. Using the method of matrix partitioning,

Yind = W11 − W12W22−1W21 =

8 0 0

0 2 + 2s −2s−10

0 −2s 16 + 2s

−

116

−8

8

−16

−8 −2 −6[ ]

=8 0 0

0 2 + 2s −2s−10

0 −2s 16 + 2s

−

−0.5

0.5

−1

−8 −2 −6[ ] =4 −1 −3

4 3 + 2s −2s − 7

−8 −2s− 2 10 + 2s

S

(b) When C is grounded,

yAB =4 −1

4 3 + 2s

S and zAB =

18s +16

3 + 2s 1

−4 4

Ω

(c) When B is grounded

yAC =4 −3

−8 10 + 2s

S

SOLUTION 20.34. (a) Writing the usual node equations we have,

IA

IB

IC

0

=W11 W12

W21 W22

VA

VB

VC

VD

=

s + 0.5 − s 0 −0.5

−s s + 0.5 0 −0.5

0 0 0.5s −0.5s

−0.5 −0.5 −0.5s 0.5s +1

VA

VB

VC

VD

Using the method of matrix partitioning,

Yind = W11 − W12W22−1W21 =

s + 0.5 − s 0

−s s + 0.5 0

0 0 0.5s

−

0.5s + 2

1

1

s

1 1 s[ ]

=s + 0.5 −s 0

− s s + 0.5 0

0 0 0.5s

−

0.5s + 2

1 1 s

1 1 s

s s s2

=1

s + 2

s2 + 2.5s + 0.5 −s2 − 2s− 0.5 −0.5s

− s2 − 2s − 0.5 s2 + 2.5s + 0.5 −0.5s

−0.5s −0.5s s

S

(b) Here

yAB =1

s + 2s2 + 2.5s + 0.5 − s2 − 2s − 0.5

−s2 − 2s− 0.5 s2 + 2.5s + 0.5

S

SOLUTION 20.35. (a) Rule 1: Consider the two networks NA (3 external nodes) and NB (4external nodes) given below:

The two indefinite admittance matrices are

YindN A=

G1 + G2 −G2 −G1

−G2 G2 + G3 −G3

−G1 −G3 G1 + G3

and YindN B

=

G1 + G2 −G2 −G1 0

−G2 G2 + G3 −G3 0

−G1 −G3 G1 + G3 0

0 0 0 0

Observe that YindN B can be obtained from YindN A

by adding a column of zeros and a row of

zeros to form a 4x4 matrix.

(b) Rule 2: Consider two networks NA and NB and a third network NC which combines

elements of NA and NB as given below:

The corresponding indefinite admittance matrices are:

YindN A=

6 −2 −4

−2 3 −1

−4 −1 5

, YindN B

=5 −3 −2

−3 5 −2

−2 −2 4

, and YindNC

=11 −5 −6

−5 8 −3

−6 −3 9

.

Clearly,YindNC

= YindN A+YindNB

(c) Rule 3: Consider the 3-terminal network

with indefinite admittance matrix

YindN A=

5 −3 −2

−3 5 −2

−2 −2 4

S

If we move node 3 inside to form a 2-terminal network and labeled as NB,

then from nodal analysis we have

I1

I2

0

=5 −3 −2

−3 5 −2

−2 −2 4

V1

V2

V3

Using the method of matrix partitioning,

YindN B=

5 −3

−3 5

−

14

−2

−2

−2 −2[ ] =

4 −4

−4 4

S

This computation is the one given by the formula in the problem. To see that this is correct, weobserve that the internal simplification of NB leads to the following:

SOLUTION 20.36.Part (a)»Yinda= [1/2 -1/4 0 -1/4 0; -1/4 1/2 0 0 -1/4; ...0 0 0 0 0; -1/4 0 0 1/2 -1/4; 0 -1/4 0 -1/4 1/2]

Yinda =

5.0000e-01 -2.5000e-01 0 -2.5000e-01 0 -2.5000e-01 5.0000e-01 0 0 -2.5000e-01 0 0 0 0 0 -2.5000e-01 0 0 5.0000e-01 -2.5000e-01 0 -2.5000e-01 0 -2.5000e-01 5.0000e-01

»Yindb= [0 0 0 0 0; 0 0 0 0 0; 0 0 3/4 -1/2 -1/4; ...0 0 -1/2 5/8 -1/8; 0 0 -1/4 -1/8 3/8]Yindb =

0 0 0 0 0 0 0 0 0 0 0 0 7.5000e-01 -5.0000e-01 -2.5000e-01 0 0 -5.0000e-01 6.2500e-01 -1.2500e-01 0 0 -2.5000e-01 -1.2500e-01 3.7500e-01

Part (b)»Yind = Yinda + YindbYind =

5.0000e-01 -2.5000e-01 0 -2.5000e-01 0 -2.5000e-01 5.0000e-01 0 0 -2.5000e-01 0 0 7.5000e-01 -5.0000e-01 -2.5000e-01 -2.5000e-01 0 -5.0000e-01 1.1250e+00 -3.7500e-01 0 -2.5000e-01 -2.5000e-01 -3.7500e-01 8.7500e-01

% To suppress nodes 4 and 5 we use the partitioned matrix formula as follows:

»W11= [Yind(1:3, 1:3)]W11 =

5.0000e-01 -2.5000e-01 0 -2.5000e-01 5.0000e-01 0 0 0 7.5000e-01

»W12=[Yind(1:3, 4:5)]W12 = -2.5000e-01 0 0 -2.5000e-01 -5.0000e-01 -2.5000e-01

»W21= [Yind(4:5, 1:3)]W21 = -2.5000e-01 0 -5.0000e-01 0 -2.5000e-01 -2.5000e-01

»W22= [Yind(4:5, 4:5)]W22 = 1.1250e+00 -3.7500e-01 -3.7500e-01 8.7500e-01

»Yind123 = W11 - W12*inv(W22)*W21Yind123 = 4.3519e-01 -2.7778e-01 -1.5741e-01 -2.7778e-01 4.1667e-01 -1.3889e-01 -1.5741e-01 -1.3889e-01 2.9630e-01

(d) For the required Y-matrix we delete row and column 3 to obtain»Ysc = Yind123(1:2,1:2)Ysc = 4.3519e-01 -2.7778e-01 -2.7778e-01 4.1667e-01

»Zoc = inv(Ysc)Zoc = 4.0000e+00 2.6667e+00 2.6667e+00 4.1778e+00

SOLUTION 20.37. Since complex roots must occur in conjugate pairs, we will only check jω0.

0 = p( j 0) = − ja3 03 − a2 0

2 + ja1 0 + a0 = a0 − a2 02 + j a1 0 − a3 0

3( )Both real and imaginary parts must be zero, i.e.,

a0 − a2 02 = 0 and a1 0 − a3 0

3 = 0 a1 − a3 02( ) = 0

From the first equation, a2 02 = a0. The second equation above must be true for arbitrary ω0

which implies that a1 = a3 02. Equivalently a1a2 = a3a2 0

2 = a3a0. Conclusion: this conditionleads to imaginary complex roots.

(b) Given the above condition, what are the resulting imaginary roots of the polynomial? Sincethe polynomial is cubic, we can assume a3 ≠ 0. In this case,

0 = p(s) = s3 +a2a3

s2 +a1a3

s +a0a3

= s s2 +a1a3

+

a2a3

s2 +a1a2

a32 = s s2 +

a1a3

+

a2a3

s2 +a1a3

= s2 +a1a3

s +

a2a3

Therefore, the roots are: s = ± ja1a3

, −a2a3

.

SOLUTION 20.38.

(a) The four 2-port equations arising from the interconnection are:

V1 = V1a −V1b I1 = I1a = −I1b V2 = V2a = V2b I2 = I2a + I2b

ThusV1 = V1a −V1b = h11aI1a + h12aV2a( ) − h11bI1b + h12bV2b( ) = h11a + h11b( )I1 + h12a − h12b( )V2

andI2 = I2a + I2b = h21aI1a + h22aV2a( ) + h21bI1b + h22bV2b( ) = h21a − h21b( )I1 + h22a + h22b( )V2

This proves that the series-parallel connection has the required h-parameters.

(b) The four 2-port equations arising from the interconnection are:

V2 = V2a −V2b I2 = I2a = −I2b V1 = V1a = V1b I1 = I1a + I1b

Thus,I1 = I1a + I1b = g11aV1a + g12aI2a( ) + g11bV1b + g12bI2b( ) = g11a + g11b( )V1 + g12a − g12b( )I2

andV2 = V2a −V2b = g21aV1a + g22aI2a( ) − g21bV1b + g22bI2b( ) = g21a − g21b( )V1 + g22a + g22b( )I2

This proves that the parallel-series connection has the required g-parameters.

SOLUTION 20.39.

(a) Refer here to Na in figure P20.39b. With reference to figure 19.28b, h11 = 11 kΩ, h12 = 0,

h21 = 95.9, and h22 = 1/105 = 10

-5 S. Similarly, by inspection with reference to equation 19.33,

h11 = 90||10 = 9 kΩ, h12 = 0.1 (reverse voltage division), h21 = –0.1, and h22 = 10-5

S.

(b) By problem 20.38, part (a),

h11 = h11a + h11b = 20 kΩ h12 = h12a – h12b = –0.1

h21 = h21a – h21b = 96 h22 = h22a + h22b = 2×10-5 S

(c) Recall from chapter 19 that Zin =V1

I1= h11 −

h12h21

h22 + YL and Yout =

I2

V2= h22 −

h12h21

h11 + Zs in

which case Zout is the reciprocal. Using our MATLAB script, we have

h = [20e3 -0.1; 96 0.02e-3 ];zL= 1e8;zs= 5e3;»twoporth(h,zL,zs)ans =twoport analysis using h-parameterszin = 4.9976e+05zout = 2.4752e+03v1tovs = 9.9009e-01v2tov1 = -9.5998e+00v2tovs = -9.5047e+00

REMARK: We have used the following m-file code for "twoporth":

»% two-port analysis in terms of h-parameters»function [zin, zout] =twoport(h, zL, zs)»['twoport analysis using h-parameters']»h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2);»zin = h11 - h12*h21/(h22+ 1/zL)»yout= h22 - h12*h21/(h11+zs);

»zout= 1/yout»v1tovs= zin/(zin+zs)»v2tov1= -h21/(zin*(h22+1/zL))»v2tovs= v1tovs*v2tov1

SOLUTION 20.40.(a) The y-parameters for Na are:

yAB =7 4

4 7

−1

=133

7 −4

−4 7

S

Hence the indefinite admittance matrix for Na is:

YindN a=

133

7 −4 −3

−4 7 −3

−3 −3 6

S

Let us consider the associate 2-port with port A grounded and B as the new port 1 input. Theproblem is then solved by computing the input impedance with port 2 open circuited. Hence, thenew y-parameters are:

yBC =133

7 −3

−3 6

S

Thus Yin = y11 −y12y21

y22=

133

7 −96

=

5.533

=16

S. Hence, Zin = 6 Ω is the unique reading.

(b) The answer is not unique as demonstrated in part (c).

(c) The following two networks have the given Z-parameters, but the meter reading for N1 is 4

Ω but for N2 it is 2 Ω.

SOLUTIONS CHAPTER 21 PROBLEMS

SOLUTION TO 21.1. (a) Low pass(b) High pass

SOLUTION TO 21.2.

SOLUTION TO 21.3.(a)»n = 0.65378;»d = [1 0.80381643 0.82306043];»w = 0:0.005:2;»h = freqs(n,d,w);»plot(w, 20*log10(abs(h)))»grid»xlabel('Frequency rads/s')»ylabel('dB Gain')»

0 0.5 1 1.5 2-15

-10

-5

0

Frequency rads/s

dB G

ain

TextEnd

(b) »poles = roots(d)poles = -4.0191e-01 + 8.1335e-01i -4.0191e-01 - 8.1335e-01i

(c)»% Poles of new transfer function»wp = 2*pi*750'wp = 4.7124e+03»wp = 2*pi*750;»polesnew = poles*wppolesnew = -1.8939e+03 + 3.8328e+03i -1.8939e+03 - 3.8328e+03i»% All zeros remain at infinity.

Further

H(s) = HNLP (s p) =( p)2

s2 + 0.80381643 ps + 0.82306043( p)2 =2.2207 ×107

s2 + 3.7879 ×103s +1.8277 ×107

SOLUTION TO 21.4. (a) The 2nd order normalized LP transfer function is HNLP (s) =1

s2 + 2s +1. This

must be frequency scaled by Kf = 1000π. Hence,

H(s) = HNLP (s K f ) =(K f )2

s2 + K f 2s + (K f )2 =9.8696 ×106

s2 + 4.4429 ×103s + 9.8696 ×106

(b) Using MATLAB,»n = (1000*pi)^2;»d = [1 sqrt(2)*pi*1e3 (1000*pi)^2];»w = 0:1:2*pi*1500;»h = freqs(n,d,w);»plot(w/(2*pi),abs(h))»grid»xlabel('Frequency in Hz')»ylabel('Magnitude')»plot(w/(2*pi),20*log10(abs(h)))»grid»xlabel('Frequency in Hz')»ylabel('Magnitude in dB')

0 500 1000 15000.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in Hz

Mag

nitu

de

TextEnd

0 500 1000 1500-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

0

Frequency in Hz

Mag

nitu

de in

dB

TextEnd

(c)»nn = 9.8696e+06»dd = 1.0000e+00 4.4429e+03 9.8696e+06»w = j*2000*pi;»mag = abs(n/(w^2 + d(2)*w + d(3)))mag = 2.4254e-01

SOLUTION TO 21.5. (a) ∈max is that value of e that places the magnitude response curve through Amax

at ω = ωp. Therefore

Amax = 10log10 H ( j p)2

=10log10 1+ ∈max2 p

p

2n

= 10log10 1+ ∈max2( )

Therefore ∈max2 =100.1Amax −1 which upon a square root yields the final answer.

(b) Similarly, ∈min puts the magnitude response curve through the Amin spec. Hence

Amin = 10log10 H ( j s)2 =10log10 1+ ∈min

2 s

p

2n

Therefore

∈min2 =

100.1Amin −1

s

p

2n

which is equivalent to the required formula.

SOLUTION TO 21.6. The relationship of ε and ωc is given by the formula: c = p

(∈)1 n . Further, ∈max

in putting the magnitude response curve through the Amax spec produces c min , and ∈min in putting the

magnitude response curve through the Amin spec produces c max . Hence, from the solution to problem5,

c min = p

(∈max )1 n = p2n

100.1Amax −1 ≤ ω ≤ c max = p

(∈min )1 n = s2n

100.1Amin −1

SOLUTION TO 21.7. (a) From above material, the second order Butterworth NLP transfer function is

HNLP2(s) =1

s2 + 2s +1and from tables, the third order is

HNLP3(s) =1

s3 + 2s2 + 2s +1 (b)n1 = 1; d1 = [1 sqrt(2) 1];n2 = 1; d2 = [1 2 2 1];w = 0:.01:5;h1 = freqs(n1,d1,w);h2 = freqs(n2,d2,w);plot(w,abs(h1))gridxlabel('Normalized Frequency')ylabel('Magnitude')holdplot(w,abs(h2),'r')hold off

0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Normalized Frequency

Mag

nitu

de

TextEnd

Notice how the 3rd order filter has a sharper transition to zero.

(c) »% The simplest way to obtain the step response is as follows:»»syms s t»StepResp1 = ilaplace(1/(s^3 + sqrt(2)*s^2 + s))StepResp1 =1-exp(-1/2*2^(1/2)*t)*cos(1/2*2^(1/2)*t)-exp(-1/2*2^(1/2)*t)*sin(1/2*2^(1/2)*t)»StepResp2 = ilaplace(1/(s^4 + 2*s^3 + 2*s^2 + s))

StepResp2 =1-exp(-t)-2/3*exp(-1/2*t)*3^(1/2)*sin(1/2*3^(1/2)*t)»Thus the step response of the second order Butterworth normalized LP filter is:

v(t) = u( t) − e−0.70711t cos(0.70711t) − sin(0.70711t)[ ]u( t)

and that of the third order Butterworth normalized LP filter is:

v(t) = 1− e−t( )u(t) −1.1547e−0.5t sin(0.86603t)u( t)

SOLUTION TO 21.8.

fp = 100; fs = 1200;Amax = 0.3; Amin = 35;n = buttord(fp,fs,Amax,Amin,'s')emax = sqrt(10^(0.1*Amax) - 1)emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^nfcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)))fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)))wcmin = 2*pi*fcminwcmax = 2*pi*fcmaxwc = wcmin;fc = fcmin;[z,p,k] = buttap(n)% Numerators are each 1. Denominators are the polynomialsd1 = poly(p(1:2))d2 = poly(p(3))zplane(p)gridpauseznew = z*wcpnew = p*wcknew = k*wc^nf = 0:fc/50:1.2*fs;h = freqs(knew*poly(znew),poly(pnew),2*pi*f);plot(f,abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(f,20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')

grid

n = 3emax = 2.6743e-01emin = 3.2538e-02fcmin = 1.5521e+02fcmax = 3.1324e+02wcmin = 9.7524e+02wcmax = 1.9681e+03z = []p = -5.0000e-01 + 8.6603e-01i -5.0000e-01 - 8.6603e-01i -1.0000e+00k = 1% Numerators are each 1. Denominators are the polynomialsd1 = 1.0000e+00 1.0000e+00 1.0000e+00d2 = 1 1

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real part

Imag

inar

y pa

rt

znew = []pnew = -4.8762e+02 + 8.4458e+02i -4.8762e+02 - 8.4458e+02i -9.7524e+02knew = 9.2753e+08

0 500 1000 15000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in Hz

Gai

n m

agni

tude

0 500 1000 1500-60

-50

-40

-30

-20

-10

0

Frequency in Hz

Gai

n in

dB

SOLUTION TO 21.9.

fp = 100; fs = 1200;Amax = 0.3; Amin = 35;n = buttord(fp,fs,Amax,Amin,'s');emax = sqrt(10^(0.1*Amax) - 1);emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n;fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)));fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)));wcmin = 2*pi*fcmin;wcmax = 2*pi*fcmax;[z,p,k] = buttap(n);wc = wcmax;fc = fcmax;znew = z*wcpnew = p*wcknew = k*wc^nf = 0:fc/50:1.2*fs;h = freqs(knew*poly(znew),poly(pnew),2*pi*f);plot(f,abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(f,20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')grid

znew = []pnew = -9.8406e+02 + 1.7044e+03i -9.8406e+02 - 1.7044e+03i -1.9681e+03knew = 7.6235e+09

0 500 1000 15000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in Hz

Gai

n m

agni

tude

0 500 1000 1500-40

-35

-30

-25

-20

-15

-10

-5

0

5

Frequency in Hz

Gai

n in

dB

SOLUTION TO 21.10.

fp = 75; fs = 450;Amax = 1; Amin = 45;n = buttord(fp,fs,Amax,Amin,'s')emax = sqrt(10^(0.1*Amax) - 1)emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^nfcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)))fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)))wcmin = 2*pi*fcminwcmax = 2*pi*fcmax[z,p,k] = buttap(n)d1 = poly(p(1:2))d2 = poly(p(3:4))zplane(p)gridpausewc = wcmin;fc = fcmin;znew = z*wcpnew = p*wcknew = k*wc^nW = 0:0.01:fs/fp;h = freqs(k*poly(z),poly(p),W);plot(W*wc/(2*pi),abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(W*wc/(2*pi),20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')grid

n = 4emax = 5.0885e-01emin = 1.3721e-01fcmin = 8.8800e+01fcmax = 1.2323e+02wcmin = 5.5795e+02wcmax = 7.7427e+02z = []p = -3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i -9.2388e-01 - 3.8268e-01ik = 1

d1 = 1.0000e+00 7.6537e-01 1.0000e+00d2 = 1.0000e+00 1.8478e+00 1.0000e+00znew = []pnew = -2.1352e+02 + 5.1548e+02i -2.1352e+02 - 5.1548e+02i -5.1548e+02 + 2.1352e+02i -5.1548e+02 - 2.1352e+02iknew = 9.6912e+10

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real part

Imag

inar

y pa

rt

0 100 200 300 400 500 6000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in Hz

Gai

n m

agni

tude

0 100 200 300 400 500 600-70

-60

-50

-40

-30

-20

-10

0

Frequency in Hz

Gai

n in

dB

SOLUTION TO 21.11.

fp = 75; fs = 450;Amax = 1; Amin = 45;

n = buttord(fp,fs,Amax,Amin,'s');% The order mfile may not be available in the student edition.emax = sqrt(10^(0.1*Amax) - 1);emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n;fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)));fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)));wcmin = 2*pi*fcmin;wcmax = 2*pi*fcmax;[z,p,k] = buttap(n);wc = wcmax;fc = fcmaxznew = z*wcpnew = p*wcknew = k*wc^nW = 0:0.01:fs/fp;h = freqs(k*poly(z),poly(p),W);plot(W*wc/(2*pi),abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(W*wc/(2*pi),20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')grid

fc = 1.2323e+02znew = []pnew = -2.9630e+02 + 7.1533e+02i -2.9630e+02 - 7.1533e+02i -7.1533e+02 + 2.9630e+02i -7.1533e+02 - 2.9630e+02iknew = 3.5940e+11

0 100 200 300 400 500 600 700 8000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in Hz

Gai

n m

agni

tude

0 100 200 300 400 500 600 700 800-70

-60

-50

-40

-30

-20

-10

0

Frequency in Hz

Gai

n in

dB

SOLUTION TO 21.12. Here we require that

s3 + 2s2 + 2s +1= s3 +C1 + C2

C1C2s2 +

C1 + C2 + LLC1C2

s +2

LC1C2Thus

LC1C2 = 2 ⇒ C1 + C2 + L = 4 and 2 =C1 + C2

C1C2×

LL

=LC1 + LC2

2

in which case, LC1 + LC2 = 4 . Therefore, L C1 + C2 + L( ) = 4L = 4 + L2. Equivalently

L2 − 4L + 4 = (L − 2)(L − 2) = 0

Hence L = 2 H is the only solution. Thus C1C2 = 1 ⇒ C1 =1

C2 ⇒

1C2

+ C2 + 2 = 4 or equivalently

C22 − 2C2 +1 = (C2 −1)(C2 −1) = 0 which implies that C1 = C2 = 1 F is the only solution, as was to be

shown.

SOLUTION TO 21.13. (a) By voltage division

H(s) =

1

Cs + G

Ls + 1

Cs + G

=

1

LC

s2 + 1

RCs + 1

LC (b) With R = 1 Ω,

H(s) =

1

LC

s2 + 1

RCs + 1

LC

=1

s2 + 2s +1

requires that C = 1/ 2 F and L = 2 H.

(c) »fc = 1000;»wc = 2*pi*fcwc = 6.2832e+03»Kf = wc;»Km = 1000;»C = 1/sqrt(2);»L = sqrt(2);»Cnew = C/(Kf*Km)Cnew = 1.1254e-07»Lnew = L*Km/KfLnew = 2.2508e-01

(d)

»Km = C/(wc*1e-6)Km = 1.1254e+02»Kf = wc;»Rnew = KmRnew = 1.1254e+02»Lnew = L*Km/wcLnew = 2.5330e-02

SOLUTION TO 21.14. (a) By voltage division

H(s) =

1

Cs

Ls + Rs + 1

Cs

=

1

LC

s2 + Rs

Ls + 1

LC (b) With R = 1 Ω,

H(s) =

1

LC

s2 + Rs

Ls + 1

LC

=1

s2 + 2s +1

requires that L = 1/ 2 F and C = 2 H.

(c)»L = 1/sqrt(2); C = 1/L;»Km = 10;»Kf = 2*pi*500;»Rnew = 10;»Cnew = C/(Km*Kf)Cnew = 4.5016e-05»Lnew = Km*L/KfLnew = 2.2508e-03

(d)»Km = C/(1e-6*Kf)Km = 4.5016e+02»Lnew = L*Km/KfLnew = 1.0132e-01»Cnew = C/(Km*Kf)Cnew = 1.0000e-06

SOLUTION TO 21.15.

(a) Since 1 = 2/(LC), L = 2/C. Since (1/L + 1/C) = (C/2 + 1/C) = sqrt(2), we have that C is a root of

the quadratic 0.5C2 – sqrt(2)C + 1 = 0. Hence»v = [0.5 -sqrt(2) 1];»r = roots(v)r = 1.4142e+00 1.4142e+00»C = r(1)C = 1.4142e+00»L = 2/CL = 1.4142e+00

(b)»Km = 1e3;»Kf = 2*pi*3500;»Cnew = C/(Km*Kf)Cnew = 6.4308e-08»Lnew = L*Km/KfLnew = 6.4308e-02

(c)»Km = C/(Kf*10e-9)Km = 6.4308e+03»Cnew = C/(Km*Kf)Cnew = 1.0000e-08»Lnew = L*Km/KfLnew = 4.1356e-01»Rs = KmRs = 6.4308e+03»RL = RsRL = 6.4308e+03

SOLUTION TO 21.16. (a) Let G = RL. Then by voltage division

H(s) =Vout

Vin=

1

Cs + G

Ls + Rs + 1

Cs + G

=

1

LC

s2 +1

CRL+

Rs

L

s +

1 + Rs RL

LC

(b) Since 1 = 1.25/(LC), L = 1.25/C. Since (Rs/L + 1/RLC) = (2C/1.25 + 1/8C) = sqrt(2), we have that

C is a root of the quadratic (16/1.25)C2 – 8sqrt(2)C + 1 = 0. Hence»C = roots([16/1.25 -8*sqrt(2) 1])C = 7.8427e-01 9.9615e-02»L = 1.25 ./CL = 1.5938e+00 1.2548e+01(c)»Km = 1e3;»Kf = 2*pi*5e3Kf = 3.1416e+04»Cnew = C/(Km*Kf)Cnew = 2.4964e-08 3.1708e-09»Lnew = L*Km/KfLnew = 5.0734e-02 3.9942e-01

SOLUTION TO 21.17. (a) Define G = 1/Rs, execute two source transformations, and apply voltage division to obtain

H(s) =Vout

Vin=

11

Cs + G+ Ls +1

×G

Cs + G=

1

RsLC

s2 + 1RsC

+ 1L

s + 1 +1 Rs

LC

(b) Since L = 1.5/C, the values of C are the roots of the quadratic, (2/1.5)C2 – 2sqrt(2)C + 1 = 0.Hence»C = roots([(2/1.5) -2*sqrt(2) 1])C = 1.6730e+00 4.4829e-01»L = 1.5 ./CL = 8.9658e-01 3.3461e+00»Km = 2e3;»Kf = 2*pi*5e3;»Cnew = C/(Km*Kf)

Cnew = 2.6627e-08 7.1347e-09»Lnew = Km*L/KfLnew = 5.7078e-02 2.1302e-01»Rsnew = 2*KmRsnew = 4000

SOLUTION TO 21.18. (a)

H( jω) =K

jω

ω p

+1

2 =K

ωω p

2

+1

The 3 dB down frequency, ωc, occurs when

1

ωcω p

2

+1

=1

2

Equivalently

ωc = ω p 2 −1 = 0.64359ω p = 6.4359 ×104 rad/sec.

(b) h(t) = Kω p2te

−ω ptu(t) since

H(s) =K

sω p

+1

2 =Kω p

2

s + ωp( )2 .

Further,

H(s)1

s=

Kω p2

s s + ωp( )2 =K

s−

K

s + ωp−

Kω p

s + ωp( )2

Hence, the step response is

K 1 − e−ω pt

− ωpte−ω pt

u(t)

SOLUTION TO 21.19.

(a) Using voltage division,

H(s) =VC

Vin×

Vout

VC=

1

Cs +1

L2s +1

1 + L1s +1

Cs +1

L2s +1

×1

L2s +1

=

1L1L2C

s3 + 1L1

+ 1L2

s

2 + L1 + L2 + CL1L2C

s + 2L1L2C

(b) Matching coefficients in

1

L1L2C

s3 +1

L1+

1

L2

s2 +

L1 + L2 + C

L1L2Cs +

2

L1L2C

=1

s3 + 2s2 + 2s +1

yields C =2

L1L2 and

L1 + L2 + CL1L2C

=L1 + L2 +

2

L1L22

= 2; equivalently, L1 + L2 +2

L1L2= 4. Further,

1L1

+1L2

=

L1 + L2L1L2

= 2 implies that L1 + L2 = 2L1L2 implies 2L1L2 +2

L1L2= 4 . This requires that

L1L2 = 1 and from earlier equations that L1 + L2 = 2L1L2 = 2 which forces L1 = L2 = 1 H and C = 2 F.The idea is to match the denominator coefficients and thus the dc gain is 0.5 instead of the desired 1. Atransformer or some amplifier device is needed to increase the gain to 1.

(c) Using MATLAB:

»Km = 1000;»Kf = 2*pi*20e3;»Lnew = Km/KfLnew = 7.9577e-03»Cnew = 2/(Km*Kf)Cnew = 1.5915e-08

»Hence, L1new = L2new = 7.96 mH and C = 15.9 nF.

(d) SPICE simulation

V0

R1K

L7.96m

L07.96m

R01K

C15.9n

IVm

MAG(V(IVM))


+0.000e+000

+100.000m

+200.000m

+300.000m

+400.000m

+500.000m

+10.000k +20.000k +30.000k +40.000k

SOLUTION TO 21.20. (a) From figure P21.19a

Vin − V1Rs

= I1, V2 = Vout

in which caseV1 = z11I1 + z12 × 0 = Vin − RsI1

Also

V2 = z21I1 + z22 × 0 = Vout

This implies that Vin = (z11 + Rs)I1 and Vout = z21I1 . Finally we conclude that

VoutVin

=z21

z11 + Rs

(b) Now from figure P21.20b, we have V1 = Vin and V2 = Vout = −RL I2 . This implies that

I2 = −VoutRL

= y21V1 + y22V2

ThusVoutVin

=−y21

y22 +1

RL

=−y21

y22 + GL

Consider here that

VoutVin

=z21

z11 +1=

1

s2 + 2s +1=

1

s

s +1

s+ 2

=

1

2ss

2+

1

2s

Hence z21 =1

2s and z11 =

s

2+

1

2s

. This leads to the circuit

with L =1

2 H and C = 2 F.

( c) Similarly,

−y21y22 +1

=

1

2ss

2+

1

2s

+1

implies y21 = −1

2s and y22 =

s

2s+

1

2s.

This yields the same circuit as above with

L = 2 H and C =1

2 F.

(d) Here Km =1000 and K f = 5000rad / s

(d-i) For (b),

L =1

2×

103

5 ×103 =1

5 2= 0.1414 H

and

C = 2 ×1

5 ⋅106 = 0.2828 µF

(d-ii) For (c)

L = 0.2828 H and C = 0.1414 µF.

SOLUTION TO 21.21.

(a) From earlier developments

H(s) = −Yin

Yout= −

1

R1

Cs + 1

R2

(b) Let C = 1 F, and R1 = R2 = 1 Ω.(c) »Kf = 2*pi*3500

Kf = 2.1991e+04»Km = 1/(Kf*1e-9)Km = 4.5473e+04

In the final design, R1 = R2 = 45.5 kΩ.

SOLUTION TO 21.22.

(a) From problem 21 and voltage division,

H(s) = −

1

R1

Cs + 1

R2

×

1

C2s

R3 + 1

C2s

= −

1

C1R1

s + 1

C1R2

×

1

R3C2

s + 1

R3C2

(b) By inspection, let C1 = 0.1 F, R1 = 1 Ω, R2 = 10 Ω, R3 = 10 Ω, and C2 = 0.1 F, in which case

H(s) = −10

s + 1×

1

s +1=

−10

s +1( )2

(c)»Kf = 1e5;»Km = 0.1/(Kf*1e-9)Km = 1000Hence, in the final design

C1 = 1 nF, R1 = 1 kΩ, R2 = 10 kΩ, R3 = 10 kΩ, and C2 = 1 nF.

(d) Cascade the circuit of figure P21.22 with another op amp section. For the first part of the design,

again set ωp = 1 and use the same values as in part (c). The extra op amp section has the same values as

the first section. As such, final values are the same as in part (c).

SOLUTION TO 21.23. The 2nd order NLP Butterworth transfer function is: HNLP (s) =1

s2 + 2s +1. The

design parameters and steps are detailed in the excel spread sheet below. An additional design called

design C is also listed. For input attenuation, the resistor R1 is replaced by the voltage divider R3-R4combination.

w0^2 w0/Q Num w0 Q KNLP KMA Kf=wp KmR KMB KMC1 1.41 1.00 1.0000 0.7071 1.00 22507.86 6283.20 10000.00 22507.86 15915.46

KMS19492.37

RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0 1 2Q 1/(2Q) 1 1 KNLP/K 1/alpha 1/(1-alpha)Design B 1 1 2 1 1/Q 1 Q KNLP/K 1/alpha 1/(1-alpha)Design C 1 1-1/Q 3-1/Q 1 1 1 1 KNLP/K 1/alpha 1/(1-alpha)Saraga RA RA/3 3-Apr rt(3)Q 1 1/Q 1/rt(3) KNLP/K 1/alpha 1/(1-alpha)

RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 1.4142 0.7071 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.4142 1.0000 0.7071 0.5000 2.0000 2.0000Design C 1.00 -0.4142 1.5858 1.0000 1.0000 1.0000 1.0000 0.6306 1.5858 2.7071Saraga 3.00 1.0000 1.3333 1.2247 1.0000 1.4142 0.5774 0.7500 1.3333 5.6569

w0 scaleDesign A ∞ 0.0000 1.0000 1.4142 0.7071 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.4142 1.0000 0.7071 0.5000 2.0000 2.0000Design C 1.00 0.1140 2.1140 1.0000 1.0000 1.0000 1.0000 0.6306 1.5858 2.7071Saraga 3.00 1.0000 1.3333 1.2247 1.0000 1.4142 0.5774 0.7500 1.3333 5.6569

wp scaleDesign A ∞ 0.000E+00 1.000E+00 2.251E-04 1.125E-04 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.592E-04 2.251E-04 1.000E+00 7.071E-01 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.592E-04 1.592E-04 1.000E+00 1.000E+00 6.306E-01 1.586E+00 2.707E+00Saraga 3.00 1.000E+00 1.333E+00 1.949E-04 1.592E-04 1.414E+00 5.774E-01 7.500E-01 1.333E+00 5.657E+00

Km scaleDesign A ∞ 0.000E+00 1.000E+00 1.000E-08 5.000E-09 2.251E+04 2.251E+04 1.000E+00 2.251E+04 #DIV/0!Design B 10000 1.000E+04 2.000E+00 7.0711E-09 1.000E-08 2.251E+04 1.592E+04 5.000E-01 4.502E+04 4.502E+04Design C 10000 1.140E+03 2.114E+00 1.000E-08 1.000E-08 1.592E+04 1.592E+04 6.306E-01 2.524E+04 4.308E+04Saraga 30000 1.000E+04 1.333E+00 1.000E-08 8.165E-09 2.757E+04 1.125E+04 7.500E-01 2.599E+04 1.103E+05

SOLUTION TO 21.24 AND 21.25. In problem 21.8, the transfer function information was computed inMATLAB as:

% Numerators are each 1. Denominators are the polynomialsd1 = 1.0000e+00 1.0000e+00 1.0000e+00d2 = 1 1

Further we know from MATLAB thatfcmin = 1.5521e+02

The Saraga design and Design A for d1, the second order section of each filter, are given by the excelspread sheet below, as well as two alternate designs labeled B and C.


KMS35521.40


RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 2.0000 0.5000 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.0000 1.0000 1.0000 0.5000 2.0000 2.0000Design C 1.00 0.0000 2.0000 1.0000 1.0000 1.0000 1.0000 0.5000 2.0000 2.0000Saraga 3.00 1.0000 1.3333 1.7321 1.0000 1.0000 0.5774 0.7500 1.3333 4.0000


wp scaleDesign A ∞ 0.000E+00 1.000E+00 2.051E-03 5.127E-04 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.025E-03 1.025E-03 1.000E+00 1.000E+00 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.025E-03 1.025E-03 1.000E+00 1.000E+00 5.000E-01 2.000E+00 2.000E+00Saraga 3.00 1.000E+00 1.333E+00 1.776E-03 1.025E-03 1.000E+00 5.774E-01 7.500E-01 1.333E+00 4.000E+00


The first order (leaky integrator) section is common to both problems. This section consists of an input

resistor (conductance) R1 (G1) connected to the inverting terminal with a parallel R2-C combination

feeding back from the output. The transfer function is: H(s) =G1

Cs + G2. For the normalized design we

set G1 = G2 = 1 S (R1 = R2 = 1 Ω) and C = 1 F. This design can be scaled independently of the S&K 2nd

order section. Hence we set Cnew = 50 nF. Thus Km = 20,508.29. Hence R1 = R2 = 20,508.29 Ω.

SOLUTION TO 21.26 AND 21.27. The relevent data from the solution of problem 21.10 is:

k = 1d1 = 1.0000e+00 7.6537e-01 1.0000e+00d2 = 1.0000e+00 1.8478e+00 1.0000e+00fcmin = 8.8800e+01wcmin = 5.5795e+02

In providing the designs, we set forth all the possible S&K designs using two excel spreadsheets, one foreach second order section.

The designs for denominator d1 with numerator equal to 1 are:


KMS40559.76


RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 2.6131 0.3827 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 0.7654 1.0000 1.3066 0.5000 2.0000 2.0000Design C 1.00 0.2346 2.2346 1.0000 1.0000 1.0000 1.0000 0.4475 2.2346 1.8100Saraga 3.00 1.0000 1.3333 2.2630 1.0000 0.7654 0.5774 0.7500 1.3333 3.0615


wp scaleDesign A ∞ 0.000E+00 1.000E+00 4.683E-03 6.859E-04 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.792E-03 1.372E-03 1.000E+00 1.307E+00 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.792E-03 1.792E-03 1.000E+00 1.000E+00 4.475E-01 2.235E+00 1.810E+00Saraga 3.00 1.000E+00 1.333E+00 4.056E-03 1.792E-03 7.654E-01 5.774E-01 7.500E-01 1.333E+00 3.061E+00


The designs for denominator d2 with numerator equal to 1 are:


KMS17922.81


RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 1.0824 0.9239 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.8478 1.0000 0.5412 0.5000 2.0000 2.0000Design C 1.00 -0.8478 1.1522 1.0000 1.0000 1.0000 1.0000 0.8679 1.1522 7.5703Saraga 3.00 1.0000 1.3333 0.9374 1.0000 1.8478 0.5774 0.7500 1.3333 7.3912


wp scaleDesign A ∞ 0.000E+00 1.000E+00 1.940E-03 1.656E-03 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.792E-03 3.312E-03 1.000E+00 5.412E-01 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.792E-03 1.792E-03 1.000E+00 1.000E+00 8.679E-01 1.152E+00 7.570E+00Saraga 3.00 1.000E+00 1.333E+00 1.680E-03 1.792E-03 1.848E+00 5.774E-01 7.500E-01 1.333E+00 7.391E+00


SOLUTION TO 21.28. For this problem we use the excel spread sheet given below. First we observe that

H(s) =ˆ K

s2 + 0

Qs + 0

2 → H ( 0s) =

ˆ K

0s( )2 + 0

Q 0s( ) + 02

=ˆ K 0

2 (= KNLP)

s2 +1

Qs +1

Thus after this type of frequency scaling, the new transfer function is:

Hnew(s) =0.7943

s2 +1

1.1286s +1

The dc gain is of course 0.7943 and the modification of the circuit to achieve the correct dc gain is givenin the spread sheet below via R3 and R4 which constitute a voltage divider that replaces R1.

w0^2 w0/Q Num w0 Q KNLP KMA Kf=wp KmR KMB KMC0.82306 0.8038 0.65378 0.9072 1.1286 0.7943 1131.42 43982.40 10000.00 501.23 501.23

KMS979.83


RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 2.2573 0.4430 1.0000 1.0000 0.7943 1.2589 4.8621Design B 1.00 1.0000 2.0000 1.0000 0.8860 1.0000 1.1286 0.3972 2.5179 1.6588Design C 1.00 0.1140 2.1140 1.0000 1.0000 1.0000 1.0000 0.3758 2.6613 1.6019Saraga 3.00 1.0000 1.3333 1.9549 1.0000 0.8860 0.5774 0.5957 1.6786 2.1917

w0 scaleDesign A ∞ 0.0000 1.0000 2.4881 0.4883 1.0000 1.0000 0.7943 1.2589 4.8621Design B 1.00 1.0000 2.0000 1.1023 0.9766 1.0000 1.1286 0.3972 2.5179 1.6588Design C 1.00 0.1140 2.1140 1.1023 1.1023 1.0000 1.0000 0.3758 2.6613 1.6019Saraga 3.00 1.0000 1.3333 2.1548 1.1023 0.8860 0.5774 0.5957 1.6786 2.1917

wp scaleDesign A ∞ 0.000E+00 1.000E+00 5.657E-05 1.110E-05 1.000E+00 1.000E+00 7.943E-01 1.259E+00 4.862E+00Design B 1.00 1.000E+00 2.000E+00 2.506E-05 2.220E-05 1.000E+00 1.129E+00 3.972E-01 2.518E+00 1.659E+00Design C 1.00 1.140E-01 2.114E+00 2.506E-05 2.506E-05 1.000E+00 1.000E+00 3.758E-01 2.661E+00 1.602E+00Saraga 3.00 1.000E+00 1.333E+00 4.899E-05 2.506E-05 8.860E-01 5.774E-01 5.957E-01 1.679E+00 2.192E+00

Km scaleDesign A ∞ 0.000E+00 1.000E+00 5.000E-08 9.813E-09 1.131E+03 1.131E+03 7.943E-01 1.424E+03 5.501E+03Design B 10000 1.000E+04 2.000E+00 5.000E-08 4.430E-08 5.012E+02 5.657E+02 3.972E-01 1.262E+03 8.315E+02Design C 10000 1.140E+03 2.114E+00 5.000E-08 5.000E-08 5.012E+02 5.012E+02 3.758E-01 1.334E+03 8.029E+02Saraga 30000 1.000E+04 1.333E+00 5.000E-08 2.558E-08 8.682E+02 5.657E+02 5.957E-01 1.645E+03 2.148E+03

SOLUTION TO 21.29. (a) HHP (s) = HNLPc

s

=

1

cs

2

+ 2 cs

+1

. At s = j p ,

HHP ( j p) = HNLPc

j p

=1

c

j p

2

+ 2 c

j p

+1

=1

−5.57

2

− j 25.57

+1

Thus in MATLAB,»Magfp = 1/abs(1 - (5.5/7)^2 -j*sqrt(2)*(5.5/7))Magfp = 8.5091e-01»Attenfp = -20*log10(Magfp)

Attenfp = 1.4023e+00»Magfs = 1/abs(1 - (5.5/1)^2 -j*sqrt(2)*(5.5/1))Magfs = 3.3040e-02»Attenfs = -20*log10(Magfs)Attenfs = 2.9619e+01

Thus the attenuation at fp is 1.4023 dB and that at fs is 29.619 dB.(b) From problem 21.15, the transfer function is

Hcir(s) =1 LC

s2 +1

C+

1

L

s +

2

LCand the values of L and C realizing the 2nd order Butterworth NLP transfer function can be computedaccording to

1C

+1L

= 2, 2

LC=1 ⇒

1C

+C2

= 2 ⇒ C2 − 2 2C + 2 = 0

(C − 2)2 = 0 ⇒ C = 2 F ⇒ L = 2 H

(c) Here Km = 1000. Ls → L C

s →

11

LKm Cs

and 1

Cs →

sC C

→ Km

C Cs . Thus in

MATLAB,

»wc = 2*pi*5.5e3wc = 3.4558e+04»Km = 1000;»C = sqrt(2); L = sqrt(2);»Lhp = Km/(C*wc)Lhp = 2.0462e-02»Chp = 1/(Km*wc*L)Chp = 2.0462e-08

Therefore, the resistors take on values of 1 kΩ, the inductor is changed to a capacitor of value of Chp =

20.46 nF and the capacitor is changed to an inductor of value Lhp = 20.46 mH.

SOLUTION TO 21.30. (a)

Ω = p, Amax = 2 dB, = max = 100.1×2 −1 = 0.76478, Ωc =

16

100.1× 2 −1= 1.0935

»Wc=1/(10^0.2-1)^(1/6)Wc = 1.0935e+00

(b)»wchp = 2*pi*5e3/Wcwchp = 2.8730e+04»fchp = wchp/(2*pi)fchp = 4.5725e+03

Thus chp = p

Ωc= 28.73 krad/s.

( c ) (i) NLP → HP transformation: scale by wchp.

Chp =1

Llp chp=17.404 F, Lhp =

1Clp chp

= 34.807 H

(ii) Magnitude scale to obtain proper value of capacitors.»Clp = 1;»Llp = 2;»Chp = 1/(Llp*wchp)Chp = 1.7404e-05»Lhp = 1/(Clp*wchp)Lhp = 3.4807e-05»Cnewhp = 100e-9;»Km = Chp/CnewhpKm = 1.7404e+02»Lnewhp = Km*LhpLnewhp = 6.0578e-03

Km =174.02, Chp,new =Chp

Km= 100 nF , Lhp,new = KmLhp = 6.06 mH

SOLUTION TO 21.31.

The 2nd order NLP Butterworth transfer function is: HNLP (s) =1

s2 + 2s +1. Using the transformation

s to 1/s, we obtain the NHP Butterworth transfer function:

HNHP(s) = HNLP1s

=

s2

s2 + 2s +1= K∞

s2

s2 + d(1) s + d(2)

INPUT: d(1) d(2) K K Kf Km KmR1.414213562 1 1.33334 1 18849.6 12247.44871 30000

NHP Params w0 Q alpha1 0.70710678 0.74999625

NHP Crt Pars C1 = Q C2 = rt(3) R1 = 1 R2 =1/(rt(3)Q) R R/30.707106781 1.73205081 1 0.816496581 1 0.333333333

HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/33.06293E-09 7.5026E-09 12247.44871 10000 30000 10000

A plot of the design without input attenuation is shown below. Notice that as predicted the gain is 4/3.

R130k

XOpAmp

Vpls15

Vminus15

Vin C07.503n

C3.063n

R012.247k

R10k

IVm

R210k

MAG(V(IVM))

Frequency (Hz)S&K HP-Small Signal AC-3

(V)

+0.000e+000

+500.000m

+1.000

+100.000 +316.228 +1.000k +3.162k +10.000k

Input attenuation requires that we replace C1 with a series combination of capacitors in which C1 = C3+ C4 and (1/C3)/(1/C4 + 1/C3) = alpha. Here then, C1 = C3 + C4 and alpha = C4/(C3 + C4) = C4/C1.Thus C4 = alpha*C1 and C3 = (1 – alpha)*C1. Thus

InputAttenuation

C3 = (1 - alpha)C1 C4 = alpha*C1

7.65744E-10 2.29719E-09

SOLUTION TO 21.32. The fourth order Butterworth NLP transfer function can be obtained from tables orfrom MATLAB as follows:

»[z,p,k] = buttap(4)z = []p = -3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i -9.2388e-01 - 3.8268e-01ik = 1»% Second Order Sections

»n1 = 1;»d1 = poly([p(1),conj(p(1))])d1 = 1.0000e+00 7.6537e-01 1.0000e+00»n2 = 1;»d2 = poly([p(3),conj(p(3))])d2 = 1.0000e+00 1.8478e+00 1.0000e+00»Thus,

HNLP (s) =1

s2 + 0.76537s +1×

1

s2 +1.8478s +1having frequency response

»w = 0:0.01:3.5;»h = freqs(k*poly(z),poly(p),w);»plot(w,abs(h))»grid»xlabel('Normalized Frequency')»ylabel('Magnitude 4th Order Butterworth')

0 0.5 1 1.5 2 2.5 3 3.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Normalized Frequency

Mag

nitu

de 4

th O

rder

But

terw

orth

TextEnd

The Saraga design parameters are given in the following Excel tables:

INPUT: d(1) d(2) KNLP/NHP w0LP/HP QNLP H(s) 0.76537 1 1 1 1.306557613NHP H(s) 0.76537 1 1 1 1.306557613

K Kf Km KmR alpha1.33334 18849.6 22630.24168 60000 0.74999625


Scale by w0HP 1.306557613 1.732050808 1 0.441886576 1 0.333333333


Input Attenuation C3 = (1 - alpha)C1 C4 = alpha*C17.65744E-10 2.29719E-09


K Kf Km KmR alpha1.33334 18849.6 10000 30000 0.74999625


Scale by w0HP 0.541184111 1.732050808 1 1.066827827 1 0.333333333

HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/32.87106E-09 9.18879E-09 10000 10668.27827 30000 10000


SOLUTION TO 21.33. Using MATLAB,

»fp = 5e3; fs = 1.5e3;»wp = 2*pi*fp; ws = 2*pi*fs;»Amax = 3; Amin = 40;»n = buttord(wp,ws,Amax,Amin,'s')n =4

»[z,p,k] = buttap(n)z =[]p =-3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i

-9.2388e-01 - 3.8268e-01ik = 1

»d1 = real(poly([p(1),p(2)]))d1 =1.0000e+00 7.6537e-01 1.0000e+00»d2 = real(poly([p(3),p(4)]))d2 =1.0000e+00 1.8478e+00 1.0000e+00

In general,

HNHP(s) = HNLP1s

=

1

1s

2

+ 0LPQ

1s

+ ( 0LP )2

=1

( 0LP )2 ×s2

s2 +1 0LP( )

Qs + 1 0LP( )2

=( 0HP )2 s2

s2 + 0HP

Qs + ( 0HP )2

The S&K Saraga design for d1 is given by the following excel spreadsheet:


K Kf Km KmR alpha1.33334 31416 45260.48336 60000 0.74999625


Scale by w0HP 1.306557613 1.732050808 1 0.441886576 1 0.333333333



The S&K Saraga design for d2 is given by the following excel spreadsheet:


K Kf Km KmR alpha1.33334 31416 20000 60000 0.74999625


Scale by w0HP 0.541184111 1.732050808 1 1.066827827 1 0.333333333

HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/38.61319E-10 2.75664E-09 20000 21336.55655 60000 20000


This completes the design.

SOLUTION TO 21.34. For the woofer,

H(s) =8

Ls + 8=

8

L

s +8

L

Thus, 8L

= 2000 × 2π ⇒ L = 636 µH.

For the tweeter,

H(s) =8

1

Cs+ 8

=8Cs

8Cs +1=

s

s +1

8C

Thus, 1

8C= 2000 × 2π ⇒ C = 9.95 µF.

SOLUTION TO 21.35. For the woofer,

HNLP (s) =1

s2 + 2s +1

The transfer function of the following circuit

is

H(s) =

1

LC

s2 +1

Cs +

1

LC

Thus 2 =1C

⇒ C = 0.70711 F and since 1

LC= 1, L = 2 = 1.4142 H. Frequency scaling the

element values by Kf = 4000π and magnitude scaling by Km = 8 yields C =0.70711KmK f

= 7.0337 µF and

L =1.4142Km

K f= 0.90032 mH:

For the tweeter we first realize the NLP Butterworth transfer function as above to obtain asabove

with Thus C = 0.70711 F and L = 2 = 1.4142 H. We now apply the frequency transformation

s → Cs

to each element (capacitors become inductors and inductors become capacitors according to

figure 21.24) and we obtain the HP circuit topology

where Cnew = 7.0337 µF and Lnew = 0.90032 mH.

SOLUTION TO 21.36. Consider figure (a). Let the current entering the RC network from Z1 be dentoted

by Ifa. Let the voltage from this point to ground be denoted Vfa. Then

Vout,a = Vfa + I faZ1 = Ha(s)VinFor figure (b) with a similar denotation of voltage and current, we have

Vout,b = Vfb + I fb +VfbZ1

k −1

Z1k

= V fb + I fbZ1k

+k −1

kVfb =

1k

(2k −1)Vfb +I fb

Z1

If(2k −1)Vfb ≅ Vfa and I fb ≅ I fa (**)

then

Vout,b ≅1k

Ha(s)Vin =Vout ,a

k

For gain enhancement, k < 1. However, for the (**) to be valid, we require that Z1

k −1 be large relative

to what it sees in the RC network. Hence, in general, k must be close to 1. Thus only small gainenhancements are possible. For such a potentially sensitive approach to gain enhancement, it might bebetter simply to add another op amp stage as op amps are comparatively inexpensive.


SOLUTION TO PROBLEM 22.1(a) For figure P22.1a, T0 = 2 and ω0 = π . Let t0 = -1 in equation 22.5b. Then f(t) = δ(t) and

cn = 0.5 ( t)e− jnπtdt−1

1

∫ = 0.5 for all n

From equation 22.6, an = 1 and bn = 0 for all n. Finally from equation 22.2

f ( t) = 0.5 + cos(nπt)n=1

∞∑

(b) For figure P22.1b, T0 = 2 and ω0 = π . Let t0 = 0 in equation 22.5b. Then f(t) = - δ(t -1) and

cn = −0.5 (t −1)e− jnπtdt−1

1

∫ = −0.5e− jnπ

From equation 22.6, bn = 0 for all n, and

an = -1 for n even

an = 1 for n odd

Finally from equation 22.2

f ( t) = −0.5 + cos(πt) − cos(2πt) + cos(3πt) − cos(4πt) +K

SOLUTION TO PROBLEM 22.2(a) T0 = 1 and ω0 = 2π . Let t0 = 0 in equation 22.5b. Then

f ( t) = e− ln (2)( )t

and from equation 22.5b

cn = e− ln (2)( )te− j2nπtdt0

1

∫ = e− ln (2)+ j2nπ( )tdt0

1

∫ =−1

ln(2) + j2nπ( ) e− ln (2)+ j2nπ( ) −1[ ] =

0.5ln(2) + j2nπ( )

(b) Using the above result for cnc0 =0.5

ln2 = 0.7213 ,

c1 = 0.5ln2+ j2π

= 0.7213e -j1.4609

c2 = 0.5ln2+ j4π

= 0.397e -j1.516

From equation 22.6

d0 = c0 = 0.7213

d1 =2c1 = 0.158,

θ1 =-1.461x180/π =-83.7o

d2 =2c2 = 0.0795

θ2 =-1.516x180/π =-86.84o

Thus, f(t) in the form of equation 22.3 is

f(t) = 0.7213 + 0.158cos(2πt - 83.7o) + 0.0795os(2πt - 86.84o)


(a) T0 = 1 and ω0 = 2π . Let t0 = 0 in equation 22.5b. Then

f ( t) = e− ln (2)( )t u(t) − u( t − 0.5)[ ]

and from equation 22.5b

cn = e− ln (2)( )te− j2nπtdt0

0.5

∫ = e− ln (2)+ j2nπ( )tdt0

0.5

∫ = −1ln(2) + j2nπ( ) e−0.5 ln (2)+ j2nπ( ) −1[ ]

=1−

(−1)n

2ln(2) + j2nπ( )

(b) Using the above result for cn, and MATLAB to evaluate the numerical result,»n= 0;»c0= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pic0 4.2256e-01»n=1;»c1= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi)c1 = 2.9612e-02- 2.6843e-01i»abs(c1)ans = 2.7006e-01»degreec1=angle(c1)*180/pidegreec1 = -8.3705e+01»n=2; »c2= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi)c2 = 1.2817e-03- 2.3237e-02i

»abs(c2)ans = 2.3272e-02degreec2= angle(c2)*180/pidegreec2 =-8.6843e+01

From equation 22.6 and equation 22.3

f(t) = 0.4226 + 0.54cos(2πt - 83.7o) + 0.04654os(2πt - 86.84o)

SOLUTION PROBLEM 22.4. (a) f(t) = cos(4t) sin(2t) = 0.5[ sin(6t) - sin (2t)] .The fundamental angular frequency of f(t) is ω0= 2 rad/s. The given f(t) can be expressed as

f(t) = -0.5 sin( ω0t) + 0.5sin(3ω0t) . Observe that b1= -0.5, b3= 0.5 and all other ai and bi are zero.

From equation 22.4 , d1= 0.5 /-90o and d3 = 0.5 /90o. From equations 22.6a and 22.6b.

c1 = 0.25j and c3 = -0.25j. All other cn are zero for n positive

(b) f(t) = sin2(4t) cos2(8t)= 0.5[1 - cos(8t)]x0.5 1+ cos(16t)

= 0.25 [ 1 - cos(8t) + cos(16t) - cos(8t) cos(16t)]

= 0.25 - 0.375 cos(8t) + 0.25cos(16t) - 0.125cos(24t) The fundamental angular frequency of f(t) is ω0= 8 rad/s. The given f(t) can be expressed as

f(t) = 0.25 -0.375 cos( ω0t) + 0.25cos(2ω0t) - 0.125sicos(3ω0t). Observe that a0= 0.5, a1=-0.375, a2=

0.25 , a3= -0.125 and all other ai and bi are zero.

Fom equation 22.4,d0= 0.25, d1= 0.375 /180o , d2 = 0.25 /0o, and d3 = 0.125 /180o,

From equations 22.6a and 22.6b.c0 = 0.25, c1 = -0.375 and c2 = 0.25, c3 = -0.125. All other cn are zero for n positive.

(c)f(t) = 2 + 1.5 sin(500t)- 2cos(2000t)]cos(106t)

=2cos(106t) + 0.75sin(1000500t) + 0.75sin(999500t)- cos(1002000t) - cos(998000t)

The fundamental angular frequency of f(t) is ω0= 500 rad/s. The given f(t) can be expressed as

f(t) =2cos( 2000ω0t) + 0.75sin(2001ω0t) + 0.75sin(1999ω0t) - cos(2004ω0t) - cos(1996ω0t)

Observe that: a1996 = -1, b1999= 0.75, a2000= 2, b2001= 0.75 , a2004= -1, and all other ai and bi are

zero. From equation 22.4 , d1996 = 1 /180o , d1999 = 0.75 /-90o, d2000 = 2 /0o ,

d2001 = 0.75/-90o, d2004 = 1 /180o , and all other di are zero.

From equations 22.6a and 22.6b.c1996 = -0.5 , c1999 = -j0.375, c2000 = 1 ,

c2001 = -j0.375o, dc2004 =-0.5 , and all other cn are zero for n positive..

SOLUTION PROBLEM 22.5

By inspesction, the derivative of f(t) is

f '( t) =AT

− A (t − nT )n=−∞

∞∑ =

AT

− f ( t)

where fδ(t) is shown in figure 22/7, with its Fourier series given by equation 22.20b, i.e.

f (t) =AT

+2AT

cos(n 0t)n=1

∞∑

Therefore

f '( t) = −2AT

cos(n 0t)n=1

∞∑

The dc component is the average value of f(t) and is given by 0.5A. Other terms in the Fourier series of

f(t) are obtained by integrating the cosine terms in the above expression. The result is

f ( t) = 0.5A −Anπ

sin(n 0t)n=1

∞∑

SOLUTION TO PROBLEM 22.6Denote by f5(t) the waveform of figure P22.5, with A= 0.5 and T =1. Then by inspection

f ( t) = f prob5(−t) + 0.5

Substituting the result of problem 22.5 into the above equation, we have

f ( t) = 0.75 +0.5nπ

sin(n 0t)n=1

∞∑


Consider the square wave g(t) shown in figure 22.4 with its Fourier series given by equation 22.13. By

inspection, the derivative of f(t) is

f '( t) = −4T

g(t) f'(t) = - 4T

g(t)

Substituting equation 22.13 into the above expression, we have

f '( t) = −8AπT

sin(n 0t)n

n=1,odd

∞∑

The dc component is the average value of f(t) and is given by 0.5A. Other terms in the Fourier series of

f(t) are obtained by integrating the sine terms in the above expression. The result is

f ( t) = 0.5A +4A

π 2cos(n 0t)

n2n=1,odd

∞∑


The method used below is simpler than that suggested in the hint.

We first sketch the waveform of f'(t) and observe that it may be expressed as the sum of two periodic rectangular

pulse trains:f'(t)= 1

αT[ fp(t + 0.5αT) +fp(t - 0.5αT)]

where fp(t) is sketched in figure 22.8. Using equation 22.23, we have

f'(t)= 1αT

[αA +2Asin(nαπ)

nπ∑n=1

∞

cos(nω0(t+0.5αT))]

-[αA +2Asin(nαπ)

nπ∑n=1

∞

cos(nω0(t-0.5αT))]

= 1αT

2Asin(nαπ)

nπ[∑

n=1

∞

cos(nω0(t+0.5αT))-cos(nω0(t-0.5αT)] \

= 1αT

[ - 4Asin2(nαπ)

nπ sin(nω0t)]∑

n=1

∞

The dc component is the average value of f(t) and is given by αA. Other terms in the Fourier series of

f(t) are obtained by integrating the sine terms in the above expression. The result is

f(t) = αA + 2Aαπ2

[sin(nαπ)

n ]2cos(nω0t)∑n=1

∞

SOLUTION TO PROBLEM 22.9

We first sketch the waveform of f'(t) and observe that it may be expressed as the sum of a periodic rectangular

pulse train and a periodic impulse train:

f'(t)= 1αT

fp(t + 0.5αT) - fδ(t )

where fp(t) is sketched in figure 22.8, and fδ(t) in figure 22.7.

Using equations 22.23 and 22.20b, we have

f'(t)= 1αT

[αA +2Asin(nαπ)

nπ∑n=1

∞

cos(nω0(t+0.5αT))

- [AT

+ 2AT ∑

n=1

∞cos(nω0t)]

= 2AT

sin(nαπ)

nαπ∑[n=1

∞

cos(nω0(t+0.5αT)) -cos(nω0t)]

The dc component is the average value of f(t) and is given by 0.5αA. Other terms in the Fourier series

of f(t) are obtained by integrating the sine terms in the above expression. The result is

f(t) =0.5αA + Anπ

[sin(nαπ)

nαπ∑n=1

∞

sin(nω0(t+0.5αT)) -sin(nω0t)]

It remains to rewrite the expression in the form of equation 22.2.To this end, let b = sin(nαπ)/(nαπ) and re-write the terms within [ ] as follows:

bsin(nαπ)cos(nω0t) + bcos(nαπ) - 1sin(nω0t)

Hence, f(t) in the form of equation 22.2 has the coefficients, for n=1,2...

an = Aαπ2n2

sin2(nαπ)

bn = Aαπ2n2

sin(nαπ)cos(nαπ) -nαπ

dn= an2 + bn2 = A

απ2n2sin4(nαπ) + sin(nαπ)cos(nαπ) -nαπ2

= Aαπ2n2

sin2(nαπ) +(nαπ)[nαπ - sin(2nαπ)]

θn= tan-1(-bnan

)

The result is item 6 of table 22.4


Observe that the present f(t) can be derived from that of problem 22.9 by (a) replacing t by -t; and (b)replacing α by β. Thus the Fourier series for f(t) is:

f(t) =0.5βA + ancos(nω0t) + bn∑[n=1

∞

sin(nω0t) ]

wherean = A

βπ2n2

sin2(nβπ)

bn = -Aβπ2n2

sin(nβπ)cos(nβπ) -nβπ

dn= an2 + bn2 = A

βπ2n2

sin4(nβπ) + sin(nβπ)cos(nβπ) -nβπ2

= A

βπ2n2

sin2(nβπ) +(nβπ)[nβπ - sin(2nβπ)]

θn= tan-1(-bnan

)

SOLUTION PROBLEM 22.11. Following the hint, we have the second derivative of f(t) given by

f ''( t) =1

(1− )Tf (t + T) −

1(1− )T

f ( t)

where fd(t) is given in example 22.5. Notice that we have focused on the part of the waveform over[–αT, (1–α)T]. By making use of equation 22.20b, we obtain

f ''( t) =2A

(1− )T 2 cos(n 0t + 2n π) − cos(n 0t)[ ]n=1

∞∑

=−4A

(1− )T 2 sin(n 0t + n π)sin(n π)[ ]n=1

∞∑

Therefore,

f ( t) = f ( t)[ ]ave +4A

(1− )T 2sin(n π)

n 0( )2 sin(n 0t + n π)

n=1

∞∑

=A2

+A sin(n π)

n2π 2 (1− )cos n 0t + (n − 0.5)π( )

n=1

∞∑

Letting T = 1 and α = 0.25 we obtain,

f ( t) =A2

+16A sin n

π4

3n2π 2 cos n2πt + (0.25n − 0.5)π( )

n=1

∞∑

Therefore, d0 = 0.5A , and

dn =16Asin n

π4

3n2π 2 .

It follows that d1 = 0.38211A and d2 = 0.13509A.


Denote by fp(t, α) the period rectangular waveform of figure 22.5, with A=1.

Then we can express the present f(t), with T= 4, as the sum of 3 terms:

f(t) = 3fp(t-0.125T, 0.25) + 4 fp(t - 0.5T, 0.5) -2

From equation 22.14b, and equation 22.12c, for n= 1,2,...

cn = 3πn

sin(0.25nπ)e-jnω0x0.125T+ 4πn

sin(0.5nπ)e-jnω0x0.5T

= 3πn

sin(0.25nπ)e-j0.25nπ+ 4πn

sin(0.5nπ)e-jnπ

The numerical values of the first few Fourier series coefficients are:

c0 = averge value of f(t) = 14

(1 + 4 - 2) = 0.75

c1= 3π

sin(0.25π)e-j0.25π+ 4π

sin(0.5π)e-jπ = -0.7958 -j 0.4775

c2= 32π

sin(0.5π)e-j0.5π+ 42π

sin(π)e-j2π =-j0.4775

Solution Problem 22.13

(a) For sinusoidal steady analysis, the transfer function is

H(jω) = YL YL +YC + YR

= 1 1 +ZLYC + ZLYR

= 1(1 - ω2LC) + jωL

R

= 1

(1 - 4×10 -5ω2) + j×10--3ω

The transfer function evaluated at various input frequencies are listed below.

H(0) = 1

H(j377) = 0.2128 /-175.4o

H(j3x377) = 0.0199/-178.7o

H(j5x377) = 0.0071/-179.2o

Using equation 15.7 and superposition, we obtain the steady state output voltage (in V):vout(t) = 200 + 200× 2×0.2128cos (377t -175.4o) +60× 2×0.0199cos (3×377t +30o -178.7o)

+80× 2×0.0071cos (5×377t +50o -17.2o)

= 200 + 2×42.55cos (377t -175.4o) + 2×1.196cos (3×377t -148.7o)

+ 2×0.5668cos (5×377t -129.2o)

(b) From equation derived in P11.39,

Vout,eff= 2002+42.552+ 1.1962 +0.56682 = 204.48 V

The average power absorbed by the 10 kΩ resistor is

Pav=204.482

104 = 4.1812 W

Solution Problem 22.14

One correction in the problem statement: in the angle expression, 5000 should be 10000.

(a) Using the identity cos(x)cos(y)= 0.5cos(x+y) + 0.5cos(x-y), we have

vin(t) = 0.1cos (998,000t) +0.2cos (999,000t) + 2cos (1,000,000t)

+ 0.2cos (1,001,000t) + 0.1cos (1,002,000t) V

(b) The transfer function has a constant magnitude of 10, and a phase shift proportional to the deviationfrom ωc. at ω = ωc + 2ωm, the phase shift is -9 degrees. From these facts, we can write directly

vout(t) = cos (998,000t +9o) +2cos (999,000t + 4.5o) + 20cos (1,000,000t)

+ 2cos (1,001,000 - 4. 5o) + cos (1,002,000t - 9o) V

(c) Using the identity

cos(x)cos(y) = 2 cos(x+y

2)cos(

x+y2

)

we can group the terms in vout(t) and re-write it as

vout(t) = 2 cos( 2ωmt - 9o)cos(ω ct) +4cos( ωmt - 4.5o)cos(ω ct) + 20cos(ω ct)= [20 + 4cos(ωmt - 4.5o) +2 cos(2ωm - 9o)]cos(ω ct) = g(t)cos( ω ct)

Thus g(t) = 20 + 4cos( ωmt - 4.5 o) +2cos(2ωmt - 90)

With td = 78.54 µs, then ωmtd = 1000 x 78.54x10-6 = 0.07854 rad, or 4.5 degrees. We have

10 f(t - td) = 20[1 + 0.2cos( ωm(t - td) +0.1cos( 2ωm(t - td)]

= 20[1 + 0.2cos(ωmt - 4.5o) + 0.1 cos( 2ωmt - 9o)] = g(t)

SOLUTION PROBLEM 22.15. (a) This proof is a special case of the general proof given in the

solution to Problem 22.16. See the solution to problem 22.16, below.

FOR THE REMAINING PARTS, WE USE THE FOLLOWING MATLAB CODE:

% chapter 22, problem 15.%part (b).Vmax= 30*pi;Vmin=0;T=4;R = 1; C = 1;voutmin= Vmin+(Vmax - Vmin)/(1+ exp(0.5*T/(R*C)))voutmax = Vmax - (Vmax -Vmin)/(1+ exp(0.5*T/(R*C)))t1= 0: 0.05: 2;vseg1= Vmax+(voutmin- Vmax)*exp((-t1/(R*C)));t2= 2:0.05: 4;vseg2= Vmin+(voutmax- Vmin)*exp(-(t2-T/2)/(R*C));t= [ t1 t2];v= [vseg1 vseg2];plot(t,v)gridxlabel('t in seconds')ylabel('Steady State Output Voltage in V')

%part (c)error1= 100*(12.235- voutmin)/voutminerror2= 100*(82.013 - voutmax)/voutmax

TO OBTAIN

(b)voutmin = 1.1235e+01voutmax = 8.3013e+01

(c)error1 = 8.9045e+00error2 = -1.2048e+00

SOLUTION PROBLEM 22.16. CORRECTION: In the problem statement, s

should be s .

For simplicity, let us consider the case when the transfer function is a voltage ratio, i.e., H(s) = Vout./Vin.

If a constant input vin(t) = Vcon is applied to the stable network, then the vout(∞) = KVcon ,

independent of the initial conditions. This is because the zero-input response for a stable network

approaches zero as t approaches infinite. To see this observe that the zero-state response is given by

vout (t) = L−1 Ks +1

×Vcon

s

= L−1 KVcons

−KVcons +1

= KVcon 1− e−t( )u( t)

from which vout(∞) = KVcon as asserted.

For the remainder of our proof we make use of the fact that in steady state, vout(t+T) = vout(t)

with t = 0 in our case. Specifically, after the first order network has reached steady state, the vout(t)

waveform will be periodic as shown in figure P22.15c, where the time reference has been chosen so thatvout,min occurs at t = 0.

Recall equation 8.19

x(t) = x(∞) + x(t0+ ) − x(∞)[ ]e−(t−t0 )

Applying this equation to the interval [0, T/2], we have

b0 = KVmax + a0 − KVmax( )e−0.5T / ≡ KVmax + a0 − KVmax( ) (1)

Note that = e−0.5T / . Similarly, applying equation 8.19 to the interval [ T/2, T] leads to

a0 = KVmin + b0 − KVmn( ) (2)

Equations (1) and (2) can be written as a single matrix equation

− 1

1 −

a0

b0

=

(1− )KVmax

(1− )KVmin

(3)

Solving equation (3) by matrix inverse (or Cramer's rule) results in

a0

b0

=

−12 −1

1

1

(1− )KVmax

(1− )KVmin

=

K+1

1

1

Vmax

Vmin

=

K+1

Vmax + Vmin

Vmax + Vmin

vout ( t)[ ]min = a0 = KVmax + Vmin

+1= K

Vmax + Vmin + Vmin − Vmin+1

= KVmin +Vmax −Vmin

+1= KVmin + K

Vmax −Vmin1+1/

= KVmin + KVmax − Vmin

1+ e0.5T /

and

vout ( t)[ ]max = b0 = KVmax + Vmin

+1= K

Vmax + Vmin + Vmax − Vmax+1

= KVmax − KVmax − Vmin

1 +1/= KVmax − K

Vmax − Vmin

1+ e0.5T /

This complete completes the derivation of the desired formulas.

SOLUTION TO PROBLEM 22.17

There is one correction in the problem statement. 1-kHz should be 5.1mHz.(a) The transfer function is, according to equation 4.3, and using the given element values R1 = 10 kΩ,

Rf = 50 kΩ and C = 20 mF:

H(s) = VoutVin

= - ZfZ1

= - Y1Yf

= -

1R1

Cs + 1Rf

= -

RfR1

1 + RfCs = -5

1 + 1000s

(b) Here we have We have K= -5, t = 1000 s, Vmax = 1 V, Vmin = -1 V and T = 1/f = 1/0.0051=

196.15 s. In using the equations derived in problem 22.16, we note that the subscripts min and max

should be switched in this case because K is negative. The following MATLAB codes perform the

needed numerical calculations.

f= 5.1e-3;Cf=20e-3;Rf=50e3;Rs= 10e3;T=1/f;K= -Rf/Rs;tau=Rf*Cf;['part (b)']Vmax=1;Vmin=-1;voutmax= K*( Vmin + (Vmax-Vmin)/(1+exp(0.5*T/tau)))voutmin= K*( Vmax - (Vmax-Vmin)/(1+exp(0.5*T/tau)))

answers from MATLAB are:

voutmax = 0.2449voutmin = -0.2449

(c)['part (c)']

t1= 0:0.005*T:0.5*T;v1= 5 + (-0.245 -5).*exp(-(t1/tau)) ;t2= 0.5*T:0.005*T:T;v2= -5 + (0.245 +5).*exp(-(t2-0.5*T)/tau) ;t=[t1,t2];v=[v1,v2];plot (t,v)xlabel('time in seconds')ylabel('vout in V')grid

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0 50 100 150 200

time in seconds

vout

in V


vi(t) = cos (ωt) = cos(6000t)

vo(t) = 10v i + vi2 + vi

3 = 10cos(ωt) + cos2(ωt) + cos3(ωt)

= 10cos(ωt) +0.5 +0.5cos(2ωt) + 0.75cos(ωt) + 0.25cos(3ωt)

= 0.5 + 10.75 cos(ωt) + 0.5cos(2ωt) 0.25cos(3ωt)

The effective values of various components of the output are:

dc fundamental 2nd harmonic 3rd harmonic total harmonic

0.5 10.75/√2 0.5/√2 0.25/√2 0.559/√2

The total harmonic distortion is0.55910.75

× 100% = 5.2%

and the average power at the fundamental frequency is

0.5× 10.752

100 = 0.5778 W


vo(t) = V0 + V1 cos(ωt) + V2cos(2ωt) + V3cos(3ωt)

We use equation 22.30 and the values of vo(t) read from the oscilloscope

to compute Vk. The total harmonic distortion is given by

H.D. = V2

2 + V32

V1×100 %

MATLAb codes:vo0= 10; vo60= 5.2; vo120= -4.6; vo180= -9.6;V0= (vo0+2*vo60+2*vo120 +vo180)/6V1= (vo0+ vo60 - vo120 -vo180)/3V2= (vo0 - vo60 - vo120 +vo180)/3V3= (vo0-2*vo60+2*vo120 -vo180)/6% total harmonic distortionHD= 100*sqrt( V2^2 +V3^2)/V1

The following answers are obtained frm MATLAB output.

V0 = 0.2667 V

V1 = 9.8000 V

V2 = -0.0667

V3 = 0 V

HD = 0.6803 (percent)


We shall follow the solution given in example 22.12, and only indicate the needed changes below. There are twocorrections in example 22.12: (1) 20cos(ωαT) should be 20cos(0.5ωαT) and (2) 1.842cos(ωt) should be 1.842

cos(3ωt).

The new input is

vi(t) = 0.9 cos(6000t) The positive peak of the output sine wave is clipped for αT.

0.9 ×20cos(0.5ωαT) = 15 → απ = cos-1 (15/18) → α = 0.1864

Equation 22.35 becomes

Equations 2.36 becomes

Using A=1, and α = 0.1864 in item #2 of table 22.4, equation 22.37 becomes

and equation 22.38 becomes

Equation 22.39 becomes

from which the harmonic distortions are:

third order 1.076/16.57 = 6.595%

fifth order 0.55/16.57 = 3.319%

SOLUTION PROBLEM 22.21. Corrections: (1) cos( 1)e

– 2− 1+πRC = cos( 2) should be

cos( 1)e− – 2 − 1+π

RC

= cos( 2) and (2) on page 943, equation 22.46 , θ2 should be –θ2, i.e., equation

22.46 should be cos( 1)e− – 2 − 1+π

RC

.

The proof is similar to that given on page 943 for the half-wave rectifier case, except for some

minor changes described below.

For the case of a full-wave rectifier, the output voltage waveform is a modification of figure 22.17

as shown below.

The exponential decay of vo(t) starts with the value Vmcos(θ1) at θ = θ1 – π (instead of θ1 – 2π, as in the

half-wave rectifier case). Therefore, in equation 22.45, change T to T/2 and 2π to π. In other words

v0 t( ) = Vm cos( 1)( )e−

t − 1 −T

2

RC

= Vm cos( 1)( )e−

t − 1−( )RC

In equation 22.46, make correction (2) above, and change 2π to π, i.e.,

cos( 1)e− – 2 − 1+π

RC

= cos( 2)

The desired proof is complete.

SOLUTION PROBLEM P22.22

Refer to figure 22.17. Assuming θ1 = 0, we have

v0(t) = Vmexp( -tRC

) = Vm 1 - ( tRC

) + ( tRC

)2 - ( tRC

)3 + ...

Assuming θ2 = 0, we compute the average value of vo(t) over the time interval [ 0, T]. For the case

RC >> T, we can approximate vo(t) over this interval by keeping only the first two terms of the infinite

series. Thus

v0(t) ≅ Vm 1 - tRC

which indicates that the plot of vo vs. t over the interval [ 0 T] is approximately a straight line.

Therefore the average of the vo(t) over [ 0 T] is equal to vo(T/2). Thus, for the case RC >> T, or

equivalently ωRC >> 2π,

Vdc ≅ voT2

= Vm 1 - T2RC

= Vm 1 - πωRC

SOLUTION PROBLEM 22.23Given values are: C = 20e-6 F, R =100kΩ, Vm = 20 V; f = 60 Hz. From equation 22.48b

Vdc ≅ (1 - πωRC

)Vm = (1 - 12fRC

)Vm = (1 - 12×60×105×20×10-6

)×20 = 19.916 V

To calculate the ripple factor, we first calculate θ2 from equation 22.47,

and then use the result in equation 22.50.

θ2 = cos-1( 2πRC

) = 0.1289 rad

ripple factor ≅ 1 - cos(θ2)

3 [ 1 +cos(θ2)] =0.2406%

For the diode average and peak currents, use equations 22.49a and 22.52

Idc = VdcR

= 19.917100,000

= 0.199 ×10-3 A

id,peak ≅ VmωCsin(θ2) = 19.4×10-3 A


Equations 22.44 - 22.52 are derived for a half-wave rectifier. For a full-wave rectifier, some of these equations will

be modified slightly as given below. The difference arises from changing T to T/2.The new equation for θ2 is derived in problem 22.21, and repeated below.

The given values are: C = 20e-6 F, R =100kΩ, Vm = 20 V; f = 60 Hz.

A modification of equation 22.48b gives

Vdc = (1 - π2ωRC

)Vm = (1 - T4RC

)Vm = (1 - 14×60×105×20×10-6

)×20 = 19.96 V

To calculate the ripple factor, we first calculate θ2 from equation derived in problem 22.21, and then use the result

in equation 22.50.

θ2 ≅ cos-1(e- πωRC) =5.23 degrees

ripple factor ≅ 1 - cos(θ2)

3 [ 1 +cos(θ2)] =0.12%

For the diode average and peak currents, use equations 22.49a and 22.52

Idc = VdcR

= 19.96100,000

= 0.1996 ×10-3 A

id,peak ≅ VmωCsin(θ2) = 13.75×10-3 A

_

SOLUTION PROBLEM 22.25. CORRECTIONS: (1) 195 Ω should be 1950 Ω. (2) 100 µF should

be 10 µF.

Since this problem only requires an estimate of the answer, we can use reasonable approximations to simply

the solution. Let H(s) = Vo(s) I(s) be the transfer function for the linear circuit to the right of the diodes. Then,

H(s) =Vin (s)I(s)

×Vo(s)Vin (s)

= Zin (s)Zpar(s)

R + Zpar(s)

where Zpar(s) is the impedance of the parallel C-RL.

The first step is to find the magnitude H ( j ) . As long as 1/RL << ωC and 1/(ωC) << R, the

following approximations are valid:

Zpar( j ) =1

j C +1

RL

≅1C

which means that the parallel impedance is essentially that of the capacitor, and

Zpar( j )

R + Zpar ( j )≅

Zpar ( j )

R≅

1RC

and again since 1/(ωC) << R,

Zin ( j ) ≅1C

Using these approximations,

H ( j ) = Zin ( j ) ×Zpar ( j )

R + Zpar ( j )≅

1C

×1RC

=R

RC( )2

It is given that Idc = 0.01 A. From the short pulse property, the input current i(t) consists of very short pulses at

120 Hz and all ac components of i(t) have peak magnitudes approximately equal to twice the average dc value.

Hence the peak magnitudes are 0.02 A. Therefore the magnitude of 120-Hz component of the output voltage is

H ( j2π ×120) × 0.02 ≅R

RC( )2 × 0.02 =1950

2π ×120 ×1950 ×10 ×10−6( )2 × 0.02 = 0.18042 V

Hence, the effective value is 0.18042 2 = 0.12757 V and the ripple factor is

0.1275730

×100 = 0.42524 %


Since this problem only requires an estimate of the answer, we can use reasonable approximations

to simply the solution. Let H(s) = Vo(s) I(s) be the transfer function for the linear circuit to the right of

the diodes. Then

H(s) =VoI

= VinI

× VoVin

Under the condition 1/(ωC) << R and 1/(ωC) << RL, we have

Zin(s) ≅ ZC(s) = 1Cs

and VoVin

≅ 1ωRC

2

Using these approximations,

H(jω) ≅Zin VoutVin

= 1ωC

( 1ωRC

)2 = R( 1ωRC

)3

For a full-wave rectifier, the fundamental frequency is 120 Hz, and ω = 240π rad/s. It is given that Idc

= 0.01 A. From the short pulse property, the input current i(t) consists of very short pulses at 120 Hz

and all ac components of i(t) have peak magnitudes approximately equal to twice the average dc value.

Hence the peak magnitudes are 0.02 A. Therefore the peak magnitude of 120-Hz component of the

output voltage is

H(j120π) x0.02 ≅ = R( 1ωRC

)3×0.02= 975×( 1120π×975×16×10-6)

3× 0.02 = 0.012 V

Hence, the effective value is 0.012/ 2 =0.0848 V and the ripple factor is

0.084830

×100% = 0.2827 %

SOLUTION PROBLEM 22.27Using H(jω) of figure P22.27 and the given formula for cn , we have

cn = 1ω s

H(jω) e-jnT0ωdω = 1ω s

e-jnT0ωdω = 1-jnT0ω s

[ e-jnT0ω]-0.25 ωs0.25ωs

-0.25 ωs

0.25ωs

-0.5 ωs

0.5ωs

= 1nπ

sin (nπ2

), n = 1,2,...

For n = 0, we have c0= d0 = a0 /2 = [Average value of H(jω) ]= 0.5 .

For n = 1,2,..., from equation 22.6

bn = 0, dn=an = 2

nπ sin (nπ2

),

The Fourier series representation of H(jω) is

H(jω) = 0.5 + 2π cos(T0ω) - - 1

3cos(3T0ω) + - 1

3cos(5T0ω) - - 1

7cos(7T0ω + ...

A plot of H(jω) vs. ω curve using the first 11 terms (n=0,1,...11) of the Fourier

series is given below together with the MATLAB codes.

-0.5

0

0.5

1

1.5

0 20 40 60 80 100

w in radians

H(j

w)

%chapter 2, problem 27.ws= 100;T0=2*pi/ws;w= 0: ws/200:ws;d0w = 0.5;d1w = (2/pi)*cos(T0*w);d3w= -(2/pi/3)*cos(3*T0*w);d5w= (2/pi/5)*cos(5*T0*w);d7w= -(2/pi/7)*cos(7*T0*w);d9w= +(2/pi/9)*cos(9*T0*w);d11w= -(2/pi/10)*cos(11*T0*w);H =d0w +d1w + d3w +d5w +d7w + d9w +d11w;plot(w,H)xlabel(' w in radians')ylabel('H(jw)')grid

R a DeCarlo and P Lin Linear Circuit Analysis s

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