ELEC 202 Electric Circuit Analysis II -...

THE CITADEL, THE MILITARY COLLEGE OF SOUTH CAROLINA

171 Moultrie Street, Charleston, SC 29409

Dr. Gregory J. Mazzaro

Spring 2016

ELEC 202 – Electric Circuit Analysis II

Lecture 10(a)

Complex Arithmetic and

Rectangular/Polar Forms

2

Real vs. Complex Numbers

1j

The product of a real number & the

operator j is an imaginary number. 2

3 , , , 5.17

j j j j

The sum of a real number & an imaginary

number is a complex number, z . 2 4 , 1j j

…where the real part is denoted Rea z

…and the imaginary part is denoted Imb z

Re 2 4 2, Im 2 4 4j j

rectangular

form

3

Addition & Subtraction

Graphical addition & subtraction

are performed like vector addition

(“tip-to-tail”).

3 1M j

2 2N j

Algebraic addition & subtraction

are performed piece-wise:

1 1

2 2

1 2 1 2

M a b j

N a b j

M N a a b b j

5 1M N j

4

Multiplication: Rectangular Form

5 3M j

2 4N j

Multiplication may be accomplished in rectangular form…

1 1 1z a b j

2 2 2z a b j

1 2 1 1 2 2

2

1 2 1 2 2 1 1 2

1 2 1 2 2 1 1 2

1 2 1 2 1 2 2 1

z z a b j a b j

a a a b j a b j b b j

a a a b a b j b b

a a b b a b a b j

2

5 3 2 4

10 20 6 12

22 14

M N j j

j j j

j

…but it is more easily accomplished in polar form.

5

Exponential Form

cos sin

cos sin

j

j

j

e j

z e z j z

z e a b j

cos

sin

a z

b z

assume |z| is

positive, real

2 22 2 2 2

2 22 2 2 2

2 2

cos sin

cos sin

a b z z

a b z z

a b z

sintan

cos

b

a

6

Rectangular Exponential Form

Re

Im

a z

b z

4 3M j

|z|

2 2

tanb

a

z a b

2 2

1

4 3 5

tan 3 4 37

z

375 jM e

|z| = magnitude of z

= phase/angle of z

7

Polar Form

jz a b j z e z

polar rectangular exponential

Polar form is a shorthand

for the exponential form.

374 3 5 5 37jM j e

|z| = magnitude of z

= phase/angle of z

8

Multiplication: Polar Form

3 4 5 53

3 33 45

2 2

M j

N j

Multiplication in polar form is carried out using exponentials…

1

2

1 1 1 1

2 2 2 2

j

j

z a b j z e

z a b j z e

1 2

1 2

1 2

1 2 1 2

1 2

1 2

j j

j j

j

z z z e z e

z z e

z z e

5 53 3 45

15 98

M N

1 1 1 2 2 1,z z z z 1 2 1 2 1 2z z z z

9

Division: Polar Form

1 2

1 2

2 2

1 1 1

2 2 2

j jj j

j j

z e z zee

z e z e z

1 1 1 2 2 1,z z z z

11 11 2

2 2 2

zz

z z

6 8 10 53

5 55 45

2 2

M j

N j

10 53 5 45

2 8

M N

10

Complex Conjugate

The complex conjugate of z is denoted z

then z a b j and if z a b j

3 1M j

3 1M j

The conjugate of z is the same

number, except that the

imaginary part is negated.

Graphically, the complex

conjugate of z is the

mirror image of z

across the Real axis.




Spring 2016


Lecture 10(b)

Complex Arithmetic Examples

and Matlab Scripts

12

Example: Complex Addition

9 2 Aj 3 1 Aj 2 6 Aj

>> i1 = 9 + 2*j;

>> i2 = -3 + j;

>> i3 = -2 + 6*j;

>> i = i1 + i2 + i3

i = 4.0000 + 9.0000i

13

9 2 Aj 3 1 Aj 2 6 Aj

4 9 Aj

>> i1 = 9 + 2*j;

>> i2 = -3 + j;

>> i3 = -2 + 6*j;

>> i = i1 + i2 + i3

i = 4.0000 + 9.0000i

Example: Complex Addition

14

Example: Complex Multiplication

Find v x i in rectangular form:

7 3 mV

5 4 mA

v j

i j

2

7 3 5 4

35 15 28 12

47 13 μW

v i j j

j j j

j

2 9 V , 3 5 Av j i j

2 9 3 5

6 27 10 45

51 17 W

v i j j

j j

j

>> V = 2 + 9*j;

>> I = -3 + 5*j;

>> p = V * I

p = -51.0000 -17.0000i

15

Find v x i in rectangular form:

7 3 mV

5 4 mA

v j

i j

2

7 3 5 4

35 15 28 12

47 13 μW

v i j j

j j j

j

2 9 V , 3 5 Av j i j

2 9 3 5

6 27 10 45

51 17 W

v i j j

j j

j

>> V = 2 + 9*j;

>> I = -3 + 5*j;

>> p = V * I

p = -51.0000 -17.0000i

Example: Complex Multiplication

16

Example: Complex Arithmetic

Determine the quantity va – vb

in polar form if vn = 0 .

va

vb

vn 110 0 V

110 120 V

110 240 V

vc

>> v_a = 110*exp(j*0);

>> v_b = 110*exp(j*-2*pi/3);

>> v = v_a - v_b;

>> abs(v)

ans = 190.5256

>> angle(v)*180/pi

ans = 30.0000

17

Determine the quantity va – vb

in polar form if vn = 0 .

va

vb

vn 110 0 V

110 120 V

110 240 V

190.5 30 V

22 1

110 0 110 120

110cos0 110sin 0

110cos 120 110sin 120

110 0 110 1 2 110 3 2

165 55 3

165 55 3 tan 55 3 165

a bv v

j

j

j j

j

vc

>> v_a = 110*exp(j*0);

>> v_b = 110*exp(j*-2*pi/3);

>> v = v_a - v_b;

>> abs(v)

ans = 190.5256

>> angle(v)*180/pi

ans = 30.0000

Example: Complex Arithmetic

18

Example: Complex Division

Determine the ratio of VL to IL:

LLVL IL

L L

VV

I I

VL

+

–

IL

3 3 3 mVj 1 3 mAj

>> V_L = -3*sqrt(3)+3*j;

>> I_L = 1+j*sqrt(3);

>> Z = V_L / I_L

Z = -0.0000 + 3.0000i

>> abs(Z)

ans = 3.0000

>> angle(Z)*180/pi

ans = 90.0000

19

2 2 1

22 1

3 3 3 mV

1 3 mA

3 3 3 tan 1 3 6 150

2 601 3 tan 3

L

L

j

j

V

I

3 90

Example: Complex Division

Determine the ratio of VL to IL:

VL

+

–

IL

3 3 3 mVj 1 3 mAj

LLVL IL

L L

VV

I I

>> V_L = -3*sqrt(3)+3*j;

>> I_L = 1+j*sqrt(3);

>> Z = V_L / I_L

Z = -0.0000 + 3.0000i

>> abs(Z)

ans = 3.0000

>> angle(Z)*180/pi

ans = 90.0000

20

Example: Complex Conjugate

Write the quantity V x I* in polar form, given 3 5 V

6 7 mA

j

j

V

I

>> V = 3 - 5*j;

>> I = 6 + 7*j;

>> S = V * conj(I)

S = -17.0000 -51.0000i

>> abs(S)

ans = 53.7587

>> angle(S)*180/pi

ans = -108.4349

21

Write the quantity V x I* in polar form, given

3 5 6 7

18 30 21 35

17 51 mW

j j

j j

j

V I

>> V = 3 - 5*j;

>> I = 6 + 7*j;

>> S = V * conj(I)

S = -17.0000 -51.0000i 2 2 117 51 tan 51 17

53.8 108 mW

V I

>> abs(S)

ans = 53.7587

>> angle(S)*180/pi

ans = -108.4349 34 59 85 49

34 85 59 49

53.8 108 mW

V I

3 5 V

6 7 mA

j

j

V

I

Example: Complex Conjugate




Spring 2016


Lecture 10(c)

Sinusoids and

Sinusoidal Steady-State

23

Alternating Current (Sinusoidal)

0 0sin , 0mv t V t

Vm = amplitude (in Volts), 0 = phase (in radians)

= frequency (in radians/second)

T = period (in seconds)

f = frequency (in cycles/second) = 1/T = / 2

0 0sin cos 2m mV t V t

+ – v t

A

B

A

B

24

Review of Sinusoids

v2 “leads” v1 by

v1 “lags” v2 by

1 2sin , sinm mv t V t v t V t

25

Sinusoids & Exponential Form

1 2sin , sinm mv t V t v t V t

cos sinj t

m m mV e V t j V t

cos sinj

e j

Im sinj t

m mV e V t

Re cos

j t

m mV e V t

26

RL Circuit with a Sinusoidal Source

i

0 cosVd R

i t i t tdt L L

0 cosV t

-- oscillates forever

-- never settles to a DC value (e.g. zero)

0 cosi t I t It’s possible that the solution is of the form

Substituting i(t) into the differential equation…

Solving for I0 and substituting back into i(t) yields

00 0sin cos cos

VRI t I t t

L L

10

2 2 2cos tan

V Li t t

RR L

amplitude scaling,

phase shift

27

RL Circuit with a Sinusoidal Source

vs vs

iL

iL

The RL circuit’s transient response

is negligible after 5t .

The remaining response is sinusoidal.

400 μH

5 5 5 1μs2 kΩ

L

Rt

tran

sien

t

28

RC Circuit with a Sinusoidal Source

vs

vs vC

+

–

tran

sien

t

vC

The RC circuit’s transient response

is negligible after 5t .


5 5

5 2 kΩ 100 pF 1μs

RCt

29

RLC Circuit with a Sinusoidal Source

is

vC

vC

+

–

tran

sien

t 5 1.6 msst

1 2 3cos sint

c d dv t e V t V t V

1 1 rad3125

2 2 10 16 μF sRC

The RLC circuit’s transient response

is negligible after ts .





Spring 2016


Lecture 10(d)

Phasors

31

2cosmI t

1cosmV t

Phasor Notation

2mI

1mV

1 1 1

1 1cos Re Rej t j jj t

m m m m mV t V e V e e V e V

2 2 2

2 2cos Re Rej t j jj t

m m m m mI t I e I e e I e I

Assume all voltages & currents

oscillate with frequency = 2 f …

Pick off the amplitude & phase for each v/i ;

write each in polar form.

phasor notation

(longer derivation provided in

textbook section 10.3)

32

Let all phasors be referenced to a cosine (zero phase).

Convert the following time-domain functions to the phasor domain :

(a) 40cos(t + 30°) mV, = 100 rad/s

(b) 25cos(t – 75°) A, = 400 rad/s

(c) 70cos(2 f + 45°) V, f = 20 MHz

(d) 36sin(2 f + 110°) mA, f = 8 MHz

sin cos 2t t

(a)

(b)

(c)

(d)

Example: Phasor vs. Time Domain

33

Let all phasors be referenced to a cosine (zero phase).

Convert the following time-domain functions to the phasor domain :

(a) 40cos(t + 30°) mV, = 100 rad/s

(b) 25cos(t – 75°) A, = 400 rad/s

(c) 70cos(2 f + 45°) V, f = 20 MHz

(d) 36sin(2 f + 110°) mA, f = 8 MHz

(a) 40 30 mV

(b) 25 75 A

(c) 70 45 V

(d) 36 20 mA sin cos 2t t

Example: Phasor vs. Time Domain

34

Example: Phasors and Voltage/Current

Let = 2000 rad/s with phasors be referenced to a cosine (zero phase).

Determine the instantaneous value, at t = 1 ms ,

of the current corresponding to this phasor: 20 + 10j A .

omega = 2000;

I = 20 + 10*j;

T = 2*pi/omega;

delta_t = T/1000;

t = 0:delta_t:2*T;

i_t = abs(I)*cos(omega*t + angle(I));

plot(t,i_t)

grid

axis([0 2*T -Inf Inf])

ylabel('Current (A)')

xlabel('Time (s)')

– 17.5 A

35

2 2 120 10 20 10 tan 10 20 22.4 26.6 Aj

22.4 cos 2000 0.464 A

1 ms 22.4cos 2.464

i t t

i

omega = 2000;

I = 20 + 10*j;

T = 2*pi/omega;

delta_t = T/1000;

t = 0:delta_t:2*T;

i_t = abs(I)*cos(omega*t + angle(I));

plot(t,i_t)

grid

axis([0 2*T -Inf Inf])

ylabel('Current (A)')

xlabel('Time (s)')

Example: Phasors and Voltage/Current

Let = 2000 rad/s with phasors be referenced to a cosine (zero phase).

Determine the instantaneous value, at t = 1 ms ,

of the current corresponding to this phasor: 20 + 10j A .




Spring 2016


Lecture 10(e)

Phasors & Ohm’s Law, KVL/KCL

37

Phasor Voltage vs. Current: R, L, C

v t R i t

For this equation to be true,

1 2andm mV I R

L L

dv t L i t

dt

1cosmv t V t 2cosmi t I t

1 2 90 , m

m

VL

I 2 1 90 , m

m

IC

V

C C

di t C v t

dt

For this equation to be true, For this equation to be true,

V = R I V = jL I

v leads i

by 90°

v and i are

in phase I = jC V i leads v

by 90°

38

Example: KCL, Phasor Domain

+

–

Determine v(t).

v t

2 nF 70

10

mH

68cos 2 10 30 mAsi t t

39

V

+

–

Determine v(t).

v t

• Convert to phasor form… +

–

IR IL IC

8 30 mAs I

2 nF 70

2 nF

10

mH 70

• Employ the appropriate Kirchhoff Law(s)…

0s j CR j L

V V

I V

6 9

6 68 30 2 10 2 10 0

70 2 10 10 10j

j

V VV

10

mH



40

+

–

Determine v(t).

v t

2 nF

10

mH 70

• Convert between rectangular & polar forms as necessary…

6 9

6 68 30 2 10 2 10 0

70 2 10 10 10j

j

V VV

1 10.0126 8 30

70 62.8j

j

V

0.0143 0.0159 0.0126 8 30j j V

0.0143 0.0033 8 30j V

0.0147 13 8 30 V



41

+

–


Determine v(t).

v t

2 nF

10

mH 70

omega = 2*pi*10^6;

I = 8*exp(j*30*pi/180);

R = 70;

L = 10e-6;

C = 2e-9;

Y = (1/R + 1/(j*omega*L) + j*omega*C);

V = I / Y;

abs(V)

ans = 545.2174

angle(V)*180/pi

ans = 43.1941


42

+

–


Determine v(t).

v t

2 nF

10

mH 70

• Convert between rectangular & polar forms as necessary…

0.0147 13 8 30 V

8 30 mA544 43 mV

0.0147 13

V

omega = 2*pi*10^6;

I = 8*exp(j*30*pi/180);

R = 70;

L = 10e-6;

C = 2e-9;

Y = (1/R + 1/(j*omega*L) + j*omega*C);

V = I / Y;

abs(V)

ans = 545.2174

angle(V)*180/pi

ans = 43.1941

• Convert from phasors to time domain…

6544cos 2 10 43 mVv t t





Spring 2016


Lecture 10(f)

Impedance

44

Impedance

Impedance, Z is the ratio of phasor voltage to phasor current

for an electrical element or network. like resistance, but it is complex

-- It is a measure of an element/network’s ability to impede current flow.

V

ZI

For a resistor, RR R V I Z

For an inductor, Lj L j L V I Z

For a capacitor, 1Cj C j C

j C

I V Z

-- current and voltage are always in-phase

-- there is no frequency dependence

-- voltage always leads current by 90°

-- at higher frequencies, less current is passed (for constant V )

-- current always leads voltage by 90°

-- at higher frequencies, more current is passed (for constant V )

45

Impedance vs. Frequency

0

R

R

R

R

R

Z

Z

Z

R = 100

90

L

L

L

j L

L

Z

Z

Z

2 f

1

1

90

C

C

C

j C

C

Z

Z

Z

L = 3 mH

C = 300 pF

46

Impedance vs. Frequency

100 , RR R Z

300 pF, 1CC j C Z

freq = 2e6:1e3:10e6;

omega = 2*pi*freq;

R = 100;

Z_R = R*ones(1,length(omega));

L = 3e-6;

Z_L = j*omega*L;

C = 300e-12;

Z_C = 1./(j*omega*C);

2 MHz 10 MHz, 2f f

3μH, LL j L Z

figure(1)

subplot(2,1,1)

plot(freq/10^6,abs(Z_R))

axis([-Inf Inf 0 300])

set(gca,'Xtick',[2:1:10])

ylabel('| Z_R | (ohms)')

title('{\bf Resistor Impedance ...

vs. Frequency}')

subplot(2,1,2)

plot(freq/10^6,phase(Z_R)*180/pi)

axis([-Inf Inf -180 180])


ylabel('phase[ Z_R ] (degrees)')

xlabel('Frequency (MHz)')

figure(2)

subplot(2,1,1)

plot(freq/10^6,abs(Z_L))



ylabel('| Z_L | (ohms)')

title('{\bf Inductor Impedance ...

vs. Frequency}')

subplot(2,1,2)

plot(freq/10^6,phase(Z_L)*180/pi)



ylabel('phase[ Z_L ] (degrees)')


figure(3)

subplot(2,1,1)

plot(freq/10^6,abs(Z_C))



ylabel('| Z_C | (ohms)')

title('{\bf Capacitor Impedance ...

vs. Frequency}')

subplot(2,1,2)

plot(freq/10^6,phase(Z_C)*180/pi)



ylabel('phase[ Z_C ] (degrees)')


47

Impedances in Series & Parallel

1

N

s n

n

R R

1

N

s n

n

Z Z

Impedances in series

are combined

like resistors in series.

Z1

Z2

Z3

ZN

V1

V2

V3

VN

Vs

I

1

s

N

n

n

VI

Z

Impedances in parallel

are combined

like resistors in parallel. 1

1

s

N

n

n

IV

Z

1

1 1N

np nR R

1

1 1N

np n

Z Z

48

Example: Equivalent Impedance

Determine the equivalent

impedance of the network at

terminals A–B if = 5 rad/s. 2

100

mF

1.6 H 33.3

mF

4

A

B

49




100

mF

2 –2j

1.6 H

8j –6j

33.3

mF

4

4

• Convert all resistances,

inductances, capacitances

into impedances…

A

B

A

B

R

L

C

R

j L

j C

Z

Z

Z


50

• Combine impedances in series &

parallel, starting away from A–B

and working towards A–B…

1

2 2 4 902 45 1

2 2 8 45

jj

j

Z

2 1 8 6

1 8 6 1

j j

j j j j

Z ZZ1


51

• Combine impedances in series &

parallel, starting away from A–B

and working towards A–B…

Z2

2

2

4 1 4

4 1 4

4 4 32 45.0

5 26 11.4

A B

j

j

j

j

ZZ

Z

1.1 33.6

If a test source were applied at terminals A–B : test 1 0 A I

then the voltage across A–B would be test 1 0 1.1 33.6 1.1 33.6 V V


52




100

mF

1.6 H 33.3

mF

4

A

B

omega = 5;

C1 = 100e-3;

C2 = 33.3e-3;

R1 = 2;

R2 = 4;

L1 = 1.6;

Z_C1 = 1/(j*omega*C1);

Z_C2 = 1/(j*omega*C2);

Z_R1 = R1;

Z_R2 = R2;

Z_L1 = j*omega*L1;

Z_eq1 = Z_C1*Z_R1/(Z_C1 + Z_R1);

Z_eq2 = Z_eq1 + Z_L1 + Z_C2;

Z_ab = Z_R2*Z_eq2 / (Z_R2 + Z_eq2)

Z_ab =

0.9217 + 0.6120i

abs(Z_ab)

ans = 1.1063

phase(Z_ab)*180/pi

ans = 33.5837


53

Example: Phasor Analysis

(a) Determine vC(t).

(b) Plot vC(t) and the source

versus time for 2 cycles. 3 cos(2×106t + 20°) V

1 k

1 nF

+

vC

54




versus time for 2 cycles.

vS

vC

3 cos(2×106t + 20°) V

1 k

1 nF

+

vC

+

55

• Convert the circuit from the time domain

to the phasor domain.

VC 6 92 10 10 160 ΩC j C j j Z–160j

1 k 3 20

V

• Use KVL/KCL to solve for V/I

in the phasor domain.

3 20 1000 160 0

160C

j

j

I I

V I

I





3 cos(2×106t + 20°) V

1 k

1 nF

+

vC

56

• Perform complex algebra to find V/I…

• Convert back to the time domain…

6472cos 2 10 60.9 mVCv t t

2 2 1

160 160 903 20 3 20

1000 160 1000 160 tan 160 1000

160 903 20 0.472 60.9

1013 9.1

C

j

j

V





+

VC –160j

1 k 3 20

V

I

57

Example: Plotting Sinusoids, Matlab




omega = 2*pi*10^6;

T = 2*pi/omega;

delta_t = T/100;

t = 0 : delta_t : 2*T - delta_t ;

v_S = 3.000 * cos( omega*t + 20.0*pi/180 );

v_C = 0.472 * cos( omega*t - 60.9*pi/180 );

figure(1)

plot(t/10^-6, v_S, 'r--', ...

t/10^-6, v_C, 'b-', 'LineWidth', 2)

ylabel('Voltage (V)')

xlabel('Time (\mus)')

legend('v_S','v_C')

grid

6472cos 2 10 60.9 mVCv t t

3 cos(2×106t + 20°) V

1 k

1 nF

+

vC

58

Example: Transient AC Circuit, PSpice

6472cos 2 10 60.9 mVCv t t

Amplitude is 472 mV

as determined by

written analysis.

59

Example: Frequency Response, Matlab

9

9 1010 ΩC

jj C j

Z

9

9

103 20 3 20

1000 1000 10

CC

C

j

j

ZV

Z

Plot the amplitude of vC(t) versus f0

for 500 kHz < f0 < 2 MHz.

V_s = 3 * exp( j*20*pi/180 );

R = 1000;

C = 1e-9;

f = 500e3 : 1e3 : 2e6 ;

omega = 2 * pi * f;

Z_C = -j ./ (omega * C);

V_C = V_s * Z_C ./ (R + Z_C);

plot(f/10^3, abs(V_C), 'LineWidth', 2)

ylabel('|V_C| (volts)')

xlabel('Frequency (kHz)')

grid

6 6

6

for 10 Hz ( 2 10 rad s) ...

472cos 2 10 60.9 mVC

f

v t t

Frequency (kHz)

3 cos(2×106t + 20°) V

1 k

1 nF

+

vC

60

Example: Frequency Response, PSpice



“VAC” part,

“SOURCE”

library

61

Example: Frequency Response, PSpice



“VAC” part,

“SOURCE”

library

6 6

6

for 10 Hz ( 2 10 rad s) ...

472cos 2 10 60.9 mVC

f

v t t




Spring 2016


Lecture 10(g)

Nodal & Mesh Analysis

in the Phasor Domain

63

Nodal Analysis with Phasors

Analysis Steps

(1) Choose a reference node (usually ground or the bottom node) to have a voltage of zero.

(2) Assign a unique voltage variable to each node that is not the reference (v1, v2, v3, … vN–1).

(3) For independent & dependent voltage sources, identify a supernode

and write the voltage across the supernode in terms of node voltages.

Write a KCL equation at all N – 1 nodes including the supernode

(and not the reference, or a supernode which includes the reference).

(4) Solve the N – 1 node equations + source equations simultaneously.

0

v1 v2

0

V1 V2 V3 v3

SAME as with DC circuits. Now use complex arithmetic.

64

Example: Nodal Analysis, Phasors

Write a valid matrix equation whose

solution includes the phasor form of v(t).

+

–

v(t)

0.5iA

iA

65



+

–

V

0

V1

V2

V3 V4

320 70 10 1.4LZ j L j j

320 250 10 0.2CZ j C j j

1.4 j

0.2 j 0.2 j

9 0 V 9 90 V

• Convert to phasor form…

• Identify supernode(s)… 1 2 0.5 AV V I

• Write KCL equations…

• Identify v sources next to the reference…

3 1 3 2 3 4 3 00.2 1.4 0.2 3

V V V V V V V

j j j

1 49, 9V V j

• Write equations governing

dep. src. ctrl…

IA

0.5IA

0

v1

v2

v3 v4

0.5iA

iA

3

3A

VI

+

–

v(t)


66



+

–

V

0

V1

V2

V3 V4

0.5IA

1.4 j

0.2 j 0.2 j

9 0 V 9 90 V IA

• Rearrange into matrix form…

3 1 3 2 3 4 3 00.2 1.4 0.2 3

V V V V V V V

j j j

1 49, 9V V j 1 2 0.5 AV V I

1 2 3 4

15 5 5 5 0

1.4 1.4 3

j jj V V j j V j V

1

4

9

9

V

V j

1 2 0.5 0AV V I

3

3A

VI

2V V

3 3 0AV I

1

3

4

105 9.3 5 0

1.4 39

1 0 0 0 09

0 0 0 1 00

1 1 0 0 0.50

0 0 1 3 0 1 A

jVj j j

V

V j

V

I

8.3 5.5 VV

8.3cos 20 5.5 Vv t t


67



+

–

V

0

V1

V2

V3 V4

0.5IA

1.4 j

0.2 j 0.2 j

9 0 V 9 90 V IA

3 1 3 2 3 4 3 00.2 1.4 0.2 3

V V V V V V V

j j j

1 49, 9V V j

3

3A

VI

>> A = [ -5*j j/1.4 9.29*j+1/3 -5*j 0 ;

1 0 0 0 0 ;

0 0 0 1 0 ;

1 -1 0 0 -0.5 ;

0 0 -1/3 0 1 ];

>> B = [0 ; 9 ; -9*j ; 0 ; 0 ];

>> x = A^-1 * B

>> V = x(2);

>> abs(V)

>> angle(V) * 180/pi

x =

9.0000 - 0.0000i

8.2703 + 0.7913i

4.3784 - 4.7477i

0.0000 - 9.0000i

1.4595 - 1.5826i

ans =

8.3080

ans =

5.4653

1 2 0.5 AV V I


8.3 5.5 VV

68

Mesh Analysis with Phasors

Analysis Steps

(1) Draw a mesh current for each mesh.

(2) Identify supermeshes.

(3) Write KVL around each supermesh,

then KVL for each mesh that is

not part of a supermesh.

(4) Express additional unknowns

(e.g. dependent-source V/I)

in terms of mesh currents.

(5) Solve the simultaneous equations.

SAME as with DC circuits.

Now use complex arithmetic.

i1 i2

I1 I2

69

Example: Mesh Analysis, Phasors



v(t)

+

–

0.5iA

iA

70

320 70 10 1.4LZ j j

320 250 10 0.2CZ j j

• Convert to phasor form…

• Identify supermesh(es)…

• Write KVL equations…

1 2 1 4

2 3 2 1

3 2 3 3 4

4 1 4 3

9 0.2 3 0

0.5 1.4 0.2 0

1.4 5 9 0.2 0

3 0.2 9 0

A

j I I I I

I j I I j I I

j I I I j j I I

I I j I I j

• Write equations governing

dep. src. ctrl… 1 4AI I I

v(t)

+

–

i2 i3

i1 i4

0.5iA

iA

I2 I3

I1 I4

+

–

V

9 0 V

9 Vj

0.5IA

IA



0.2 j 0.2 j

1.4 j


71




1

2

3

4

0.2 3 0.2 0 3 0 0 9

0.2 1.2 1.4 0 0.5 0 0

0 1.4 1.2 5 0.2 0 0 9

3 0 0.2 3 0.2 0 0 9

1 0 0 1 1 0 0

0 0 5 0 0 1 0

A

j j I

j j j I

j j j I j

j j I j

I

V

8.3 5.5 VV 8.3cos 20 5.5 Vv t t

I2 I3

I1 I4

+

–

V

9 0 V

9 Vj

0.5IA

IA

0.2 j 0.2 j

1.4 j

1 2 1 4

2 3 2 1

3 2 3 3 4

4 1 4 3

9 0.2 3 0

0.5 1.4 0.2 0

1.4 5 9 0.2 0

3 0.2 9 0

A

j I I I I

I j I I j I I

j I I I j j I I

I I j I I j

1 4AI I I 35V I

1 2 4

1 2 3

2 3 4

1 3 4

1 4

3

0.2 3 0.2 3 9

0.2 1.2 1.4 0.5 0

1.4 1.2 5 0.2 9

3 0.2 3 0.2 9

0

5 0

A

A

j I j I I

j I j I j I I

j I j I j I j

I j I j I j

I I I

I V


72



I2 I3

I1 I4

+

–

V

9 0 V

9 Vj

0.5IA

IA

0.2 j 0.2 j

1.4 j

1 2 1 4

2 3 2 1

3 2 3 3 4

4 1 4 3

9 0.2 3 0

0.5 1.4 0.2 0

1.4 5 9 0.2 0

3 0.2 9 0

A

j I I I I

I j I I j I I

j I I I j j I I

I I j I I j

1 4AI I I 35V I

A = [ -0.2*j+3 0.2*j 0 -3 0 0 ;

0.2*j 1.2*j -1.4*j 0 0.5 0 ;

0 -1.4*j 1.2*j+5 0.2*j 0 0 ;

-3 0 0.2*j 3-0.2*j 0 0 ;

1 0 0 -1 -1 0 ;

0 0 -5 0 0 1 ];

B = [ 9 ; 0 ; -9*j ; 9*j ; 0 ; 0 ];

x = A^-1 * B

V = x(6);

abs(V)

angle(V) * 180/pi

x =

-18.1368 +20.4776i

5.6121 - 2.6198i

1.6540 + 0.1583i

-19.5970 +22.0609i

1.4602 - 1.5833i

8.2699 + 0.7916i

ans =

8.3077

ans =

5.4679


8.3 5.5 VV




Spring 2016


Lecture 10(h)

Thevenin Equivalence

in AC Circuits

74

Review of Thevenin Equivalents

• allows us to replace a large, complicated circuit with a much

simpler 2-element series/parallel circuit

• the simpler circuit allows for rapid calculations of V, I, P that the

original circuit can deliver to a load

• helps us to choose the best value of load resistance to maximize

the power delivered (e.g. from an amplifier, to a speaker)

75


only ind.

src.?

To determine VTH ...

To determine ZTH ...

• source transformation

• find VTH = VOC

• find VTH = VOC

YES NO

only ind.

src.?

• deactivate src., find ZTH = Zeq

• source transformation

• find ISC, use ZTH = VOC/ISC

• insert test source, find ZTH = Vtest/Itest

YES NO

VTH = 0 ?

• find ISC, use ZTH = VOC/ISC

• insert test source, find ZTH = Vtest/Itest

NO

• insert test source,

find ZTH = Vtest/Itest

YES

76

Determine the Thevenin

equivalent of Network A using

open-circuit voltage and

short-circuit current.

OC TH

6 612 12 8 V =

6 3 9V

V V

+

–

VOC

SC

1 7 84 A

1 2 1 7 9

I

OCTH

SC

89

8 9

VZ

I

ISC


77

Determine the Thevenin

equivalent of Network A

by using a test source.

TH8 V V

+

–

VOC

Itest

Vtest

Vtest

Itest

2

testtest

2 7

VI

testTH

test

9 V

ZI


78

Example: Thevenin & Sinusoids

Determine the phasor voltage

difference V1 – V2 (with –j10 ) .

Use the Thevenin equivalent

at V1 , V2 (without –j10 ) .

79

TH OC A B 1 0 4 2 0.5 90 2 4

4 2 0.5 2 4 4 2 2 6 3 V

j j

j j j j j j

V V V V

VA VB

TH 4 2 2 4 6 2j j j Z

1 2

106 3

6 2 10

3 6 6.7 63.4 V

jj

j j

j

V V

Example: Thevenin & Sinusoids

Determine the phasor voltage

difference V1 – V2 (with –j10 ) .

Use the Thevenin equivalent

at V1 , V2 (without –j10 ) .




Spring 2016


Lecture 10(i)

Phasors &

Superposition

81

Linearity

linear

network

N

1x t 1y t 2x t 2y t

1 1x t t 1 1y t t

1 1 1 2 2 2A x t t A x t t

linear

network

N

N

1 1 1A x t t 1 1 1A y t tN

1 1 1 2 2 2A y t t A y t t

x = input or

source or

stimulus

y = output or

response

N

N

82

Superposition: Voltage Sources

2 k

2 k

4 V

8 V

16 V

32 V

Determine the current i using superposition.

16 4 4 mAi 32 4 8 mAi

8 4 2 mAi 4 4 1mAi

deactivate all but 1 solve sum

60 V

60 4 15 mAi

4 8 2 1

15 mA

i i i i i

83

Determine the voltage v

using superposition.

deactivate all but 1

solve sum

3 6 A 6 12 A 24 A

6 0 12 V3 6

v vv

12 0 24 V3 6

v vv

24 0 48 V3 6

v vv

12 24 48

60 Vv

Superposition: Current Sources

84

Example: Superposition & Sinusoids

Determine the phasor

voltage difference V1 – V2 .

Use superposition.

85

Example: Superposition & Sinusoids

1V 2

V 2V

1V

1 2 1 2 2

1V V V V V V

1 2

4 21 10 2 4 V

4 2 2 6

jj j

j j

V V

2

2 40.5 10 1 2 V

2 4 4 12

jj j j

j j

1

V V

1

1 2 3 6 9 36 tan 2

6.7 63.4 V

j

V V



Use superposition.




Spring 2016


Lecture 10(x,1)

Thevenin Equivalent

Example

87

Determine the Thevenin equivalent

circuit with respect to terminals a–b.

Example: Thevenin & Sinusoids, #2

88



TH eq

5 10 55 || 10 5

5 10 5

25 502.5 5

10

j jj j

j j

jj

Z Z

TH OC

1 10 53 30 10 5

1 10 5 1 5

13.0 10.6 V

RLs RL

RL C

jj

j j

j

YV V I Z

Y Y


89



TH OC SC

13.0 10.6 3 30

2.5 5 j

Z V I

2.5 5 j

13.0 10.6 Vj


90


impedance with respect to terminals a–b.

TH 2.5 5 j Z





Spring 2016


Lecture 10(j)

Phasors &

Source Transformation

92

Review of Source Transformation

If these two circuits provide the same v/i characteristics at their

outputs (vL, iL), the two circuits are equivalent.

sV

max s sI V R

max sV V

max sI I

max s pV I R

ss p

s

VR R

I

condition

for equivalence

sI

93

Source Transformation & Phasors

If these two circuits provide the same V/I characteristics at their

outputs (VL, IL), the two circuits are equivalent.

sV

max s sI V Z

max sV V

max sI I

max s p V I Z

ss p

s

V

Z ZI

condition

for equivalence

sZ LI

LV

LI

LVpZsILZ LZ

94

Example: Src. Transform. & Sinusoids



Use source transformation(s).

95

Example: Src. Transform. & Sinusoids

VS1 VS2

Z2 Z1

1 2 S1 S2

1 2

10

10

j

j

V V V V

Z Z

1

2

4 2

2 4

j

j

Z

Z

S1 1 0 4 2

4 2 V

j

j

V

S1 0.5 90 2 4

0.5 2 4 2 V

j

j j j

V

1 2

104 2 2

10 4 2 2 4

106 3 3 6

8 6

6.7 63.4 V

jj j

j j j

jj j

j

V V



Use source transformation(s).




Spring 2016


Lecture 10(k,1)

PSpice for

AC Circuits

97

• From nodal analysis…

3 1 3 3 4 3 00.2 1.4 0.2 3

V V V V V V V

j j j

1 49, 9V V j 1 2 0.5 AV V I

3

3A

VI

1

3

4

105 9.3 5 0

1.4 39

1 0 0 0 09

0 0 0 1 00

1 1 0 0 0.50

0 0 1 3 0 1 A

jVj j j

V

V j

V

I

8.3 5.5 VV

8.3cos 20 5.5 Vv t t

Example #1: Written Analysis

+

–

V

0

V1

V2

V3 V4

1.4 j

0.2 j 0.2 j

9 0 V 9 90 V IA

0.5IA

0.5iA

iA

+

–

v(t)

Solve for v(t) .

98

+

–

V V1

V2

V3 V4

0.5IA

1.4 j

0.2 j 0.2 j

9 0 V 9 90 V IA


3 1 3 2 3 4 3 00.2 1.4 0.2 3

V V V V V V V

j j j

1 49, 9V V j

3

3A

VI

1 2 0.5 AV V I

8.3 5.5 V V

8.3cos 20 5.5 Vv t t

Example #1: Matlab

Solve for v(t) .

>> A = [ -5*j j/1.4 9.29*j+1/3 -5*j 0 ;

1 0 0 0 0 ;

0 0 0 1 0 ;

1 -1 0 0 -0.5 ;

0 0 -1/3 0 1 ];

>> B = [0 ; 9 ; -9*j ; 0 ; 0 ];

>> x = A^-1 * B

>> V = x(2);

>> abs(V)

>> angle(V) * 180/pi

x =

9.0000 - 0.0000i

8.2703 + 0.7913i

4.3784 - 4.7477i

0.0000 - 9.0000i

1.4595 - 1.5826i

ans =

8.3080

ans =

5.4653

99

Example #1: PSpice

Amplitude is 8.4 V

as determined by

written analysis.

Plot v(t) using PSpice.

8.3 5.5 V V

8.3cos 20 5.5 Vv t t

100



Confirm this answer

using PSpice.

VS1 VS2

Z2 Z1

1 2 S1 S2

1 2

10

10

j

j

V V V V

Z Z

1

2

4 2

2 4

j

j

Z

Z

S1 1 0 4 2

4 2 V

j

j

V

S1 0.5 90 2 4

0.5 2 4 2 V

j

j j j

V

1 2

104 2 2

10 4 2 2 4

106 3 3 6

8 6

6.7 63.4 V

jj j

j j j

jj j

j

V V

Example #2: Written Analysis

+

–

+

–

101

1 2 6.7 63.4 V V V

Example #2: PSpice

“IAC” part,

“SOURCE”

library



Confirm this answer

using PSpice.




Spring 2016


Lecture 10(k,2)

Op Amps

in AC Circuits

103

Example #3: Phasors & Op Amps

200

2 k

125 nFSketch |Vout/Vin| vs. for 1 rad/s < < 100 krad/s .

Determine vout(t) for vin(t) = cos(4x104t) V .

104

200

out

in

||1 1

1

1 400010

1 4000

f f f f f

i i i f f

f f f

i f f

R j C R j C

R R R j C

R R C

R j R C j

ZV

V Z

2 k

125 nF

out

2 2in

400010

4000

V

V

4000

Sketch |Vout/Vin| vs. for 1 rad/s < < 100 krad/s .

Determine vout(t) for vin(t) = cos(4x104t) V .

4

4

out in3

4 10 104 10 7.1 135 V

4 10 4000 1j j

V V

4

out 7.1cos 4 10 135 Vv t t

Example #3: Phasors & Op Amps




Spring 2016


Lecture 10(k,3)

AC Thevenin Equivalent

w/ a Dependent Source

106

Example #4: Thevenin, Dependent Src

A

B


of this circuit at terminals A–B .

107

Example #4: Thevenin, Dependent Src

A

B


of this circuit at terminals A–B .

OC

10 30 30 5 25 0

25

a a a

a

j

j

I I I

I V

OC

10 300.28 75

25 25

25 0.28 75 7.07 15 V

aj

j

I

V

SC

SC

10 30 30 5 25 0

25 50 0

a a a

a

j

j

I I I I

I I

SC

25 25 30 10 30

25 50 0

aj

j

I

I SC 106 2 mA I7.07 15 V

TH

7.07 15

0.106 2

66.7 13

Z

A

B

ELEC 202 Electric Circuit Analysis II -...

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