Linear Circuit Analysis

Warming up snacks in a microwave oven

Second-Order Circuit1. What is second-order circuit?A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.

Typical examples of second-order circuits:

C

R L

US

+

_

CUS

+

_LR

(a) series RLC circuit (b) parallel RLC circuit

2. Discharging a capacitor through an inductor (§8.1)

CU0

+

_L

A BS

+uL_

+uC_

iC

iL

Solution

Step 1. Write down the u-i relationship for the capacitor and inductor and apply KCL and KVL, respectively.

CC L

dui C i

dt

LL C

diu L u

dt

Step 2. Obtain differential equations in the capacitor voltage uC and inductor current iL.

2

2

1CC

d uu

dt LC

2

2

1LL

d ii

dt LC

Step 3. Construct the waveforms uC(t) and iL(t).

Specifically recall the differentiation properties of the sine and cosine functions:

sin( ) cos( )d

t tdt

cos( ) sin( )d

t tdt

22

2sin( ) sin( )

dt t

dt

22

2cos( ) cos( )

dt t

dt

Thus one reasonable assumes that the solutions of differential equations have the general forms:

( ) cos( )Cu t K t

( ) cos( )Li t K t

Note : General solution has the equivalent form:

sin( ) cos( )2

K t K t

cos( ) sin( )2

K t K t

Step 4. Find ω, K and θ for the capacitor voltage and inductor current.

2

2

1CC

d uu

dt LC ( ) cos( )Cu t K t

sin( )CduK t

dt

22 2

2cos( )C

C

d uK t u

dt

Based on and

2 1

LC or

1

LC

The values of K and θ depend on the initial conditions.

CU0

+

_L

A BS

+uL_

+uC_

iC

iL 0(0 ) cosCu K U 1 1

(0 ) sin (0 ) (0 ) 0C C Lu K i iC C

0 0K Uand

0( ) cos( )C

tu t U

LC

0( ) sin( )L

C ti t U

L LCObtain iL(t) directly by differentiating

Several interesting and significant facts.• For the source-free LC circuit, the voltage and current responses are sinusoidal waveforms with an angular frequency equal to1/ LC

• The frequency ω depends on the values of L and C only, whereas the amplitude K and the phase angleθdepend on the initial values of the capacitor voltage and inductor current .

• Although the energy stored in the capacitor, WC(t), and the energy stored in the inductor, WL(t), both vary with time, their sum in constant. • In theory, the simplest circuit that generates sinusoidal waveforms is LC circuit. Such an electronic circuit is an idealized oscillator circuit. Oscillator circuits play an important role in many communication and instrumentation systems.

.Since the peak amplitude of the sinusoidal oscillations dose not damp out, the circuit is said to be undamped.

3. Source-free second-order linear networks (§8.2, §8.4)

C

R L

(a) series RLC circuit

CLR

(b) parallel RLC circuit

• Development of differential equation models for series an parallel RLC circuits

Solution

R L Ci i i

+uC_

Step 1. Apply KVL to the series RLC. 0R L Cu u u

Step 2. Chouse uC as circuit variable and express uR, uL, and uC in terms of uC.

0CC C

diRi L u

dt

2

2

10C C

C

d u duRu

dt L dt LC

Step 3. Chouse iL as circuit variable and express uR, uL, and uC in terms of iL.

1( ) 0

tL

L L

diRi L i d

dt C

2

2

10L L

L

d i diRi

dt L dt LC

Step 2. Chouse iL as circuit variable and express iR, iL, and iC in terms of iL.

Step 1. Apply KCL to the parallel RLC.

0R L Ci i i

Li

C L Ru u u +

_

Solution

CLR

(b) parallel RLC circuit

0L LL

u dui C

R dt

2

2

1 10L L

L

d i dii

dt RC dt LC

Step 3. Chouse uC as circuit variable and express iR, iL, and i

C in terms of uC.1

( ) 0t

C CC

u duu d C

R L dt

2

2

1 10C C

C

d u duu

dt RC dt LC

• Solution of the general second-order differential equation model

All four differential equations have the general form:2

20

d x dxb cx

dt dt

For appropriate constants b and c, where x is either uC or iL.

SolutionStep 1. Substitute Kest

22

2( ) 0

st stst std e de

K bK cKe Ke s bs cdt dt

2 0s bs c characteristic equation

Step 2. Solve characteristic equation, the roots are: 2

1 2

4,

2

b b cs s

natural frequencies

series RLC circuit / 1 /b R L c LC parallel RLC circuit 1/ 1 /b RC c LC

Step 3. Discuss three distinct possibilities

Case 1. b2 — 4c>0 , two distinct real roots. 1 21 2( ) s t s tx t K e K e

The constants K1 and K2 depend on the initial conditions of the differential equation, which depend on the initial voltages and currents in the circuit.

1 21 2 1 2

0(0 ) s t s t

tx K e K e K K

1 21 2 1 1 2 2

0

(0 ) s t s t

t

x K e K e s K s K

Series RLC circuit / 1 /b R L c LC 2

22

44 0

Rb c

L LC 2

LR

C

Parallel RLC circuit 1/ 1 /b RC c LC

22

1 44 0

( )b c

RC LC 2

CG

L

overdamped

Case 2. b2 — 4c<0 , two distinct conjugate complex roots.2

1 2

4,

2 d

b b cs s j

where / 2b and 2 4 / 2d b c

( ) [ cos( ) sin( )] cos( )t td d dx t e A t t Ke t solution

where 2 2 1, tan ( )B

K A BA

As before, A and B depend on the initial conditions as follows:

0(0 ) [ cos( ) sin( )]t

d d tx e A t B t A

0

(0 ) [ cos( ) sin( )]

[ sincos( ) cos( )]

td d

td d d d t

d

x e A t B t

e A t B t

A B

series RLC circuit 2L

RC

parallel RLC circuit 2C

GL

underdamped

Case 3. b2 — 4c=0 , two repeated (equal) roots. 11 2( ) ( ) s tx t K K t e

initial conditions: 1(0 )x K

1 1 2(0 )x s K K

series RLC circuit 2L

RC

parallel RLC circuit 2C

GL

If s1=s2 are negative, the response decays to zero and is said to the critically damped. Critically damped defines the boundary between overdamped and underdamped. This means that with a slight change in circuit parameters, the response would almost always change into either overdamped or underdamped.

Undamped Response Underdamped Response

Overdamped Response Critically damped Response

Generic waveforms corresponding to the four cases of damping.

• Response calculation of series and parallel RLC circuits

Procedure for solving second-order RLC circuits

Step 1. Determine the differential equation model of the circuit.

Step 2. From the differential equation model, construct the characteristic equation and find its roots using the quadratic root formula.

Step 3. From the nature (real distinct, real equal, or complex) of the roots of the characteristic equation, determine the general form of the solution. The solution form will contain two unknown parameters.

Step 4. Find the two unknown parameters using the initial conditions on the circuit.

+

_

Practical inductor

+

_

2010mH

L1R

2R

1C F

10

10V

A B

S

Cu

Example 9.5. Suppose the switch S in figure has remained in position A for a long time and moves to position B at t=0. The 1μF capacitor is assumed to be ideal, whereas the practical inductor is modeled by a 10mH ideal inductor in series with a 20Ω resistor. Find and plot uC(t) for t≥0 for the following three cases: case 1: R

2=405 Ω; case 2: R2=0; and case 3: R2=180Ω. Each of these cases represents a different type of response.

Solution

The first step in the calculation of the circuit response is to find a second-order differential equation in the unknown uC.

For the series RLC.2

2

10C C

C

d u duRu

dt L dt LC

where 1 2R R R

+

_

Practical inductor

+

_

2010mH

L1R

2R

1C F

10

10V

A B

S

Cu

With L and C fixed, the series RLC characteristic equation is

2 2 2 82

1(20 )10 10 0

Rs s s R s

L LCCase 1. R2=405 Ω

2 842500 10 0s s

1,2 21250 18750

2500, 40000

s

Real distinct roots imply an overdamped response of the form

2500 40001 2( ) 0t t

Cu t K e K e t

Step 1. Find the characteristic equation and the general form of the response.

Step 2. Find K1 and K2 using initial conditions on the circuit.

1 2(0 ) (0 ) 10C Cu u K K

1 2

(0 )(0 ) 0 2500 40000C

C

iu K K

C

1 210.667 0.667K K

Step 3. Set forth the solution for uC(t). 2500 4000( ) 10.667 0.667 0t t

Cu t e e V t

Case 2. R2=0ΩStep 1. Find the characteristic equation and the general form of the response.

1 1000 9950 ds j j Since b2 — 4c= － 396000000<0, the roots are complex .

2 82000 10 0s s

2 1000 9950 ds j j andThe underdamped response form is

1000[ cos( ) s( ) [ cos(9950 ) sin(9950 )]in( )]tdC d

tu t ee A t BA t t tB

Step 2. Find A and B using initial conditions on the circuit.(0 ) 10Cu A (0 ) 1000 9950 0C du A B A B

10 1.005A B Step 3. Set forth the solution for uC(t).

1000

1000

( ) [10cos(9950 ) 1.005sin(9950 )]

10.05 cos(9950 5.7 ) 0

tC

t o

u t e t t

e t V t

Case 3. R2=180Ω

Step 1. Find the characteristic equation and the general form of the response.

2 820000 10 0s s

Whose roots are s1=s2= － 104, implying a critically damped response.

4101 2( ) ( ) 0t

Cu t K K t e t

Step 2. Find K1 and K2 using initial conditions on the circuit.

1 2(0 ) 10Cu K K 4

1 1 2 1 2(0 ) 10 0Cu s K K K K 5

1 210 10K K

Step 3. Set forth the solution for uC(t). 44 10( ) 10(1 10 ) 0t

Cu t t e V t

Waveforms of uC(t) for three different degrees of damping. Critical damping represents the boundary between overdamped and the oscillatory behavior of underdamped.

• Application to selected second-order circuits

Not all second-order circuits are RLC. Some are only RC but with two capacitors, and some are RL with two inductors.

Passive RC or RL cannot have an oscillatory response.

However, with controlled sources a second-order RC or RL circuit can have an oscillatory response, which is not a phenomenon of a first-order circuit, but of second or higher order circuits.

The next example illustrates the analysis of an oscillatory second-order RC circuit containing controlled sources.

Example 9.6. This example illustrate the analysis of the second-order RC circuit shown in figure. The objective is to find u1C(t) and u2C(t) for t>0 given the initial conditions u1C(0)= 10V and u2C

(0) =0.

Solution

Step 1. Write a differential equation in u1C(t) .

6 311 2 110 0.1 10C

C C C

dui u u

dt

1k

1Ci

1 F

+

_ 1Cu20.1 Cu 10.1 Cu

+

_ 2Cu2Ci

1 F

6 5 311 2 110 10 10C

C C C

dui u u

dt

By considering the right node, we have6 2

2 110 0.1CC C

dui u

dt 52

110CC

duu

dt

Differentiating both sides of equation one

25 3 10 31 2 1 1

1210 10 10 10C C C C

C

d u du du duu

dt dt dt dt 2

3 101 112

10 10 0C CC

d u duu

dt dt

Step 2. Determine the characteristic equation, its roots, and the form of the response .

characteristic equation 2 3 1010 10 0s s

characteristic roots 1,2 500 99998.75 ds j j

underdamped response

1

500

( ) [ cos( ) sin( )]

[ cos(99998.75 ) sin(99998.75 )]

tC d d

t

u t e A t B t

e A t B t

Step 3. Find A and B.

1 2(0) 10, (0) 0C Cu u

1(0) 10Cu A

5 3 412 1

(0)10 (0) 10 (0) 10C

C C d

duu u A B

dt 25.0001 10B

From the initial conditions,

Step 4. Set forth the final form of uC1(t). 500 2

1( ) [10cos(99998.75 ) 5.0001 10 sin(99998.75 )] 0tCu t e t t V t

Underdamped (oscillatory) response showing envelope of exponential decay.

It is important to observe here that example 2 achieves a second-order RLC response without the use of an inductor, which is of significance for integrated circuit technology .

Step 5. Plot the response.

4. Second-order linear networks with constant inputs (§8.3)

The preceding section studied source-free second-order linear networks. When independent sources are present, the network differential equations are similar to the source-free case except for an additional term, which accounts for the effect of the input.

2

2( )

d x dxb cx f x

dt dt

2

2

d x dxb cx

dt dtF

解的一般形式为 :

( ) ( )n Fx t x t X F

FX

c

The interpretation XF=F/c is a mathematical one. When the differential equation describes a linear circuit with constant inputs, there is a physical interpretation of XF and a circuit theoretic method for computing its value, even without writing the differential equation.

( ) ( )n Fx t x t X

If Re[s1] and Re[s2]<0, then xn(t) tends to zero for large t, and hence x(t) tends to XF for large t. Consequently XF is termed the final value of the response.

( ) ( )n F Fx x X X a constant XF is either a constant capacitor voltage or a constant inductor current. If a capacitor voltage is constant, its current is zero; this interprets as an open circuit. Similarly, if an inductor current is constant, its voltage is zero; this interprets as a short circuit.

The value of XF can be obtained by analyzing the resistive network resulting when all capacitors are open-circuited and all inductor are short-circuited. Physically speaking then, XF equals either uC(∞) or iL(∞) when Re[s1] and Re[s2]<0.

Example 3. A step current input iin(t)=u(t) A excites the parallel RLC circuit of the figure whose initial conditions satisfy iL(0)=0 and uC(0)=0. This simply means that the current source turns on with a value of 1A at t=0 and maintains this constant current excitation for all time. The objective is to find the inductor current iL(t) for t≥0 for three values of R: 500Ω, 25Ω, and 20Ω.

R( )ini t0.25

L

mH 0.1

C

F

Solution由于这是一个并联 RLC 电路 , 因此特征方程为

72 12 01 1 10

4 10 0ss sLC

sRRC

For all positive values of R, the roots of the circuit’s characteristic equation have negative real parts. Thus for large t or ideally, at t=∞, the inductor looks like a short circuit and the capacitor like an open circuit. Hence for all cases of this example, XF=iL(∞)=1A.

Case 1. R=500 Ω.

characteristic equation 2 1020000 4 10 0s s

characteristic roots4 5

1,2 1.0 10 1.9975 10 ds j j

underdamped response ( ) [ cos( ) sin( )]tL d d Fi t e A t B t X

(0) 0, (0) 0L Ci u From the initial conditions,

(0 ) 1 0L Fi A X A 1A

(0 )(0 ) (0 )0CL L

d

udi uA B

dt L L

25.0063 10

d

AB

Hence for t≥0, 10000 5 2 5( ) [cos(1.9975 10 ) 5.0063 10 sin(1.9975 10 )] 1t

Li t e t t A

Case 2. R=25 Ω.

characteristic equation 2 5 104 10 4 10 0s s

characteristic roots5

1,2 2.0 10s

critically damped response 11 2( ) ( ) s t

L Fi t K K t e X

(0) 0, (0) 0L Ci u From the initial conditions,

1 1(0 ) 1 0L Fi K X K 1 1K

1 1 2

(0 )(0 ) (0 )0CL L udi u

s K Kdt L L

2 200000K

Hence for t≥0, 200000( ) (1 200000 ) 1t

Li t t e A

Case 3. R=20 Ω.

characteristic equation 2 5 105 10 4 10 0s s

characteristic roots 51 1.0 10s

overdamped response 1 21 2( ) s t s t

L Fi t K e K e X

(0) 0, (0) 0L Ci u From the initial conditions,

1 2 1 2(0 ) 1 0L Fi K K X K K

1 1 2 2

(0 )(0 ) (0 )0CL L udi u

s K s Kdt L L

1

4

3K

Hence for t≥0, 100000 400000( ) 1.3333 0.3333 1t t

Li t e e A

and 52 4.0 10s

2

1

3K and

In a linear circuit or system, the response to a step input often indicates the quality of the system performance. The problem of measuring a battery voltage using a voltmeter is illustrative of this indicator.In the previous examples observe that the characteristic equations are independent of the source values. This is a general property of linear circuits with constant parameters. Hence when constructing the characteristic equation we may without loss of generality set independent source values to zero, i.e., independent voltage sources become short circuits and independent current sources become open circuits. With this operation, some circuits that appear to be non-series/parallel become series parallel. This allows us to easily compute the characteristic equation and then the previous methods to obtain the solution without having to construct the differential equation explicitly.

Example 4. The circuit of figure b is driven by the input of figure a, i.e., us(t)= － 60u( － t)+60u(t)+60u(t － 1)V, up to 2.5s, the switch opens, changing the circuit characteristics to those of two uncoupled first-order circuits. Our goal is to find the response uC(t) for t≥0.

1

120

60

60,t s

( ),su t V

( )a

2H

1

8F

6

3

+

_( )su t

2.5t s

+

_( )Cu t

( )Li t

( )b

Solution

Step 1. Analysis at 0 － .

1

120

60

60,t s

( ),su t V

The capacitor looks like an open circuit and the inductor like a short circuit.

(0 ) 60Cu V 60

(0 ) 106Li A

Step 2. Analysis at 0 ＋ .

The equivalent circuit at 0 ＋ :

6

3

+

_60V

+

_

(0 ) (0 )

60C Cu u

V

(0 ) (0 )

10L Li i

A

1Ri

2Ri

(0 )Ci

+

_

(0 )Lu

(0 ) 60 ( 60) 120Lu V

2 1(0 ) (0 )

120 60( 10)

3 640

C R L Ri i i i

A

Step 3. Find the characteristic equation and the form of the response.To find the characteristic equation, we set the independent voltage source to zero. The resulting circuit is a parallel RLC with characteristic equation.

2 2 21 14 4 ( 2) 0s s s s s

RC LC

2H

1

8F

6

3

+

_( )su t

2.5t s

+

_( )Cu t

( )Li t

1,2 2s

Which corresponds to a critically damped response of form.

11 2( ) ( ) s t

C Fu t K K t e X

Step 4. Find constants in the response form for 0≤t<1.

1

120

60

60,t s

( ),su t VThe input is constant for 0≤t<1 but changes its value to 120V at t=1s.

2H

1

8F

6

3

+

_( )su t

2.5t s

+

_( )Cu t

( )Li t

( ) 60F CX u V 1(0 ) 60C Fu K X 1 120K

1 1 2

(0 )(0 ) 320C

C

iu s K K

C2 80K

Thus the response of the circuit for 0≤t<1 is 2( ) ( 120 80 ) 60t

Cu t t e V

Similarly, one can compute for 0≤t<1, 2( ) ( 20 20 ) 10t

Li t t e A

1

120

60

60,t s

( ),su t V

2H

1

8F

6

3

+

_( )su t

2.5t s

+

_( )Cu t

( )Li t

(1 ) (1 ) 54.59C Cu u V

(1 ) (1 ) 10L Li i A

6

3

+

_120V

+

_ (1 ) 54.59Cu

(1 ) 10Li A

1Ri

2Ri

(1 )Ci

+

_

(1 )Lu

(1 ) 120 54.59 65.41Lu V

2 1(1 ) (1 )

65.41 54.5910

3 622.71

C R L Ri i i i

A

Step 5. Analysis at t=1 － .

Step 6. Analysis at t=1 ＋ .

The equivalent circuit at 1 ＋ :

Step 7. Computation of the response for 1≤t<2.5.Because the characteristic equation is independent of the input excitation, the form of the response is almost the same as in step3, except for the replacement of t by t － 1.

1 ( 1)1 2( ) [ ( 1)] s t

C Fu t K K t e X

Similarly, ( ) 120F CX u V 1(1 ) 54.59C Fu K X 1 65.41K

1 1 2

(1 )(1 ) 181.7C

C

iu s K K

C2 50.9K

2( 1)( ) [ 65.41 50.9( 1)] 120tCu t t e V

Thus, for 1≤t<2.5,

Step 8. Computation of the response for t>2.5s.

2H

1

8F

6

3

+

_( )su t

2.5t s

+

_( )Cu t

( )Li t

At t=2.5s the switch opens.

Because the switch opens, the parallel RC part of the circuit becomes independent of the parallel RL part of the circuit.

The capacitor then discharges through the 6Ω resistor with a time constant τ=RC=6/8=0.75s.

(2.5) 120.545Cu V

( 2.5) ( 2.5)

0.75( ) (2.5) 120.545t t

C Cu t u e e V

Thus, the capacitor voltage for t≥2.5 satisfies

5. Formulation of a single second-order differential equation (§8.5)

The examples of second-order circuits given in prior sections are relatively simple; general second-order circuits have a more complex structure, which necessitates a systematic method for constructing a differential equation model.

Such general circuits will have arbitrary interconnections of resistors, energy storage elements, (L,L), (C,C), or (L, C), and, possibly, controlled sources.

本节介绍了一种针对一般二阶电路构建简单特征方程模型的方法 .

构建过程运用了一个新的概念 : 状态方程 (state equations).

For a second-order system, the state equations have the form of a pair of coupled first order differential equations:

111 1 12 2 1( )

dxa x a x u t

dt

221 1 22 2 2 ( )

dxa x a x u t

dt

x1(t) 和 x2(t) 为电容电压 , 电感电流 , 或者一个电容电压和电感电流的组合 .

The uk(t) are state model inputs, and each uk(t) is a scaled sum of the circuit source inputs.列写状态方程的步骤 :Step 1. Write the state equations, i.e., write two first-order differential equations using the capacitor voltages and/or the inductor currents as the unknowns. Manipulate the equations to have the general form.Step 2. Compute the desired second-order differential equation in either x1(t) or x2(t) by eliminating the unwanted variable from the state equations.

• Writing the state equations

Writing state equations has three basic steps.

1.The first step sets forth the defining differential u-i relationships for C and/or L.

2.The second step is the elimination of the unwanted variables.

3.In the third step one then substitutes the results of the second step into the LC defining equations written in step 1.

在步骤 2中 ,要找出 iC 和 uL 的表达式 , 变量 uC 和 iL 作为已知量 . 因此我们可以用一个独立电压源置换电容 ,由 uC 表示 ,用一个独立电流源置换电感 ,由 iL 表示 .

使用节点分析法和网孔分析法 ,或者其他方法 , 都可以解出联立的电阻电路方程 ,并且用 uC, iL 和输入 来置换 iC 和 uL.

Example 5. Find the state equations for the circuit of the figure.

SolutionStep 1. 根据题意写出 L 和 C 的相关方程 .

CC

dui C

dt L

L

diu L

dtStep 2. 用 uC, iL, 和 uin 来表示 iC 和 uL.

用一个电压源置换电容 C, 用一个电流源置换电感 L, 得到一个电阻电路如图 b.

4

( )a

+

_inu

+

_

2

8

0.2H4

0.1F Cu

Li

4

( )b

+

_inu

+

_

2

8

4

Cu

Li

Ci

+ _Lu

4

( )b

+

_inu

+

_

2

8

4

Cu

Li

Ci

+ _Lu

Li

1i Ci

1(2 8) 8L Li i u

18 (8 4) 4L C ini i i u

14 (4 4) C Ci i u

Using mesh analysis:

3 8 14 2L C L inu i i u

16 2 10L C C Lu i u i

14 23 8

2 101 16L L in

C C L

u i u

i u i

14 216 812 1040 1 3

L L in

C C L

u i u

i u i

0.4 3.6 0.8L C L inu u i u

0.15 0.4 0.05C C L ini u i u

Step 3. Construct the state equations.

10 1.5 4 0.5CC C L in

dui u i u

dt

5 2 18 4LL C L in

diu u i u

dt

• Reducing the state equations to a single second-order differential equation

111 1 12 2 1( )

dxa x a x u t

dt

221 1 22 2 2 ( )

dxa x a x u t

dt

我们的目的是消除其中一个状态变量 ,例如 x2, 得到一个仅含 x1 的简单二阶状态方程 .

11 1 12 2 1( ) ( )d

a x a x u tdt

21 1 22 2 2( ) ( )d

a x a x u tdt

22 11 1 22 12 2 22 1( )( ) ( ) ( ) ( )d d d d

a a x a a x a u tdt dt dt dt

12 21 1 12 22 2 12 2( ) ( )d

a a x a a x a u tdt

2

1 1 111 22 11 22 12 21 1 22 1 12 22

( ) ( )d x dx du

a a a a a a x a u a udt dt dt

2

2 2 211 22 11 22 12 21 2 11 2 21 12

( ) ( )d x dx du

a a a a a a x a u a udt dt dt

A high-accuracy pressure sensor application

Window Heat ControlsElectric Power Management

Communication and Navigation SystemsEngine vibration Monitor

Cabin Pressure/Temperature Controls

Sinusoidal Steady-State Analysis by Phasor Methods1. Notion of sinusoidal steady-state (§7.10)

• Transient( 瞬态 ) means the circuit behavior is changing from one type to another; it represents a circuit’s response from initial turn-on to an equilibrium behavior called steady state.

USC

+

_u(t)

+

_

R

K

( ) /u t V

/t s0 1t 2t 3t

US

Steady state Steady state Steady state

Transient state

Transient state

Switch

Switch

0t

• Steady state( 稳态 ) means that all transient behavior of the stable circuit has died out, i.e., decayed to zero.

US

－ US

Output response: cos( )mU t

Um

－ Um

• Sinusoidal means that source excitations have the form:

cos( )S Su U t or

sin( )S Su U t

上图所示为一个正弦稳态线性电路 .(US 和 Um 可以换为 IS 和 I

m)

Period( 周期 )

2T

Frequency 1

2f

T

Angular frequency

2 f

( 频率 )

( 角频率 )

Deducible from trigonometric identities,

cos( ) cos( )cos( ) sin( )sin( )m m m mu U t U t U t cos( ) sin( )A t B t

where cosmA U and sinmB U

Conversely, by summing the squares of A and B one obtains

2 2mU A B

By taking the inverse tangent of the ratio of B and A, one obtains

1tan ( )B

A

向量法是建立在数学中复数和基本电路分析法的基础上的一种分析方法 .

2. Brief review of complex numbers (§9.2)

An arbitrary complex number z a jb where 1j

Where

Re[ ]

2

z za z Imaginary part( 虚部 )

Im[ ]

2

z zb z

j z a jb is the complex conjugate of z

The magnitude or modulus of z( 模 ):

Real part( 实部 )

2 2 2z a b

Im[ ]z

Re[ ]z

Imaginary axis( 虚轴 )

Real axis( 实轴 )

jb

a

sinb

cosa

2 2a b

Diagram showing relationship between polar and rectangular coordinates of complex number.

2 2a b z

1tan ( )b

a

Hence z = a + jb is completely specified by its magnitude ρ and angle θ, i.e.,

z a jb cos sinj [cos sin ]j je

ρ ∠ θ 是 ρe j θ 的简单记法 .

and cos sinje is the famous Euler identity ( 欧拉公式 )指数的性质 : 1 2 1 2( )j j je e e

11 1 1 1

jz a jb eSuppose and 2

2 2 2 2jz c jd e

then 1 2 ( )( ) ( ) ( )a jb c jd ab bd j adz bcz 1 2 1 2(

1 2)

1 2 12 21 ( )j j je e e

1

22 2

( )( ) ( ) ( )

( )( )

a jb a jb c jd ac bd j bc ad

c jd c jd c jd c

z

bz

1

1 2

2

( 11 2

2

)1 1

22

( )j

jj

ee

e

2 21 1

2 2[tan ( ) tan ( )]

a b b d

a cc d

Example 1. Suppose 1 3 4 5 53.13oz j and 2 8 6 10 36.87oz j

(a) Compute z1z2 in rectangular coordinates.

(b) Compute z1z2 in polar coordinates.

1 2 (3 4)(8 6) (24 24) (18 32) 48 14z z j j j jSolution

Equivalently, 1 2 5 53.13 10 36.87 50 16.26o o oz z

16.2650 50cos16.26 50sin16.26 48 14oj o oe j j

Example 2. Suppose ( )( ) Re[ ]j tu t Ue

and 455 6 3 20oj j j jUe j Ue Ue e find U, φand u(t).

SolutionFactoring Uejφ out to the left and dividing by ( － 8+j6) yields

4598.1320

28 6

o

oj

j jeUe e

j

Hence, 2, 98.13 ,oU and ( )( ) Re[ ] 2cos( 98.13 )j t ou t Ue t

Summary of properties of complex numbers

cos( ) cos( ) sin( )mU t A t B t 2 2 1, tan ( )m

BU A B

A

Euler identity cos sinje B

Real part of sum 1 2 1 2Re[ ] Re[ ] Re[ ]z z z z

Proportionality property 1 1Re[ ] Re[ ]z z For all real scalars α

Linearity property 1 1 2 2 1 1 2 2Re[ ] Re[ ] Re[ ]z z z z

Differentiation property Re[ ] Re ( ) Re[ ]j t j t j td dAe e j Ae

dt dt

Equality propertyFor all t if and only if A=B

Single-frequency property:

Sum of complex exponentials , or their derivatives, or their indefinite integrals of any order is a complex exponential of the same frequency ω

Re[ ] Re[ ]j t j tAe Be

3. Naïve technique for computing the sinusoidal steady state (§7.10)

A first-order circuit differential equation model with a sinusoidal excitation has the form

( )( ) cos( )m

dx tax t U t

dt

Or in the second-order case,2

2

( ) ( )( ) cos( )m

d x t dx tb cx t U t

dt dt

x(t) 是待求的电压或电流 , 例如 uC(t) 或 iL(t).

An important assumption: if the zero-input response consists of decaying exponentials or exponentially decaying sinusoids, there is a well-defined sinusoidal steady-state (SSS) response.

Example 3. Let the source excitation to the circuit of the figure be ( ) cos( )S Si t I t .Compute the SSS response iL(t).

R( )Si t

( )Ri t ( )Li t

L

SolutionStep 1. 写出电路的特征方程式 :

R L Si i i 由 KCL

LL S

diLi i

R dt L

L S

di R Ri i

dt L L

Step 2. 写出响应公式 ,因为输入是一个正弦波 ,因此 SSS响应 也应该是正弦形式 :

( ) cos( ) sin( )Li t A t B t

Step 3. 把以上式子代入特征方程式 :

[ cos( ) sin( )] [ cos( ) sin( )]d R

A t B t A t B tdt L

sin( ) cos( ) cos( ) sin( ) cos( )S

RA RB RA t B t t t I t

L L L

Step 4. 合并同类项并且解出 A 和 B.

cos( ) sin( ) 0S

RA R RBB I t A t

L L L （ ） （ ）

0S

RA RB I

L L

0RB

AL

2

2 2 2SR I

AR L

2 2 2SRLI

BR L

Step 5. 写出正弦稳态响应 . 2

2 2 2 2 2 2( ) cos( ) sin( )S S

L

R I RLIi t t t

R L R L

化简成更简单的形式 : ( ) cos( )L mi t I t

2 2 2( ) cos( )S

L

RIi t t

R L

where 1tan ( )

L

R

4. Complex exponential forcing functions in sinusoidal steady-state computationComplex exponential forcing functions are simply complex exponential input excitations of the form ( ) ( )j t j t

S SU e or I e

From the properties of complex numbers, we can replace the input excitation and the assumed circuit response cos( ) SU t

by their complex counterpartscos( ) mU t ( ) ( ) j t j tS mU e and U e

respectively, without penalty.

cos( ) SU t cos( ) mU t

( ) j tSU e ( ) j t

mU e( )Re[ ] cos( ) j t

S SU e U t ( )Re[ ] cos( ) j tm mU e U t

Substitution

这种代换和取实部的过程实际上简化了计算 ,这是由指数函数的性质和特征决定的 .

Example 3. For the series RC circuit of the figure, let ( ) cos( )S Su t U t .Compute the steady-state response uC(t).

+

_

+

_

R

( )Su t ( )Cu tC

( )Ci t

SolutionStep 1. 写出电路的特征方程 .

CC S

duRC u u

dt

Step 2. 把输入和响应的复数形式代换到特征方程中 .

Set ( ) j tS Su t U e and agree that ( )( ) j t

C mu t U e

( ) ( ) j t j t j tm m Sj RCU e U e U e

1

j S

m

UU e

j RCStep 3. 计算 Um 和 φ.

2 2 21

S

m

UU

R Cand 1tan ( ) RC

Step 4. 写出稳态响应 :1[ tan ( )]

2 2 2( ) Re

1

j t RCS

C

Uu t e

R C

1

2 2 2cos[ tan ( )]

1

SU

t RCR C

As we can see, the use of complex exponential does indeed lead to a more direct calculation of the SSS response.

However, this method and the method of section 3 require a differential equation model of the circuit.

For multiple sources, dependent sources, and large interconnections of circuit elements, finding the differential equation model is often a nontrivial task.

In the next section we eliminate the need to find a differential equation model of the circuit by introducing the phasor concept.

5. Phasor representations of sinusoidal signals (§9.3,§9.5)

If the frequency ω is known, then the complex number A∠φ completely determines the complex exponential Aej (ωt+φ). In turn, if ω is known, then A∠φ completely specifies Acos(ωt+φ)=Re[Aej(ωt+φ )]

This means that the complex number A can represent a sinusoidal function Acos(ωt+φ) whenever ω is known. Complex number representations that denote sinusoidal signals at a fixed frequency are called phasors.

A typical voltage phasor is &mU U

A typical voltage phasor is &mI I

( ) 25cos( 45 ) oi t tFor example 25 45 & oI

( ) 15sin( 30 ) ou t t ( ) 15cos( 30 90 180 ) o o ou t t

( ) 15cos( 120 ) ou t t 15 120 & oU

As all voltages and current satisfy KVL and KCL, respectively, phasor voltages and currents do likewise. For example, consider the circuit node drawn in the figure.

3( ) 8sin( )i t t A

1( ) 10cos( )i t t A

2( ) 5.043cos( 7.52 ) oi t t A

4( )i t

4 1 2 3( ) ( ) ( ) ( ) i t i t i t i t

10cos( ) 5.043cos( 7.52 ) 8cos( 90 ) o ot t t

10cos( 60 ) ot4 10 60 & oI

1 2 3 10 0 5.043 7.52 8 90 & & & o o oI I I10 (5 0.66) ( 8) 5 8.66 j j j 410 60 &o I

4 1 2 3 & & & &I I I I satisfy KCL

Example 4. Determine the voltage across the resistor in the circuit of the figure using the phasor concept.

+

_

1( )u t20cos( 53.13 ) ot V

2( )u t19.68sin( 152.8 ) ot V

2( )u t4.215cos( 71.61 ) ot V

3( )u t

+

_( )Ru t

+_

+_

Solution

From KCL, 1 2 3( ) ( ) ( ) ( ) Ru t u t u t u t

Since voltage phasors must satisfy KVL,

1 2 3 & & & &RU U U U 20 53.13 19.68 62.8 4.215 71.6 o o o

12 16 (9 17.5) 1.33 4 j j j 4.33 2.5 j

Equivalently, 5 30 & oRU and ( ) 5cos( 30 ) o

Ru t t

6. Elementary impedance concepts: phasor relationships for RS, LS, and CS (§9.3,§9.5)

( ) ( )( )( ) j t j t

RR R t Ru t Ri RI e U e

,j jR RR RI I e andU U e

From Ohm’s law,

In terms of the phasors

This relationship reduces to

( )RR R RU R I Z j I