Linear Circuit Analysis

In an oscilloscope a timing signal called a horizontal sweep acts as a time base, which allows one to view measured input signals as a function of time.

In practice, the linear increase in voltage is approximated by the “linear” part of an exponential response of an RC circuit.

First-Order RL and RC Circuits1. What is a first-order circuit?A first-order circuit is characterized by a first-order differential equation. It consists of resistors and the equivalent of one energy storage element.

Typical examples of first-order circuits:

(a) First-Order RL circuit (b) First-Order RC circuit

USL

R

+

_

C

R

US

+

_

Interconnections of sources, resistors, capacitors, and inductors lead to new and fascinating circuit behavior.

C

R

US

+

_

i+

_Cu

First-Order RC circuit

A loop equation leads to S CU Ri u

since Cdui C

dt

CS C

duU RC u

dt

or , equivalently,1 1C

C S

duu U

dt RC RC

This equation is called Constant-coefficient first-order linear differential equation

Apply duality principle,1 1L

C S

dii I

dt GL GL

2. Some mathematical preliminaries

The first-order RL and RC circuits have differential equation models of the form

1 1CC S

duu U

dt RC RC

1 1LC S

dii I

dt GL GL

RC circuit first-order differential equation

RL circuit first-order differential equation

0 0

( )( ) ( ), ( )

dx tx t f t x t x

dt

or , equivalently,0 0

( )( ) ( ), ( )

dx tx t f t x t x

dt

valid for t≥t0, where x(t0)=x0 is the initial condition. The term f(t) denotes a forcing function. Usually, f(t) is a linear function of the input excitations to the circuit.The parameter λdenotes a natural frequency of the circuit.

0 0

( )( ) ( ), ( )

dx tx t f t x t x

dt

0 0

( )( ) ( ), ( )

dx tx t f t x t x

dt

The main purpose of this chapter is to find a solution to the differential equation.

or

The solution to the equation for t≥t0 has the form

0

0

( ) ( )0( ) ( ) ( )

tt t t

tx t e x t e f d

(1) Satisfies the differential equation

(2) Satisfies the correct initial condition, x(t0)=x0

Integrating factor method

0 0

( )( ) ( ), ( )

dx tx t f t x t x

dt

First step, multiply both sides of the equation by a so-called integrating factor e － λt. ( )

( ) ( )t t tdx te e x t e f t

dt By the product rule f

or differentiation,

[ ( )]tde x t

dt ( )

( )t tdx te e x t

dt ( )te f t

Integrate both sides of the equation from t0 to t as follows

0

[ ( )]t

t

de x d

d

0

( )t

te x 0

0( ) ( )tte x t e x t 0

( )t

te f d

0

00( ) ( ) ( )

ttt

te x t e x t e f d

0

0

( ) ( )0( ) ( ) ( )

tt t t

tx t e x t e f d

3. Source-free or zero-input response

LR

( )a

R C

( )b

A parallel connection of a resistor with an inductor or capacitor without a source.In these circuits, one assumes the presence of an initial inductor current or initial capacitor voltage.

(a) KCL implies (b) KVL implies

R Li i

Ri Li+

_Lu

Ci+

_Cu

+

_Ru

L LR

u diLi

R R dt

LL

di Ri

dt L

R Cu u

CR C

duu i R RC

dt

1CC

duu

dt RC

Both differential equation models have the same general form,

( ) 1( ) ( )

dx tx t x t

dt

( ) 1( ) ( )

dx tx t x t

dt

The solution to the equation for t≥t0 has the form

0( )0( ) ( )t tx t e x t

0

0( )t t

e x t

Where τ is a special constant called the time constant of the circuit. The response for t≥t0 of the undriven RL and RC circuit are, respectively, given by

0( )

0( ) ( )R

t tL

L Li t e i t

01

( )

0( ) ( )t t

RCC Cu t e u t

RC circuit

RL circuit

RC

LGL

R

The time constant of the circuit is the time it takes for the source-free circuit response to drop to e － 1=0.368 of its initial value.

For more general circuits, those containing multiple resistors and dependent sources, it is necessary to use the Thevenin equivalent resistance seen by the inductor or capacitor in place of the R.

LinearResistivecircuit

Noindependent

sources

Li

L

LinearResistivecircuit

Noindependent

sources

C+

_Cu

Thevenin equivalent Li

LthR

+

_Cu C

thR

Thevenin equivalent

0( )

0( ) ( )th t t

LL L

R

i t e i t

01

( )

0( ) ( )th

t tC

C CRu t e u t

Example 1. For the circuit of the figure, find iL(t) and uL(t) for t ≥0 given that iL(0 － )=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5Ω resistor over the time interval [0.4, ∞).

8H+

_Lu20

Li

5

0.4t sS

USC

+

_u(t)

+

_

R

K

( ) /u t V

/t s0

US

过渡期为零

USR

+

_u(t)

K

+

_

（ 1 ）问题的提出

( ) /u t V

/t s0 1t 2t 3t

US

第一个稳定状态 第二个稳定状态 第三个稳定状态

过渡状态 过渡状态

瞬态（ transient state ）

稳态（ steady state ）

换路

换路

代数方程描述微分方程描述

0t

一阶电路的初始条件

（ 2 ）电路的初始条件① t ＝ 0 － 和 t ＝ 0 ＋ 的概念

认为换路在 t ＝ 0 时刻进行

换路前一瞬间

换路后一瞬间t ＝ 0＋

t ＝ 0－

00

(0 ) lim ( )tt

f f t

00

(0 ) lim ( )tt

f f t

初始条件为 t ＝ 0 ＋ 时 u 、 i 及其各阶导数的值② 电容的初始条件

C+

_Cu

i1

( ) ( )t

Cu t i dC

0

0

1 1( ) ( )

ti d i d

C C

0

1(0 ) ( )

t

Cu i dC

0

0

1(0 ) (0 ) ( )C Cu u i d

C

t ＝ 0 ＋ 时

当 ( )i 为有限值 (0 ) (0 )C Cu u

Cq Cu(0 ) (0 )q q 电荷守恒

换路瞬间，若电容电流保持为有限值，则电容电压（电荷）换路前后保持不变。

③ 电感的初始条件

L+

_u

Li1

( ) ( )t

Li t u dL

0

0

1 1( ) ( )

tu d u d

L L

0

1(0 ) ( )

t

Li u dL

0

0

1(0 ) (0 ) ( )L Li i u d

L

t ＝ 0 ＋ 时

当 ( )u 为有限值 (0 ) (0 )L Li i

LLi (0 ) (0 ) 磁链守恒

换路瞬间，若电感电压保持为有限值，则电感电流（磁链）换路前后保持不变。④ 换路定律

(0 ) (0 )C Cu u

(0 ) (0 )q q

(0 ) (0 )L Li i

(0 ) (0 )

换路瞬间，若电容电流保持为有限值，

则电容电压（电荷）换路前后保持不变。

换路瞬间，若电感电压保持为有限值，

则电感电流（磁链）换路前后保持不变。

反映了能量不能跃变

例 1 电路如图所示，试求 。

(0 )Ci

解：+

_10V

10k

C40k

K

+

_Cu

Ci

+

_10V

10k

40k+

_Cu

+

_10V

10k

+

_

Ci

8V

② 由换路定律

① 画出 电路，求出 0 (0 )Cu

电路0

电路0

③ 画出 电路，求出 0 (0 )Ci

40(0 ) 10 8

10 40Cu V

(0 ) (0 ) 8C Cu u V 3

10 8(0 ) 0.2

10 10Ci mA

(0 ) 0Ci A

(0 ) (0 )C Ci i 可见，

电容开路

电容用电压源替代

Ci

+

_10V

1 4

K L+

_Lu

Li

+

_10V

1 4

+

_Lu

Li

+

_10V

1 4

+

_Lu2A

例 2 电路如图所示，试求 。

(0 )Lu

解：

② 由换路定律

① 画出 电路，求出 0 (0 )Li

电路0

电路0

③ 画出 电路，求出 0 (0 )Lu

10(0 ) 2

1 4Li A

(0 ) (0 ) 2L Li i A (0 ) 4 2 8Lu V

(0 ) 0Lu V

(0 ) (0 )L Lu u 可见，

电感短路 电感用

电流源替代

求解初始条件的步骤

① 画 0 － 等效电路，即换路前电路（稳定状态），求 uC(0 － ) 和 iL(0 － ) 。

② 由换路定律得 uC(0+) 和 iL(0+) 。

③ 画 0+ 等效电路，即换路后的电路。

④ 由 0+ 电路求所需各变量的 0+ 值。

电容用电压源来替代，大小为 uC(0 ＋ )

电感用电流源来替代，大小为 iL(0 ＋ ) 。

电压源和电流源的方向均与原来的电压、电流方向一致。

其中

电容相当于开路

电感相当于短路其中

Si

L

R C0t

K

+_

LuLi

Ci+

_Cu

Si R

Li

+

_Cu

+_

LuR

Si

+

_ SRi

例 3 电路如图所示，试求 ， 。

(0 ) (0 )C Li u

解：

② 由换路定律

① 画出 电路，求出 ，

0 (0 ) (0 )C Lu i

电路0

电路0

③ 画出 电路，求出 ，

0 (0 ) (0 )C Li u

(0 )L Si i A

(0 ) (0 )C C Su u Ri V (0 ) 0S

C S

Rii i A

R

(0 )C Su Ri V Ci

(0 ) (0 )L L Si i i A (0 )L Su Ri V

+

_48V

2

L

2

3

C

K

+

_Lu

Li

+

_Cu

Ci

i

例 4 电路如图所示，试求开关 K 闭合瞬间各支路的电流和电感上的电压。

解： 电路0

电路0

48(0 ) (0 ) 12

2 2L Li i A

2(0 ) (0 ) 48 24

2 2C Cu u V

(0 ) 48 2 12 24Lu V

48 24(0 ) 8

3Ci A

(0 ) (0 ) (0 ) 20L Ci i i A

+

_48V

2

23Li

+

_Cu

+

_Lu+

_48V

2

312A

+

_24V

Li

i

Ci

100

L C

100+

_200V

100K

+_

Cu

Li

100 100+

_200V

100

+_100V1A

+_

Lu Ci

Ki

例 5 电路如图所示，试求开关 K 闭合瞬间流过它的电流值。 解： 电路0

电路0200(0 ) (0 ) 1

100 100L Li i A

(0 ) (0 ) 100 100C C Lu u i V

100100+

_200V

100

+_

CuLi

200 100(0 ) 1 2

100 100Ki A

单位阶跃函数0 0

( )1 0

tu t

t

( )u t

t0

1

延时（ delayed ）单位阶跃函数

00

0

0( )

1

t tu t t

t t

( )u t

t0

1

0t分段常量信号（ piecewise-constant signal ）

( )f t

t0 0t

( )f t

t0

1

0t 02t 03t（矩形）脉冲（ pulse ） 脉冲串（ pulse train ）

1

单位阶跃函数及分段常量信号

运用阶跃函数和延时阶跃函数，分段常量信号可以表示为一系列阶跃信号之和。

( )f t

t0 0t

1( )f t

t0 0t 02t 03t

1

0( ) ( ) ( )f t u t u t t

( )f t

t0 0t 02t

A

A

0 0 0( ) ( ) ( ) ( 2 ) ( 3 )f t u t u t t u t t u t t

0 0( ) ( ) 2 ( ) ( 2 )f t Au t Au t t Au t t

Solution

Example 8.2. For the circuit of the figure, find iL(t) and uL(t) for t ≥0 given that iL(0 － )=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5Ω resistor over the time interval [0.4, ∞).

8H+

_Lu20

Li

5

0.4t sS

Step 1. With switch S open, compute the response for 0≤t ≤0.4s.

From the continuity property of the inductor current, (0 ) (0 ) 10L Li i A

( ) (0 )thR

tL

L Li t e i 2.510 te A

Step 2. With switch S closed, compute the response for t ≥0.4s.

20 // 5 4thR

0( )

0( ) ( )thR

t tL

L Li t e i t

0.5( 0.4)(0.4) t

Li e 0.5( 0.4)3.679 te A

Step 3. Write the complete response as a single expression.2.5 0.5( 0.4)( ) 10 [ ( ) ( 0.4)] 3.679 ( 0.4)t t

Li t e u t u t e u t A

Step 4. Plot the complete response.

( ),Li t A

,t s

0.4s

2s

The 0.4s time constant has a much faster rate of decay than the lengthy 2s time constant.

Step 5. Compute uL(t).

8H+

_Lu20

Li

5

0.4t sS

( ) (0 )thR

tL

L Lu t e u for 0≤t ≤0.4s,

in particular, (0 ) 20 (0 ) 200L Lu i A

hence,2.5( ) 200 t

Lu t e V

for t ≥0.4s,( 0.4)

( ) (0.4)thR

tL

L Lu t e u

in particular, (0.4) 4 (0.4 ) 14.716L Lu i V

0.5( 0.4)14.716 te V

Step 6. Compute energy dissipated in the 5Ω resistor over the time interval [0.4, ∞).

2 0.5( 0.4) 2( 0.4)

5

( ) [ 14.716 ]( ) 43.3

5 5

ttLu t e

p t e W

( 0.4)5 0.4

(0.4, ) 43.3 43.3tW e dt J

Example 8.3. Find uC(t) for t ≥0 for the circuit of the figure given that uC(0)=9V. Solution

72 4.59 0.1F

+

_Cu

S

0t 1t Step 1. Compute the response for 0≤t ≤1s. By the continuity of the capacitor voltage,

(0 ) (0 ) 9C Cu u V

hence,1

( ) (0 )th

tR C

C Cu t e u

1.259 te V

Step 2. Compute the response for t ≥ 1s. 0

1( )

0( ) ( )th

t tR C

C Cu t e u t

1

0.32.58t

e V

Step 3. Use step functions to specify the complete response.1

1.25 0.3( ) 9 [ ( ) ( 1)] 2.58 ( 1)t

tCu t e u t u t e u t V

Step 4. Obtain a plot of the response.

Here the part of the response with the 0.3s time constant shows a greater rate of decay than the longer 0.8s time constant.

( ),Cu t V

,t s

0.8s

0.3s

Exercise. Suppose that in example 2 the switch moves to the 4.5Ω resistor at t=0.5s instead of 1s. Compute the value uC(t) at t=1.2s.

72 4.59 0.1F

+

_Cu

S

0t 0.5t

Example 8.4. Find uC(t) for the circuit of the figure, assuming that gm=0.75S and uC(0 － )=10V.

Solution

It is straightforward to show that the Thevenin equivalent seen by the capacitor is a negative resistance,

m xg u 4+

_xu 0.25F

+

_Cu

a

bThevenin equivalent

2 0.25F+

_Cu

a

b

2thR

hence,1

( ) (0 )th

tR C

C Cu t e u

210 ( )te u t V

( ),Cu t V

,t s

Negative resistance causes capacitor voltage to increase without bound.

Because of the negative resistance, this response grows exponentially, as shown in the figure.

A circuit having a response that increases without bound is said to be unstable.

Exercise. In example 3, let gm=0.125S. Find the equivalent resistance seen by the capacitor and uC(t), t ≥0.

m xg u 4+

_xu 0.25F

+

_Cu

a

b

5F

2

3 6

K1i

3i2i+

_Cu

5F 4

1i

+

_Cu

示例 已知图示电路中的电容原本充有 24V 电压，求 K 闭合后，电容电压和各支路电流随时间变化的规律。

解：

0 0t

RCCu U e t

本题为求解一阶 RC 电路零输入响应问题

等效电路 t > 0

则有

0 24U V又由已知条件

4 5 20RC s

2024 0t

Cu e V t

201 6 0

4

tCu

i e A t

利用并联分流，得

202 1

24 0

3

t

i i e A t

20

3 1

12 0

3

t

i i e A t

+

_24V

4 4

2

3

6H

6

( 0)K t

Li

12

+

_Lu

0t

6H

R

+

_Lu

Li

示例 t = 0 时 ，开关 K 由 1→2 ，求电感电压和电流。

解： RL 电路零输入响应问题

0 0R

tL

Li I e A t

0 (0 ) (0 )L LI i i

24 62

2 3 // 6 4 3 6A

(2 4) // 6 3 6R

61

6

Ls

R

2 0tLi e A t

12 0tLL

diu L e V t

dt

① 一阶电路的零输入响应是由储能元件的初值引起的响应 , 都是由初始值衰减为零的指数衰减函数。

② 衰减快慢取决于时间常数 。

③ 同一电路中所有响应具有相同的时间常数。

④ 一阶电路的零输入响应和初始值成正比，称为零输入线性。

小结

( ) (0 ) 0t

x t x e t

RC 电路 (0 ) (0 )C Cu u

RL 电路 (0 ) (0 )L Li i

RC 电路 RC

RL 电路L

GLR

R 为与动态元件相连的一端口电路的等效电阻。

4. DC or step response of first-order circuitsThis section takes up the calculation of voltage and current responses when constant voltage or constant current sources are present.

LinearResistivecircuitWith

independentsources

Li

L

LinearResistivecircuit With

independentsources

C+

_Cu

Thevenin equivalent

Thevenin equivalent

C+

_

thR

ocU

+

_Cu

Ci

thR

L+

_ocU

Li+

_Lu

C+

_

thR

ocU

+

_Cu

CithR

L+

_ocU

Li+

_Lu

Deriving the differential equation models characterizing each circuit’s voltage and current responses.

By KVL and Ohm’ law,

L oc th Lu U R i

LdiL

dt

1thLL oc

Rdii U

dt L L

By KCL and Ohm’ law,

oc CC

th

U ui

R

Cdu

Cdt

1 1CC oc

th th

duu U

dt R C R C

Exercise. Constant differential equation models for the parallel RL and RC circuits of the figure. Note that these circuits are Norton equivalents of those in the figure. Again choose iL(t) as the response for the RL circuit and uC(t) as the response for the RC circuit.

Li

LthR+

_Lu

scI thRscI

+

_Cu C

Ci

Answers:

th thLL sc

R Rdii I

dt L L

1 1CC sc

th

duu I

dt R C C

1thLL oc

Rdii U

dt L L

1 1CC oc

th th

duu U

dt R C R C

th thLL sc

R Rdii I

dt L L

1 1CC sc

th

duu I

dt R C C

Observe that four differential equations have the same structure: ( ) 1

( )dx t

x t Fdt

where,th

th

LG L

R

,thR C

oc sc thU I RF

L L

sc oc

th

I UF

C R C

And the general formula for solving such a differential equation:

0

0

( ) ( )0( ) ( ) ( )

tt t t

tx t e x t e f d

for RC case

for RL case

0

0

( ) ( )0( ) ( ) ( )

tt t t

tx t e x t e f d

where 1

,

(0 ) (0 )x x as long as x(t) is a capacitor

voltage or inductor current, and f(τ)=F is a constant (nonimpulsive) forcing function.

0

00( ) ( )

t t t qt

tx t e x t e Fdq

0 0

0( ) (1 )t t t t

e x t F e

0

0[ ( ) ]t t

F x t F e

Which is valid for t≥t0. After some interpretation, this formula will serve as a basis for computing the response to RL and RC circuits driven by constant sources.

0

0( ) [ ( ) ]t t

x t F x t F e

if 0, then0

0( ) lim ( ) lim [ ( ) ] t t

t tx x t F x t F e

Foc

scth

UI

R

ocU

for RC case

for RL case

Hence, the solution of the differential equation given constant or dc excitation becomes

0

0( ) ( ) [ ( ) ( )]t t

x t x x t x e

0( )

0( ) ( ) [ ( ) ( )]thR

t tL

L L L Li t i i t i e

0

0( ) ( ) [ ( ) ( )] th

t t

R CC C C Cu t u u t u e

for RC case

for RL case

Example 8.5. For the circuit of the figure, suppose a 10V unit step excitation is applied at t=1 when it is found that the inductor current is iL(1 － )=1A. The 10V excitation is represented mathematically as uin(t)=10u(t － 1)V for t≥1. Find iL(t) and uL

(t) for t≥1.

5R

+

_

Li +

_Lu( )inu t 2L H

SolutionStep 1. Determine the circuit’s differential equation model.

1thLL oc

Rdii U

dt L L

2.5 5 ( 1)Li u t 1t where the time constant 0.4s

Step 2. Determine the form of the response.1

( ) ( ) [ (1 ) ( )] ( 1)t

L L L Li t i i i e u t A

5R

+

_

Li +

_Lu( )inu t 2L H

Step 3. Compute iL(∞) and set forth the final expression for iL(t).

replace the inductor by a short circuit,

( ) 2inL

ui A

R scI

It follows that, 2.5( 1)( ) [2 (1 2) ] ( 1)tLi t e u t A

and since (1 ) (1 ) 1L Li i A

Step 4. Plot iL(t).2.5( 1)[2 ] ( 1)te u t A

Step 5. Compute uL(t). 5R

+

_

Li +

_Lu( )inu t 2L H( ) L

L

diu t L

dt

2.5( 1)5 ( 1)te u t V

1

[ (1 ) ( )] ( 1)t

L L

Li i e u t

Exercise. Verify that in example 4, uL(t) can be obtained without differentiation by ( ) ( )L oc th Lu t U R i t

Exercise. In example 4, suppose R is changed to 4Ω. Find iL

(t) at t=2s.

Specifically, we need only compute , , and the time constant or .

0( )x t ( )x

th

L

R thR C

Example 8.6. The source in the circuit of the figure furnishes a 12V excitation for t<0 and a 24V excitation for t≥0, denoted by uin(t)=12u( － t)+24u(t)V. The switch in the circuit closes at t=10s. First determine the value of the capacitor voltage at t=0 － , which by continuity equals uC(0+). Next determine uC

(t) for all t≥0. 1 6R k

+

_( )inu t

2 3R k

10t sC

0.5mF

+

_Cu

Solution

Step 1. Compute initial capacitor voltage

(0 ) (0 ) 12C Cu u V

Step 2. Obtain uC(t) for 0≤t ≤ 10s.

1 3 ,R C s ( ) 24Cu V

( ) ( ) [ (0 ) ( )] th

t

R CC C C Cu t u u u e

324 12

t

e V

Step 3. Compute the initial condition for the interval t>10.

1 6R k

+

_( )inu t

2 3R k

10t sC

0.5mF

+

_Cu

10

3(10 ) (10 ) 24 12C Cu u e

Step 4. Find uC(t) for t>10.

23.57V

Thevenin equivalent

2thR k

+

_8ocU V C

0.5mF

+

_Cu

1 ,thR C s ( ) 8Cu V 10

( ) ( ) [ (10 ) ( )] th

t

R CC C C Cu t u u u e

( 10)8 15.57 te V

Step 5. Set forth the complete response using step functions.

10( 10)3( ) (24 12 )[ ( ) ( 10)] [8 15.57 ] ( 10)t

Cu t e u t u t e u t

Step 6. Plot uC(t).

3s

1s

Exercise. Suppose the switch in example 5 opens again at t=20s. Find uC(t) at t=25s.

+

_100V

500

10 F+

_Cu

K i

示例 t = 0 时 , 开关 K 闭合，已知 uC(0 － ) = 0V ，求（ 1 ）电容电压和电流；（ 2 ） uC ＝ 80V 时的充电时间 t 。

解：（ 1 ） RC 电路零状态响应问题

5 3500 10 5 10RC s

200(1 ) 100(1 ) 0t

tC Su U e e V t

2000.2 0t

tC Sdu Ui C e e A t

dt R

（ 2 ）设经过 t 秒， uC ＝ 80V

200( ) 100(1 ) 80tCu t e

8.045t ms

10A 2H

1R

200 300

80

+

_ LuK

Li

10A 2H R+

_ Lu

Li

例 1 t = 0 时 , 开关 K 打开，求 t > 0 后 iL ， uL 的变化规律 。

解： RL 电路零状态响应问题，先化简电路

80 200 // 300 200R

0t

20.01

200

Ls

R

( ) 10SS L

UI i A

R

10010(1 ) 0tLi e A t

100 10010 200 2000 0t tLu e e V t

2H

R

+

_Lu

Li

+

_SU

2A 2H

5

10

10

+

_LuK

Li

+

_u

例 2 t = 0 时 , 开关 K 打开，求 t > 0 后 iL ， uL 以及电流源两端的电压 u 。

解： RL 电路零状态响应问题，先化简电路

10 10 20R

0t

2 10 20SU V

20.1

20

Ls

R

( ) 1SL

Ui A

R

10(1 )tLi e A

10 1020t tL Su U e e V

105 10 20 10 tS L Lu I i u e V

全响应的两种分解方式

① 根据电路的两种工作状态

0( ) 0t

C S Su U U U e t

稳态分量 暂态分量 全响应

/Cu V

/t s0

US

U0 － U

S

稳态分量

暂态分量

U0

物理概念清晰

② 根据激励与响应的因果关系

0 (1 ) 0t t

C Su U e U e t

零输入响应 零状态响应 全响应

便于叠加计算

/Cu V

/t s0

US

U0

零输入响应

零状态响应

全响应

全响应

+

_24V

4

0.6H

8

( 0)K t Li+

_Lu

例 1 t = 0 时 , 开关 K 打开，求 t > 0 后的 iL 。

解： RL 电路全响应问题24

(0 ) (0 ) 64L Li i A

0 6 1

12 20

L .s

R

零输入响应 206 tLi e A

零状态响应 20 2024(1 ) 2(1 )

12t t

Li e e A

全响应 20 20 206 2(1 ) 2 4t t tL L Li i i e e e A

+

_10V

1 1

1F+

_ Cu1A

1

+

_u

K

例 2 t = 0 时 , 开关 K 闭合，求 t > 0 后的 iC ， uC 以及电流源两端的电压 u ，已知 uC(0+) = 1V 。

解： RC 电路全响应问题

稳态分量 ( ) 10 1 11Cu V

(1 1) 1 2RC s

全响应 0.511 tCu Ke V

(0 ) ( ) 10C CK u u

0.511 10 tCu e V

0.55 tCC

dui C e A

dt

0.51 1 1 12 5 tC Cu i u e V

故

例3

t = 0 时 , 开关闭合，求换路后的 uC(t) 。

1A 2 3F 1+

_Cu

解： (0 ) (0 ) 2C Cu u V

Cu2

( ) (2 // 1) 1 V3

23 2

3RC s

( ) ( ) [ (0 ) ( )]t

C C C Cu t u u u e

te 0.52 2(2 )

3 3

te t0.52 4 0

3 3

小结

状态变量

uC

iL

零输入响应 零状态响应 全响应

0

t

RCCu U e

(1 )

t

RCC Su U e

0( )

t

RCC S Su U U U e

0

0

Rt

LL

Rt

L

Ui e

R

I e

(1 )

(1 )

Rt

S LL

Rt

LS

Ui e

R

I e

0

0( )

Rt

S S LL

Rt

LS S

U U Ui e

R R

I I I e

5. Superposition and linearity

Provided one properly accounts for initial conditions, superposition still apply when capacitors and inductors are added to the circuit.

CC

dui C

dt

1 21 2,C C

C C

du dui C i C

dt dt

1 2( )C C C

di C u u

dt

suppose and 1 2C C Cu u u

for a capacitor

1 2C Cdu duC C

dt dt 1 2C Ci i

By the same arguments, the current due to the input excitation

1 1 2 2C C Cu a u a u is .1 1 2 2C C Ci a i a i

One the other hand, suppose

1 1 2 2

1 1( ) , ( )

t t

C C C Cu i d u i dC C

and 1 1 2 2C C Ci a i a i

1 1 2 2

1[ ( ) ( )]

t

C C Cu a i a i dC

1 1 2 2

1 1( ) ( )

t t

C Ca i d a i dC C

1 1 2 2C Ca u a u

Thus linearity and, hence, superposition hold.

Arguments analogous to the preceding imply that a relaxed inductor satisfies a linear relationship, and thus superposition is valid, whether the inductor is excited by currents or by voltages.

For a general linear circuit, one can view each initial condition as being set up by an input which shuts off the moment the initial condition is established.

This mean that when using superposition on a circuit, one first looks at the effect of each independent source on a circuit having no initial conditions.

Then one sets all independent sources to zero and computes the response due to each initial condition with all other initial conditions set to zero.

The sum of all the responses to each of the independent sources plus the individual initial condition responses yields the complete circuit response, by the principle of superposition.

Example 8.8. The linear circuit of the figure has two source excitations applied at t=0, as indicated by the presence of the step functions. The initial condition on the inductor current is i

L(0 － )= － 1A. Determine the response iL(t) for t≥0 using superposition.

SolutionStep 1. Compute the part of the circuit response due only to the initial condition, with all independent sources set to zero.

2( ) (0 ) ( )thR

t tLL Li t e i e u t A

+

_10 ( )SU u t V

1 10R Li

2L H

+ _1Ru

2

20

3R

2H 4Li

2 ( )SI u t A

Step 2. Determine the response due only to US=10u(t)V.

+

_10 ( )SU u t V

1 10R Li

2L H

+ _1Ru

2

20

3R 2 ( )SI u t A

1

( ) 1 ,SL

Ui A

R 1 2// 4thR R R (0 ) 0 ,Li A

( ) ( ) [ (0 ) ( )] ( )thR

tL

L L L Li t i i i e u t 2(1 ) ( )te u t A

Step 3. Compute the response due only to the current source IS=2u(t)A.

( ) 2 ,Li A 1 2// 4thR R R (0 ) 0 ,Li A

( ) ( ) [ (0 ) ( )] ( )thR

tL

L L L Li t i i i e u t 22(1 ) ( )te u t A

Step 4. Apply the principle of superposition.

( ) ( ) ( ) ( )L L L Li t i t i t i t 2 2 2( ) (1 ) ( ) 2(1 ) ( )t t te u t e u t e u t A

due to initial condition

due to source US

due to source IS

2(3 4 ) ( )te u t A

Question 1. What is the new response if the initial condition is changed to iL(0 － )=5A?

Question 2. What is the new response if the voltage source US is changed to 5u(t)V, with all other parameters held constant at their original values?

Question 3. What is the new response if the initial condition is changed to 5A, the voltage source US is changed to 5u(t)V, and the current source IS is changed to 8u(t)A?

The question still remains as to why the superposition principle holds an advantage over the Thevenin equivalent method.

This allows one to explore easily a circuit’s behavior over a wide range of excitations and initial conditions.

6. Response classifications

Zero-input response The zero-input response of a circuit is the response to the initial conditions when the input is set to zero.

Zero-state response The zero-state response of a circuit is the response to a specified input signal or set of input signals given that the initial conditions are all set to zero.

Complete response By linearity, the sum of the zero-input and zero-state responses is the complete response of the circuit.

Natural response The natural response is that portion of the complete response that has the same exponents as the zero-input response.

Forced response The forced response is that portion of the complete response that has the same exponents as the input excitation.

7. Further points of analysis and theory

Not only a capacitor voltage or an inductor current, it turns that any voltage or current in an RC or RL first-order linear circuit with constant input has the form

0

0( ) ( ) [ ( ) ( )]t t

x t x x t x e

by linearity, any voltage or current in the circuit has the form

1 2( ) ( )Cx t K K u t

and0

3 4( )t t

Cu t K K e

which implies that0

5 6( )t t

x t K K e

Example 8.9. For the circuit of the figure, let uin(t)= － 18u( － t )+9u(t)V. Find iin(t) for t>0.Solution

Step 1. Compute iin(0+).3

(0 ) (0 ) ( 18) 69C Cu u V

+

_( )inu t

( )ini t

3k

2k

0.5mA

6kCi

+

_Cu

0+ circuit

+

_( )inu t

( )ini t

3k

2k6kCi

+

_6V

9 6 3(0 )

6 3 // 2 2 3 // 6 3 6ini mA

1.75mA

Step 2. Computeτand iin(∞).

2 3 // 6 4thR k

2thR C s 9

( ) 16 3ini mA mA

Step 3. Set forth the complete response iin(t).

( ) ( ) [ (0 ) ( )]t

in in in ini t i i i e

0.51 0.75 te mA

Exercise. In example 7, find iC(0 － ), iC(0+), and iC(t) for t>0.

+

_( )inu t

( )ini t

3k

2k

0.5mA

6kCi

+

_Cu

例 1 t = 0 时开关闭合，求 t > 0 后 iL ， i1 和 i2 。

解法 1 ： 10(0 ) (0 ) 2

5L Li i A 10 20

( ) 65 5Li A

0.60.2

5 // 5

Ls

R

+

_10V

1i5 5

+

_20V2A

2i

0电路

+

_10V

1i5 5

0.5H

Li+

_20V

2i

1

10 20(0 ) 1 0

10i A

1

10( ) 2

5i A

2

20 10(0 ) 1 2

10i A

2

20( ) 4

5i A

5 5( ) 6 (2 6) 6 4 0t tLi t e e t

5 51( ) 2 (0 2) 2 2 0t ti t e e A t

5 52( ) 4 (2 4) 4 2 0t ti t e e A t

例 1 t = 0 时开关闭合，求 t > 0 后 iL ， i1 和 i2 。

解法 2 ： 10(0 ) (0 ) 2

5L Li i A 10 20

( ) 65 5Li A

0.60.2

5 // 5

Ls

R

( ) ( ) [ (0 ) ( )]t

L L L Li t i i i e 56 (2 6) te 56 4 0te t

( ) LL

diu t L

dt 5 50.5 ( 4 ) ( 5) 10 0t te e V t

1

10( )

5Lu

i t

52 2 0te A t

1

20( )

5Lu

i t

54 2 0te A t

+

_10V

1i5 5

0.5H

Li+

_20V

2i

2A 4

4

1 0.1F

2

1

+

_

8V

1i

+ _12i

+

_Cu

例 2 t = 0 时开关由 1→2 ，求换路后的 uC(t) 。

解：(0 ) (0 ) 8C Cu u V

1 1 1( ) 4 2 6 12Cu i i i V

10 0.1 1RC s

( ) ( ) [ (0 ) ( )]t

C C C Cu t u u u e

戴维宁等效10

+

_12V

0.1F+

_Cu

12 ( 8 12) te

12 20 0te V t

例 3 t = 0 时 , 开关闭合，求换路后的 i(t) 。

+

_10V

1H 5

2i

0.25H

解： (0 ) (0 ) 10C Cu u V

1 2 0.25 0.5RC s

( ) 0Cu

(0 ) (0 ) 0L Li i

( ) 10 / 5 2Li A

2

10.2

5

Ls

R

2( ) ( ) [ (0 ) ( )] 10 0t

tC C C Cu t u u u e e V t

5( ) ( ) [ (0 ) ( )] 2(1 ) 0t

tL L L Li t i i i e e A t

5 2( )( ) ( ) 2(1 ) 5 0

2t tC

L

u ti t i t e e A t

例 4 已知：电感无初始储能， t = 0 时闭合 K1 ， t =0.2s 时闭合 K2 ， 求两次换路后的电感电流。

解： 0 0.2t s

1

10.2

5

Ls

R

(0 ) (0 ) 0i i A 10

( ) 25

i A

5( ) 2 2 0 0.2ti t e A t

0.2t s

2

10.5

2

Ls

R

5 0.2(0.2 ) (0.2 ) 2 2 1.26i i e A 10

( ) 52

i A

2( 0.2)( ) 5 3.74 0.2ti t e A t

2

+

_10V

1H

3

1( 0)K t

2( 0.2 )K t s

i