Download for t>0.

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 Circuits
1. 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:
R
R
+
US
+
L
_
(a) First-Order RL circuit
US
C
_
(b) First-Order RC circuit
Interconnections of sources, resistors, capacitors, and inductors
lead to new and fascinating circuit behavior.
R
A loop equation leads to U S  Ri  uC
+
+
US
uC
_
i
C
_
duC
since i  C
dt
duC
 uC
dt
First-Order RC circuit
duC
1
1

uC 
US
or , equivalently,
dt
RC
RC
U S  RC
This equation is called Constant-coefficient first-order linear
differential equation
Apply duality principle,
diL
1
1

iC 
IS
dt GL
GL
2. Some mathematical preliminaries
duC
1
1

uC 
US
dt
RC
RC
diL
1
1

iC 
IS
dt GL
GL
RC circuit first-order differential equation
RL circuit first-order differential equation
The first-order RL and RC circuits have differential equation
models of the form dx( t )
  x( t )  f ( t ),
or , equivalently,
dt
dx( t )
  x( t )  f ( t ),
dt
x ( t 0 )  x0
x ( t 0 )  x0
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.
or
dx( t )
  x( t )  f ( t ), x( t 0 )  x0
dt
dx( t )
  x( t )  f ( t ), x( t 0 )  x0
dt
The main purpose of this chapter is to find a solution to the
differential equation.
The solution to the equation for t≥t0 has the form
x( t )  e
 ( t  t0 )
t
x( t 0 )   e  ( t  ) f ( )d
t0
(1) Satisfies the differential equation
(2) Satisfies the correct initial condition, x(t0)=x0
dx( t )
  x( t )  f ( t ),
dt
x ( t 0 )  x0
Integrating factor method
First step, multiply both sides of the equation by a so-called
integrating factor e－λt.
  t dx ( t )
t
t
e


e
x
(
t
)

e
f (t )
By the product rule
dt
for differentiation,
dx( t )
d t
  e   t x(t )  e t f (t )
[e x ( t )]  e   t
dt
dt
Integrate both sides of the equation from t0 to t as follows
t
t
d  
 
 
  t0
t

e

e
x
(

)
[
e
x
(

)]
d


e
x
(
t
)

e
x
(
t
)
0
t0 f ( )d
t0 d
t0
t
e
t
x( t )  e
 x( t )  e
  t0
 ( t  t0 )
t
x ( t 0 )   e   f ( )d
t0
t
x ( t 0 )   e  ( t  ) f ( )d
t0
3. Source-free or zero-input response
iR
R
+
uL
_
iL
L
+
R
(a )
(a) KCL implies
+
uR uC
_
_
(b)
iC
C
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.
(b) KVL implies
i R  i L
uR  uC
uL L diL
iR 

R R dt
diL
R

  iL
dt
L
duC
uR  iC R   RC
dt
du
1
 C 
uC
dt
RC
Both differential equation models dx( t )
1
  x(t )   x(t )
have the same general form,
dt

dx( t )
1
  x(t )   x(t )
dt

The solution to the equation for t≥t0 has the form
x ( t )  e  ( t  t0 ) x ( t 0 )  e
 t  t0 


  
x( t0 )
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
iL (t )  e
R
 ( t  t0 )
L
uC ( t )  e

iL (t0 )
1
( t  t0 )
RC
uC ( t0 )
RL circuit
L
  GL 
R
RC circuit
  RC
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.
Linear
Resistive
circuit
No
independent
sources
Linear
Resistive
circuit
No
independent
sources
iL
i L Thevenin equivalent
iL (t )  e
+
uC
L
Rth
L

Rth
( t  t0 )
L
iL (t0 )
Thevenin equivalent
C
Rth
_
uC (t )  e
+

1
RthC
( t  t0 )
uC (t0 )
uC
_
C
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, ∞).
S
t  0.4s
+
5
20
uL
_
iL
8H
一阶电路的初始条件
稳态（steady state）
代数方程描述
瞬态（transient state） 微分方程描述
（1）问题的提出
u( t ) / V
+
US
_
US
+
u(t)
R
_
K
0
过渡期为零
t/s
R
u( t ) / V
K
+
+
US
u(t)
_
_
换路
过渡状态
C
过渡状态
US
换路
0 t0
第一个稳定状态
t1
t2
第二个稳定状态
t3
t/s
第三个稳定状态
（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
t ＝0－ 换路前一瞬间
f (0 )  lim f ( t )
t ＝0＋ 换路后一瞬间
f (0 )  lim 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
i
+
uC
_
1
uC ( t ) 
C
1
i
(

)
d



C
t 0
t 0
t 0
t 0
0
1 t
 i ( )d  C 0 i ( )d
1 t

 uC (0 )    i ( )d
C
C 0
0
1


t＝0＋ 时 uC (0 )  uC (0 )    i ( )d
C 0


当 i ( ) 为有限值 uC (0 )  uC (0 )
q  CuC
q(0 )  q(0 ) 电荷守恒
t
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
iL
+
u
_
1 t
1 0
1 t
i L ( t )   u( )d   u( )d    u( )d
L 
L 
L 0
1 t

 i L (0 )    u( )d
L
L 0
0
1


t＝0＋ 时 i L (0 )  i L (0 )    u( )d
L 0


当 u( ) 为有限值 i L (0 )  i L (0 )
  LiL
 (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
反映了能量不能跃变
q(0 )  q(0 )
换路瞬间，若电容电流保持为有限值，
uC (0 )  uC (0 )
则电容电压（电荷）换路前后保持不变。
 (0 )   (0 )
换路瞬间，若电感电压保持为有限值，
i L (0 )  i L (0 )
则电感电流（磁链）换路前后保持不变。
例1

电路如图所示，试求 iC (0 ) 。
10k
解： ① 画出 0  电路，求出 uC (0  )
0  电路
+
10k
iC
+
10V
_
40k
iC (0 )  0 A
40
uC (0 ) 
 10  8V
10  40
② 由换路定律
uC (0 )  uC (0 )  8V


③ 画出 0 电路，求出 iC (0 )
+
电容
开路
iC
40k
C
10V
_
0  电路
电容用
电压源
替代
10k
iC
+
10V
+
8V
_
iC (0 ) 
uC
_
K
uC
_
+
_
10  8
 0.2mA
3
10  10


可见，iC (0 )  iC (0 )
1

电路如图所示，试求 uL (0 ) 。
例2
解： ① 画出 0  电路，求出 i L (0  )
0  电路
1
10V
4
iL
+
+
10V
+
uL
_
_
uL (0 )  0V
10

i L (0 ) 
 2A
1 4
电感
短路
4
iL
+
L
uL
K
_
0  电路
_
1
电感用
电流源
替代
4
+
+
2A
10V
_
_
② 由换路定律

uL (0 )  4  2  8V

i L (0 )  i L (0 )  2 A
③ 画出 0


电路，求出 uL (0 )
uL


可见，uL (0 )  uL (0 )
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－)。
电容相当于开路
其中
电感相当于短路
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。
其中
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。
④ 由 0+ 电路求所需各变量的 0+ 值。


电路如图所示，试求 iC (0 ) ，uL (0 ) 。
例3
L iL
_
+ uL
解： ① 画出 0  电路，求出 uC (0  ) ，i L (0  )

0 电路
iS
iL
R
+
uC
_

uC (0 )  Ri S V
R
K
t 0
iS
iC
uC
C
_
iS
0  电路
_
+
uL

i L (0 )  i S A
iC
R
i L (0 )  i L (0 )  i S A



i
(0
)
u
(0
)
0
③ 画出
电路，求出 C
，L
+
Ri S
_
② 由换路定律
uC (0 )  uC (0 )  Ri S V
+
RiS
iC (0 )  iS 
 0A
R
uL (0 )   Ri S V

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： 0 电路
2
2
iL
+
48V
_
iL +
L uL
i
2
3
+
uC
+
K
48V
_
_
2
_
2
 48  24V
22
48


i L (0 )  i L (0 ) 
 12 A
22
48  24

iC (0 ) 
 8A
3
uC (0 )  uC (0 ) 
i (0 )  i L (0 )  iC (0 )  20 A
uL (0 )  48  2  12  24V
3
iC
+
uC
C
_
0  电路
i
+
+
48V
_
12 A
iL
2
uL
_
3
iC
+
24V
_
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： 0 电路
L
iL
+ uC
100
100
+
iL
_
+ uC
100
100
K
200V
_
0  电路

uC (0 )  uC (0 )  100i L  100V
200 100
 i K (0 )  1 

 2A
100 100
100
+
_
200
i L (0 )  i L (0 ) 
 1A
100  100
_
100
200V

C
1A
+ uL
100
_
+ 100V
iC
_
iK
100
+
200V
_
100
单位阶跃函数及分段常量信号
u( t )
单位阶跃函数
1
0 t  0
u( t )  
1 t  0
0
延时（delayed）单位阶跃函数
u( t )
0 t  t 0
u( t  t0 )  
1 t  t0
1
分段常量信号（piecewise-constant signal）
f (t )
t0
0
f (t )
1
1
0
t
t0
t
（矩形）脉冲（pulse）
0
t0
2t 0
3t 0
脉冲串（pulse train）
t
t
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 )
f (t )
1
1
t0
0
t
f (t )  u(t )  u(t  t0 )
0
t0
2t 0
3t 0
t
f (t )  u(t )  u(t  t0 )  u(t  2t0 )  u(t  3t0 )
f (t )
A
0
A
t0
2t 0 t
f (t )  Au(t )  2 Au(t  t0 )  Au(t  2t0 )
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, ∞).
t  0.4s
S
Solution
iL
+
Step 1. With switch S open, compute
uL 8H
20
5
the response for 0≤t ≤0.4s.
_
From the continuity property of the


i
(0
)

i
(0
)  10 A
inductor current,
L
L
 iL (t )  e

Rth
t
L
i L (0 )  10e 2.5 t A
Step 2. With switch S closed, compute the response for t ≥0.4s.
Rth  20 // 5  4
iL (t )  e

Rth
( t  t0 )
L
0.5( t  0.4)
0.5( t  0.4)

3.679
e
A

i
(0.4)
e
iL (t0 )
L
Step 3. Write the complete response as a single expression.
i L ( t )  10e 2.5 t [u( t )  u( t  0.4)]  3.679e 0.5( t 0.4) u( t  0.4) A
Step 4. Plot the complete response.
iL ( t ), A
  0.4s
The 0.4s time constant has a
much faster rate of decay than
the lengthy 2s time constant.
  2s
t, s
for 0≤t ≤0.4s,
uL ( t )  e
t  0.4s
S
Step 5. Compute uL(t).

Rth
t
L

uL (0 )


in particular, uL (0 )  20  iL (0 )  200 A
+
5
20
uL
iL
8H
_
2.5 t
hence, uL ( t )  200e V
for t ≥0.4s,
uL ( t )  e

Rth
( t  0.4)
L
uL (0.4)  14.716e 0.5( t 0.4)V
in particular, uL (0.4)  4  iL (0.4 )  14.716V
Step 6. Compute energy dissipated in the 5Ω resistor over the
time interval [0.4, ∞).
uL2 ( t ) [14.716e 0.5( t 0.4) ]2
p5 ( t ) 

 43.3e ( t 0.4)W
5
5

 W5 (0.4, )  43.3 e ( t 0.4)dt  43.3J
0.4
Example 8.3. Find uC(t) for t ≥0 for the circuit of the figure
given that uC(0)=9V.
Solution
S
t 0
72
t 1
4.5
+
9 0.1F
uC
Step 1. Compute the response
for 0≤t ≤1s. By the continuity of
the capacitor voltage,
_
uC (0 )  uC (0 )  9V
hence, uC (t )  e

1
RthC
t
uC (0 )  9e 1.25 tV
Step 2. Compute the response for t ≥ 1s.
uC (t )  e

1
RthC
( t  t0 )
uC (t0 )  2.58e
 t 1 


 0.3 
V
Step 3. Use step functions to specify the complete response.
uC (t )  9e 1.25 t [u(t )  u(t  1)]  2.58e
 t 1 


 0.3 
u(t  1)V
Step 4. Obtain a plot of the response.
uC ( t ),V
  0.8s
  0.3s
t, s
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.
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.
S
t 0
72
t  0.5
4.5
+
9 0.1F
uC
_
Example 8.4. Find uC(t) for the circuit of the figure, assuming
that gm=0.75S and uC(0－)=10V.
a
+
gm ux
Solution
+
ux 4 0.25F
uC
_
_
b
It is straightforward to show
that the Thevenin equivalent
seen by the capacitor is a
negative resistance,
Rth  2
Thevenin equivalent
a
+
2
0.25F
uC
_
b
hence,
uC (t )  e

1
RthC
t
uC (0 )
 10e 2 t u( t )V
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.
uC ( t ),V
Negative resistance causes
capacitor voltage to
increase without bound.
t, s
Exercise. In example 3, let gm=0.125S. Find the equivalent
resistance seen by the capacitor and uC(t), t ≥0.
a
+
gm ux
+
ux 4 0.25F
uC
_
_
b
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：本题为求解一阶 RC 电路零输入响应问题
则有
uC  U 0e
又由已知条件

t
RC
K
t 0
5F
uC
 uC  24 e
V
t

uC
 i1 
 6 e 20 A
4
3
_
U 0  24V
t
20
i2
+
  RC  4  5  20s

2 i1
6
等效电路 t > 0
t 0
i1
t 0
+
t

20
2
利用并联分流，得 i2  i1  4e
A
3
t

1
i3  i1  2e 20 A
3
5F
t 0
t 0
uC
_
4
i3
t = 0 时 ，开关K由1→2，求电感电压和电流。
示例
K ( t  0)
解： RL电路零输入响应问题
iL  I 0e
R
 t
L
+
A t 0

24V
2
3
iL
6H
_

I 0  i L (0 )  i L (0 )

1
2
24
6

 2A
2  3 // 6  4 3  6
4
4
_
R
L 6
     1s
R 6
+
i L  2e  t A t  0
diL
 12e  tV
dt
uL
t 0
R  (2  4) // 6  3  6
uL  L
+
t 0
6H uL
_
iL
6
小结
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。


x(t )  x(0 )e

t

t 0

RC 电路 uC (0 )  uC (0 )
RL 电路 i L (0 )  i L (0 )
② 衰减快慢取决于时间常数  。
RC 电路
  RC
RL 电路
  GL 
L
R
R 为与动态元件相连的一端口电路的等效电阻。
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。
4. DC or step response of first-order circuits
This section takes up the calculation of voltage and current
responses when constant voltage or constant current sources
are present.
Rth
Linear
Resistive
circuit
With
independent
sources
Linear
Resistive
circuit
With
independent
sources
i L Thevenin equivalent
L
+
+
U oc
uL
iL
L
_
_
Rth
+
uC
_
Thevenin equivalent
C
+
U oc
_
+
uC
_
iC
C
Deriving the differential equation models characterizing
each circuit’s voltage and current responses.
Rth
Rth
+
+
U oc
uL
_
iL
L
_
By KVL and Ohm’ law,
uL  U oc  Rth iL
diL
L
dt
Rth
diL
1

iL  Uoc
dt
L
L
+
U oc
_
+
uC
iC
C
_
By KCL and Ohm’ law,
iC 
du
U oc  uC
C C
Rth
dt
duC
1
1

uC 
U oc
dt
RthC
RthC
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.
+
I sc
Answers:
Rth
uL
_
iL
L
+
I sc
Rth
Rth
diL

iL 
I sc
dt
L
L
duC
1
1

uC  I sc
dt
RthC
C
Rth
uC
_
iC
C
Rth
diL
1

iL  Uoc
dt
L
L
duC
1
1

uC 
U oc
dt
RthC
RthC
Rth
Rth
diL

iL 
I sc
dt
L
L
duC
1
1

uC  I sc
dt
RthC
C
Observe that four differential equations have the same
structure:
dx ( t )
1
dt
where
  x( t )  F

L
Uoc I sc Rth
  Gth L 
, F

Rth
L
L
I
U
  RthC ,
F  sc  oc
C RthC
for RL case
for RC case
And the general formula for solving such a differential
equation:
t
 ( t  t0 )

x( t )  e
x( t0 )   e  ( t  ) f ( )d
t0
x( t )  e
 ( t  t0 )
t
x( t )   e  ( t  ) f ( )d

0
t0
1
where    , x(0 )  x(0 ) as long as x(t) is a capacitor

voltage or inductor current, and f(τ)=F is a constant
(nonimpulsive) forcing function.
x( t )  e
e


t  t0

t
x( t0 )   e

t q

t0
t  t0

Fdq
x ( t0 )  F (1  e
 F  [ x( t0 )  F ]e


t  t0

)
t  t0

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.
x( t )  F  [ x( t 0 )  F ]e

t  t0

if   0, then

t  t0
x()  lim x(t )  lim{F  [ x(t )  F ]e  }
t 
t 
U
I sc  oc for RL case
Rth
 F 
U oc
for RC case

0
Hence, the solution of the differential equation given
constant or dc excitation becomes
x( t )  x( )  [ x( t 0 )  x()]e

t  t0
i L ( t )  i L ( )  [i L ( t 0 )  i L ()]e


Rth
( t  t0 )
L
uC (t )  uC ()  [uC (t0 )  uC ()]e

t  t0
RthC
for RL case
for RC 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.
Solution
R  5
Step 1. Determine the circuit’s
differential equation model.
iL +
+
uin ( t )
_
L  2H
uL
_
R
diL
1
  th iL  Uoc
dt
L
L
 2.5iL  5u( t  1)
t 1
where the time constant   0.4s
Step 2. Determine the form of the response.
iL ( t )  {iL ()  [iL (1 )  iL ()]e

t 1

}u( t  1) A
Step 3. Compute iL(∞) and set forth the
final expression for iL(t).
R  5
+
replace the inductor by a short circuit, uin ( t )
uin
 i L () 
 2 A  I sc
R
and since i L (1 )  i L (1 )  1 A
_
2.5( t 1)
]u( t  1) A
It follows that, iL ( t )  [2  (1  2)e
Step 4. Plot iL(t).
 [2  e 2.5( t 1) ]u(t  1) A
iL +
uL
L  2H
_
Step 5. Compute uL(t).
di
+
uL (t )  L L
uin ( t )
dt
_
t 1

L
  [iL (1 )  iL ()]e  u( t  1)

R  5
iL +
uL
L  2H
_
 5e 2.5( t 1) u(t  1)V
Exercise. Verify that in example 4, uL(t) can be obtained
without differentiation by uL (t )  U oc  RthiL (t )
Exercise. In example 4, suppose R is changed to 4Ω. Find
iL(t) at t=2s.
Specifically, we need only compute x ( t0 ) , x( ) , and the time
L
constant  
or RthC .
Rth
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.
R1  6k 
Solution
+
t  10s
+
Step 1. Compute initial
C
uC
uin ( t )
capacitor voltage
0.5mF
_

_

uC (0 )  uC (0 )  12V
R2  3k 
Step 2. Obtain uC(t) for 0≤t ≤ 10s.
  R1C  3s, uC ()  24V
 uC (t )  uC ()  [uC (0 )  uC ()]e

t
RthC

t
3
 24  12e V
R1  6k 
Step 3. Compute the initial condition
for the interval t>10.
uC (10 )  uC (10 )  24  12e

10
3
 23.57V
+
uin ( t )
t  10s
+
R2  3k 
_
C
0.5mF
uC
_
Step 4. Find uC(t) for t>10.
  RthC  1s,
Thevenin equivalent
uC ()  8V
 uC (t )  uC ()  [uC (10 )  uC ()]e
 8  15.57e

t 10
RthC
 ( t 10)
V
Step 5. Set forth the complete response
using step functions.
uC ( t )  (24  12e

10
3
Rth  2k 
+
U oc  8V
_
+
C
0.5mF
)[u( t )  u( t  10)]  [8  15.57e ( t 10) ]u( t  10)
uC
_
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.
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 。
－
K
解：（1） RC 电路零状态响应问题
+
100V
  RC  500  10  5  10 s
5
uC  U S (1  e

t

3
)  100(1  e 200 t )V
duC U S  t
iC

e  0.2e 200 t A
dt
R
（2）设经过 t 秒，uC＝80V
uC ( t )  100(1  e 200 t )  80
 t  8.045ms
_
t0
t0
i
10  F
500
+
uC
_
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
R1
+
R  80  200// 300  200
L
2
 
 0.01s
R 200
80
10A
K
2H uL
iL
t 0
US
 I S  iL ()  10 A
R
+
i L  10(1  e 100 t ) A t  0
uL  10  200e 100 t  2000e 100 tV
200 300
_
10A
t0
2H uL
_
iL
R
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
R  10  10  20
U S  2  10  20V
5
10
+
+
2A
u
10 2H uL
_
iL
K
_
L 2
 
 0.1s
R 20
US
 iL ( )  1 A
R
i L  (1  e
10 t
)A
uL  U S e 10 t  20e 10 tV
u  5 I S  10i L  uL  20  10e 10 tV
t 0
R
+
US
_
+
2H uL
iL
_
全响应的两种分解方式
US
① 根据电路的两种工作状态
uC  U S  (U 0  U S )e
稳态分量

t
全响应
t 0

U0
0
暂态分量
暂态分量
U0－ US
uC / V
② 根据激励与响应的因果关系
uC  U 0e
t


t
 U S (1  e )
零输入响应
t/s
全响应
物理概念清晰

uC / V 稳态分量

零状态响应
t 0
全响应
US
U0
零状态响应
全响应
零输入响应
便于叠加计算
0
t/s
例1
t = 0 时 , 开关K打开，求 t > 0 后的 iL 。
8
解： RL 电路全响应问题
i L (0 )  i L (0 ) 
24
 6A
4
L 0.6 1
 

s
R 12 20
4
+
K ( t  0)
24V
_
零输入响应
i L  6e 20 t A
零状态响应
i L 
全响应
i L  i L  i L  6e 20 t  2(1  e 20 t )  2  4e 20 t A
24
(1  e 20 t )  2(1  e 20 t ) A
12
+
iL
uL 0.6H
_
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。
例2
解： RC电路全响应问题
稳态分量
uC ()  10  1  11V
1
1
+
  RC  (1  1)  1  2 s 10V
_
全响应
uC  11  Ke 0.5 tV
 K  uC (0 )  uC ()  10
0.5 t
u

11

10
e
V
故
C
duC
iC  C
 5e 0.5 t A
dt
u  1  1  1  iC  uC  12  5e 0.5 tV
1F
1
K
+ 1A
u
_C
+
u
_
例3
t = 0 时 , 开关闭合，求换路后的uC(t) 。
解： uC (0 )  uC (0 )  2V
2
uC ( )  (2 // 1)  1  V
3
2
  RC   3  2s
3
uC ( t )  uC ()  [uC (0 )  uC ()]e
2
2
 (2  )e 0.5 t
3
3
2 4
  e 0.5 t t  0
3 3

2
1A
3F
+
uC
_

t

1
小结
状态变量
uC
iL
零输入响应
uC  U 0e

t
RC
U 0  RL t
iL 
e
R
 I 0e
R
 t
L
零状态响应
uC  U S (1  e

全响应
t
RC
) uC  U S  (U 0  U S )e

t
RC
R
R
 t
 t
US
U
U

U
S
iL 
(1  e L ) i L  S  0
e L
R
R
R
 I S (1  e
R
 t
L
)
 I S  ( I 0  I S )e
R
 t
L
5. Superposition and linearity
Provided one properly accounts for initial conditions,
superposition still apply when capacitors and inductors are
added to the circuit.
duC
for a capacitor iC  C
dt
duC 1
duC 2
i

C
,
i

C
suppose C 1
C2
dt
dt
and uC  uC 1  uC 2
d
duC 1
duC 2
 iC 1  iC 2
 iC  C ( uC 1  uC 2 )  C
C
dt
dt
dt
By the same arguments, the current due to the input excitation
uC  a1uC 1  a2 uC 2 is iC  a1iC 1  a2 iC 2 .
One the other hand, suppose
uC 1
1

C

t
i ( )d , uC 2
 C 1
1

C

t
i ( )d and iC  a1iC 1  a2 iC 2
 C 2
1 t
 uC   [a1iC 1 ( )  a2 iC 2 ( )]d
C 
1 t
1 t
  a1iC 1 ( )d   a2 iC 2 ( )d  a1uC 1  a2 uC 2
C 
C 
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
iL(0－)=－1A. Determine the response iL(t) for t≥0 using
u _
superposition.
+ R1
+
U S  10u(t )V
_
R1  10
L  2H
iL
R2 
20

3
I S  2u( t ) A
Solution
Step 1. Compute the part of the circuit response due only to the
initial condition, with all independent sources set to zero.
iL
2H
4
i L ( t )  e

Rth
t
L
i L (0 )  e 2 t u( t ) A
+
+
U S  10u(t )V
_
uR1 _
R1  10
L  2H
iL
R2 
20

3
I S  2u( t ) A
Step 2. Determine the response due only to US=10u(t)V.
US
 1 A, Rth  R1 // R2  4
R1
R
 th t

 i L ( t )  i L ()  [i L (0 )  i L ()]e L u( t )  (1  e 2 t )u( t ) A
i L (0 )  0 A, iL () 
Step 3. Compute the response due only to the current source
IS=2u(t)A.

i L(0 )  0 A, iL()  2 A, Rth  R1 // R2  4
 i L( t )  i L()  [i L(0 )  i L()]e

Rth
t
L
u( t )  2(1  e 2 t )u( t ) A
Step 4. Apply the principle of superposition.
iL (t )  iL (t )  iL (t )  iL(t )
 e 2 t u(t )  (1  e 2 t )u(t )  2(1  e 2 t )u(t ) A
due to
initial
condition
due to
source US
 (3  4e 2 t )u( t ) A
due to
source IS
The question still remains as to why the superposition
principle holds an advantage over the Thevenin equivalent
method.
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?
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 zeroinput 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
x( t )  x( )  [ x( t 0 )  x()]e

t  t0

by linearity, any voltage or current in the circuit has the
form
x(t )  K1  K 2 uC (t )
and
uC ( t )  K 3  K 4e
which implies that

t  t0

x( t )  K 5  K 6e

t  t0

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+).
iin ( t ) 6k
+
+
3
uC (0 )  uC (0 )   ( 18)  6V uin ( t )
_
9
9
6
3

iin (0 ) 


mA
6  3 // 2 2  3 // 6 3  6
3k
Step 2. Computeτand iin(∞).
  RthC  2s
iin () 
9
mA  1mA
63
0.5mA
uC
_
0+ circuit
 1.75mA
Rth  2  3 // 6  4k
2k iC
iin ( t ) 6k
+
uin ( t )
_
2k iC
+
3k
6V
_
Step 3. Set forth the complete response iin(t).

iin ( t )  iin ()  [iin (0 )  iin ()]e
t

 1  0.75e 0.5 t mA
Exercise. In example 7, find iC(0－), iC(0+), and iC(t) for t>0.
iin ( t ) 6k
2k iC
+
+
uin ( t )
_
3k
0.5mA
uC
_
例1
t = 0 时开关闭合，求 t > 0 后 iL，i1 和 i2 。
10
解法1：i L (0 )  i L (0 ) 
 2A
i1
5
10 20
+
i L ()  
 6A
10V
5
5
_
L 0.6
 
 0.2 s
R 5 // 5
10  20
10

i1 (0 ) 
 1  0 A i1 ( ) 
 2A
10
5
20  10
20
i2 (0 ) 
 1  2 A i2 (  ) 
 4 A i1
10
5
+
10V
i L ( t )  6  (2  6)e 5 t  6  4e 5 t t  0


_
i1 ( t )  2  (0  2)e 5 t  2  2e 5 t A t  0
i2 ( t )  4  (2  4)e 5 t  4  2e 5 t A t  0
5
5
i2
iL
+
0.5H 20V
_
0电路
5
5
i2
+
2A
20V
_
例1
t = 0 时开关闭合，求 t > 0 后 iL，i1 和 i2 。
5
10
解法2： i L (0 )  i L (0 ) 
 2A
5
10 20
i L ()  
 6A
5
5
L 0.6
 
 0.2 s
R 5 // 5


iL ( t )  iL ( )  [iL (0 )  i L ()]e
i1
+
5
iL
10V
_

t

i2
+
0.5H 20V
_
 6  (2  6)e 5 t  6  4e 5 t t  0
diL
uL (t )  L
 0.5  (4e 5 t )  (5)  10e 5 tV t  0
dt
10  uL
 2  2e 5 t A t  0
i1 ( t ) 
5
20  uL
i1 (t ) 
 4  2e 5 t A t  0
5
例2
t = 0 时开关由1→2，求换路后的uC(t) 。
4
2
解：
i1
uC (0 )  uC (0 )  8V
uC ()  4i1  2i1  6i1  12V
2A
4
  RC  10  0.1  1s
uC ( t )  uC ()  [uC (0 )  uC ()]e
 12  (8  12)e
 12  20e V
t

+
0.1F
_
8V
2i _
+ 1

t
1
1
+
t

戴维宁等效
10
t 0
+
12V
_
+
0.1F
uC
_
uC
_
例3
t = 0 时 , 开关闭合，求换路后的i(t) 。

1H

解： uC (0 )  uC (0 )  10V
uC ()  0
+
10V
 1  RC  2  0.25  0.5s
5
i
2
_
0.25H
i L (0 )  i L (0 )  0
iL ()  10 / 5  2 A
L 1
 2    0.2 s
R 5
t

uC ( t )  uC ()  [uC (0 )  uC ()]e   10e 2 tV
iL ( t )  iL ()  [iL (0 )  iL ()]e

t

t 0
 2(1  e 5 t ) A t  0
uC ( t )
i (t )  iL (t ) 
 2(1  e 5 t )  5e 2 t A t  0
2
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  t  0.2s
K1 ( t  0)
+
10V
_
2 i
i (0 )  i (0 )  0 A
10
1H
i () 
 2A
5
L 1


  0.2 s
1
3
R 5
K 2 ( t  0.2 s )
i (t )  2  2e 5 t A 0  t  0.2
t  0.2s i (0.2 )  i (0.2 )  2  2e 50.2  1.26 A
10
i () 
 5A
2
L 1
 2    0.5 s
R 2
i (t )  5  3.74e 2( t 0.2) A t  0.2