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11.1 The Laplace Transform
11.2 Applications of Laplace Transform
Chapter11 The Laplace Transform 拉普拉斯变
换
1. Definition of the Laplace Transform
)()( sFtf
dtetftfLsF st
0
)()]([)(
js
The Laplace transform is an integral transformation of a function f(t) from the time domain into the complex frequency domain, giving F(s).
11.1 The Laplace Transform 拉普拉斯变换
The inverse Laplace transform 拉普拉斯反变换
dsesFj
tfsFLj
j
st
1
1
)(2
1)()]([1
Laplace transform pair 拉普拉斯变换对
complex frequency 复频率
se
sdtetuLsF stst 11
1)]([)( 0
0
ase
as
dtedteetueLsF
tas
tasstatat
11
)]([)(
0)(
0
)(
0
1)()]([)( 0
0
edtettLsF st
(1)
(3)
(2)
Example 11.1 Determine the Laplace transform of each of the following functions:(1) u(t), (2) , and (3) )(tue at )(tSolution:
2. Properties of the Laplace Transform
(1) Linearity 线性性质
)()()]()([ 22112211 sFasFatfatfaL
,)]([)( 11 tfLsF )]([)( 22 tfLsF If ,)]([)( tfLsF
(2) Time Shift 时移)()]()([ sFeatuatfL as
(3) Frequency Shift 频移)()]([ asFtfeL at
(4) Time Differentiation 微分
)0()(])(
[ fssFdt
tdfL
(5) Time Integration 积分
s
sFdttfL
t )(])([
0
3. The Inverse Laplace Transform 拉普拉斯反变换
The inverse Laplace transform
dsesFj
tfsFLj
j
st
1
1
)(2
1)()]([1
The general form of
)(
)()(
sD
sNsF
(1) Decompose F(s)into simple terms using partial fraction expansion 部分分式展开 .
Steps to find the inverse Laplace transform:
(2) Find the inverse of each term by matching entries in table 14.2
)())((
)()(
21 npspsps
sNsF
)()()()(
2
2
1
1
n
n
ps
k
ps
k
ps
ksF
ipsii sFpsk )()(
If
when pi is Simple Poles
the residue method: 留数法
(1) Simple Poles 单根Partial fraction expansion 部分分式展开
Example 11.2 Find the inverse Laplace transform of
)4)(3)(1(
)2(6)(
sss
ssF
)4()3()1()( 321
s
k
s
k
s
ksF
1)4)(3(
)2(6)()1( 111
ss ss
ssFsk
3)4)(1(
)2(6)()3( 332
ss ss
ssFsk
4)1)(3(
)2(6)()4( 443
ss ss
ssFsk
)4(
4
)3(
3
)1(
1)(
sss
sF
043)( 43 teeetf ttt
By the residue method
Solution:
)()(
)()(
qsps
sNsF
n p is repeated poles
)()()()(
)( 1221 sF
ps
k
ps
k
ps
ksF
nn
)()!1(
!2)(
11
23
21
tfetn
k
etktekektf
ptnn
ptptpt
The inverse transform
(2) Repeated Poles 重根
psn
n sFpsk )()(
psn
n sFpsds
dk )]()[(1
psn
n sFpsds
dk )]()[(
!2
12
2
2
Example 11.3 Obtain g(t) if )3()1(
62)(
2
3
sss
sssG
1)1()3()(
2
s
D
s
C
s
B
s
AsG
2)3()1(
62)( 02
3
0
ss ss
ssssGA
25.2)1(
62)()3(
2
3
3
ss
sssGsB s
5.1)3(
62)()1( 1
3
12
ss ss
sssGsC
25.3)]()1[( 12 ssGs
ds
dD
Solution:
1
25.3
)1(
5.1
)3(
25.22)(
2
sssssG
025.35.125.2)(2)( 3 teteetutg ttt
The inverse transform
When )()( 1221 sFbass
AsAsF
02 bass
jbaa
p 22,1 )
2(
2
2)2
(,2
ab
a
))((
)( 112
21
jsjs
BsA
bass
AsA
)()()(
)(
)()(
)()(
1221
221
12211
sFs
B
s
sA
sFs
BsAsF
The poles
(3) Complex Poles 共轭复根
let
)(sincos)( 111 tfteBteAtf tt
)()()(
)()( 122
122
1 sFs
B
s
sAsF
11.2 Applications of Laplace Transform
(3) Take the inverse transform of the solution and thus obtain the solution in the time domain.
Steps in applying the Laplace transform:
(1) Transform the circuit from the time domain to the s domain.
(2) Solve the circuit using any circuit analysis technique with which we are familiar.
(1) For a resistor
In the time domain )()( tRitv
In the s domain )()( sRIsV
1. Circuits Element Models
Rv(t) R
i(t)
RV(s) R
I(s)
(2) For an inductor
In the time domaindt
tdiLtv
)()(
In the s domain )0()()( LissLIsV
v(t)
i(t)
LL V(s)
I(s)
)(Li
sL
)0-(Li
sL
(3) For a capacitor
In the time domaindt
tdvCti
)()(
In the s domain
Cv(t)
i(t)
V(s)
I(s)
sC
1
sC
1
s
v )0(
s
v )0(
s
vsI
sCsV
)0()(
1)(
(4) For the impedance
Under zero initial conditions:
)(
)()(
sI
sVsZ
)(
)(
)(
1)(
sV
sI
sZsY
Example 11.4 Find vo(t) in the circuit. Assume vo(0-)=5V
10
+0.1Fv0(t)
V)(10 tue t
-A)(2 t10
Solution:
s
10
10 V0(s)
A210V
5
s
V1
10
s
Transform the circuit to the s-domain
Apply nodal analysis. At the top node
2/10
5)(
10
)(
101
10)( 0
00
ss
sVsVssV
)2)(1(
3525)(0
ss
ssV
2)1(
s
B
s
A
10|)()1( 10 ssVsA
15|)()2( 20 ssVsB
Thus 2
15
)1(
10)(0
ss
sV V)()1510()( 20 tueetv tt
部分电路图和内容参考了： 电路基础（第 3 版），清华大学出版社 电路（第 5 版），高等教育出版社 特此感谢！
