Introduction to Automatic Control
The Laplace Transform
The Laplace Transform
Li Huifeng
Tel:82339276
Email:Lihuifeng@buaa.edu.cn
The Laplace Transform
Module objectives
Introduction to Automatic Control
• When you have completed this module you should be able to:
– Apply the Laplace transform to differential
equations.
– Solve linear differential equations.
– Apply the main theorems of the Laplace
transform.
– Know how useful this techniques is to handle
dynamical systems
The Laplace Transform
Introduction to Automatic Control
Subsections
•
•
•
•
•
Definition
Correspondences of the Laplace transform
Main theorems of the Laplace transform
The inverse Laplace transform
Solving linear differential equations using
the Laplace transform
The Laplace Transform
Introduction to Automatic Control
How to understand LT
• The Laplace transform is conceptually
similar to multiplication via logarithms
• log(axb) = log(a) + log(b)
• To multiply a by b
– Compute logarithms of a and b
– Add these logarithms
– Inverse logarithm of sum gives product of a and
b.
The Laplace Transform
t
t
o
o
Introduction to Automatic Control
f1 (t )  f 2 (t )   f1 ( ) f 2 (t   )d   f1 (t   ) f 2 ( )d
Linear
differential
equation
Convolution
Time domain
solution
Laplace
Transform
Time
Domain
Inverse Laplace
Transform
Laplace
transformed
Equation
Laplace
Solution
Algebraic
manipulation
Laplace
Domain
Introduction to Automatic Control
The Laplace Transform
The Laplace Transform
Definition
Introduction to Automatic Control
函数f(t)，t为实变量，如果线性积分


f (t )e  st dt ( s    jw为复变量）
0
存在，则称其为函数f(t)的拉普拉斯变换。变换后的
函数是复变量s的函数，记作F(s)或L[f(t)]，即：

L[ f (t )]  F ( s)   f (t )e  st dt
0
称F(s)为f(t)的变换函数或象函数，而f(t)为F(s)的原
函数。
The Laplace Transform
Introduction to Automatic Control
Correspondences of the Laplace
transform
f (t )  F ( s)
F ( s)  L[ f (t )]
1
f (t )  L [ F ( s)]
• Table 2.1: Corresponding elements of the Laplace
transform or Textbook Page 21
The Laplace Transform
几种典型函数的拉氏变换
Introduction to Automatic Control
1. 单位阶跃函数1(t)
其数学表达式为：
t 0
1
f ( t )  1( t )  
t 0
0
其拉氏变换为：

1
L [ f ( t )]  F ( s )   1  e dt 
s
0
 st
f(t)
0
t
The Laplace Transform
Introduction to Automatic Control
2. 单位斜坡函数
f(t)
其数学表达式为：
t
f ( t )  t  1( t )  
0
t 0
t 0
其拉氏变换为：

1
L [ f ( t )]  F ( s )   t  e dt  2
s
0
 st
0
t
The Laplace Transform
Introduction to Automatic Control
3. 等加速函数
其数学表达式为：
 1 2
t 0
t
f ( t )  2
t 0
 0
其拉氏变换为：

1 2  st
1
L [ f ( t )]  F ( s )   t  e dt  3
s
0 2
Introduction to Automatic Control
The Laplace Transform
The Laplace Transform
4. 指数函数
Introduction to Automatic Control
其数学表达式为：
t
e

f(t ) 
0
t 0
t 0
其拉氏变换为：

1
L [ f ( t )]  F ( s )   e  e dt 
s 
0
t
 st
The Laplace Transform
5. 正弦函数
Introduction to Automatic Control
其数学表达式为：
 sin wt
f(t ) 
 0
t 0
t 0
其拉氏变换为：

w
L [ f ( t )]  F ( s )   sin wt  e dt  2
2
s

w
0
 st
The Laplace Transform
Introduction to Automatic Control
6. 余弦函数
其数学表达式为：
cos wt
f(t ) 
0
t 0
t 0
其拉氏变换为：

s
L [ f ( t )]  F ( s )   cos wt  e dt  2
2
s

w
0
 st
The Laplace Transform
Introduction to Automatic Control
7. 单位脉冲函数
 (t )
其数学表达式为：
t 0
0
f(t )(t ) 
t 0

0
其拉氏变换为：
L [ f ( t )]  F ( s )  1

定义：
  ( t )dt  1

t
The Laplace Transform
jt
1
 cos t  j sin t  F ( s) 
( s  j )
 jt
1
 cos t  j sin t  F ( s) 
( s  j )
Introduction to Automatic Control
f (t )  e
f (t )  e
The Laplace Transform
Introduction to Automatic Control
Euler equation
e j  t  e  j t
cos t 
2
sint 
e
jt
e
2j
 jt
The Laplace Transform
Introduction to Automatic Control
Equating real & imaginary parts yields:
s
cos t  2
(s   2 )
sin t 

(s   )
2
2
Introduction to Automatic Control
The Laplace Transform
e
e
t
t
s 
cos t 
2
2
(( s   )   )
sin t 

(( s   ) 2   2 )
Introduction to Automatic Control
The Laplace Transform
典型函数的拉氏变换形式
The Laplace Transform
Introduction to Automatic Control
Main theorems of the Laplace
transform
① Superposition theorem(叠加定理):
L{a1 f1 (t )  a2 f 2 (t )}  a1F1 (s)  a2 F2 (s)
各函数和的拉氏变换＝各函数拉氏变换的和
Introduction to Automatic Control
The Laplace Transform
② Similarity theorem: (比例尺改变)
1
s
L{ f (at )}  F ( )
a a
a0
Introduction to Automatic Control
The Laplace Transform
③ Real Shifting theorem(延时定理):
L{ f (t  a)}  eas F (s)
a0
The Laplace Transform
例1
求 f ( t )  ( t   )  1( t   )的拉氏变换。
Introduction to Automatic Control
f (t )
t  1( t )
0

解:
( t   )  1( t   )
提示：
f(t) 相当于t· 1(t)在
时间上延迟了一个
值。
t
F ( s)  L[(t   )  1(t   )]
1 s
 2e
s
Introduction to Automatic Control
The Laplace Transform
④ Complex Shifting theorem(衰减定理）:
L{e
 at
f (t )}  F (s  a)
a0
The Laplace Transform
例2
Introduction to Automatic Control
求 f (t )e
解:
 at
sinwt的拉氏变换。
F ( s)  L[e sinwt]
w

2
2
(s  a)  w
 at
The Laplace Transform
⑤ Derivative theorem:
df ( t )
]  sF ( s )  f ( 0 )
dt
d2 f (t )
2
'



L[
]
s
F
(
s
)
sf
(
0
)
f
(0 )
2
dt
dn f (t )
n
n1
n 2 '
( n 1 )





L[
]
s
F
(
s
)
s
f
(
0
)
s
f
(
0
)
f
(0 )
L
n
dt
Introduction to Automatic Control
L[
拉氏变换将原函数求导数的运算转换为“象函数
乘s后减初值”的代数运算。
The Laplace Transform
⑥ Complex differentiation theorem:
Introduction to Automatic Control
(不要求掌握)
k
d
F ( s)
k
k
L{t f (t )}  (1)
k
ds
The Laplace Transform
Introduction to Automatic Control
⑦ Integral theorem:
1
1
L [  f ( t )dt ]  F ( s )  f ( 1 ) ( 0 )
s
s
1
1 ( 1 )
1 ( 2 )
2
L [  f ( t )dt ]  2 F ( s )  2 f
(0 ) f
(0 )
s
s
s
1
1
1
L [  L  f ( t )dt n ]  n F ( s )  n f ( 1 ) ( 0 )  L  f (  n ) ( 0 )
s
s
s
拉氏变换将原函数求积分的运算转换为“象函数除
以s后加初值”的代数运算。
The Laplace Transform
Introduction to Automatic Control
⑧ Initial value theorems:
lim sF ( s)  lim f (t )  f (0)
s 
t 0
The Laplace Transform
Given:
Introduction to Automatic Control
F ( s) 
( s  2)
( s  1)2 52
Find f(0)
解:


s 2  2s
f (0)  lim sF ( s )  lim s
 lim  2

2
2
s
s
s
( s  1)  5
 s  2 s  1  25 
( s  2)
s2 s2  2 s s2
 lim
s 2
s
s  2 s s  ( 26 s )
2
2
2
1
The Laplace Transform
Introduction to Automatic Control
⑨ final value theorems:
lim sF ( s )  lim f (t )  f ()
s 0
t 
The Laplace Transform
Given:
Introduction to Automatic Control
F ( s) 
( s  2) 2  3 2
( s  2)
2
 32

note F 1 ( s )  te  2 t cos 3t
Find f () .
解:
f ( )  lim sF ( s )  lim s
s0
s0
( s  2) 2  3 2
( s  2)
2
3
2

0
The Laplace Transform
⑩ Convolution in the time domain:
Introduction to Automatic Control
(不要求掌握)
t
t
o
o
f1 (t )  f 2 (t )   f1 ( ) f 2 (t   )d   f1 (t   ) f 2 ( )d
L{ f1 (t )  f 2 (t )}  F1 (s) F2 (s)
The Laplace Transform
Introduction to Automatic Control
⑪ Convolution in the frequency domain:
(不要求掌握)
c  j
1
L{ f1 (t ) f 2 (t )} 
F1 ( p) F2 ( s  p )dp

2j c  j
拉氏变换性质的证明
The Laplace Transform
Introduction to Automatic Control
The inverse Laplace transform
定义：
由象函数 F(s) 求其原函数 f(t) 的运算称
为拉氏反变换。
L [ F ( s )]  f ( t ) 
1
1
  j
st
F
(
s
)
e
dt

2 j   j
The Laplace Transform
Introduction to Automatic Control
Partial Fraction Expansion
思路：
将 F(s) 分解成一些简单的有理分式函数之和，
然后由拉氏变换表一一查出对应的反变换函
数，即得所求的原函数 f(t) 。
F ( s )  F1 ( s )  F2 ( s )  L  Fn ( s )

L1{F ( s )}  L1{F 1( s )}  L1{F 2( s )}  L  L1{F n( s )}
 f1 (t )  f 2 (t )  L f n (t )  f (t )
The Laplace Transform
Introduction to Automatic Control
F ( s ) 的一般形式为：
B( s ) b0 s m  b1 s m 1  L  bm 1 s  bm
F( s ) 

n
n1
A( s )
s  a1 s  L a n  1 s  a n
式中，a1 , a2 ,Lan 及 b0 , b1 ,Lbm 均为实数，
m , n 为正数，且 m  n。
Introduction to Automatic Control
The Laplace Transform
将F(s)的分母多项式A(s)进行因式分解，
即写为：
A(s) = (s - s1)(s - s2)…(s - sn)
式中，s1, s2, …sn为A(s) = 0的根。
分两种情况讨论：
1. A(s) = 0无重根
2. A(s) = 0有重根
The Laplace Transform
Introduction to Automatic Control
1. A(s)=0无重根情况
可将F(s)换写为n个部分分式之和，每个分式
的分母都是A(s)的一个因式。
C1
C2
Ci
Cn
F ( s) 

L
L
s  s1 s  s2
s  si
s  sn
Ci
F ( s)  
i 1 s  si
n
关键问题: 确定每个部分分式中的待定常数Ci。
The Laplace Transform
Introduction to Automatic Control
确定待定常数Ci
C i  lim( s  si )  F ( s )
s  si
或
B( s )
Ci  '
A ( s) s s
i
The Laplace Transform
Introduction to Automatic Control
代入Ci即可求得f(t):
Ci
L [ F ( s )]  f ( t )  L [
]
i 1 s  si
1
1
n
  Ci e
i 1
si t
n
The Laplace Transform
Introduction to Automatic Control
例3
求F ( s )的拉氏反变换
s2
F ( s)  2
s  4s  3
确定待定系数
所以
Introduction to Automatic Control
The Laplace Transform
要点：
明确四个步骤，语言表述清晰
The Laplace Transform
分子分母同阶的例子：
Introduction to Automatic Control
例4
所以
The Laplace Transform
分母有复数根的例子：
Introduction to Automatic Control
例5
Introduction to Automatic Control
The Laplace Transform
求Ci
所以
欧拉方程
Introduction to Automatic Control
The Laplace Transform
欧拉方程
Introduction to Automatic Control
The Laplace Transform
The Laplace Transform
2. A(s)=0有重根情况
Introduction to Automatic Control
设s1为m阶重根，sm 1 , sm  2 ,L, sn为单根。
则F ( s )可展成如下部分分式之和 :
Cm
C m 1
C1
F ( s) 

L
m
m 1
( s  s1 )
( s  s1 )
( s  s1 )
C m 1
Cn

L
( s  sm  1 )
( s  sn )
The Laplace Transform
确定待定常数Ci
Introduction to Automatic Control
m
Cm  lim
(
s

s
)
F
(
s
)
1
ss
1
d
m
Cm 1  lim
[( s  s1 ) F ( s )]
ss
ds
1
L
Cm j
1
dj
 lim j [( s  s1 )m F ( s )]
j! s  s ds
1
L
1
d ( m 1 )
C1 
lim m 1 [( s  s1 )m F ( s )]
( m  1)! s  s ds
1
The Laplace Transform
Introduction to Automatic Control
例6
求F ( s )的原函数。
s2
F ( s) 
s ( s  1) 2 ( s  3)
Introduction to Automatic Control
The Laplace Transform
The Laplace Transform
Introduction to Automatic Control
Solving linear differential
equations using the Laplace
transform
Linear
differential
equation
Laplace
Transform
Convolution
3
1
Laplace
transformed
Equation
Time domain
solution
2
Algebraic
manipulation
Laplace
Solution
Time
Domain
Inverse Laplace
Transform
Laplace
Domain
Introduction to Automatic Control
The Laplace Transform
1
2
3
The Laplace Transform
Introduction to Automatic Control
三个步骤:
① 代入初始条件对微分方程进行拉氏变换；
② 解变换方程(代数方程)，求出响应函数的拉
氏变换式；
③ 用部分分式法求拉氏反变换，得到微分方程
的解。
The Laplace Transform
例7
d 2 y( t )
dy ( t )

5
 6 y( t )  u( t )
2
dt
dt

Introduction to Automatic Control
设u( t )  6  1( t )，初始条件为y(0)  2, y(0)  2,
试用拉氏变换法求解该方程。
分析:
d 2 y( t )
dy(t )
5
 6 y(t )  u(t )
2
dt
dt
L[ y(t )]  Y ( s)
Introduction to Automatic Control
The Laplace Transform
d 2 y( t )
dy(t )
5
 6 y(t )  u(t )
2
dt
dt
dy(t )
L[
]  sY ( s )  y(0)
dt
Introduction to Automatic Control
The Laplace Transform
d 2 y( t )
dy(t )
5
 6 y(t )  u(t )
2
dt
dt
2
d y( t )
2
L[
]

s
Y ( s )  sy(0)  y (0)
2
dt
The Laplace Transform
Introduction to Automatic Control
2
d y( t )
dy(t )
5
 6 y(t )  u(t )
2
dt
dt
6
L[u(t )]  L[6  1(t )] 
s
The Laplace Transform
解:
Step1: For a ODE w/initial conditions,
apply Laplace transform to each term.
6
s Y ( s )  sy (0)  y (0)  5sY ( s )  5 y(0)  6Y ( s ) 
s
Introduction to Automatic Control
2
6
s Y ( s )  5sY ( s )  6Y ( s )  2s  12 
s
2
Step2: Solve for Y(s)
2 s 2  12 s  6
Y ( s) 
s( s 2  5 s  6)
The Laplace Transform
Introduction to Automatic Control
Step3: Apply partial fraction
expansion to obtain inverse Laplace
transform. 1
4
5
Y ( s)  

s ss s2
y(t )  1  4e  5e
3 t
2 t
(t  0)
The Laplace Transform
例8
Introduction to Automatic Control
d2y
dy

6
 8 y  2 y(0)  y' (0)  0
2
dt
dt
s Y (s)  6s Y (s)  8Y (s)  2 / s
2
2
Y ( s) 
s (s  2) (s  4)
1
1
1
Y ( s) 


4s 2 (s  2) 4 (s  4)
1 e 2t e 4t
y (t )  

4 2
4





ODE w/initial conditions
Apply Laplace transform
to each term
Solve for Y(s)
Apply partial fraction
expansion
Apply inverse Laplace
transform to each term
The Laplace Transform
Introduction to Automatic Control
OVERVIEW
• This module is a mathematical section to establish
a base for the theory of control systems.
• This is a tool and it is indispensable as most of
linear system dynamics are described in a mapped
space that can only be understood when the main
theorems of the Laplace transform are known.
• Special focus is put on the solution of differential
equations using the Laplace transform.
The Laplace Transform
Introduction to Automatic Control
Homework
• 见WORD文件：拉氏变换作业