introduction to automatic control the laplace transform li huifeng tel:82339276...

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The Laplace Transform


Li Huifeng

Tel:82339276

Email:[email protected]

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Module objectives

• 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

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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

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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.

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Linear differential equation

Time domain solution

Laplace Solution

Laplace transformed Equation

Convolution

Time Domain

Laplace Domain

Laplace Transform

Inverse Laplace Transform

Algebraic manipulation

t

o

t

o

dftfdtfftftf )()()()()()( 212121

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Definition函数 f(t)， t 为实变量，如果线性积分

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：

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。

为复变量）jwsdtetf st

()(0

0

)()()]([ dtetfsFtfL st

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Correspondences of the Laplace transform

• Table 2.1: Corresponding elements of the Laplace transform or Textbook Page 21

)]([)(

)]([)(

)()(

1 sFLtf

tfLsF

sFtf

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几种典型函数的拉氏变换

t

1. 1(t)单位阶跃函数

0t0

0t1)t(1)t(f

其拉氏变换为：

s

1dte1)s(F)]t(f[L

0

st

f(t)

0

其数学表达式为：

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t

2. 单位斜坡函数

0t

0t

0

t)t(1t)t(f


20

st

s

1dtet)s(F)]t(f[L

f(t)

0


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3. 等加速函数

0t

0t

0

t2

1)t(f

2


30

st2

s

1dtet

2

1)s(F)]t(f[L


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4. 指数函数

0t

0t

0

e)t(f

t


s

1dtee)s(F)]t(f[L

0

stt


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5. 正弦函数

0t

0t

0

wtsin)t(f


220

st

ws

wdtewtsin)s(F)]t(f[L


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6. 余弦函数

0t0

0twtcos)t(f


220

st

ws

sdtewtcos)s(F)]t(f[L


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7. 单位脉冲函数

0t

0t0)t()t(f



1)s(F)]t(f[L

1dt)t(定义：

t

)(t

0

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)(

1)(sincos)(

jssFtjtetf tj

)(

1)(sincos)(

jssFtjtetf tj

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Euler equation

2cos

tjtj eet

j

eet

tjtj

2sin

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)(cos

22

s

st

)(sin

22

s

t

Equating real & imaginary parts yields:

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))((cos

22

s

ste t

))((sin

22

ste t

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典型函数的拉氏变换形式

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Main theorems of the Laplace transform

① Superposition theorem( 叠加定理 ):

)()()}()({ 22112211 sFasFatfatfaL

各函数和的拉氏变换＝各函数拉氏变换的和

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② Similarity theorem: ( 比例尺改变 )

0)(1

)}({ aa

sF

aatfL

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③ Real Shifting theorem( 延时定理 ):

0)()}({ asFeatfL as

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提示： f(t) 相当于 t· 1(t) 在时间上延迟了一个值。

)(tf

t0

)(1 tt )(1)( tt

的拉氏变换。求 )t(1)t()t(f

)](1)[()( ttLsF

ses

2

1

解:

例 1

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④ Complex Shifting theorem( 衰减定理） :

0)()}({ aasFtfeL at

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的拉氏变换。求 wtsine)t(f at

]sin[)( wteLsF at

22)( was

w

解:

例 2

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⑤ Derivative theorem:

)0(f)0(fs)0(fs)s(Fs]dt

)t(fd[L

)0(f)0(sf)s(Fs]dt

)t(fd[L

)0(f)s(sF]dt

)t(df[L

)1n('2n1nn

n

n

'2

2

2

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。

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⑥ Complex differentiation theorem: ( 不要求掌握 )

k

kkk

ds

sFdtftL

)()1()}({

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⑦ Integral theorem:

)0(fs

1)0(f

s

1)s(F

s

1]dt)t(f[L

)0(fs

1)0(f

s

1)s(F

s

1]dt)t(f[L

)0(fs

1)s(F

s

1]dt)t(f[L

)n()1(

nn

n

)2()1(

22

2

)1(

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。

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⑧ Initial value theorems:

)0()(lim)(lim0

ftfssFts

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Given:

22 5)1(

)2()(

s

ssF

Find f(0)

1)26(2

2lim

2512

2lim

5)1(

)2(lim)(lim)0(

2222

222

2

2

22

sssss

ssss

ss

ss

s

ssssFf

ss s

s

解:

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⑨ final value theorems:

)()(lim)(lim0

ftfssFts

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Given:

ttesFnotes

ssF t 3cos)(

3)2(

3)2()( 21

22

22

Find )(f .

03)2(

3)2(lim)(lim)(

22

22

s

ssssFf

0s0s

解:

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⑩ Convolution in the time domain: ( 不要求掌握 )

t

o

t

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dftfdtfftftf )()()()()()( 212121

)()()}()({ 2121 sFsFtftfL

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⑪ Convolution in the frequency domain: ( 不要求掌握 )

jc

jc

dppsFpFj

tftfL )()(2

1)}()({ 2121

拉氏变换性质的证明

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The inverse Laplace transform

j

j

ts1 dte)s(Fj2

1)t(f)]s(F[L

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。

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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) 。

)}({)}({)}({)}({

)()()()(

12

11

11

21

sFLsFLsFLsFL

sFsFsFsF

n

n

)()()()( 21 tftftftf n

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。为正数，且均为实数，及式中，

的一般形式为：

nmn,m

b,b,ba,a,a

asasas

bsbsbsb

)s(A

)s(B)s(F

)s(F

m10n21

n1n

1n

1

n

m1m

1m

1

m

0

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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

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1. A(s)=0 无重根情况

n

n

i

i

ss

C

ss

C

ss

C

ss

CsF

2

2

1

1)(

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。

n

i i

i

ss

CsF

1

)(

关键问题 : 确定每个部分分式中的待定常数 Ci 。

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确定待定常数 Ci

)()(lim sFssC issii

iss

i sA

sBC

)(

)('

或

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n

i i

i

ss

CLtfsFL

1

11 ][)()]([

n

i

ts

iieC

1

代入 Ci 即可求得 f(t):

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34

2)(

)(

2

ss

ssF

sF 的拉氏反变换求

确定待定系数

所以

例 3

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要点：

明确四个步骤，语言表述清晰

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分子分母同阶的例子：

所以

例 4

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分母有复数根的例子：

例 5

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求 Ci

所以欧拉方程

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欧拉方程

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2. A(s)=0 有重根情况

:)(

,,, 211

和可展成如下部分分式之则为单根。阶重根，为设

sF

sssms nmm

)()(

)()()()(

1

1

1

1

1

1

1

1

n

n

m

m

m

m

m

m

ss

C

ss

C

ss

C

ss

C

ss

CsF

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确定待定常数 Ci

)()(lim 11

sFssC m

ssm

)]()[(lim 111

sFssds

dC m

ssm

)]()[(lim

!

11

1

sFssds

d

jC m

j

j

ssjm

)]()[(lim)!1(

111

)1(

11

sFssds

d

mC m

m

m

ss

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)3()1(

2)(

)(

2

sss

ssF

sF 的原函数。求例 6

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Solving linear differential equations using the Laplace

transform

Linear differential equation

Time domain solution

Laplace Solution

Laplace transformed Equation

Convolution

Time Domain

Laplace Domain

Laplace Transform

Inverse Laplace Transform

Algebraic manipulation

1

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1

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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。

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方程。试用拉氏变换法求解该，初始条件为设 ,2)0(,2)0()(16)(

)()(6)(

5)(

2

2

yyttu

tutydt

tdy

dt

tyd

分析 :

)()(6)(

5)(

2

2

tutydt

tdy

dt

tyd

)()]([ sYtyL

例 7

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)()(6)(

5)(

2

2

tutydt

tdy

dt

tyd

)0()(])(

[ yssYdt

tdyL

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)()(6)(

5)(

2

2

tutydt

tdy

dt

tyd

)0()0()(])(

[ 2

2

2

ysysYsdt

tydL

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)()(6)(

5)(

2

2

tutydt

tdy

dt

tyd

stLtuL

6)](16[)]([

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解:

Step1: For a ODE w/initial conditions, apply Laplace transform to each term.

ssYyssYysysYs

6)(6)0(5)(5)0()0()(2

sssYssYsYs

6122)(6)(5)(2

Step2: Solve for Y(s)

)65(

6122)(

2

2

sss

sssY

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Step3: Apply partial fraction expansion to obtain inverse Laplace transform.

2

541)(

sssssY

)0(541)( 23 teety tt

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0)0(')0(2862

2

yyydt

dy

dt

yd 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

ssYsYssYs /2)(8)(6)(2

)4()2(

2)(

ssssY

)4(4

1

)2(2

1

4

1)(

sss

sY

424

1)(

42 tt eety

例 8

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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.

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Homework

• 见WORD 文件：拉氏变换作业

introduction to automatic control the laplace transform li huifeng tel:82339276...

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