Chapter 9 Laplace Transform §9.1 Definition of Laplace Transform §9.2 Properties of Laplace...

Chapter 9 Laplace Transform

§9.1 Definition of Laplace Transform §9.2 Properties of Laplace Transform §9.3 Convolution §9.4 Inverse Laplace Transform §9.5 Application of Laplace Transform

§9.1 Definition of Laplace Transform

Definition

For ( ),t 0,if ( ) converge for certain

(complexvariable), then ( ) ( ) is called

f t f t e dt

S F s f t e dt

the Laplace Transform of ( ),denoted as ( ) [ ( )],

( )is called the inverse Laplace Transform of ( ),denoted

as ( ) [ ( )].

f t F s L f t

f t F s

f t L F s

Unit Step: 1, 0,( )

The properties of ( ) are as followings.t

Unit Impulse

( ) is smooth function.f t

( ) 1 ( ) ( ) (0)t dt t f t dt f

0 0( ) ( ) ( )t f t t dt f t

0 0( ) ( ) ( )t t f t dt f t

0, 0,( ) and ( ) 1

tt t dt

( ) ( ) ( )

( ) ( ) (0)

( ) ( ) ( 1) (0)

δ t f t dt f ,

t f t dt f

Ex:0 0

(1). ; (2). ; (3). sin1 0

kt, tu(t) f(t) e f(t) kt

1(1).L[ ( )]

let then (cos sin )

when Re(s) 0, 0 lim 0

1L[ ( )] .

st st st

st t ti t

αt st

u t u(t)e dt e dt es

s α βi, e e e t i t

α e (t ) e

1(2). [ ] , (Re( ) ) Re( - ) 0kt kt st (s-k)tL e e e dt e dt s k s k

(3). [sin ] sin (sin cos )0

( Re(s) 0).

stst e

L kt kte dt kt k kts k

TH.9.1.1 Existence Theorem

Support that

1.f(t) continous or piecewise continous on every finite

interval [0,b];

2.there are positive constants 0, ,such that

,(0 ),then the Laplace Transformc t

f t Me t

of ( ) ,i.e. ( ) ( ) exists for Re( ) ,

( ) absolutely converges and uniformly

converges for Re( ) .And ( )analytic on Re( ) .

f t F s f t e dt s c

f t e dt

s c c F s s c

The conditions in the theorem are sufficient, not necessary.

1. ( ) , ( ) ( 0),Ex f t f t t

1( )1 2however, , (Re( ) 0)

, ( function)

s te dt st s

Ex:9.1.5 ( ) [ ( )]t L t -0 0

[ ( )] ( ) ( ) ( )

st st st

L t t e dt t e dt t e dt

( ) ( ) ( 0), ( )is continuous or piecewise continuous

1on a period,prove [ ( )] ( ) Re( ) 0 .

T ss T

f t T f t t f t

L f t f t e dt se

- - -2 ( 1)

( 1) ( 1)- - - ( )

- - - -

[ ( )] ( )

( ) ( ) ( )

[ ( ) ] ( ) - ( )

( ) ( )

st st stT T k T

k T k T Tst st s u kT

kT kTk

T TskT su skT st

L f t f t e dt

f t e dt f t e dt f t e dt

f t e dt f t e dt t kT u f u kT e du

e f u e du e f t e

dt- - -

0 00 0

[ ( )] [ ( ) ] ( )

1( ) (Re( ) 0).

T TskT st st skT

T stsT

L f t e f t e dt f t e dt e

f t e dt se

Ex:9.1.6

§9.2 Properties of Laplace Transform1.Linearity

1 1 2 2 1 1 2 2

11 1 2 2 1 1 2 2

( ) ( ) ( 1,2), then

( ) ( ) ( ) ( ),

( ) ( ) ( ) ( ).

i iL f t F s i

L a f t a f t a F s a F s

L b F s b F s b f t b f t

1( ) [sh ] [ ] [ [ ] [ ]]

2 21 1 1

[ ] , (Re( ) ).2

kt ktkt kte e

F s L kt L L e L e

s k s k s k

Ex.9.2.1

Similar:

[ch ] , (Re( ) ).s

L kt s ks k

Let [ ( )] ( ), Re(s) C , then [ ( )] ( ) (0).L f t F s L f' t sF s f

Pf: [ ( )] '( ) ( ) | ( )

(0) ( ) ( ) (0).

s t s t s t

L f' t f t e dt f t e s f t e dt

f s f t e dt sF s f

( ) 1 2 ( 1)

Corollary:

Let [ ( )] ( ), then

( ) ( ) (0) (0) (0)

1,2, Re( )

n n n n n

L f t F s

L f t s F s s f s f f

2.Derivation

Special case: 0 0 0 0, then

( ) ( ) ( ).

L f t s L f t s F s

Ex.9.2.2 ( ) cos ( ).f t kt F s

2 2 2 2

1 1(1). [cos ] [ (sin ) '] [(sin ) ']

1{ [sin ] sin | }

1, Re( ) 0 .

L kt L kt L ktk k

sL kt ktk

s k ss

k s k s k

(2). ( ) cos , ''( ) cos

[ cos ] [ ''( )]

[cos ] [ ( )] (0) '(0)

[cos ] .

[cos ] , Re( ) 0

f t kt f t k kt

L k kt L f t

k L kt s L f t sf f

s L kt s

sL kt s

Let [ ( )] ( ), then ( ) [ ( )], Re( ) .L f t F s F s L tf t s c

Pf: ( ) [ ( ) ] ' ( )( )

[ ( )( )]

[ ( )( )], Re( ) .

s t s ts

F s f t e dt f t t e dt

f t t e dt

L f t t s c

In general:

( ) [( ) ( )], Re( ) .n nF s L t f t s c

Ex. ( ) sh ( )f t t kt F s

2 2 2 2 2

Solution: [sh ] , Re( ) .

2( sh ) [ ] , Re( ) .

kL kt s k

s kd k ks

L t kt s kds s k s k

Similar:

( sh ) , Re( ) .( )

s kL t kt s k

3.Integration

( )Let ( ) ( ), then ( ) .

t F sL f t F s L f t dt

Pf: Let ( ) ( )d , '( ) ( ),

[ '( )] [ ( )] [ ( )d ]

( )( ) .

g t f t t g t f t

L g t sL g t sL f t t

F sL f t dt

0 0 0{ }

In general:

1d d ( )d ( )

nL t t f t t F s

( )Let ( ) ( ), then ( )

or ( ) [ ( ) ].

f tL f t F s F s ds L

f t tL F s ds

Pf: ( ) { ( ) }

1( ){ } ( )

( ) ( )

( )( ) , Re( ) .

F s ds f t e dt ds

f t e ds dt f t e dtt

f t f te dt L

f tL F s ds s c

In general: ( ) .n s s s

f tL ds ds F s ds

Ex.9.2.4sin

( ) ( )t

f t F st

1Solution: [sin ] ,

1sin 1

[ ] arctan cot .1 2s

tL ds s arc s

Ex. ( ) sin ( )f t t F s

( ) sin cos ,

1 1[sin ] [ cos ] [cos ] .

f t t tdt

L t L tdt L ts s

Homework

P217:2.(1)(3)(5) 3 4 5(1)(2)(3)(4)

f(t) f(t)

4.Delay

( ) ( ) 0, 0 ,

( ) ( ) ( ),( 0)s s

L f t F s t f t

L f t e L f t e F s

Ex: 1[ ( )] [ ( )] .s sL u t e L u t e

Ex:9.2.8 ( ) [ ( )]f t L f t

( ) [ ( ) ( - ) ( - 2 ) ] [ ( )]

1 1 1[ ( )] ( )

f t A u t u t u t L f t

L f t A e es s s

- -2 2 2 2

when Re( ) 0, then | | 1

1( ( ))

( - ) ( )

(1 coth ) (Re( )2 2

s s s s

AL f t

A e e e e

We can get ( ) ( ) ( ),

( ( )) [ ( )] [ ( )]

2 2[sin ( )] [sin ( - ) ( - )]

(1 ).2

f t f t f t

L f t L f t L f t

T TEL t u t EL t u t

Ex:9.2.9

5.Displacement

( ) ( ) Re( ) , thenL f t F s s c

( ), Re( ) ,

L e f t F s s c

F s e f t

sin ( )tf t e kt F s

[sin ] ,

[e sin ]( )

L kts a k

6.Initial ＆ Terminal Value Theorems

(1).Initial Value Theorem

Let [ ( )] ( ), lim ( ) exists,

lim ( ) lim ( ), (0) lim ( ).s

L f t F s sF s

f t sF s or f sF s

Re( ) Re( )

Pf: [ '( )] ( ) - (0)and lim ( ) exists,

lim ( ) lim ( )

lim [ '( )] lim [ ( ) - (0)]

lim ( ) (0).

L f t sF s f sF s

sF s sF s

L f t sF s f

sF s f

0Re( ) Re( )

0 Re( )

lim [ '( )] lim '( )

lim '( ) 0

lim ( ) (0) lim ( ).

L f t f t e dt

f t e dt

sF s f f t

(2).Terminal Value Theorem

Let [ ( )] ( ),all singularities of ( ) lie in the left-half plane,

lim ( ) lim ( ), ( ) lim ( ).t s s

L f t F s sF s

f t sF s or f sF s

Pf: [ '( )] ( ) - (0)

lim [ '( )] lim[ ( ) - (0)] lim ( ) (0).s s s

L f t sF s f

L f t sF s f sF s f

lim [ '( )] lim '( )

lim '( ) '( )

lim ( ) (0).

lim ( ) lim ( ) ( ) lim ( ).

L f t f t e dt

f t e dt f t dt

f t sF s or f sF s

Ex:9.2.112

1[ ( )] (0) ( ).

( 1) 4

sL f t f and f

( 1)(0) lim ( ) lim 1,

( 1) 4

( 1)( ) lim ( ) lim 0.

( 1) 4

s sf sF s

Satisfying the conditions of the theorem, then you can use the theorem.

1( )= , the singularities are , not satisfying the conditions,

lim ( ) lim 0,but ( ) lim ( ) lim sin not exists.1s s t t

F s s js

ssF s f f t t

Ex:9.2.12 -4( ) cos5 ( )tf t te t F s 2

'2 2 2 2

- 25[cos 5 ] [ cos5 ] -( )

25 25 ( 25)

( 4) - 25( ) [ cos5 ]

[( 4) 25]t

s s sL t L t t

sF s L te t

-3 sin 2

( ) ( ).te t

f t F st

2[sin 2 ]

4sin 2 2

[ ] arctan | - arctan4 2 2 2

sin 2 3cot ( ) [ ] cot

t s sL ds

s e t sarc F s L arc

Ex:9.2.14

Table for properties on P201

§9.3 Convolution

1.Definition

1 2 1 20Let ( ) ( ) 0, ( 0), ( ) ( )d

tf t f t t f f t

1 2( ) ( )f t and f t 1 2( ) ( )f t f t

1 2 1 20( ) ( ) ( ) ( )d

tf t f t f f t

1 2 1 2

1 2 1 20

( ) ( ) ( ) ( )d

( ) ( )d ( ) ( )d

( ) ( )d .

f t f t f f t

f f t f f t

1 2( ( ) ( ) 0, 0)f t f t t

is called the convolution of ,denoted as , i.e. .

Note: Convolution in Fourier transform is same to that in Laplace transform.

Properties:

1.Commutative Law

2.Associative Law

3.Distributive Law

1 2 2 1( ) ( ) ( ) ( )f t f t f t f t

1 2 3 1 2 3( ) [ ( ) ( )] [ ( ) ( )] ( )f t f t f t f t f t f t

1 2 3 1 2 1 3( ) [ ( ) ( )] ( ) ( ) ( ) ( )f t f t f t f t f t f t f t

1 2 1 2( ) ( ) ( ) ( )f t f t f t f t

Ex: eatt

Solution:

e e d e e d

1 ee de e e d

t tat a t at a

att ttat a a a

e 1e e

e 1e (e 1)

1(e 1)

attat a

atat at

2.Convolution TheoremTH.9.3.1

1 2 1 1 2 2

1 2 1 2 1 2

Let ( ) ( ) 0, ( 0),and [ ] , [ ]

then [ ] [ ] [ ] ( ) ( ),

f t f t t L f t F s L f t F s

L f t f t L f t L f t F s F s

11 2 1 2 1 2 1 2or [ ( ) ( )] . , . . ( ) ( ).

F s F s f t f t i e f t f t F s F s L

1 2 1 20 0 0

Pf: [ ]

[ ( ) ( )d ] d ( )[ ( ) d ]d

t st st

L f t f t

f t f t e t

f f t e t f f t e t

1 20 0

1 2 1 2 1 20 0

( )[ ( ) d ]d

( ) ( )d ( ) d ( ) ( ) ( ).

u tsu s

f f u e e u

f e F s f e F s F s F s

Ex.9.3.2 12 2

1( ) ( ) ( )

( 1)F s f t F F s

1 1 12 2

Solution:

1 1(1). ( ) sin

1 1(2). ( ) sin

sin( ) cos( )

cos( ) cos( )

sin( ) sin

F F s L L t ts s

F F s t ts s

t d d t

t t s ds

t t s t t

Homework:

P217:5.(5)-(13) 7.(1)(3)(5) 8

§9.4 Inverse Laplace Transform1.Inverse Integral FormulaFrom the inverse Fourier transform, we have the inverse Laplace transform formula.

[ ( ) ( ) ] ( ) ( )

( ) ( )

t t j t

j t st

F f t u t e f t u t e e dt

f t e dt s j f t e dt F s

j ( j )

1( ) ( ) e ( ) ( ) e e d e d

e d ( )e d21

( j ) e d , ( Re( ) , 0).2

f t u t f u

F s c t

( j )1 1

( ) ( j ) e d , 0,Let , ,2

1then ( ) ( ) , ( Re( ) , 0).

f t F t j s d dsj

f t F s e ds s c tj

RO Real axis

Imaginary axis

singularities

TH.9.4.1

1Let ( )has only finite number of singularities , , ,

(lie in the left side of Re( ) ) and lim ( ) 0,thenn

F s s s

1( ) ( ) Re ( ) , , ( 0).

nj st stkj

f t F s e ds s F s e s tj

2.Evaluation(1).Using integral formula

1( ) ( ).

( 1)F s f t

0 is pole of order 1,and 1is pole of order 2,s s

( ) Re ,0 Re ,1

1 d 1e lim e

( 1) d

11 lim e e

1 ( e e ) 1 e ( 1) ( 0).

f t s F s e s F s e

(2).Using convolution theorem

11 2 1 2[ ( ) ( )]L F s F s f t f t

1( ) ( ) .

( 2 5)F s L F s

1 12 2 2 2

1 1 12 2 2 2

[( 1) 2 ]

1 1 2 1sin 2

( 1) 2 2 2 2

1 1( )

( 1) 2 ( 1) 2

1 1 1sin 2 sin 2 ( sin 2 )( sin 2( )

2 2 41

(cos(4 2 ) cos 2 )8

tt t t

L e L e ts s

L F s L Ls s

e t e t e e t d

1sin 2 2 cos 2 .

t td e t t t

(3).Using partial fraction

1( ) ( ).

( 1)F s f t

1 1 1 1( )

( 1) 1

1 e ( 0).t

F ss s s s s

f t Ls s

-1 -1 -12

1 -1 1=

s s s+1

(4).Using properties1 1 1

1 2 1 2

1. [ ( ) ( )] [ ( )] [ ( )].

2 . [ ( )] [ ( )].

3 . [ ( )] ( ).

4 . [ '( )] ( ).

( )5 . [ ( ) ] .

L F s F s L F s L F s

L F s e L F s

L e F s f t

L F s tf t

f tL F s ds

1( ) ( ) .

sF s L F s

1 12 2

1 1( ) [ ] [ ]

2 2 ( 1) 1

[ ] cos .1

s sf t L L

se L e t

(5).Using L-transform table

§9.5 Application of Laplace Transform

(1). ( ) ( )

( ) (0), ( ) ( ).

f t e dt F s

f t dt F f t e dt F s

3Ex: cos 2 | .

ste dt

( ) ( )(2). ( ) ( ) ,Let 0, ( ) ,

( ), ( ) .

f t f tL F s ds s dt F s ds

t tf t

s s e dt F s dst

1.Evaluating the improper integral

sin 1Ex: arctan | .

1 4te dt ds s

0( ) ( )00 0

3 . ( ) ( 1) (0). ( ) ( 1) ( ).s tn n n n n nt f t dt F t f t e dt F s

1 1Ex: sin [ ]'| .

st t e dts

Using Laplace transform solves the differential equation:

The block diagram shows the details.

Solution of

Differential equation

Algebra equation of

2.Solving Differential Equation

( )x t

( )X s

Ex:( ) 2 ( ) 2 ( ) 2 cos

Solve(0) (0) 0

tx t x t x t e t

Let ( ) ( ) ,differential equation equation of ( )

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

( 1) 1

2( 1) 2( 1)( ) 2 ( ) 2 ( ) ( )

( 1) 1 ( 1) 1

2( 1)( )

( 1) 1

X s L x t X s

ss X s sx x sX s x X s

s ss X s sX s X s X s

sx t L

1 12 2

1 1sin

e L te L te ts s

( ) 2 ( ) 2 ( ) 10

Solve 2 ( ) ( ) 3 ( ) 13 .

(0) 1, (0) 3

x t x t y t e

x t y t y t e

Let ( ) ( ) , ( ) ( )

differential equation equation of ( ) and ( ).

10( ) 1 2 ( ) 2 ( )

2 ( ) ( ) 3 3 ( )2

( )23 ( ) 3( )

X s L x t Y s L y t

X s Y s

sX s X s Y ss

X s sY s Y ss

X sx t es

y t eY ss

Homework:

P218: 9.(1)(3)(5) 10.(1)(3) 11.(1)(3)

1. The properties of Laplace Transform.2. Application in solving differential equations.

The key points and difficulties of the chapter.

Chapter 9 Laplace Transform §9.1 Definition of Laplace Transform §9.2 Properties of Laplace...

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