Chapter 9 Laplace Transform §9.1 Definition of Laplace Transform §9.2 Properties of Laplace...
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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
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Transcript of 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
0
0
For ( ),t 0,if ( ) converge for certain
(complexvariable), then ( ) ( ) is called
s t
s t
f t f t e dt
S F s f t e dt
1
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,( )
0, 0.
tu t
t
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
, 0.
tt t dt
t
( ) ( ) ( )
( ) ( ) (0)
( ) ( ) ( 1) (0)
' '
n n n
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
, t
0 0
-
1(1).L[ ( )]
0
let then (cos sin )
when Re(s) 0, 0 lim 0
1L[ ( )] .
st st st
st t ti t
αt st
t
u t u(t)e dt e dt es
s α βi, e e e t i t
α e (t ) e
u ts
0 0
1(2). [ ] , (Re( ) ) Re( - ) 0kt kt st (s-k)tL e e e dt e dt s k s k
s k
2 20
2 2
(3). [sin ] sin (sin cos )0
( Re(s) 0).
stst e
L kt kte dt kt k kts k
k ,
s 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
M c R
f t Me t
0
0
1
of ( ) ,i.e. ( ) ( ) exists for Re( ) ,
( ) absolutely converges and uniformly
converges for Re( ) .And ( )analytic on Re( ) .
s t
s t
f t F s f t e dt s c
f t e dt
s c c F s s c
Note:
The conditions in the theorem are sufficient, not necessary.
1. ( ) , ( ) ( 0),Ex f t f t t
t
10 2
12
1( )1 2however, , (Re( ) 0)
, ( function)
s te dt st s
s
Ex:9.1.5 ( ) [ ( )]t L t -0 0
0
[ ( )] ( ) ( ) ( )
1
st st st
stt
L t t e dt t e dt t e dt
e
t 0
( ) ( ) ( 0), ( )is continuous or piecewise continuous
1on a period,prove [ ( )] ( ) Re( ) 0 .
1
T ss T
f t T f t t f t
L f t f t e dt se
-
0
- - -2 ( 1)
0
( 1) ( 1)- - - ( )
00
- - - -
0 0
[ ( )] ( )
( ) ( ) ( )
[ ( ) ] ( ) - ( )
( ) ( )
st
st st stT T k T
T kT
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
-- 0
[ ( )] [ ( ) ] ( )
1( ) (Re( ) 0).
1
T TskT st st skT
k k
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
2 2
1( ) [sh ] [ ] [ [ ] [ ]]
2 21 1 1
[ ] , (Re( ) ).2
kt ktkt kte e
F s L kt L L e L e
ks k
s k s k s k
Ex.9.2.1
2 2
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
00 0
0
Pf: [ ( )] '( ) ( ) | ( )
(0) ( ) ( ) (0).
s t 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
n s c
2.Derivation
1
( )
Special case: 0 0 0 0, then
( ) ( ) ( ).
n
n n n
f f f
L f t s L f t s F s
Ex.9.2.2 ( ) cos ( ).f t kt F s
0
2 2 2 2
1 1(1). [cos ] [ (sin ) '] [(sin ) ']
1{ [sin ] sin | }
1, Re( ) 0 .
t
L kt L kt L ktk k
sL kt ktk
s k ss
k s k s k
2
2
2 2
2
2 2
(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
s k
Let [ ( )] ( ), then ( ) [ ( )], Re( ) .L f t F s F s L tf t s c
0 0
0
Pf: ( ) [ ( ) ] ' ( )( )
[ ( )( )]
[ ( )( )], Re( ) .
s t s ts
s t
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 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
2 2
2 2 2
Similar:
( sh ) , Re( ) .( )
s kL t kt s k
s k
3.Integration
0
( )Let ( ) ( ), then ( ) .
t F sL f t F s L f t dt
s
0
0
0
Pf: Let ( ) ( )d , '( ) ( ),
[ '( )] [ ( )] [ ( )d ]
( )( ) .
t
t
t
g t f t t g t f t
L g t sL g t sL f t t
F sL f t dt
s
0 0 0{ }
In general:
1d d ( )d ( )
t t t
nL t t f t t F s
sn
1
( )Let ( ) ( ), then ( )
or ( ) [ ( ) ].
s
s
f tL f t F s F s ds L
t
f t tL F s ds
0
0 0
0
Pf: ( ) { ( ) }
1( ){ } ( )
( ) ( )
( )( ) , Re( ) .
st
s s
st st
ss
st
s
F s ds f t e dt ds
f t e ds dt f t e dtt
f t f te dt L
t t
f tL F s ds s c
t
( )
In general: ( ) .n s s s
n
f tL ds ds F s ds
t
Ex.9.2.4sin
( ) ( )t
f t F st
2
2
1Solution: [sin ] ,
1sin 1
[ ] arctan cot .1 2s
L ts
tL ds s arc s
t s
Ex. ( ) sin ( )f t t F s
0
20
( ) sin cos ,
1 1[sin ] [ cos ] [cos ] .
1
t
t
f t t tdt
L t L tdt L ts s
Homework
P217:2.(1)(3)(5) 3 4 5(1)(2)(3)(4)
O t
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
s
1
u(t)
tO
Ex:9.2.8 ( ) [ ( )]f t L f t
- -2
- -2
( ) [ ( ) ( - ) ( - 2 ) ] [ ( )]
1 1 1[ ( )] ( )
(1 ).
s s
s s
f t A u t u t u t L f t
L f t A e es s s
Ae e
s
-
-
- -2 2 2 2
-2 2
when Re( ) 0, then | | 1
1( ( ))
1-
( - ) ( )
2-
(1 coth ) (Re( )2 2
s
s
s s s s
s s
s e
AL f t
s e
A e e e e
se e
A ss
s
1 2
1 2
-2
2 2
We can get ( ) ( ) ( ),
( ( )) [ ( )] [ ( )]
2 2[sin ( )] [sin ( - ) ( - )]
2 22
(1 ).2
( )
Ts
f t f t f t
L f t L f t L f t
T TEL t u t EL t u t
T T
ET e
sT
Ex:9.2.9
5.Displacement
( ) ( ) Re( ) , thenL f t F s s c
1
( ), Re( ) ,
( )
t
t
L e f t F s s c
F s e f t
L
sin ( )tf t e kt F s
2 2
2 2
[sin ] ,
[e sin ]( )
at
kL kt
s kk
L kts a k
Ex:
6.Initial ＆ Terminal Value Theorems
(1).Initial Value Theorem
0
Let [ ( )] ( ), lim ( ) exists,
lim ( ) lim ( ), (0) lim ( ).s
t s s
L f t F s sF s
f t sF s or f sF s
Re( )
Re( ) Re( )
Pf: [ '( )] ( ) - (0)and lim ( ) exists,
lim ( ) lim ( )
lim [ '( )] lim [ ( ) - (0)]
lim ( ) (0).
s
s s
s s
s
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( )
0
lim [ '( )] lim '( )
lim '( ) 0
lim ( ) (0) lim ( ).
st
s s
st
s
s t
L f t f t e dt
f t e dt
sF s f f t
(2).Terminal Value Theorem
0 0
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
0 0 0
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
00 0
0 00
0 0
lim [ '( )] lim '( )
lim '( ) '( )
lim ( ) (0).
lim ( ) lim ( ) ( ) lim ( ).
st
s s
st
s
t
t s s
L f t f t e dt
f t e dt f t dt
f t f
f t sF s or f sF s
Ex:9.2.112
1[ ( )] (0) ( ).
( 1) 4
sL f t f and f
s
2
20 0
( 1)(0) lim ( ) lim 1,
( 1) 4
( 1)( ) lim ( ) lim 0.
( 1) 4
s s
s s
s sf sF s
s
s sf sF s
s
Satisfying the conditions of the theorem, then you can use the theorem.
2
20 0
1( )= , the singularities are , not satisfying the conditions,
1
lim ( ) lim 0,but ( ) lim ( ) lim sin not exists.1s s t t
F s s js
ssF s f f t t
s
Ex:9.2.12 -4( ) cos5 ( )tf t te t F s 2
'2 2 2 2
2-4
2 2
- 25[cos 5 ] [ cos5 ] -( )
25 25 ( 25)
( 4) - 25( ) [ cos5 ]
[( 4) 25]t
s s sL t L t t
s s s
sF s L te t
s
-3 sin 2
( ) ( ).te t
f t F st
2
20
-3
2[sin 2 ]
4sin 2 2
[ ] arctan | - arctan4 2 2 2
sin 2 3cot ( ) [ ] cot
2 2
s
t
L ts
t s sL ds
t s
s e t sarc F s L arc
t
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
0
1 2 1 2
1 2 1 20
1 20
( ) ( ) ( ) ( )d
( ) ( )d ( ) ( )d
( ) ( )d .
t
t
t
f t f t f 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
4.
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
( )
0 0
00 0
Solution:
e e d e e d
1 ee de e e d
t tat a t at a
att ttat a a a
t
a a
0
2
e 1e e
e 1e (e 1)
1(e 1)
attat a
atat at
at
ta a
ta a
t
a a
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 [ ( ) ( )] . , . . ( ) ( ).
L
F s F s f t f t i e f t f t F s F s L
1 2
1 20
1 2 1 20 0 0
Pf: [ ]
[ ] d
[ ( ) ( )d ] d ( )[ ( ) d ]d
st
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
s 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
s s
1 1 12 2
1 1 12 2
0 0
0 0
0
Solution:
1 1(1). ( ) sin
1
1 1(2). ( ) sin
1
sin( ) cos( )
cos( ) cos( )
sin( ) sin
t t
tt
t
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
L L
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.
( )
0 0
[ ( ) ( ) ] ( ) ( )
( ) ( )
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
j ( j )
0
j
1( ) ( ) e ( ) ( ) e e d e d
21
e d ( )e d21
( j ) e d , ( Re( ) , 0).2
t t
t
t
f t u t f u
f
F s c t
( j )1 1
( ) ( j ) e d , 0,Let , ,2
1then ( ) ( ) , ( Re( ) , 0).
2
t
j st
j
f t F t j s d dsj
f t F s e ds s c tj
RO Real axis
Imaginary axis
LCR
+jR
jR
singularities
analy
TH.9.4.1
1Let ( )has only finite number of singularities , , ,
(lie in the left side of Re( ) ) and lim ( ) 0,thenn
s
F s s s
s F s
，
1
1( ) ( ) Re ( ) , , ( 0).
2
nj st stkj
k
f t F s e ds s F s e s tj
2.Evaluation(1).Using integral formula
2
1( ) ( ).
( 1)F s f t
s s
0 is pole of order 1,and 1is pole of order 2,s s
2 10
21
( ) Re ,0 Re ,1
1 d 1e lim e
( 1) d
11 lim e e
1 ( e e ) 1 e ( 1) ( 0).
st st
st st
ss
st st
s
t t t
f t s F s e s F s e
s s s
t
s s
t t t
Ex:
(2).Using convolution theorem
11 2 1 2[ ( ) ( )]L F s F s f t f t
12 2
1( ) ( ) .
( 2 5)F s L F s
s s
Ex:
2 2 2
1 12 2 2 2
1 1 12 2 2 2
( )
0
1( )
[( 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
t t
tt t t
t
F ss
L e L e ts s
L F s L Ls s
e t e t e e t d
e t t
0
1sin 2 2 cos 2 .
16
t td e t t t
(3).Using partial fraction
2
1( ) ( ).
( 1)F s f t
s s
2 2
12
1 1 1 1( )
( 1) 1
1( )
( 1)
1 e ( 0).t
F ss s s s s
f t Ls s
L L L
t t
-1 -1 -12
1 -1 1=
s s s+1
Ex:
(4).Using properties1 1 1
1 2 1 2
1 1
1
1
1
1. [ ( ) ( )] [ ( )] [ ( )].
2 . [ ( )] [ ( )].
3 . [ ( )] ( ).
4 . [ '( )] ( ).
( )5 . [ ( ) ] .
t
s
s
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
t
。
。
。
。
。
12
1( ) ( ) .
2 2
sF s L F s
s s
1 12 2
12
1 1( ) [ ] [ ]
2 2 ( 1) 1
[ ] cos .1
t t
s sf t L L
s s s
se L e t
s
Ex:
(5).Using L-transform table
§9.5 Application of Laplace Transform
0
0
00 0
(1). ( ) ( )
( ) (0), ( ) ( ).
s t
s t
f t e dt F s
f t dt F f t e dt F s
3
32 20
3Ex: cos 2 | .
2 13t
s
ste dt
s
0
0
0 0
0 0
( ) ( )(2). ( ) ( ) ,Let 0, ( ) ,
( ), ( ) .
s
s t
s
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
120 1
sin 1Ex: arctan | .
1 4te dt ds s
t 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
120
1 1Ex: sin [ ]'| .
1 2t
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
1L
( )X s
( )X s
L
Ex:( ) 2 ( ) 2 ( ) 2 cos
Solve(0) (0) 0
tx t x t x t e t
x x
22
222 2
122
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
s ss X s sX s X s X s
s s
sx t L
s
122
1 12 2
2
1
1 1sin
1 1
t
t t t
se L
s
e L te L te ts s
Ex:
2
2
( ) 2 ( ) 2 ( ) 10
Solve 2 ( ) ( ) 3 ( ) 13 .
(0) 1, (0) 3
t
t
x t x t y t e
x t y t y t e
x y
2
2
Let ( ) ( ) , ( ) ( )
differential equation equation of ( ) and ( ).
10( ) 1 2 ( ) 2 ( )
213
2 ( ) ( ) 3 3 ( )2
1( )
( )23 ( ) 3( )
2
t
t
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.
