inverse laplace transform
DESCRIPTION
Inverse Laplace Transform. Chairul Hudaya, ST, M.Sc. Electric Power & Energy Studies (EPES) Department of Electrical Engineering University of Indonesia http://www.ee.ui.ac.id/epes. - PowerPoint PPT PresentationTRANSCRIPT
Depok, October, 2009 Laplace Transform Electric Circuit
Inverse Laplace Transform
Electric Power & Energy Studies (EPES)Department of Electrical Engineering
University of Indonesiahttp://www.ee.ui.ac.id/epes
Chairul Hudaya, ST, M.Sc
Depok, October, 2009 Electric Circuit
Depok, October, 2009 Laplace Transform Electric Circuit
Inverse Laplace transform (ILT)
The inverse Laplace transform of F(s) is f(t), i.e.
)()( 1 sFtf L
where L−1 is the inverse Laplace transform operator.
Depok, October, 2009 Laplace Transform Electric Circuit
Example 4
Find the inverse Laplace transform of
3
2
s(a) (b)
4
2
s
(d)9
652
s
s
(c)25
12 s
(e)4)1(
12
s
s (f)4)1( 2 s
s
Depok, October, 2009 Laplace Transform Electric Circuit
Solution
From the table of Laplace transform,
31 2
sL(a)
31 !2
sL 2t
41 2
sL(b)
41 !3
!3
2
sL 3
3
1t
25
12
1
sL(c)
221
5
5
5
1
sL t5sin
5
1
Depok, October, 2009 Laplace Transform Electric Circuit
9
652
1
s
sL
(d)22222 3
32
35
9
65
ss
s
s
s
221
221
3
32
35
ss
sLL
tt 3sin23cos5
4)1(
12
1
s
sL(e) te t 2cos
Write
Depok, October, 2009 Laplace Transform Electric Circuit
4)1( 21
s
sL
(f)
22
122
1
2)1(
2
2
1
2)1(
1
ss
sLL
tete tt 2sin2
12cos
Since the ILT of the term cannot be found directly from the table, we need to rewriteit as the following
2222
2222
2)1(
2
2
1
2)1(
1
4)1(
1
4)1(
1
4)1(
1)1(
4)1(
ss
s
ss
s
s
s
s
s
Depok, October, 2009 Laplace Transform Electric Circuit
Example 5
Find the inverse Laplace transform of
8
( 2)
s
s s
(a) (b)2
9
2 7 4s s
(d)2
7 20
( 4 20)
s
s s s
(c)3 2
4 1
2
s
s s s
(e)2
2 5 6
s
s s
Depok, October, 2009 Laplace Transform Electric Circuit
Solution
We use the partial fractions technique:
1 8
( 2)
s
s s
(a)
12
9
2 7 4s s
(b)
1 4 3
2s s
24 3 te
1 2 1
2 1 4s s
/ 2 4t te e
1
12
1 1
4s s
=LL
L =L
=L
Depok, October, 2009 Laplace Transform Electric Circuit
13 2
4 1
2
s
s s s
(c) 12
4 1
( 1)
s
s s
12
1 3 1
( 1) 1s s s
1 3 t te t e
where, if we let2
1( )F s
s , then ( ) .f t t Hence,
1 12
1( 1) ( )
( 1)t tF s e f t e t
s
=LL
=L
L =L
Depok, October, 2009 Laplace Transform Electric Circuit
12
7 20
( 4 20)
s
s s s
(d) 12
1 3
4 20
s
s s s
12 2
1 2 5
( 2) 16 ( 2) 16
s
s s s
12
1 3
( 2) 16
s
s s
12
1 ( 2) 5
( 2) 16
s
s s
2 2541 cos 4 sin 4t te t e t
=LL
=L
=L
=L
Depok, October, 2009 Laplace Transform Electric Circuit
21
2 5 6
s
s s
(e) 12
5 61
5 6
s
s s
1 5 61( 2)( 3)
s
s s
1 4 91
2 3s s
2 3( ) 4 9t tt e e
L =L
=L
=L=L
Depok, October, 2009 Laplace Transform Electric Circuit
The convolution theorem
)()()()(1 tgtfsGsF L
)()( tgtf where is called as the convolution of
f(t) and g(t),
t
dgtftgtf0
)()()()(
Convolution property: )()()()( tftgtgtf
Therefore, tt
dtgfdgtftgtf00
)()()()()()(
Sometimes, )()( tgtf denoted as
))(( tgf or simply .gf
defined by
Depok, October, 2009 Laplace Transform Electric Circuit
Example 6
(a))2)(1(
1
ss
Use the convolution theorem to find the inverseLaplace transforms of the following:
(b))9(
122 ss
(c))5(
72 ss
Depok, October, 2009 Laplace Transform Electric Circuit
Solution
)2)(1(
11
ssL(a)
2
1
1
1 11
ssLL
tt ee 2
t
t dee0
2 t
t de0
3
tte
0
3
3
33
22 tttt eeee
Depok, October, 2009 Laplace Transform Electric Circuit
)9(
122
1
ssL(b)
9
314
21
ssL
9
314
211
ssLL
)3sin1(4 t
t
d0
3sin14 t
03
3cos4
)3cos1(
3
4t
Depok, October, 2009 Laplace Transform Electric Circuit
)5(
72
1
ssL(c)
5
117
21
ssL
5
117 1
21
ssLL
tet 57 t
t de0
)(57 t
t de0
)(57
t ttt
dee
0
)(5
0
)(5
57
57
ttete
0
)(50
257
5
07
25
)1(7
5
7 5tet )15(
25
7 5 tet