Inverse Laplace Transform

A review of integral Transforms - Inverse Laplace Transformations and sample problems with solutions.

1. The Laplce Transforms - II
Review of Integral Transforms - Inverse Laplace Transforms
Vishnu V
Assistant professor, Department of Mathematics
Sree Ayyappa College, Eramallikkara
1 / 28

2. Definition of Inverse Laplace Transform
Inverse Laplace Transform
If the laplace transform of a functionF(t) isf(s), i.e., if
L(F(t)) =f(s), thenF(t) is called an inverse laplace transform
off(s) and we write aymbolicallyF(t) =L
−1
f(s).
In other words, inverse transform of a given functionf(s) is that
functionF(t) whose, Laplace transform isf(s).
2 / 28

3. Properties of Inverse Laplace Transforms
First property
L
−1
(kf(t)) =kL
−1
(f(t)), where k is a constant.
2
nd
property - Linearity property
Ifc1andc2are any two constants whilef1(s) &f2(s) are the
functions with Inverse Laplace transformsF1(t) &F2(t)
respectively, then
L
−1
(c1f1(s) +c2f2(s)) =c1L
−1
(f1(s)) +c2L
−1
(f2(s))
=c1F1(t) +c2F2(t)
3 / 28

4. Properties of Inverse Laplace Transforms
3
rd
property -Translation or Shifting property
IfL
−1
(f(s)) =F(t) thenL
−1
(f(s−a)) =e
at
F(t)
4
th
property - Multiplication by power of t
IfL
−1
(f(s)) =F(t), andF(0) =F
′
(0) =. . .F
(
n−1)(0) = 0
thenL
−1
Γ
d
n
ds
nf(s)
∆
= (−1)
n
t
n
F(t), wheren= 1,2,3...
5
th
property
ifL
−1
(f(s)) =F(t), then
L(e
−as
f(s)) =



F(t−a),t>a
0, t<a− − − −(1)
4 / 28

5. Remark
It should be noted that the right side of (1) can be written in a
different way using the Unit step function.
Letua(t) =



1,t>a
0,t<a− − − − −(2)
From equations (1) and (2) we can wriiten in a different form
L
−1
(e
−as
f(s)) =F(t−a)ua(t)
.
5 / 28

6. Method of Convolution
Convolution Property
IfF(t) andG(t) are the inverse transforms off(s) andg(s),
respectively, the inverse transform of the productf(s)g(s) is the
convolution ofF(t) andG(t), written (F∗G)(t) and defined by
(F∗G)(t) =
Z
1
0
F(t−u)G(u)du
i.e.,
L
−1
(f(S)g(s)) = (F∗G)(t) =
Z
1
0
F(t−u)G(u)du
6 / 28

7. Method of Convolution
Corrollary
Puttingt−u=vin the above integral, we obtain,
(F∗G)(t) =−
Z
0
t
F(v)G(t−v)du
=
Z
t
0
G(t−v)F(v)du
= (G∗F)(t)
7 / 28

8. Remark
Properties of convolution
(F∗(G+H))(t) = (F∗G)(t) + (F∗H)(t)
((F∗G)∗H)(t) = (F∗(G∗H))(t)
(F∗0)(t) = (0∗F)(t)
8 / 28

9. Examples : - Partial Fractions
Use partial fractions to decompose :
s+3
(s−2)(s+1)
To the linear factorss−2 ands+ 1, we associate respectively the
fractions
A
(s−2)
and
B
(s+1)
. We set
s+ 3
(s−2)(s+ 1)
=
A
(s−2)
+
B
(s+ 1)
i.e.,s+ 3 =A(s+ 1) +B(s−2)
To find A and B, we substitutes=−1 ands= 2 into the above
equation,⇒A=
5
3
andB=
−2
3
.
∴
s+ 3
(s−2)(s+ 1)
=
5
3
(s−2)
−
2
3
(s−2)
9 / 28

10. Examples :
Find theL
−1
n
s+3
(s−2)(s+1)
o
L
−1
æ
s+ 3
(s−2)(s+ 1)
œ
=
5
3
L
−1
æ
1
s−2
œ
−
2
3
L
−1
æ
1
s+ 1
œ
=
5
3
e
2x
−
2
3
e
−x
10 / 28

11. Example:
Find the inverse transforms of
1.
3s+7
s
2
−2s−3
2.
5s
2
−15s−11
(s+1)(s−2)
3
3.
3s+1
(s−1)(s
2
+1)
In the problems of the above kind, we use the method ofpartial
fractions.
1. Resolving into partial fractions, we have
3s+ 7
s
2
−2s−3
= 4.
1
s−3
−
1
s+ 1
.
∴L
−1
`
3s+ 7
s
2
−2s−3
´
= 4L
−1
`
1
s−3
´
−L
−1
`
1
s+ 1
´
= 4.e
3t
−e
−t
11 / 28

12. Examples
2. Resolving into partial fraction, we have,
5s
2
−15s−11
(s+ 1)(s−2)
3
=−
1
3
`
1
s+ 1
´
+
1
3
`
1
s−2
´
+
4
(s−2)
2
−
7
(s−2)
3
∴L
−1
`
5s
2
−15s−11
(s+ 1)(s−2)
3
´
=−
1
3
L
−1
`
1
s+ 1
´
+
1
3
L
−1
`
1
s−2
´
+
+ 4L
−1
`
1
(s−2)
2
´
−7L
−1
`
1
(s−2)
3
´
=−
1
3
e
−t
+
1
3
e
2t
+ 4te
2t
−
7
2
t
2
e
2t
.
12 / 28

13. Examples :
3. Resolving into partial fractions, we have,
3s+ 1
(s−1)(s
2
+ 1)
=
2
s−1
+
−2s+ 1
s
2
+ 1
∴L
−1
æ
3s+ 1
(s−1)(s
2
+ 1)
œ
= 2L
−1
æ
1
s−1
œ
−2L
−1
æ
s
s
2
+ 1
œ
+L
−1
æ
1
s
2
+ 1
œ
= 2e
t
−2 cost+ sint.
13 / 28

14. Examples :
Find the inverse tranforms of,
1.
6s−4
s
2
−4s+20
2.
s
(s−2)
4
3.
s
2
+2s+3
(s
2
+2s+2)(s
2
+2s+5)
1. Here the denominator cannot be factorised into rational factors
and hence the partial fraction method is in-applicable. However
completing the square in the denominator, we obtain,
6s−4
s
2
−4s+ 20
=
6(s−2) + 8
(s−2)
2
+ 16
= 6.
(s−2)
(s−2)
2
+ 16
+
8
(s−2)
2
+ 16
∴L
−1
æ
6s−4
s
2
−4s+ 20
œ
=L
−1
æ
s−2
(s−2)
2
+ 16
œ
+ 2L
−1
æ
4
(s−2)
2
+ 16
œ
= 6e
2t
cos 4t+ 2e
2t
sin 4t
14 / 28

15. Examples :
2. Since the denominator contains (s−2) as unit, rewrite the
numerator also in terms ofs−2. Thus
s
(s−2)
4
=
(s−2) + 2
(s−2)
4
=
1
(s−2)
3
+
2
(s−2)
4
∴L
−1
æ
s
(s−2)
4
œ
=L
−1
æ
1
(s−2)
3
œ
+ 2L
−1
æ
2
(s−2)
4
œ
=e
2t
.
t
2
2!
+ 2e
2t
.
t
3
3!
=
1
3!
e
2t
(3t
2
+ 2
3
).
15 / 28

16. Examples :
3. By the method of partial frations
s
2
+ 2s+ 3
(s
2
+ 2s+ 2)(s
2
+ 2s+ 5)
=
1
3
1
s
2
+ 2s+ 2
+
2
3
1
s
2
+ 2s+ 5
=
1
3
1
(s+ 1)
2
+ 1
+
2
3
1
(s+ 1)
2
+ 4
∴L
−1
æ
s
2
+ 2s+ 3
(s
2
+ 2s+ 2)(s
2
+ 2s+ 5)
œ
=
1
3
L
−1
æ
1
(s+ 1)
2
+ 1
œ
+
1
3
L
−1
æ
1
(s+ 1)
2
+ 4
œ
=
1
3
e
−t
sint+
1
3
e
−t
sin 2t
=
1
3
e
−t
(sint+sin2t)
16 / 28

17. Examples :
Find the inverse Laplace transfroms of
1.
1
(s+a)
n
2.
1
(s
2
+a
2
)
2
3.
s
2
(s
2
+a
2
)
2
1. By applying 3
rd
property -Translation or Shifting property, we
have
L
−1
æ
1
(s+a)
n
œ
=e
−at
L
−1
æ
1
s
n
œ
=e
−at
t
n−1
(n−1)!
2. Since we know the inverse transforms of
1
s
2
+a
2and
s
2
−a
2
(s
2
+a
2
)
2, we
shall rewrite the given expression in terms of these.Thus :
17 / 28

18. Examples :
1
(s
2
+a
2
)
2
=
1
2a
2
(s
2
+a
2
)−(s
2
−a
2
)
(s
2
+a
2
)
2
=
1
2a
2
æ
1
s
2
+a
2
−
s
2
−a
2
(s
2
+a
2
)
2
œ
∴L
−1
æ
1
(s
2
+a
2
)
2
œ
=
1
2a
2
`
1
a
sinat−tcosat
´
=
1
2a
3
[sinat−atcosat]
18 / 28

19. Examples :
3.
s
3
(s
2
+a
2
)
2
=
s[(s
2
+a
2
)−a
2
(s
2
+a
2
)
2
=
s
s
2
+a
2
−a
2
s
(s
2
+a
2
)
2
∴L
−1
æ
s
3
(s
2
+a
2
)
2
œ
= cosat−a
2
.
1
2a
tsinat
= cosat−
1
2
atsinat.
19 / 28

20. Examples :
FindL
−1
n
e
−3s
(s−1)
4
o
We have
L
−1
æ
e
−3s
(s−1)
4
œ
=e
t
L
−1
æ
1
s
4
œ
=e
t
t
3
3!
; 3
rd
property -Translation or Shifting property
=
1
6
t
3
e
t
L
−1
æ
e
−3s
(s−1)
4
œ
=



1
6
(t−3)
3
e
t−3
;t>3.Using 6
th
property
0; t<3
Using the remark;
=
1
6
(t−3)
3
e
t−3
u3(t);
20 / 28

21. Examples :
FindL
−1
n
e
πs
s
2
+2s+2
o
We have
L
−1
æ
e
πs
s
2
+ 2s+ 2
œ
=L
−1
æ
1
(s+ 1)
2
+ 1
œ
=e
−t
L
−1
æ
1
s
2
+ 1
œ
=e
−t
sint.Hence using property 6, we have
L
−1
æ
e
πs
s
2
+ 2s+ 2
œ
=



e
−t−π
sin(t−π);t> π.
0;t≤π
21 / 28

22. Remark
Remark
We known that
L{tF(t)}=−
d
ds
L{F(t)}
∴tF(t) =L
−1
æ
−
d
ds
L{F(t)}
œ
22 / 28

23. Examples :
Find the inverse transforms of the log
s+1
s−1
Solution: LetF(t) =L
−1
n
log
s+1
s−1
o
, ThenL[F(t)] = log
s+1
s−1
Hence from above remark,
tL
−1
æ
log
`
s+ 1
s−1
´œ
=L
−1
æ
−
d
ds
[log(s+ 1)−log(s−1)]
œ
=−L
−1
æ
d
ds
log(s+ 1)
œ
+L
−1
æ
d
ds
log(s−1)
œ
=−L
−1
æ
1
s+ 1
œ
+L
−1
æ
1
s−1
œ
=−e
−t
+e
t
= 2 sinht
∴L
−1
æ
log
`
s+ 1
s−1
´œ
=
2 sinht
t
23 / 28

24. Examples :
Find the inverse Laplace transforms of tan
−1
(
2
s
2)
Solution: LetF(t) =L
−1
Φ
tan
−12
s
2

, ThenL[F(t)] = tan
−12
s
2
Hence from above remark,
tL
−1
æ
tan
−1
(
2
s
2
)
œ
=L
−1
æ
−
d
ds
[tan
−1
(
2
s
2
)
œ
=L
−1
æ
4s
s
2
+ 4
œ
=L
−1
æ
4s
(s
2
+ 2)
2
−4s
2
œ
=L
−1
æ
4s
(s
2
+ 2 + 2s)(s
2
+ 2−2s)
œ
=L
−1
æ
1
s
2
−2s+ 2
−
1
s
2
+ 2s+ 2
œ
=L
−1
æ
1
(s−1)
2
+ 1
−
1
(s+ 1)
2
+ 1
œ
24 / 28

25. Examples :
=e
t
sint−e
−t
sint
= 2 sinhtsint.
∴L
−1
æ
tan
−1
(
2
s
2
)
œ
=
2 sinhtsint
t
Using Convolution property, FindL
−1
n
1
s(s
2
+a
2
)
o
Solution: We have already solved this type of problems by partial
fraction method. However convolution method the work easy.
Letf(s) =
1
s
andg(s) =
1
s
2
+a
2
ThenF(t) =L
−1
Φ
1
s

= 1 andG(t) =L
−1
n
1
s
2
+a
2
o
=
1
a
sinat.
25 / 28

26. Examples :
∴L
−
æ
1
s(s
2
+a
2
)
œ
=
Z
t
0
F(t−u)G(u)du
=
Z
t
0
sinau
a
du
=
ˇ
−cosau
a
2
˘
t
0
=
1
a
2
(1−cosat)
Using Convolution property, FindL
−1
n
1
s
2
(s−a)
o
Solution: Letf(s) =
1
s
2andg(s) =
1
s−a
ThenF(t) =L
−1
Φ
1
s
2

=tandG(t) =L
−1
n
1
s−a
o
=e
at
26 / 28

27. Examples :
∴L
−1
æ
1
s
2
(s−a)
œ
=
Z
t
0
F(u)G(t−u)du
=
Z
t
0
ue
a(t−u)
du
=e
at
Z
t
0
ue
−au
du
=
1
a
2
(e
at
−at−1)
27 / 28

28. Thank You
28 / 28