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
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2. Definition of Inverse Laplace Transform
Inverse Laplace Transform
If the laplace transform of a function F(t) is f (s), i.e., if
L(F(t)) = f (s), then F(t) is called an inverse laplace transform
of f (s) and we write aymbolically F(t) = L−1f (s).
In other words, inverse transform of a given function f (s) is that
function F(t) whose, Laplace transform is f (s).
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3. Properties of Inverse Laplace Transforms
First property
L−1(kf (t)) = kL−1(f (t)), where k is a constant.
2nd property - Linearity property
If c1 and c2 are any two constants while f1(s) & f2(s) are the
functions with Inverse Laplace transforms F1(t) & F2(t)
respectively, then
L−1
(c1f1(s) + c2f2(s)) = c1L−1
(f1(s)) + c2L−1
(f2(s))
= c1F1(t) + c2F2(t)
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4. Properties of Inverse Laplace Transforms
3rd property -Translation or Shifting property
If L−1(f (s)) = F(t) then L−1(f (s − a)) = eatF(t)
4th property - Multiplication by power of t
If L−1(f (s)) = F(t), and F(0) = F′(0) = . . . F(n − 1)(0) = 0
then L−1 dn
dsn f (s)
= (−1)ntnF(t), where n = 1, 2, 3...
5th property
if L−1(f (s)) = F(t), then
L(e−asf (s)) =



F(t − a), t a
0, t a − − − −(1)
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5. Remark
It should be noted that the right side of (1) can be written in a
different way using the Unit step function.
Let ua(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)
.
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6. Method of Convolution
Convolution Property
If F(t) andG(t) are the inverse transforms of f (s) and g(s),
respectively, the inverse transform of the product f (s)g(s) is the
convolution of F(t) and G(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
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7. Method of Convolution
Corrollary
Putting t − u = v in 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)
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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)
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9. Examples : - Partial Fractions
Use partial fractions to decompose : s+3
(s−2)(s+1)
To the linear factors s − 2 and s + 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 substitute s = −1 and s = 2 into the above
equation, ⇒ A = 5
3 and B = −2
3 .
∴
s + 3
(s − 2)(s + 1)
=
5
3
(s − 2)
−
2
3
(s − 2)
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10. Examples :
Find the L−1
n
s+3
(s−2)(s+1)
o
L−1
s + 3
(s − 2)(s + 1)
=
5
3
L−1
1
s − 2
−
2
3
L−1
1
s + 1
=
5
3
e2x
−
2
3
e−x
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11. Example:
Find the inverse transforms of
1. 3s+7
s2−2s−3
2. 5s2−15s−11
(s+1)(s−2)3
3. 3s+1
(s−1)(s2+1)
In the problems of the above kind, we use the method of partial
fractions.
1. Resolving into partial fractions, we have
3s + 7
s2 − 2s − 3
= 4.
1
s − 3
−
1
s + 1
.
∴ L−1
3s + 7
s2 − 2s − 3
= 4L−1
1
s − 3
− L−1
1
s + 1
= 4.e3t
− e−t
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12. Examples
2. Resolving into partial fraction, we have,
5s2 − 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
5s2 − 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
e2t
+ 4te2t
−
7
2
t2
e2t
.
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13. Examples :
3. Resolving into partial fractions, we have,
3s + 1
(s − 1)(s2 + 1)
=
2
s − 1
+
−2s + 1
s2 + 1
∴ L−1
3s + 1
(s − 1)(s2 + 1)
= 2L−1
1
s − 1
− 2L−1
s
s2 + 1
+ L−1
1
s2 + 1
= 2et
− 2 cos t + sin t.
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14. Examples :
Find the inverse tranforms of,
1. 6s−4
s2−4s+20
2. s
(s−2)4
3. s2+2s+3
(s2+2s+2)(s2+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
s2 − 4s + 20
=
6(s − 2) + 8
(s − 2)2 + 16
= 6.
(s − 2)
(s − 2)2 + 16
+
8
(s − 2)2 + 16
∴ L−1
6s − 4
s2 − 4s + 20
= L−1
s − 2
(s − 2)2 + 16
+ 2L−1
4
(s − 2)2 + 16
= 6e2t
cos 4t + 2e2t
sin 4t 14 / 28

15. Examples :
2. Since the denominator contains (s − 2) as unit, rewrite the
numerator also in terms of s − 2. Thus
s
(s − 2)4
=
(s − 2) + 2
(s − 2)4
=
1
(s − 2)3
+
2
(s − 2)4
∴ L−1
s
(s − 2)4
= L−1
1
(s − 2)3
+ 2L−1
2
(s − 2)4
= e2t
.
t2
2!
+ 2e2t
.
t3
3!
=
1
3!
e2t
(3t2
+ 23
).
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16. Examples :
3. By the method of partial frations
s2 + 2s + 3
(s2 + 2s + 2)(s2 + 2s + 5)
=
1
3
1
s2 + 2s + 2
+
2
3
1
s2 + 2s + 5
=
1
3
1
(s + 1)2 + 1
+
2
3
1
(s + 1)2 + 4
∴ L−1
s2 + 2s + 3
(s2 + 2s + 2)(s2 + 2s + 5)
=
1
3
L−1
1
(s + 1)2 + 1
+
1
3
L−1
1
(s + 1)2 + 4
=
1
3
e−t
sin t +
1
3
e−t
sin 2t
=
1
3
e−t
(sin t + sin2t)
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17. Examples :
Find the inverse Laplace transfroms of
1. 1
(s+a)n
2. 1
(s2+a2)2
3. s2
(s2+a2)2
1. By applying 3rd property -Translation or Shifting property, we
have
L−1
1
(s + a)n
= e−at
L−1
1
sn
= e−at tn−1
(n − 1)!
2. Since we know the inverse transforms of 1
s2+a2 and s2−a2
(s2+a2)2 , we
shall rewrite the given expression in terms of these.Thus :
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18. Examples :
1
(s2 + a2)2
=
1
2a2
(s2 + a2) − (s2 − a2)
(s2 + a2)2
=
1
2a2
1
s2 + a2
−
s2 − a2
(s2 + a2)2
∴ L−1
1
(s2 + a2)2
=
1
2a2
1
a
sin at − t cos at
=
1
2a3
[sin at − at cos at]
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19. Examples :
3.
s3
(s2 + a2)2
=
s[(s2 + a2) − a2
(s2 + a2)2
=
s
s2 + a2
− a2 s
(s2 + a2)2
∴ L−1
s3
(s2 + a2)2
= cos at − a2
.
1
2a
t sin at
= cos at −
1
2
at sin at.
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20. Examples :
Find L−1
n
e−3s
(s−1)4
o
We have
L−1
e−3s
(s − 1)4
= et
L−1
1
s4
= et t3
3!
; 3rd
property -Translation or Shifting property
=
1
6
t3
et
L−1
e−3s
(s − 1)4
=



1
6(t − 3)3et−3; t 3. Using 6th property
0; t 3
Using the remark;
= 1
6(t − 3)3et−3u3(t);
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21. Examples :
Find L−1
n
eπs
s2+2s+2
o
We have
L−1
eπs
s2 + 2s + 2
= L−1
1
(s + 1)2 + 1
= e−t
L−1
1
s2 + 1
= e−t
sin t. Hence using property 6, we have
L−1
eπs
s2 + 2s + 2
=



e−t−π sin(t − π); t π.
0; t ≤ π
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22. Remark
Remark
We known that
L {tF(t)} = −
d
ds
L {F(t)}
∴ tF(t) = L−1
−
d
ds
L {F(t)}
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23. Examples :
Find the inverse transforms of the log s+1
s−1
Solution : Let F(t) = L−1
n
log s+1
s−1
o
, Then L[F(t)] = log s+1
s−1
Hence from above remark,
tL−1
log
s + 1
s − 1
= L−1
−
d
ds
[log(s + 1) − log(s − 1)]
= −L−1
d
ds
log(s + 1)
+ L−1
d
ds
log(s − 1)
= −L−1
1
s + 1
+ L−1
1
s − 1
= −e−t
+ et
= 2 sinh t
∴ L−1
log
s + 1
s − 1
=
2 sinh t
t
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24. Examples :
Find the inverse Laplace transforms of tan−1( 2
s2 )
Solution : Let F(t) = L−1
tan−1 2
s2
, Then L[F(t)] = tan−1 2
s2
Hence from above remark,
tL−1
tan−1
(
2
s2
)
= L−1
−
d
ds
[tan−1
(
2
s2
)
= L−1
4s
s2 + 4
= L−1
4s
(s2 + 2)2 − 4s2
= L−1
4s
(s2 + 2 + 2s)(s2 + 2 − 2s)
= L−1
1
s2 − 2s + 2
−
1
s2 + 2s + 2
= L−1
1
(s − 1)2 + 1
−
1
(s + 1)2 + 1
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25. Examples :
= et
sin t − e−t
sin t
= 2 sin ht sin t.
∴ L−1
tan−1
(
2
s2
)
=
2 sin ht sin t
t
Using Convolution property, Find L−1
n
1
s(s2+a2)
o
Solution : We have already solved this type of problems by partial
fraction method. However convolution method the work easy.
Let f (s) = 1
s and g(s) = 1
s2+a2
Then F(t) = L−1
1
s
= 1 and G(t) = L−1
n
1
s2+a2
o
= 1
a sin at.
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26. Examples :
∴ L−
1
s(s2 + a2)
=
Z t
0
F(t − u)G(u) du
=
Z t
0
sin au
a
du
=
− cos au
a2
t
0
=
1
a2
(1 − cos at)
Using Convolution property, Find L−1
n
1
s2(s−a)
o
Solution : Let f (s) = 1
s2 and g(s) = 1
s−a
Then F(t) = L−1
1
s2
= t and G(t) = L−1
n
1
s−a
o
= eat
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27. Examples :
∴ L−1
1
s2(s − a)
=
Z t
0
F(u)G(t − u) du
=
Z t
0
uea(t−u)
du
= eat
Z t
0
ue−au
du
=
1
a2
(eat
− at − 1)
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28. Thank You
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