The Laplace Transform
Objective
To study the Laplace transform and inverse Laplace transform.
Modules
Module I- Laplace transform of some elementary functions
Module II- Properties of Laplace transform
Module III- Existence conditions
Module IV- Inverse Laplace transform
Introduction
A Transformation is a mathematical device which converts one function into another. For
example, when the differential operator D operates on f(x) = sin x, it gives a new function g(x) =
D f(x) = cos x.
Laplace transform or Laplace transformation is widely used by scientists and engineers. It is
particularly effective in solving linear differential equations. It is very useful in system analysis
where initial conditions can be easily included to give system response. We begin this session by
the definition of Laplace transform.
Definitions
There are two types of Laplace transforms. The transform defined by
 (s) 

 f (t )e
 st
dt ,

where s is a parameter which may be real or complex is known as two sided or bilateral Laplace
transform of the function f(t), provided the integral exists.
1
For some functions the above transform cause problem of convergence. This can be almost
avoided by restricting the range of integration to between 0 and  and considering f(t) = 0 for t <
0. Thus the transformation defined by

 ( s)   e st . f (t )dt
0
where t > 0 and s is a parameter which may be real or complex is known as unilateral or simply,
Laplace transform of the function f(t), provided the integral exists. It is also denoted as L{f(t)} or
f (s ) .
In this session we will be used this second definition.
Module I- Laplace transform of some elementary functions
I. f(t) = k

Therefore
 ( s)   e st .kdt
0

 e st 
k
 k
 ,s  0.

  s 0 s
II. f(t) = eat

Therefore
 ( s)   e st .e at dt
0

  e ( sa )t dt
0

 e ( s  a ) t 
1


,s  a .

  ( s  a)  0 s  a
III. f(t) = tn, (n > -1)
2

Therefore
 ( s)   e st .t n dt
0

n
dk
 k  dk
  e .  . , by putting st  k , dt 
s
s s
0



k
1
n1
s

. e k .k n dk
0
1
s

. e k .k ( n1)1dk
n1 
0
(n  1)
, if n > -1 and s > 0.
s n1
Note. If n is a positive integer, we have (n  1)  n!
Therefore, Lt n  
n!
s n1
IV. f(t) = cos at

Therefore
 (s)   e st . cos atdt
0

 e at

 2
( s cos at  a sin at )
2
s  a
0

s
.
s  a2
2
V. f(t) = sin at

Therefore
 ( s)   e st . sin atdt
0
3

 e at

 2
(a sin at  a cos at )
2
s  a
0

1
.
s  a2
2
Module II- Properties of Laplace transform
I. Linearity property
If c1, c2 are two constants and f1, f2 are two functions of t, then
Lc1 f1 (t )  c2 f 2 (t )  c1L f1 (t ) c2 L f 2 (t ) .
Proof.
By definition, we have

Lc1 f1 (t )  c2 f 2 (t )   e st c1 f1 (t )  c2 f 2 (t )dt
0

 c1  e
 st

f1 (t )dt  c2  e st f 2 (t )dt
0
0
 c1L f1 (t ) c2 L f 2 (t ).
This result may be generalized as
Lc1 f1 (t )  c2 f 2 (t )  ...  cn f n (t )
,
 c1 L f1 (t )  c2 L f 2 (t )  ...  cn L f n (t )
for n constants c1, c2, …, cn and n functions f1, f2, …, fn.
Examples
We have cosh at 
e at  e  at
.
2
4
 e at  e  at 
L{cosh at}  L

2


Therefore
    

1
L e at  L e  at
2

1 1
1 


2  s  a s  a 

s
.
s  a2
2
 e at  e  at 
L{sinh at}  L

2


Similarly,
    

1
L e at  L e at
2

1 1
1 


2  s  a s  a 

a
.
s  a2
2
II. First shifting property
If L{ f (t )}   ( s) , then L{e at . f (t )}   ( s  a) .
Proof.
By definition, we have

L{e . f (t )}   e st .e at . f (t )dt
at
0

  e ( sa )t . f (t )dt
0
5

  e rt . f (t )dt , where r = s – a
0
  (r )   ( s  a) .
Examples.
1. If n is positive integer, we know that

L tn 
n!
.
s n1
Therefore by first shifting property,
 
L e at t n 
n!
.
( s  a) n1
2. We know that Lcos 3t 
s
.
s  32
2
By shifting property,


L e 2t cos 3t 
( s  2)
.
( s  2) 2  32
III. Change of scale property
1 s
If L{ f (t )}   ( s) , then L{ f (at )}     .
a a
Proof.
By definition, we have

L{ f (at )}   e st . f (at )dt
0
6

 e

sr
a
. f (r )
0
du
dr
, where at = r and dt 
a
a

s
1
  e  pr . f (r )dr , where p 
a
a0
1
1 s
  ( p)     .
a
a a
Examples.
1. We know that L{sinh t} 
1
.
s 1
2
Therefore by change of scale property, we have
1
1
2
L{sinh 2t}  .
 2
.
2
2 s
s 4
  1
2
2. Let L f (t ) 
s 1
, then
s2  2
s
1
1
s3
.
L f (3t )  . 3 2
 2
3 s
s  18
  2
 3
IV. Change of scale shifting property
1  s b
If L{ f (t )}   ( s) , then L{ebt . f (at )}   
.
a  a 
Proof is similar to the proof of above property.
Example. Let L{cos t} 
s
, then
s 1
2
7
 s2


1  3 
2t
L e cos 3t  .
3  s2

 1
 3 



s2
.
( s  2) 2  32
Module III- Existence conditions
The Laplace transform does not exists for all functions. If it exists, it is uniquely determined. The
following conditions are to be satisfied:
Let f(t) be the given function. If
1. f(t) is piecewise continuous on every finite interval
2. f(t) satisfies the inequality f (t )  b.eat for all t  0 and for some constants a and b,
and
then L{f(t)} exists.
The function which satisfies the condition 2 is known as exponential order.
For example, cos ht < et for all t > 0,
tn < n! et (n = 0, 1, 2, …) for all t > 0.
  does not exist. Similarly, 1t does not have
But et  beat , whatever may be a and b. So L et
2
Laplace transform.
Example. Find the Laplace transforms of
(i) sin 2t cos 3t
(i) Here
sin 2t cos 3t 
(ii) sin
 t
1
(2 cos 3t sin 2t )
2
8
2

Therefore
1
(sin 5t  sin t ) .
2
1

Lsin 2t cos 3t  L  (sin 5t  sin t )
2


1
Lsin 5t  Lsin t
2

1 5
1 
 2 2
2
2

2 s  5
s 1 

2( s 2  5)
.
( s 2  25)( s 2  1)
 t   t
sin  t   t 
3!
5!
3
(ii) We have
Therefore
5
   
3
5


t
t


t  L t 

 ...
3!
5!




  
L sin
 ...
 
 L t 1/ 2 

 
 
1
1
L t 3 / 2  L t 5 / 2  ...
3!
5!
(3 / 2) 1 (5 / 2) 1 (7 / 2)


 ...
s3/ 2
3! s 5 / 2
5! s 7 / 2
1
3 1
5 3 1
. 
. . 
. . . 
1 2 2
1 2 2 2
2
 3/ 2  .

 ...
s
6
s5/ 2
120
s7/ 2


 
2s 3 / 2
2
3

 1  1 1  1 1 
1           ...

  4s  2!  4s  3!  4s 
 .e 1/ 4 s
2s 3 / 2
.
9
Example. Find the Laplace transform of t e-4t sin3t.
Solution.
We know that Lt 
1
.
t2
Therefore by first shifting theorem, we have
 
L te3it 

or
1
( s  3i) 2
( s  3i ) 2
(s  3i)( s  3i)2
(s
Lt (cos 3t  i sin 3t ) 
 9)  6is
(s 2  9) 2
2
Equating the imaginary parts on both sides, we get
Lt sin 3t ) 
6s
.
( s  9) 2
2
Again applying the first shifting theorem, we have


L e 4t .t sin 3t ) 

6( s  4)
(s  4)
s
2
9

2
6( s  4)
2
 8s  25

2
.
Module IV- Inverse Laplace transform
If L f (t )   (s) , then f(t) is called the inverse Laplace transform of  (s ) and is denoted by
L1 (s)  f (t ) .
Here L1 denotes the inverse Laplace transform.
10
For example, since Le 2t  
1
 1  2t
, we have L1 
e .
s2
s  2
Inverse Laplace transform follows all the properties of Laplace transform.
From the results of Laplace transforms, we have
k 
(1) L1    k , k being constant.
s
 1  at
(2) L1 
e
s  a
t n1
t n1
1
(3) L1  n  
if n is positive integer. Otherwise 
.
(n)
 s  (n  1)!
 1  at t n1
(4) L1 
e
n
(n  1)!
 ( s  a) 
 1  1
 sin at
(5) L1  2
2
s  a  a

 1 bt
1
(6) L1 
 e sin at
2
2
 ( s  b)  a  a
 s 
 cos at
(7) L1  2
2
s  a 

 bt
s b
(8) L1 
 e cos at
2
2
 ( s  b)  a 
 1  1
 sinh at
(9) L1  2
2
s  a  a
 s 
 cosh at
(10) L1  2
2
s  a 
11
Example. Find the inverse Laplace transform of


3 s2 1
(i)
2s 5
2
4s  15
16s 2  25
(ii)
Solution.
(i) We have


2
3 s2 1
3s 4  6s 2  3

2s 5
2s 5
3 1
1 3 1
 .  3. 3  . 5
2 s
s
2 s
Therefore


2

3 1  1 
3 s 2  1 
 3 1  1 
1  1 
L 
  .L    3.L  3   .L  5 
5
s
s  2
s 


 2s
 2
1
t 2  3 t 4 
3
 (1)  3    
2
 2!  2  4! 

(ii) We have
4s  15

16s 2  25
4s  15
25 

16 s 2  
16 

1
 .
4
Therefore
3 3 2 1 4
 t  t .
2 2
16
s
5
s2   
4
2

15
.
16
1
5
s2   
4
2
.








1 1 
s
1
 15 1 

1  4 s  15 
L  2
 .L 
  .L 
2
2
16s  25  4
 s 2   5   16
 s2   5  
  
  


4 
4 
12

1
 5  15 1
5 
cosh  t   .
sinh  t 
4
 4  16 5 / 4
4 

1
5  3
5 
cosh  t   sinh  t  .
4
4  4
4 
Inverse Laplace transforms using method of partial fractions
If  (s ) is rational algebraic function, then we have to express  (s ) in terms of partial fractions
in order to find the inverse Laplace transform.
Example. Find the inverse Laplace transform of
s2 1
s 3  3s 2  2s
Solution.
Here the denominator can be written as
s 3  3s 2  2s  s(s 2  3s  2)
 s ( s  1)( s  2)
Let
s2 1
A
B
C
.
 

s( s  1)( s  2) s s  1 s  2
Then
s 2  1  A(s  1)( s  2)  Bs (s  2)  Cs(s  1) .
Putting s = 0,
we get
1 = 2A  A = ½
Putting s = -1,
we get
2 = -B  B = -2
Putting s = -2,
we get
5 = 2C  C = 5/2.
Therefore
s2 1
1 1
1
5 1
 .  2.
 .
s(s  1)( s  2) 2 s
s 1 2 s  2
13
and

 1 1 1 
s2 1
5 1  1 
1  1 
L1 
  .L    2.L 
  .L 

s 
 s  1 2
s  2
 s( s  1)( s  2)  2

1
5
 2e t  e 2t .
2
2
14