case study on Laplace transform
By
Panchal Kaushik M.
Enrollment No:13040006035
Department: Civil Engineering Department
The Laplace Transforms
TOPIC :
1.
2.
3.
4.
5.
6.
7.
8.
9.
Definition & Laplace transform of some elementary functions
First shifting theorem
Inverse Laplace transform
Laplace transform of Integrals
Multiplication by power of t
Inverse Laplace transform derivation
Division by t
Convolution theorem
Application to Differential equation
Topic -1
Definition : Let f(t) be a function of t defined by for all t 0 ; then the laplace
transform of f(t) is denoted by L{ f (t )} ; and is defined as

L{ f (t )} 
e
0
 st
f (t )dt
L{1} 
1
s
1
L{e at } 
;s  a
Provided that the integral exists ‘S’ is parameter which may be raels or
 a complex.
s
L{cos at}  2 2
s a
Note : (1) L{ f (t )}  is clearly a function of s and is also written as
s
L{sin at}  2 2
s a
i.e. L{ f (t )}  f s 
n!
L{t n }  n1 ; nN
s
(2) L{af (t )  bg (t )}  aL{ f (t )}  bL{g (t )}
(n  1)
L{t n }   n1 ; n 
s
Laplace transform of some elementary functions
1
L{1} 
s
1
L{e } 
;s  a
sa
s
L{cos at}  2
s  a2
s
L{sin at}  2
2
s a
at
n!
L{t }  n 1 ; nN
s
(n  1)
n
L{t }   n 1 ; n  1
s
s
L{cosh at}  2
s  a2
a
L{sinh at}  2
s  a2
n
2 First shifting theorem
If
L{ f (t )}  f s  then prove that L{e at f (t )}  f ( s  a )
Formula :
L{e at f (t )}  f ( s  a )
L{e at t n } 
n!
( s  a ) n 1
b
(s  a) 2  b 2
(s  a)
cos bt} 
(s  a) 2  b 2
b
sinh bt} 
(s  a) 2  b 2
L{e at sin bt} 
L{e at
L{e at
3 Inverse Laplace transform
If L{ f (t )}  f s  then f (t )
and is denoted by
is called the inverse laplace transform f s  of
L1{ f ( s )}  f t 
Formula :
 ( s  a )  at
L 
 e cos at
2
2
 (s  a)  a 
1 
L1    1
s
1
 1  sin at
L1  2

2
s

a
a


1  sinh at
1 
L  2

2
a
s  a 
 s 
L1  2
 cos at
2
s

a


s 
1 
L
 cosh at


at
2
2
 e sin at
s a 



2

1
L1 
2
 (s  a)  a 
a
1  1 
at
L

e




(
s

a
)
L1 
 e at cos at
s  a
2
2
(s  a)  a


n 1

1
t

L1  n  
 s  (n  1)!
 1  at t n 1
L 
e
n
(
s

a
)
(n  1)!


1
 s 2  3s  4 
E.x. Find inverse Laplace transform of L 

3
s


1
2

s
 3s  4 
1
L 

3
s


3 4
1  1
 L   2  3
s 
s s
 1  1  3  1  4 
 L   L  2  L  3 
s
s 
s 
t2
 1  3t  4
2!
1
4 Laplace transform of integrals
Theorem
: if L{ f (t )}  f s  then
t


f s 

L  f (u ) du  
s


0
t


f s 

L  f (u ) du  
s


0
f ( s)
L{ (t )} 
s
t
e
u
cos udu
0
E.x. Find the Laplace transform of
solution
 t u
 1
 L   e cos udu   L e u cos u
0
 s



1
s 1
s ( s  1) 2  12
1  s 1 
  2
s  s  2 s  2 
5 Multiplication by power of t
 
dn
f ( s)
Theorem : If L{ f (t )}  f s  then L{t f (t )}  (1)
n
ds
n
e.x.
FIND Lt sin t
Lsin t 
1
 f (s)
2
s 1
 Lt sin t  (1)


d  1 
ds  s 2  1


d 2
( s  1) 1  (1)( s 2  1)  2 (2s)
ds
2s
( s 2  1) 2
n
6 Inverse Laplace transform
derivation


 dn

n n
n n 1
L
f
(
s
)

(

1
)
t
f
(
t
)

(

1
)
t L f ( s)
Theorem :
 n

 ds

1
e.x.
FIND
  s  1 
L1 log 

s

1



 s 1
f ( s )  log 
  log( s  1)  log( s  1)
s

1



1
1
f (s) 

s 1 s 1
 1 

 

1  1
 L1  f ( s )   L1 

L





 s  1
 s  1


 ( 1)tL1 f ( s )  e t  e t
t
t

2 sinh t
e t  e t
 s  1  e  e
 L log 

 

t
t
 s  1 

t
1
7 Division by t
f (t 
Theorem : if L{ f (t )}  f s  then L


 t 
e.x.

 f (s)ds
s
1  e 2t 
FIND L  t 


1
1
L 1  e 2t  
 f (s)
s s2

1  e 2t  
1 
1
 L
   f ( s)ds    
 ds
t
s
s

2


 s
s 


 log s  log( s  2)

s
 s 
 log    log 

s

2


 s2
 log 

 s 

  s 
 log 

s

2

 s

8 Convolution theorem
Theorem : if L1 f ( s) f (t )
1
and L g ( s) g (t )
 Apply convolution theorem
 1

L1  2

s
(
s

1
)


Let
f (s) 
1 and
1
g
(
s
)

s2
s 1
then




  1 
 f (t )  L1 f ( s )  L1  2   t
  s 
  1 
t
 g (t )  L1 g ( s )  L1 
  e
  s  1 
By convolution theorem

1
1 
L  2

 s ( s  1) 
t
( t u )
ue
du

0
t
 e t  ue u du
o
 e  ue
t

u
e

u t
0

 et  tet  et  1
 e  t 1
t
9 Application to Differential equation
• Solution of ordinary linear differential equation:
the Laplace transform can also be used to solve ordinary as well as
partial differential equation. We shall apply this method to solve only
Ly(tequation
)  s y ( s)  y (0)with constant coefficients. The
ordinary linear differential
Lyis
(t )that
 s y ( sit
)  sy
(0)  y(the
0)
advantage of this method
gives
particular solution directly ;
y(t )  s y ( s)  s y (0)  sy(0)  y(0)
without the necessity ofLfirst
finding the general solution and evaluating
the arbitrary constants.
2
3
2
• Note
Ly(t )  s y ( s)  y (0)
Ly(t )  s 2 y ( s)  sy (0)  y(0)
Ly(t )  s 3 y ( s)  s 2 y (0)  sy(0)  y(0)
Method :
1. take Laplace transform on both side of the given differential
equation using initial conditions. This gives an algebraic
equation.
2. Solve the algebraic equation to get y in term of S i.e.
divide by the coefficient of y ; getting y as known
function of S.
3. Resolve this function of S in to partial fraction and take the
inverse Laplace transform of both sides. This gives Y as a
function of t which is the required solution satisfying the
given conditions.
E.x.1: use transform method to solve ; dy  2 y  10e3t
dt
y
(
0
)

6
given that
.
Here given equation is y  2 y  10e3t
Taking the Laplace transform on both sides;
 
Ly 2 Ly  10 L e3t


 s y ( s )  y (0)  2 y ( s ) 
y  2 y
Ly 
10
s 3
Using initial conditions ; we get
10
s 3
10
6s  8
 s  2y ( s ) 
6
s 3
s 3
6s  8
4
2
 y ( s) 


( s  2)( s  3) s  2 s  3
s y( s)  6  2 y( s) 

 s y(
s y(s) 
 s 
 y( s)
now taking Inverse Laplace transform on both sides;
 
 4 
1  2 
 L1 y ( s )  L1 

L



s

2
s

3




 y(t )  4e
2t
 2e
Which is required solution.
3t