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Introduction
Module-2
find the Laplace transform of the
following functions
CONTENTS
 SECOND SHIFTING THROEM and its example.
 DERIVATIVE PROPERTY (multiplication by t property) with examples.
 INTEGRAL PROPERTY (division by t property) with examples.
If 𝐿{𝑓(𝑡)}=𝐹(𝑠) and 𝑢(𝑡−𝑎)=
then we define a new function
𝐿{𝑓(𝑡−𝑎)𝑢(𝑡−𝑎)} = e−sa F(s)
And its Inverse Laplace is
𝐿−1{e−sa F(s)} = 𝑓(𝑡−𝑎)𝑢(𝑡−𝑎)
0
1
{ if t a
if t a


Second shifting throem
Example :-
𝐿{𝑓(𝑡−𝑎) 𝑢(𝑡−𝑎)}
Comparing with second shift throem
𝑢(𝑡−𝑎) = 𝑢(𝑡−2) 𝑖.𝑒 𝑎 = 2
𝑓(𝑡−𝑎) = 𝑓(𝑡−2) = (𝑡−1)2
Put 𝑡=𝑡+2 then 𝑓(𝑡) = (𝑡+1)2 = 𝑡2+2𝑡+1
𝐹(𝑠)=𝐿{𝑡2+2𝑡+1 }= Hence by the second shifting theorem, we get
𝐿 {𝑓(𝑡−𝑎)𝑢(𝑡−𝑎)}=e−sa F(s)
𝐿 {(𝑡−1)2𝑢(𝑡−2)}=e−2s
3 2
2 2 1
s s s
 
 
 
 
3 2
2 2 1
s s s
 
 
 
 
𝐿−1{e−sa F(s)} = 𝑓(𝑡−𝑎)𝑢(𝑡−𝑎)
here e-as= e-3s i.e. a=3
i.e u(t-3)
And f(s) =
f(t)= L-1 = sint
f(t-3) = sin (t-3)
i.e ans = sin(t-3)u(t-3)
Example of L-1 using
second shifting throem
3
1
2
[ ]
1
s
e
L
s



2
1
1
s 
2
1
1
s 
DERIVATIVE PROPERTY (multiplication by t property)
{ ( )} ( )
L f t F s

if
then
{ ( )} ( 1) { ( )}
n
n n
n
d
L t f t F s
ds
 
{ ( )} ( 1) { ( )}
d
L t f t F s
ds
 
2
2 2
2
{ ( )} ( 1) { ( )}
d
L t f t F s
ds
 
Example of derivitive property (multiplication by t property)
Find the laplace of
2 3
( )
t
L t e
{ ( )} ( 1) { ( )}
n
n n
n
d
L t f t F s
ds
 
Comparing with derivative property
2
1 ( 3)
( 3)
d s
s ds
 


1
( ) ( )
3
d d
f s
ds ds s


3
2 ( 3)
( )
( 3)
d s
s ds



2
1
( 1)
s


3
2
( 3)
s


2 2
3
2
{ ( )} ( 1)
( 3)
L t f t
s
 

2
2 2
1
{ ( )}
( 3)
d d
f s
ds ds s



3
{ ( )} { } ( )
t
L f t L e F s

 
1
( )
3
F s
s


2
3
2
{ ( )}
( 3)
L t f t
s


INTEGRAL PROPERTY (DIVISION by t property)
{ ( )} ( )
L f t F s

( )
{ } ( )
s
f t
L F s ds
t

 
2
( )
{ } ( )
s
s
f t
L F s ds ds
t


  
if
then
Example of integral property (division by t property)
Find the laplace of
comparing it with equation we get
1 1
{ ( )} ( )
4 6
L f t F s
s s
  
 
4 6
( ) ( ) ( )
t t
f t L e L e
 
 
4 6
( ) ( )
t t
f t e e
 
 
4 6
{ ( )} ( )
t t
L f t L e e
 
 
4 4
limln( ) ln( )
6 6
s
s s
s s

 
 
 
4
[ln( )]
6
s
s
s




[ln( 4) ln( 6)]s
s s 
  
1 1
( ) ( )
4 6
s s
F s ds
s s
 
 
 
 
4
0 ln
6
s
s

 

4
1
4
limln( ) ln( )
6 6
1
s
s
s
s
s



 


1
4
ln( )
6
s
s




6
ln( )
4
s
s



4 6
t t
e e
t
 

4 6
{ } ( )
t t
s
e e
L F s ds
t

 

 
The knowledge of Laplace transform has in recent years become an
essential part of mathematical background required of engineers and
scientists. This is because the transform method an easy and effective
means for the solution of many problems arising in engineering.
The properties presented are very useful to solve specfic find of
laplace transform problems in a easy and more enhanced way.
conclusion
THANK YOU

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PROPERTIES OF LAPLACE TRANSFORM part 2

  • 1.
  • 2. Introduction Module-2 find the Laplace transform of the following functions
  • 3. CONTENTS  SECOND SHIFTING THROEM and its example.  DERIVATIVE PROPERTY (multiplication by t property) with examples.  INTEGRAL PROPERTY (division by t property) with examples.
  • 4. If 𝐿{𝑓(𝑡)}=𝐹(𝑠) and 𝑢(𝑡−𝑎)= then we define a new function 𝐿{𝑓(𝑡−𝑎)𝑢(𝑡−𝑎)} = e−sa F(s) And its Inverse Laplace is 𝐿−1{e−sa F(s)} = 𝑓(𝑡−𝑎)𝑢(𝑡−𝑎) 0 1 { if t a if t a  
  • 5. Second shifting throem Example :- 𝐿{𝑓(𝑡−𝑎) 𝑢(𝑡−𝑎)} Comparing with second shift throem 𝑢(𝑡−𝑎) = 𝑢(𝑡−2) 𝑖.𝑒 𝑎 = 2 𝑓(𝑡−𝑎) = 𝑓(𝑡−2) = (𝑡−1)2 Put 𝑡=𝑡+2 then 𝑓(𝑡) = (𝑡+1)2 = 𝑡2+2𝑡+1 𝐹(𝑠)=𝐿{𝑡2+2𝑡+1 }= Hence by the second shifting theorem, we get 𝐿 {𝑓(𝑡−𝑎)𝑢(𝑡−𝑎)}=e−sa F(s) 𝐿 {(𝑡−1)2𝑢(𝑡−2)}=e−2s 3 2 2 2 1 s s s         3 2 2 2 1 s s s        
  • 6. 𝐿−1{e−sa F(s)} = 𝑓(𝑡−𝑎)𝑢(𝑡−𝑎) here e-as= e-3s i.e. a=3 i.e u(t-3) And f(s) = f(t)= L-1 = sint f(t-3) = sin (t-3) i.e ans = sin(t-3)u(t-3) Example of L-1 using second shifting throem 3 1 2 [ ] 1 s e L s    2 1 1 s  2 1 1 s 
  • 7. DERIVATIVE PROPERTY (multiplication by t property) { ( )} ( ) L f t F s  if then { ( )} ( 1) { ( )} n n n n d L t f t F s ds   { ( )} ( 1) { ( )} d L t f t F s ds   2 2 2 2 { ( )} ( 1) { ( )} d L t f t F s ds  
  • 8. Example of derivitive property (multiplication by t property) Find the laplace of 2 3 ( ) t L t e { ( )} ( 1) { ( )} n n n n d L t f t F s ds   Comparing with derivative property 2 1 ( 3) ( 3) d s s ds     1 ( ) ( ) 3 d d f s ds ds s   3 2 ( 3) ( ) ( 3) d s s ds    2 1 ( 1) s   3 2 ( 3) s   2 2 3 2 { ( )} ( 1) ( 3) L t f t s    2 2 2 1 { ( )} ( 3) d d f s ds ds s    3 { ( )} { } ( ) t L f t L e F s    1 ( ) 3 F s s   2 3 2 { ( )} ( 3) L t f t s  
  • 9. INTEGRAL PROPERTY (DIVISION by t property) { ( )} ( ) L f t F s  ( ) { } ( ) s f t L F s ds t    2 ( ) { } ( ) s s f t L F s ds ds t      if then
  • 10. Example of integral property (division by t property) Find the laplace of comparing it with equation we get 1 1 { ( )} ( ) 4 6 L f t F s s s      4 6 ( ) ( ) ( ) t t f t L e L e     4 6 ( ) ( ) t t f t e e     4 6 { ( )} ( ) t t L f t L e e     4 4 limln( ) ln( ) 6 6 s s s s s        4 [ln( )] 6 s s s     [ln( 4) ln( 6)]s s s     1 1 ( ) ( ) 4 6 s s F s ds s s         4 0 ln 6 s s     4 1 4 limln( ) ln( ) 6 6 1 s s s s s        1 4 ln( ) 6 s s     6 ln( ) 4 s s    4 6 t t e e t    4 6 { } ( ) t t s e e L F s ds t      
  • 11. The knowledge of Laplace transform has in recent years become an essential part of mathematical background required of engineers and scientists. This is because the transform method an easy and effective means for the solution of many problems arising in engineering. The properties presented are very useful to solve specfic find of laplace transform problems in a easy and more enhanced way. conclusion