Page 66 - LAPLACE TRANSFORM

Example 2

Apply the Partial Fraction method to find the Inverse Laplace Transform
of the following.

s
7 − 1
s
( + 1 )( + 2 )( − ) 3
s
s
7 − 1 = A + B +
s
C
( + 1 )( + 2 )( − ) 3 ( + ) 1 ( + ) 2 ( − ) 3
s
s
s
s
s
s

7 − = A + B + C
s 1
s 3
s 3
s 1
( + )( s + 2 )( − ) ( + ) s ( + ) 2 ( − )
s 1

7 s 1 =− A( s + 2 )( s 3 +− ) B( + )( s 3 +− ) C( + )( s + 2 )....... eq 1
s 1
s 1
if s = − 1
( 7 − ) 1 − 1 = A 1(− + 2 )(− 1− ) 3 + B )0( + C )0(

− 8 = A 4(− )
A = 2
if s = − 2
( 7 − ) 2 − 1 = A )0( + B 2(− + 1 )(− 2 − ) 3
− 15 = B )5(
B = − 3
if s = 3
) 3 ( 7 − 1 = A )0( + B )0( + C 3( + 1 3)( + ) 2
20 = C 20( )
C = 1
2 3 1
∴ Partial Fraction = − +
( + ) s ( + ) 2 ( − )
s 1
s 3
 2   3   1 
∴ Inverse Laplace Transform = L 1 −   − L 1 −   + L 1 −  
s 3
s 1
 ( + )   s ( + ) 2   ( − ) 
= e 2 t − − e 3 − t 2 + e t 3

59