Page 464 - Mechanical Engineers' Handbook (Volume 2)

2 Laplace Transforms 455

B(s)
F(s) 1 2 r
A(s) s p 1 (s p ) 2 (s p ) r
1
1
j (24)
n
j r 1 s p j
The ’s can be evaluated as before using Eq. (22). The ’s are given by
j
i
B(s) (s p )
r
r
A(s) 1 s p 1
B(s) (s p )
d

r
r 1
ds A(s) 1
s p 1

B(s) (s p )
1
d
j
r j 1 r
j! ds j A(s)
s p 1

r 1
d
1 B(s) (s p )
r
1 r 1 1
(r 1)! ds A(s)
s p 1
The inverse Laplace transform of Eq. (24) can then be obtained as follows:
ƒ(t) r r 1 r 1 r 2 t e pt 1
(r 1)! t (r 2)! t 2 1
e pt 1 t 0 (25)
n
j r 1 j
Example 7 Find the inverse Laplace transform of the function
2
s 2s 3
F(s)
(s 1) 3
Expanding F(s) into partial fractions, we obtain
B(s)
F(s) 1 2 3
A(s) s 1 (s 1) 2 (s 1) 3
where , , and are determined as already shown:
1
2
3
B(s) (s 1)
3
3
A(s) s 1 2
similarly 0 and 1. Thus we get
2
1
ƒ(t) L L 2
1
1
1
s 1 (s 1) 3
e t te t 0
2 t