082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

21.6 Inverse Laplace transforms 639
Example21.12 Find the inverse Laplace transformsof the following functions:
10 (s+1)
(a) (b) (c)
(s +2) 4 (s+1) 2 +4
{ }
Solution (a) L −1 10
= 10 { }
6
(s +2) 4 6 L−1 (s +2) 4
15
(s−1) 2 −9
= 5t3 e −2t
3
{ } { }
(b) L −1 s +1
= L −1 s +1
= e −t cos2t
(s+1) 2 +4 (s+1) 2 +2 2
{ } { }
(c) L −1 15
= 5L −1 3
= 5e t sinh3t
(s−1) 2 −9 (s−1) 2 −3 2
The function is written to match exactly the standard forms given in Table 21.1, with
possibly a constant factor being present. Often the denominator needs to be written in
standard form asillustratedinthe next example.
Example21.13 Findthe inverse Laplace transformsofthe following functions:
(a)
s +3
s 2 +6s+13
(b)
2s+3
s 2 +6s+13
Solution (a) By completing the square we can write
(c)
s 2 +6s+13=(s+3) 2 +4=(s+3) 2 +2 2
s −1
2s 2 +8s+11
(b)
Hence we may write
{ } { }
L −1 s +3
= L −1 s +3
= e −3t cos2t
s 2 +6s+13 (s+3) 2 +2 2
2s+3
s 2 +6s+13 = 2s+3
(s+3) 2 +2 = 2s+6
2 (s+3) 2 +2 − 3
2 (s+3) 2 +2 2
(
(
= 2
s +3
(s+3) 2 +2 2 )
− 3 2
)
2
(s+3) 2 +2 2
The expressions in brackets are standard forms so their inverse Laplace transforms
can befound fromTable 21.1.
{ }
L −1 2s+3
= 2e −3t cos2t − 3e−3t sin2t
s 2 +6s+13
2
(c) We write the expression usingstandard forms:
s −1
2s 2 +8s+11 = 1 s −1
2s 2 +4s+5.5 = 1 s −1
2 (s +2) 2 +1.5
= 1 (
)
s +2
2 (s +2) 2 +1.5 − 3
(s +2) 2 +1.5
= 1 (
s +2
2 (s +2) 2 +1.5 − √ 3
√ ) 1.5
1.5 (s +2) 2 +1.5