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

642 Chapter 21 The Laplace transform
and so
( ) 6s+8
L −1 s 2 +3s+2
=2e −t +4e −2t
Example21.15 Findthe inverse Laplace transformof 3s2 +6s+2
s 3 +3s 2 +2s .
Solution Following fromthe work inSection 1.7.1,we have
3s 2 +6s+2
s 3 +3s 2 +2s = 1 s + 1
s +1 + 1
s +2
The inverse Laplace transform of the partialfractions iseasily found:
{ } 3s
L −1 2 +6s+2
=1+e −t +e −2t
s 3 +3s 2 +2s
Example21.16 Findthe inverse Laplace transformof
Solution From Example 1.33 wehave
3s 2 +11s +14
s 3 +2s 2 −11s−52 .
3s 2 +11s +14
s 3 +2s 2 −11s−52 = 2
s −4 + s +3
s 2 +6s+13
We find the inverse Laplace transformsof the partialfractions:
{ } 2
L −1 =2e 4t
s −4
{ }
L −1 s +3
= e −3t cos2t
s 2 +6s+13
So
{ } 3s
L −1 2 +11s +14
= 2e 4t +e −3t cos2t
s 3 +2s 2 −11s−52
EXERCISES21.7
1 Expressthe followingaspartialfractions andhence
findtheir inverse Laplacetransforms:
(a)
(b)
(c)
5s+2
(s +1)(s +2)
3s+4
(s +2)(s +3)
4s+1
(s +3)(s +4)
(d)
(e)
(f)
(g)
6s−5
(s +5)(s +3)
4s+1
s(s +2)(s +3)
7s+3
s(s +3)(s +4)
6s+7
s(s +2)(s +4)