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

648 Chapter 21 The Laplace transform
For any functions f (t)andg(t)
f∗g=g∗f
21.9.1 Theconvolutiontheorem
Let f (t) and g(t) be piecewise continuous functions, with L{f (t)} = F(s) and
L{g(t)} = G(s). The convolution theorem allows us to find the inverse Laplace transform
of a product oftransforms,F(s)G(s):
L −1 {F(s)G(s)} = (f ∗g)(t)
Example21.20 Use the convolution theorem to find the inverse Laplace transforms of the following
functions:
1 3
(a) (b)
(s +2)(s +3) s(s 2 +4)
Solution (a) LetF(s) = 1
s +2 ,G(s) = 1
s +3 .
Then f(t) = L −1 {F(s)} = e −2t ,g(t) = L −1 {G(s)} = e −3t .
{ }
L −1 1
= L −1 {F(s)G(s)} = (f ∗g)(t)
(s +2)(s +3)
=
=
∫ t
0
∫ t
0
e −2(t−v) e −3v dv =
e −2t e −v dv
∫ t
0
e −2t e 2v e −3v dv
= e −2t [−e −v ] t 0 = e−2t (−e −t +1) = e −2t −e −3t
(b) LetF(s) = 3 s ,G(s) = 1
s 2 +4 .Thenf(t) =3,g(t) = 1 sin2t. So,
2
{ }
L −1 3
= L −1 {F(s)G(s)}
s(s 2 +4)
=(f ∗g)(t)
∫ t
∫ t
= 3 sin2v dv = 3 sin2vdv
0 2 2 0
= 3 [ −cos2v
= 3 2 2 4 (1−cos2t)
] t
0