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

21.9 The convolution theorem 647
21.9 THECONVOLUTIONTHEOREM
Let f (t) andg(t) be two piecewise continuous functions. The convolution of f (t)
andg(t), denoted (f ∗g)(t), isdefined by
(f ∗g)(t)=
∫ t
0
f(t −v)g(v)dv
Example21.18 Find the convolution of2t andt 3 .
Solution f (t) = 2t,g(t) =t 3 , f(t −v) =2(t −v),g(v) = v 3 .Then
2t∗t 3 =
∫ t
0
2(t −v)v 3 dv =2
∫ t
0
tv 3 −v 4 dv
[ tv
4
= 2
4 − v5
5
= t5
10
] t
0
[ ] t
5
= 2
4 − t5 5
Itcan beshown that
f∗g=g∗f
butthe proof isomitted. Instead, thispropertyisillustratedby anexample.
Example21.19 Show that f ∗g=g ∗ f where f (t) = 2t andg(t) =t 3 .
Solution f ∗g= 2t ∗t 3 =t 5 /10 by Example 21.18. Fromthe definitionofconvolution
(g∗f)(t)=
∫ t
0
g(t −v)f(v)dv
We haveg(t) =t 3 ,sog(t −v) = (t −v) 3 ,and f(v) =2v.Therefore
g∗f=t 3 ∗2t=
=
= 2
∫ t
0
∫ t
0
∫ t
(t −v) 3 2vdv
(t 3 −3t 2 v +3tv 2 −v 3 )2vdv
0
t 3 v −3t 2 v 2 +3tv 3 −v 4 dv
[ t 3 v 2
] t
= 2
2 −t2 v 3 + 3tv4
4 − v5
= t5
5
0
10