INTEGRAL TRANSFORMS
Figure 13.7 Two representations of the Laplace transform convolution (see
text).
where the integral in the brackets on the LHS is theconvolutionoffandg,
denoted byf∗g. As in the case of Fourier transforms, the convolution defined
above is commutative, i.e.f∗g=g∗f, and is associative and distributive. From
(13.64) we also see that
L−^1
[
f ̄(s) ̄g(s)
]
=
∫t
0
f(u)g(t−u)du=f∗g.
Prove the convolution theorem (13.64) for Laplace transforms.
From the definition (13.64),
f ̄(s) ̄g(s)=
∫∞
0
e−suf(u)du
∫∞
0
e−svg(v)dv
=
∫∞
0
du
∫∞
0
dv e−s(u+v)f(u)g(v).
Now lettingu+v=tchanges the limits on the integrals, with the result that
f ̄(s) ̄g(s)=
∫∞
0
du f(u)
∫∞
u
dt g(t−u)e−st.
As shown in figure 13.7(a) the shaded area of integration may be considered as the sum
of vertical strips. However, we may instead integrate over this area by summing over
horizontal strips as shown in figure 13.7(b). Then the integral can be written as
f ̄(s) ̄g(s)=
∫t
0
du f(u)
∫∞
0
dt g(t−u)e−st
=
∫∞
0
dt e−st
{∫t
0
f(u)g(t−u)du
}
=L
[∫t
0
f(u)g(t−u)du