Let
f
(
t
)
and
g
(
t
)
be two functions of
t
. The
convolution
of
f
(
t
)
and
g
(
t
)
is also a function of
t
, denoted by
(
f
∗
g
)
(
t
)
and is defined by the relation
(
f
∗
g
)
(
t
)
=
∫
−
∞
∞
f
(
t
−
x
)
g
(
x
)
d
x
However if
f
and
g
are both
causal
functions then (strictly)
f
(
t
)
,
g
(
t
)
are written
f
(
t
)
u
(
t
)
and
g
(
t
)
u
(
t
)
respectively, so that
(
f
∗
g
)
(
t
)
=
∫
−
∞
∞
f
(
t
−
x
)
u
(
t
−
x
)
g
(
x
)
u
(
x
)
d
x
=
∫
0
t
f
(
t
−
x
)
g
(
x
)
d
x
because of the properties of the step functions:
u
(
t
−
x
)
=
0
if
x
>
t
and
u
(
x
)
=
0
if
x
<
0
.
Convolution
If
f
(
t
)
and
g
(
t
)
are causal functions then their convolution is defined by:
(
f
∗
g
)
(
t
)
=
∫
0
t
f
(
t
−
x
)
g
(
x
)
d
x
This is an odd looking definition but it turns out to have considerable use both in Laplace transform theory and in the modelling of linear engineering systems. The reader should note that the variable of integration is
x
. As far as the integration process is concerned the
t
-variable is (temporarily) regarded as a constant.
Find the convolution of
f
and
g
if
f
(
t
)
=
t
u
(
t
)
and
g
(
t
)
=
t
2
u
(
t
)
.
f
(
t
−
x
)
=
(
t
−
x
)
u
(
t
−
x
)
and
g
(
x
)
=
x
2
u
(
x
)
Therefore
(
f
∗
g
)
(
t
)
=
∫
0
t
(
t
−
x
)
x
2
d
x
=
1
3
x
3
t
−
1
4
x
4
0
t
=
1
3
t
4
−
1
4
t
4
=
1
12
t
4
Find the convolution of
f
(
t
)
=
t
.
u
(
t
)
and
g
(
t
)
=
sin
t
.
u
(
t
)
.
Here
f
(
t
−
x
)
=
(
t
−
x
)
u
(
t
−
x
)
and
g
(
x
)
=
sin
x
.
u
(
x
)
and so
(
f
∗
g
)
(
t
)
=
∫
0
t
(
t
−
x
)
sin
x
d
x
We need to integrate by parts. We find, remembering again that
t
is a constant in the integration process,
∫
0
t
(
t
−
x
)
sin
x
d
x
=
[
−
(
t
−
x
)
cos
x
]
0
t
−
∫
0
t
(
−
1
)
(
−
cos
x
)
d
x
=
0
+
t
−
∫
0
t
cos
x
d
x
=
t
−
[
sin
x
]
0
t
=
t
−
sin
t
so that
(
f
∗
g
)
(
t
)
=
t
−
sin
t
or, equivalently, in this case
(
t
∗
sin
t
)
(
t
)
=
t
−
sin
t
In Example 4 we found the convolution of
f
(
t
)
=
t
.
u
(
t
)
and
g
(
t
)
=
sin
t
.
u
(
t
)
. In this Task you are asked to find the convolution
(
g
∗
f
)
(
t
)
that is, to reverse the order of
f
and
g
.
Begin by writing
(
g
∗
f
)
(
t
)
as an appropriate integral:
Answer
g
(
t
−
x
)
=
sin
(
t
−
x
)
.
u
(
t
−
x
)
and
f
(
x
)
=
x
u
(
x
)
, so
(
g
∗
f
)
(
t
)
=
∫
0
t
sin
(
t
−
x
)
.
x
d
x
Now evaluate the convolution integral:
Answer
(
g
∗
f
)
(
t
)
=
∫
0
t
sin
(
t
−
x
)
.
x
d
x
=
[
x
cos
(
t
−
x
)
]
0
t
−
∫
0
t
cos
(
t
−
x
)
d
x
=
t
−
0
+
sin
(
t
−
x
)
0
t
=
t
−
sin
t
This Task illustrates the general result in the following Key Point:
Commutativity Property of Convolution
(
f
∗
g
)
(
t
)
=
(
g
∗
f
)
(
t
)
In words: the convolution of
f
(
t
)
with
g
(
t
)
is the same as the convolution of
g
(
t
)
with
f
(
t
)
.
Obtain the Laplace transforms of
f
(
t
)
=
t
.
u
(
t
)
and
g
(
t
)
=
sin
t
.
u
(
t
)
and
(
f
∗
g
)
(
t
)
.
Begin by finding
L
{
f
(
t
)
}
,
L
{
g
(
t
)
}
:
Answer
L
{
f
(
t
)
}
=
1
s
2
L
{
g
(
t
)
}
=
1
s
2
+
1
(from Table 1)
Now find
L
{
(
f
∗
g
)
(
t
)
}
:
Answer
From Example 4
(
f
∗
g
)
(
t
)
=
t
−
sin
t
and so
L
{
(
f
∗
g
)
(
t
)
}
=
L
{
t
−
sin
t
}
=
1
s
2
−
1
s
2
+
1
Now compare
L
{
f
(
t
)
}
×
L
{
g
(
t
)
}
with
L
{
f
∗
g
(
t
)
}
. What do you observe?
Answer
L
{
(
f
∗
g
)
(
t
)
}
=
1
s
2
−
1
s
2
+
1
=
1
s
2
1
s
2
+
1
=
L
{
f
(
t
)
}
L
{
g
(
t
)
}
=
F
(
s
)
G
(
s
)
We see that the Laplace transform of the convolution of
f
(
t
)
and
g
(
t
)
is the product of their separate Laplace transforms. This, in fact, is a general result which is expressed in the statement of the
convolution theorem
which we discuss in the next subsection.