The key to evaluating a convolution integral such as
x
(
t
)
*
h
(
t
)
=
∫
-
∞
∞
x
(
τ
)
h
(
t
-
τ
)
d
τ
x
(
t
)
*
h
(
t
)
=
∫
-
∞
∞
x
(
τ
)
h
(
t
-
τ
)
d
τ
(1)
is to realize that as far as the integral is concerned, the variable tt is a constant and the integral is over the variable ττ. Therefore, for each tt, we are finding the area of the product x(τ)h(t-τ)x(τ)h(t-τ). Let's look at an example that illustrates how this works.
Example 3.1
Find the convolution of x(t)=u(t)x(t)=u(t) and h(t)=e-tu(t)h(t)=e-tu(t). The convolution integral is given by
h
(
t
)
*
x
(
t
)
=
∫
-
∞
∞
e
-
τ
u
(
τ
)
u
(
t
-
τ
)
d
τ
h
(
t
)
*
x
(
t
)
=
∫
-
∞
∞
e
-
τ
u
(
τ
)
u
(
t
-
τ
)
d
τ
(2)
Figure 1 shows the graph of e-τu(τ)e-τu(τ), e-tu(t)e-tu(t), and their product. From the graph of the product, it is easy to see the the convolution integral becomes
∫
0
t
e
-
τ
d
τ
=
1
-
e
-
t
,
t
≥
0
0
,
t
<
0
∫
0
t
e
-
τ
d
τ
=
1
-
e
-
t
,
t
≥
0
0
,
t
<
0
(3)
Signals which can be expressed in functional form should be convolved as in the above example. Other signals may not have an easy functional representation but rather may be piece-wise linear. In order to convolve such signals, one must evaluate the convolution integral over different intervals on the tt-axis so that each distinct interval corresponds to a different expression for x(t)*h(t)x(t)*h(t). The following example illustrates this:
Example 3.2
Suppose we attempt to convolve the unit step function x(t)=u(t)x(t)=u(t) with the trapezoidal function
h
(
t
)
=
t
,
0
≤
t
<
1
1
,
1
≤
t
<
2
0
,
elsewhere
h
(
t
)
=
t
,
0
≤
t
<
1
1
,
1
≤
t
<
2
0
,
elsewhere
(4)
From Figure 2, it can be seen that on the interval 0≤t<10≤t<1, the product x(t-τ)h(τ)x(t-τ)h(τ) is an equilateral triangle with area t2/2t2/2. On the interval 1≤t<21≤t<2, the area of x(t-τ)h(τ)x(t-τ)h(τ) is t-1/2t-1/2. This latter area results by adding the area of an equilateral triangle having a base of 1, and the area of a rectangle having a base of t-1t-1 and a height of 1. For all values of tt greater than 2, the convolution is 1.5 since x(t-τ)h(τ)=h(τ)x(t-τ)h(τ)=h(τ) and h(τ)h(τ) is a trapezoid having an area of 1.5. Finally, for t<0t<0, the convolution is zero since x(t-τ)h(τ)=0x(t-τ)h(τ)=0.
