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(tτ)x(τ)h(tτ). 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)=etu(t)h(t)=etu(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 eτu(τ)eτu(τ), etu(t)etu(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 piecewise linear. In order to convolve such signals, one must evaluate the convolution integral over different intervals on the ttaxis 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(tτ)h(τ)x(tτ)h(τ) is an equilateral triangle with area t2/2t2/2. On the interval 1≤t<21≤t<2, the area of x(tτ)h(τ)x(tτ)h(τ) is t1/2t1/2. 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 t1t1 and a height of 1. For all values of tt greater than 2, the convolution is 1.5 since x(tτ)h(τ)=h(τ)x(tτ)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(tτ)h(τ)=0x(tτ)h(τ)=0.