Digital Signal Processing Reference
In-Depth Information
This result is easily verified from the system equation, (3.20):
t
t
t
t
2
2
u(t).
y(t) =
L
x(t) dt =
L
tu(t) dt =
L
t dt =
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-
q
-
q
0
Example 3.2 illustrates that the convolution integral can be used to find the
output of an LTI system, given its input. The example also illustrates that as a prac-
tical matter, other methods are used; the convolution integral is seldom the most
efficient procedure for finding the response of a system. Instead, the convolution in-
tegral is normally used in developing the properties of LTI systems and in develop-
ments involving the use of LTI systems. In practice, computer solutions of the
system equations, called
system simulations,
are used to find system responses. Four
additional examples in convolution will now be given.
Convolution for the system of Example 3.1
EXAMPLE 3.3
Consider the system of Example 3.1 for the case that the time increment between impulses
approaches zero. If we let
¢=0.1,
the input signal in Example 3.1 can be written as
x(t) =
q
k= 0
q
k=-
q
0.1d(t - 0.1k) =
u(k¢)d(t - k¢) (¢),
because
u(k¢) = 0
for
k 6 0
and
u(k¢) = 1
for
k Ú 1.
From (3.12), as
¢ : 0,
the input sig-
nal becomes
q
x(t) =
3
u(t)d(t - t)dt = u(t).
-
q
From (3.18), the output signal is calculated from
q
q
u(t)e
-(t -t)
u(t - t)dt.
y(t) =
3
x(t)h(t - t)dt =
3
-
q
-
q
Because
u(t) = 0
for
t 6 0,
and
u(t - t) = 0
for
t 7 t,
this convolution integral can be
rewritten as
t
t
e
-(t -t)
dt = e
-t
e
t
dt = (1 - e
-t
)u(t).
y(t) =
3
3
0
0
The output signal is plotted in Figure 3.4. Compare this result with the summation result
shown in Figure 3.2(d).
y
(
t
)
1
0.5
Figure 3.4
System output signal for
0
0
2
4
6
8
10
t
Example 3.3.
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