Digital Signal Processing Reference
In-Depth Information
2.
The second integration applies for
0 F t F 2
and is illustrated in Figure 3.9(b):
q
0
y
2
(t) =
L
x
2
(t)h(t - t) dt =
L
(0)h(t - t) dt
-
q
-
q
q
t
3e
-0.5t
dt +
L
+
L
x
2
(t)(0) dt
0
t
3e
-0.5t
-0.5
t
= 6(1 - e
-0.5t
),
=
0 F t F 2.
0
2 F t F
q
:
3.
Figure 3.9(c) applies for
t
3e
-0.5t
-0.5
t
3e
-0.5t
dt =
= 6(e
-0.5(t - 2)
- e
-0.5t
)
y
2
(t) =
L
t - 2
t - 2
= 6e
-0.5t
(e
1
- 1) = 10.31e
-0.5t
,
2 F t 6
q
.
The output
y(t)
is plotted in Figure 3.10.
In Example 3.5, the time axis is divided into three ranges. Over each range, the
convolution integral reduces to the form
t
b
y(t) =
L
x(t)h(t - t) dt,
t
i
F t F t
j
,
(3.21)
t
a
where the limits
t
a
and
t
b
are either constants or functions of time
t
. The integral
applies for the range
t
i
F t F t
j
, where
t
i
and
t
j
are constants. Hence, three different
integrals of the form of (3.21) are evaluated.
y
(
t
)
3.793
4
y
2
(
t
)
y
2
(
t
)
3
2
y
1
(
t
)
1
3
1
0
2
4
6
t
Figure 3.10
Output signal for Example 3.5.
■
EXAMPLE 3.6
A system with a time-delayed exponential impulse response
h(t) = e
-t
u(t - 1)
Consider a system with an impulse response of and an input signal
The system's impulse response and the input signal are shown in Fig-
ure 3.11(a) and (b), respectively. The system's output is
x(t) = e
t
u(-1 - t).
y(t) = x(t)
*
h(t)
, from (3.16).
