Digital Signal Processing Reference
In-Depth Information
Fig. 3.2. Output response of the
system considered in Example
3.3.
6
5
4
3
w
(
t
)
2
w
zs
(
t
)
1
0
w
zi
(
t
)
t
−1
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Eq. (3.19) and solving the resulting equation yields
K
=
4. The zero-state
response of the system is, therefore, given by
−
4
t
−
3
t
−
t
)
u
(
t
)
.
w
zs
(
t
)
=
(
C
1
e
+
C
2
e
+
4e
To compute the values of constants
C
1
and
C
2
, we use the initial conditions
w
(0
−
−
)
=
0. Substituting the initial conditions in
w
zs
(
t
) leads to
the following simultaneous equations:
)
=
0 and
w
(0
+
4
=
0
,
−
4
C
1
−
3
C
2
−
4
=
0
,
C
1
+
C
2
with solutions
C
1
=
8 and
C
2
=−
12. The zero-state solution of Eq. (3.18) is,
therefore, given by
−
4
t
−
3
t
−
t
)
u
(
t
)
.
w
zs
(
t
)
=
(8e
−
12e
+
4e
(iii) Overall response of the system
The overall response of the system can
be obtained by summing up the zero-input and zero-state responses, and can be
expressed as
−
4
t
+
8e
−
3
t
+
4e
−
t
)
u
(
t
)
.
w
(
t
)
=
(
−
7e
The zero-input, zero-state, and overall responses of the system are plotted in
Fig. 3.2.
Section 3.1 presented the procedure for calculating the output response of a
LTIC system by directly solving its input-output relationship expressed in the
form of a differential equation. However, there is an alternative and more con-
venient approach to calculate the output based on the impulse response of a
system. This approach is developed in the following sections.
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