Digital Signal Processing Reference
In-Depth Information
Solving these two equations, we get
A
1
=
9
.
903 and
A
2
=−
8
.
4. So the total
response is
=
9
.
903
(
0
.
3
)
n
−
8
.
4
(
0
.
2
)
n
(
0
.
1
)
n
y(n)
+
Example 2.13
Let us reconsider Example 2.10. The zero input response and the zero state
response in this example were found to be
[0
.
376
(
0
.
2
)
n
0
.
088
(
0
.
1
)
n
]
u(n)
y
0
i
(n)
=
−
−
0
.
2
)
n
(
0
.
2
)
n
]
u(n)
=
[0
.
5
(
0
.
2
)
n
−
0
.
2
)
n
]
u(n)
y
0
s
(n)
=
0
.
5[
−
(
+
−
0
.
5
(
The characteristic polynomial for the system given in Example 2.10 is easily
seen as [1
0
.
02
z
−
2
]. After multiplying it by
z
2
,wefindthenatu-
ral frequencies as the zeros of [
z
2
0
.
3
z
−
1
−
+
−
0
.
3
z
+
0
.
02]
=
[
(z
−
0
.
2
)(z
−
0
.
1
)
]tobe
(c
1
)
(
0
.
1
)
. Note that the zero input response
y
0
i
(n)
has a
term 0
.
376
(
0
.
2
)
n
u(n)
, which has the natural frequency equal to
(
0
.
2
)
and the
term
=
(
0
.
2
)
and
(c
2
)
=
0
.
088
(
0
.
1
)
n
u(n)
with the natural frequency of
(
0
.
1
)
, while the zero state
response
y
0
s
(n)
also contains the term 0
.
5
(
0
.
2
)
n
u(n)
with the natural frequency
of
(
0
.
2
)
. We also noticed that the pole of
Y
0
s
(z)
at
z
−
=
0
.
1 was canceled by a zero
at
z
=
0
.
1. Therefore there is no term in the zero state response
y
0
s
(n)
with the
natural frequency of
(
0
.
1
)
. So the term containing the natural frequency of
(
0
.
2
)
is
the sum 0
.
5
(
0
.
2
)
n
u(n)
=
0
.
876
(
0
.
2
)
n
u(n)
, whereas the other
term with the natural frequency of
(
0
.
1
)
is
−
0
.
088
(
0
.
1
)
n
u(n)
. Consequently, the
natural response of the system is 0
.
876
(
0
.
2
)
n
u(n)
+
0
.
376
(
0
.
2
)
n
u(n)
−
0
.
088
(
0
.
1
)
n
u(n)
.
−
0
.
2
)
n
u(n)
is the forced response with the fre-
The remaining term
−
0
.
5
(
quency
(
−
0
.
2
)
, which is found in the forcing function or the input function
−
0
.
2
)
n
u(n)
. Thus the total response of the system is now expressed as
the sum of its natural response 0
.
876
(
0
.
2
)
n
u(n)
x(n)
=
(
−
0
.
088
(
0
.
1
)
n
u(n)
and forced
0
.
2
)
n
u(n)
. We repeat that in the zero state response, there are
terms with natural frequencies of the system, besides terms with input frequen-
cies; hence it is erroneous to state that the zero input response is equal to the
natural response or that the zero state response is the forced response.
−
0
.
5
(
−
response
Example 2.14
As another example, let us analyze the discrete-time system model shown in
Figure 2.6. Assuming the initial states are zero, we get the equations for the
outputs of the two adders as
y
1
(n)
=
x(n
−
1
)
−
0
.
2
y
1
(n
−
1
)
−
0
.
4
y
1
(n
−
2
)
y
2
(n)
=
2
y
1
(n
−
1
)
−
0
.
1
y
2
(n
−
1
)
When a discrete-time system is described by several linear difference equations
like the equations above, it is difficult to derive a single-input, single-output
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