Digital Signal Processing Reference
In-Depth Information
When the input
x(n)
is zero,
X(z)
=
0; hence the second term on the right side
is zero, leaving only the first term due to initial conditions given. It is the
z
transform of the zero input response
y
0
i
(n)
.
The inverse
z
transform of this first term on the right side
−
0
.
02
z
−
1
y(
[0
.
3
y(
−
1
)
−
1
)
−
0
.
02
y(
−
2
)
]
Y
0
i
(z)
=
[1
−
0
.
3
z
−
1
+
0
.
02
z
−
2
]
gives the response when the input is zero, and so it is the zero input response
y
0
i
(n)
. The inverse
z
transform of
X(z)
[1
−
0
.
1
z
−
1
]
0
.
02
z
−
2
]
=
Y
0
s
(z)
[1
−
0
.
3
z
−
1
+
gives the response when the initial conditions (also called the initial states) are
zero, and hence it is the zero state response
y
0
s
(n)
.
Substituting the values of the initial states and for
X(z)
, we obtain
[0
.
288
−
0
.
02
z
−
1
]
[1
−
0
.
3
z
−
1
[0
.
288
z
2
−
0
.
02
z
]
z
[0
.
288
z
−
0
.
02]
Y
0
i
(z)
=
+
0
.
02
z
−
2
]
=
+
0
.
02
=
z
2
−
0
.
3
z
(z
−
0
.
1
)(z
−
0
.
2
)
and
X(z)
[1
−
0
.
1
z
−
1
]
[1
−
0
.
1
z
−
1
]
z
Y
0
s
(z)
=
0
.
02
z
−
2
]
=
[1
−
0
.
3
z
−
1
+
z
+
0
.
2
[1
−
0
.
3
z
−
1
+
0
.
02
z
−
2
]
z
[
z
2
z
2
(z
−
0
.
1
z
]
−
0
.
1
)
=
0
.
02
)
=
(z
+
0
.
2
)(z
2
−
0
.
3
z
+
(z
+
0
.
2
)(z
−
0
.
1
)(z
−
0
.
2
)
We notice that there is a pole and a zero at
z
=
0
.
1 in the second term on the
right, which cancel each other, and
Y
0
s
(z)
reduces to
z
2
/
[
(z
+
0
.
2
)(z
−
0
.
2
)
]. We
divide
Y
0
i
(z)
by
z
, expand it into its normal partial fraction form
Y
0
i
(z)
z
[0
.
288
z
−
0
.
02]
0
.
376
(z
0
.
088
(z
=
−
0
.
2
)
=
−
0
.
2
)
−
(z
−
0
.
1
)(z
−
0
.
1
)
and multiply by
z
to get
0
.
376
z
(z
0
.
088
z
(z
Y
0
i
(z)
=
−
0
.
2
)
−
−
0
.
1
)
Similarly, we expand
Y
0
s
(z)/z
=
z/
[
(z
+
0
.
2
)(z
−
0
.
2
)
] in the form
−
0
.
5
/(z
+
0
.
2
)
+
0
.
5
/(z
−
0
.
2
)
and get
z
2
−
0
.
5
z
(z
0
.
5
z
Y
0
s
(z)
=
−
0
.
2
)
=
+
0
.
2
)
+
(z
+
0
.
2
)(z
(z
−
0
.
2
)
=
[0
.
376
(
0
.
2
)
n
−
0
.
088
(
0
.
1
)
n
]
u(n)
Therefore, the zero input response is
y
0
i
(n)
−
0
.
2
)
n
(
0
.
2
)
n
]
u(n)
.
and the zero state response is
y
0
s
(n)
=
0
.
5[
−
(
+
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