Digital Signal Processing Reference
In-Depth Information
function
G(z)
, which has a simple pole at
z
=
1 and a triple pole at
z
=
2:
z(
2
z
2
−
11
z
+
12
)
G(z)
=
(2.44)
(z
−
1
)(z
−
2
)
3
(
2
z
2
G(z)
z
−
11
z
+
12
)
C
0
C
1
C
2
k
=
=
−
2
)
3
+
−
2
)
2
+
−
2
)
+
−
2
)
3
(z
−
1
)(z
(z
(z
(z
(z
−
1
)
z
=
1
=−
3
(
2
z
2
−
11
z
+
12
)
k
=
(z
−
2
)
3
z
=
2
=−
2
(
2
z
2
−
11
z
+
12
)
C
0
=
(z
−
1
)
z
=
2
=
z
=
2
=−
1
(
2
z
2
2
z
2
d
dz
−
11
z
+
12
)
−
4
z
−
1
C
1
=
(z
−
1
)
(z
−
1
)
2
z
=
2
=
z
=
2
=
3
dz
2
(
2
z
2
2
z
2
d
2
1
2
−
11
z
+
12
)
1
2
d
dz
−
4
z
−
1
C
2
=
(z
−
1
)
(z
−
1
)
2
Therefore we have
−
2
z
−
z
3
z
−
3
z
(z
G(z)
=
−
2
)
3
+
−
2
)
2
+
−
2
)
+
(2.45)
(z
(z
(z
−
1
)
a)
2
is easily obtained from
Table 2.1, as
na
n
u(n)
. We now have to reduce the term
−
Now note that the inverse
z
transform of
az/(z
−
(
2
)
2
z/
−
2
)
2
to
−
z/(z
(
2
)n
2
n
u(n)
.From
the transform pair 6 in Table 2.1, we get the inverse
z
transform of
z/(z
2
)
2
so that its inverse
z
transform is correctly written as
(z
−
−
a)
3
−
−
1
)/
2!
a
n
−
2
u(n)
.
Therefore the inverse
z
transform of
−
2
z/(z
as
n(n
−
2
)
3
is obtained as
−
2
n(n
−
1
)
=
−
n(n
−
1
)
(
2
)
n
−
2
u(n)
(
2
)
n
u(n)
2!
4
Finally, we get the inverse
z
transform of
G(z)
as
−
−
3
u(n)
n(n
−
1
)
n
2
(
2
)
n
(
2
)
n
−
3
(
2
)
n
−
4
2.3.1 More Applications of
z
Transform
In this section, we consider the circuit shown in Figure 2.5 and model it by
equations in the
z
domain, instead of the equivalent model given by equations
(2.1) in the time domain. This example is chosen to illustrate the analysis of a
discrete-time system that has a large number of adders and hence gives rise to
a large number of difference equations in the
z
domain. Writing the
z
transform
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