Digital Signal Processing Reference
In-Depth Information
p
=
2
.
0000
+
0
.
0000
i
2
.
0000
−
0
.
0000
i
2
.
0000
1
.
0000
k
=
[]
From these data, we construct the partial fraction expansion as
3
(
1
−
2
z
−
1
)
+
0
.
5
(
1
−
2
z
−
1
)
2
−
0
.
5
(
1
−
2
z
−
1
)
3
−
3
(
1
−
G
1
(z)
=
z
−
1
)
which can be reduced to the equivalent expression
0
.
5
z
2
(z
0
.
5
z
3
(z
3
z
3
z
G(z)
=
2
)
+
2
)
2
−
2
)
3
−
(z
−
−
−
(z
−
1
)
which differs from the partial fraction expansion shown in (2.45) or (2.79). But
let us expand
0
.
5
z
2
(z
z
0
.
5
z
(z
=
−
2
)
2
+
−
2
)
2
(z
−
2
)
and
0
.
5
z
3
(z
−
2
z
(z
2
z
0
.
5
z
(z
−
=
−
2
)
3
−
−
2
)
2
−
−
2
)
3
(z
−
2
)
Substituting these expressions in the preceding form for
G(z)
,weget
−
2
z
(z
z
3
z
3
z
G(z)
=
−
2
)
3
−
−
2
)
2
+
−
2
)
−
(z
(z
(z
−
1
)
which is exactly the same as the form obtained in (2.79).
Example 2.31
We can use a MATLAB function
deconv(b,a)
to find a few values in the inverse
z
transform of a transfer function, and it is based on the recursive formula given
by (2.65). Let us select the transfer function (2.67) to illustrate this function.
%MATLAB
program to find a few samples of the inverse
z
transform
b = [0.1 0.25 0];
a = [1 0.4 0.5];
n=5;
b= [b zeros(1, n-1)];
[x,r] = deconv(b,a)
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