Digital Signal Processing Reference
In-Depth Information
Row
Coefficients
1
5
4
3
1
1
1
2
1
1
1
3
4
5
3
24
19
14
2
1
4
1
2
14
19
24
5
575
454
322
29
6
29
322
454
575
7
329784
251712
171984
Check whether the three test criteria are satisfied:
−
1
)
5
R(
D(
1
)
=
D(z)
|
z
=
1
=
15
(
−
1
)
=
3
a
0
=
5
;
a
5
=
1
;
a
0
>
|
a
5
|
c
0
=
24
;
c
4
=
1
;
|
c
0
|
>
|
c
4
|
d
0
=
575
;
d
3
=
29
;
|
d
0
|
>
|
d
3
|
r
0
=
329784
;
r
2
=
171984
;
|
r
0
|
>
|
r
2
|
All criteria are satisfied, and therefore the
D(z)
above has its five zeros inside
the unit circle.
Example 2.20
3
z
4
5
z
3
3
z
2
Now consider another example:
D(z)
=
+
+
+
2
z
+
1. The Jury-
Marden array is constructed as shown below:
JURY-MARDEN ARRAY
Row 1
3
5
3
2
1
2
1
2
3
5
3
3
8
13
6
1
4
1
6
13
8
5
63
98
35
Although we have calculated all the entries in the array, we find that the second
criterion is not satisfied because
(
1
)
4
D(
0. We conclude that there is at
least one zero of
D(z)
that is not
inside
the unit circle. Indeed, it is found that
there is one zero at
z
−
−
1
)
=
=−
1
.
000. It is a good idea to check at the beginning,
whether the first two criteria are satisfied, because if one or both of these two
criteria (which are easy to check) fail, there is no need to compute the entries in
the rows after the first two rows of the Jury-Marden array.
2.9 SOLUTION USING MATLAB FUNCTIONS
In the previous sections, we have described many models for the discrete-time
system and discussed three methods of finding the output of the system when
the input sequence is given, along with initial conditions in some cases.
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