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236 Chapter 6 IMPLEMENTATION OF DISCRETE-TIME FILTERS
A practical problem with the structure in Figure 6.16 is that it has
poles on the unit circle, which makes this filter critically unstable. If the
filter is not excited by one of the pole frequencies, then the output is
bounded. We can avoid this problem by sampling H (z) onacircle |z| = r,
where the radius r is very close to 1 but is less than 1 (e.g., r =0.99),
which results in
H (z) = 1 − rM z −M
M
M−1
∑
k=0
H (k)
(
1 − rW −k ; H (k) =H re j2πk/M)
M z−k (6.15)
Now approximating H ( re j2πk/M) ≈ H ( e j2πk/M) for r ≈ 1, we can obtain
a stable structure similar to the one in Figure 6.16 containing real values.
This is explored in Problem P6.20.
□ EXAMPLE 6.6 Let h (n) = 1 9 {1 ↑, 2, 3, 2, 1}. Determine and draw the frequency sampling form.
Solution
MATLAB script:
>> h = [1,2,3,2,1]/9; [C,B,A] = dir2fs(h)
C =
0.5818
0.0849
1.0000
B =
-0.8090 0.8090
0.3090 -0.3090
A =
1.0000 -0.6180 1.0000
1.0000 1.6180 1.0000
1.0000 -1.0000 0
Since M =5is odd, there is only one 1st-order section. Hence
[
H (z) = 1 − z−5 −0.809 + 0.809z−1
0.5818
5
1 − 0.618z −1 + z −2
]
0.309 − 0.309z−1
+0.0848
1+1.618z −1 + z + 1
−2 1 − z −1
The frequency-sampling form is shown in Figure 6.17.
□
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