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320 Chapter 7 FIR FILTER DESIGN
The plots and the zero locations are shown in Figure 7.6. From these plots, we
observe that H r (ω) =0atω =0and at ω = π. There is one zero-quadruplet
constellation, two zero pairs, and zeros at ω =0and ω = π as expected. □
10
Impulse Response
30
Type–3 Amplitude Response
5
20
h(n)
0
Hr
10
−5
0
c(n)
10
5
0
−5
0 5 10
n
c(n) coefficients
−10
0 0.5 1
frequency in π units
Pole–Zero Plot
z–plane
imaginary axis
1
0
−1
0 5 10
n
FIGURE 7.6 Plots in Example 7.6
−1 0 1
real axis
□ EXAMPLE 7.7 Let h(n) = {−4, 1, −1, −2, 5, 6, −6, −5, 2, 1, −1, 4}. Determine the amplitude
↑
response H r (ω) and the locations of the zeros of H (z).
Solution
This is a Type-4 linear-phase FIR filter since M =12and since h (n) isantisymmetric
with respect to α = (12 − 1) /2 =5.5. From (7.16) we have
d(1) =2h ( 12
− 1) =12, d(2) =2h ( 12
− 2) =10,d(3) = 2h ( 12
− 3) = −4
2 2 2
d(4) =2h ( 12
− 4) = −2, d(5) =2h ( 12
− 5) =2, d(6) = 2h ( 12
− 6) = −8
2 2 2
Hence from (7.17) we obtain
[ (
H r(ω) =d(1) sin ω 1 − 1 )] [ (
+ d(2) sin ω 2 − 1 )] [ (
+ d(3) sin ω 3 − 1 )]
2
2
2
[ (
+d(4) sin ω 4 − 1 )] [ (
+ d(5) sin ω 5 − 1 )] [ (
+ d(6) sin ω 6 − 1 )]
2
2
2
( ) ( ) ( ) ( )
ω 3ω 5ω 7ω
=12sin +10sin − 4 sin − 2 sin
2 2
2
2
( ) ( )
9ω 11ω
+2 sin − 8 sin
2
2
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