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318 Chapter 7 FIR FILTER DESIGN
Solution
This is a Type-2 linear-phase FIR filter since M =12and since h (n) issymmetric
with respect to α = (12 − 1) /2 =5.5. From (7.10) we have
b(1) =2h ( 12
2 − 1) =12, b(2) = 2h ( 12
2 − 2) =10,b(3) = 2h ( 12
2 − 3) = −4
b(4) =2h ( 12
2 − 4) = −2, b(5) = 2h ( 12
2 − 5) =2, b(6) = 2h ( 12
2 − 6) = −8
Hence from (7.11) we obtain
H r(ω) =b(1) cos [ ω ( [ ( )] [ ( )]
1 − 2)] 1 + b(2) cos ω 2 −
1
2 + b(3) cos ω 3 −
1
2
)] [ ( + b(6) cos ω 6 −
1
+ b(4) cos [ ω ( [ (
4 − 2)] 1 + b(5) cos ω 5 −
1
2
( ) ( ) ( ω 3ω 5ω
=12cos +10cos − 4 cos
2 2
2
( ) ( )
9ω 11ω
+2cos − 8 cos
2
2
MATLAB script:
)
− 2 cos
( 7ω
2
)
2
)]
>> h = [-4,1,-1,-2,5,6,6,5,-2,-1,1,-4];
>> M = length(h); n = 0:M-1; [Hr,w,a,L] = Hr_Type2(h);
>> b,L
b = 12 10 -4 -2 2 -8
L = 6
>> bmax = max(b)+1; bmin = min(b)-1;
>> subplot(2,2,1); stem(n,h); axis([-1 2*L+1 bmin bmax])
>> xlabel(’n’); ylabel(’h(n)’); title(’Impulse Response’)
>> subplot(2,2,3); stem(1:L,b); axis([-1 2*L+1 bmin bmax])
>> xlabel(’n’); ylabel(’b(n)’); title(’b(n) coefficients’)
>> subplot(2,2,2); plot(w/pi,Hr);grid
>> xlabel(’frequency in pi units’); ylabel(’Hr’)
>> title(’Type-1 Amplitude Response’)
>> subplot(2,2,4); pzplotz(h,1)
The plots and the zero locations are shown in Figure 7.5. From these plots, we
observe that H r (ω) iszero at ω = π. There is one zero-quadruplet constellation,
three zero pairs, and one zero at ω = π as expected.
□
□ EXAMPLE 7.6 Let h(n) = {−4, 1, −1, −2, 5, 0, −5, 2, 1, −1, 4}. Determine the amplitude response
↑
H r (ω) and the locations of the zeros of H
(z).
Solution Since M = 11, which is odd, and since h(n) is antisymmetric about α =
(11 − 1)/2 =5, this is a Type-3 linear-phase FIR filter. From (7.13) we have
c(0) = h (α) =h(5) =0,c(1) = 2h(5 − 1) = 10, c(2) =2h(2 − 2) = −4
c (3) =2h (5 − 3) = −2, c(4) =2h (5 − 4) = 2, c(5) = 2h (5 − 5) = −8
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