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Properties of Linear-phase FIR Filters 319
10
Impulse Response
30
20
Type–2 Amplitude Response
h(n)
5
0
Hr
10
0
−5
−10
b(n)
10
5
0
−5
0 5 10
n
b(n) coefficients
−20
0 0.5 1
frequency in π units
imaginary axis
1
0
−1
Pole–Zero Plot
z–plane
0 5 10
n
FIGURE 7.5 Plots in Example 7.5
−1 0 1
real axis
From (7.14) we obtain
H r(ω) =c(0) + c(1) sin ω + c(2) sin 2ω + c(3) sin 3ω + c(4) sin 4ω + c(5) sin 5ω
=0+10sin ω − 4 sin 2ω − 2 sin 3ω +2sin 4ω − 8 sin 5ω
MATLAB script:
>> h = [-4,1,-1,-2,5,0,-5,2,1,-1,4];
>> M = length(h); n = 0:M-1; [Hr,w,c,L] = Hr_Type3(h);
>> c,L
a = 0 10 -4 -2 2 -8
L = 5
>> cmax = max(c)+1; cmin = min(c)-1;
>> subplot(2,2,1); stem(n,h); axis([-1 2*L+1 cmin cmax])
>> xlabel(’n’); ylabel(’h(n)’); title(’Impulse Response’)
>> subplot(2,2,3); stem(0:L,c); axis([-1 2*L+1 cmin cmax])
>> xlabel(’n’); ylabel(’c(n)’); title(’c(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)
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