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372 Chapter 7 FIR FILTER DESIGN
Solution
The following MATLAB script describes the design procedure.
>> w1 = 0; w2 = 0.3*pi; delta1 = 0.01;
>> w3 = 0.4*pi; w4 = 0.7*pi; delta2 = 0.005;
>> w5 = 0.8*pi; w6 = pi; delta3 = 0.001;
>> weights = [delta3/delta1 delta3/delta2 1];
>> Dw = min((w3-w2), (w5-w3));
>> M = ceil((-20*log10((delta1*delta2*delta3)^(1/3))-13)/(2.285*Dw)+1)
>> M = 51
>> f = [0 w2/pi w3/pi w4/pi w5/pi 1];
>> m = [1 1 0.5 0.5 0 0];
>> h = firpm(M-1,f,m,weights);
>> [db,mag,pha,grd,w] = freqz_m(h,[1]);
>> delta_w = 2*pi/1000;
>> w1i=floor(w1/delta_w)+1; w2i = floor(w2/delta_w)+1;
>> w3i=floor(w3/delta_w)+1; w4i = floor(w4/delta_w)+1;
>> w5i=floor(w5/delta_w)+1; w6i = floor(w6/delta_w)+1;
>> Asd = -max(db(w5i:w6i))
Asd = 62.0745
>> M = M-1; h = firpm(M-1,f,m,weights);
>> [db,mag,pha,grd,w] = freqz_m(h,[1]);
>> Asd = -max(db(w5i:w6i))
Asd = 60.0299
>> M = M-1; h = firpm(M-1,f,m,weights);
>> [db,mag,pha,grd,w] = freqz_m(h,[1]);
>> Asd = -max(db(w5i:w6i))
Asd = 60.6068
>> M
M = 49
The time- and the frequency-domain plots of the designed filter are shown in
Figure 7.38.
□
□ EXAMPLE 7.27 In this example we will design a digital differentiator with different slopes in
each band. The specifications are
Band-1:
0 ≤ ω ≤ 0.2π, Slope = 1 sam/cycle
Band-2: 0.4π ≤ ω ≤ 0.6π, Slope = 2 sam/cycle
Band-3: 0.8π ≤ ω ≤ π,
Slope = 3 sam/cycle
Solution
We need desired magnitude response values in each band. These can be obtained
by multiplying band-edge frequencies in cycles/sam by the slope values
in sam/cycle
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