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FIR Filter Designs for Sampling Rate Conversion 501
Amplitude Response: alpha = 1
Amplitude Response: alpha = 0.75
4
4
Amplitude
Amplitude
0
0
0 0.25 1
Frequency in π units
Log–mag Response: alpha = 1
0 0.25 1
Frequency in π units
Log–mag Response: alpha = 0.75
0
0
Decibels
−22
Decibels
−40
−50
0 0.25 1
−50
0 0.25 1
Frequency in π units
Frequency in π units
FIGURE 9.19 FIR interpolation filter design plots for I = 4 and L = 5
that has 0.1 dBripple in the passband and is down by at least 30 dB in the
stopband. Choose reasonable values for band-edge frequencies.
Solution
The passband cutoff frequency should be ω p = π/I =0.2π. Toget a reasonable
value for the filter length we choose the transition width of 0.12π, which gives
stopband cutoff frequency of ω s =0.32π. Note that the nominal gain of the
filter in the passband should be equal to I =5,which means that the ripple
values computed using the decibel values are scaled by 5. A filter of length
M =31achieves the design specifications given above. The details are given in
the following MATLAB script.
I = 5; Rp = 0.1; As = 30; wp = pi/I; ws = wp+pi*0.12;
[delta1,delta2] = db2delta(Rp,As); weights = [delta2/delta1,1];
F = [0,wp,ws,pi]/pi; A = [I,I,0,0];
h = firpm(30,F,A,weights); n = [0:length(h)-1];
[Hr,w,a,L] = Hr_Type1(h); Hr_min = min(Hr); w_min = find(Hr == Hr_min);
H = abs(freqz(h,1,w)); Hdb = 20*log10(H/max(H)); min_attn = Hdb(w_min);
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