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FIR Filter Designs for Sampling Rate Conversion 509
x(n)
FIR Decimator
FIR LPF
H(ω)
v(n)
↓D
y(m)
FIGURE 9.24
Rate: F x
F x
An FIR integer decimator
F x
= F y
D
which is nothing but the aliased sum of the H(ω)X(ω). Thus, the condition
necessary to avoid aliasing is
π
H(ω)X(ω) =0 for ≤|ω| ≤π (9.54)
D
Then,
Y (ω y )= 1 X(ω)H(ω) (9.55)
D
as in (9.25), where the ideal filtering was accomplished with H D (ω) as
given in (9.20).
□ EXAMPLE 9.11 Design a decimator that downsamples an input signal x(n) byafactor D =2.
Use the firpm algorithm to determine the coefficients of the FIR filter that has
a0.1 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/D =0.5π. Toget a reasonable
value for the filter length we choose the transition width of 0.1π, which gives
stopband a cutoff frequency of ω s =0.3π. Afilter of length M =37achieves the
preceding design specifications. The details are given in the following MATLAB
script.
% Filter Design
D = 2; Rp = 0.1; As = 30; wp = pi/D; ws = wp+0.1*pi;
[delta1,delta2] = db2delta(Rp,As);
[N,F,A,weights] = firpmord([wp,ws]/pi,[1,0],[delta1,delta2],2);
h = firpm(N,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);
Hf1 = figure(’units’,’inches’,’position’,[1,1,6,4],...
’paperunits’,’inches’,’paperposition’,[0,0,6,4]);
subplot(2,2,1); Hs1 = stem(n,h,’filled’); set(Hs1,’markersize’,2);
axis([-1,length(n),-0.15,0.6]); ylabel(’Amplitude’,’vertical’,’cap’);
xlabel(’n’,’vertical’,’bottom’);set(gca,’xtick’,[n(1),n(end)],’ytick’,[0,0.5]);
Title(’Impulse Response’,’fontsize’,TF,’vertical’,’baseline’);
subplot(2,2,3); plot(w/pi,Hr,’m’,’linewidth’,1.0); axis([0,1,-0.1,1.1]);
set(gca,’xtick’,[0,wp/pi,ws/pi,1],’ytick’,[0,1]); grid;
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