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Frequency Sampling Design Techniques 347
7.4.1 NAIVE DESIGN METHOD
In this method we set H(k) =H d (e j2πk/M ), k =0,...,M − 1 and use
(7.35) through (7.39) to obtain the impulse response h(n).
□ EXAMPLE 7.14 Consider the lowpass filter specifications from Example 7.8.
ω p =0.2π, R p =0.25 dB
ω s =0.3π, A s =50dB
Design an FIR filter using the frequency sampling approach.
Solution
Let us choose M =20sothat we have a frequency sample at ω p, that is, at
k =2:
ω p =0.2π = 2π
20 2
and the next sample at ω s, that is, at k =3:
ω s =0.3π = 2π
20 3
Thus we have 3 samples in the passband [0 ≤ ω ≤ ω p] and 7 samples in the
stopband [ω s ≤ ω ≤ π]. From (7.36) we have
H r (k) =[1, 1, 1, 0,...,0 , 1, 1]
} {{ }
15 zeros
Since M = 20, α = 20−1 =9.5 and since this is a Type-2 linear-phase filter,
2
from (7.37) we have
⎧
⎨−9.5 2π
̸ H (k) = 20 k = −0.95πk, 0 ≤ k ≤ 9
⎩
+0.95π (20 − k) , 10 ≤ k ≤ 19
Now from (7.35) we assemble H (k) and from (7.39) determine the impulse
response h (n). The MATLAB script follows:
>> M = 20; alpha = (M-1)/2; l = 0:M-1; wl = (2*pi/M)*l;
>> Hrs = [1,1,1,zeros(1,15),1,1]; %Ideal Amp Res sampled
>> Hdr = [1,1,0,0]; wdl = [0,0.25,0.25,1]; %Ideal Amp Res for plotting
>> k1 = 0:floor((M-1)/2); k2 = floor((M-1)/2)+1:M-1;
>> angH = [-alpha*(2*pi)/M*k1, alpha*(2*pi)/M*(M-k2)];
>> H = Hrs.*exp(j*angH); h = real(ifft(H,M));
>> [db,mag,pha,grd,w] = freqz_m(h,1); [Hr,ww,a,L] = Hr_Type2(h);
>> subplot(2,2,1);plot(wl(1:11)/pi,Hrs(1:11),’o’,wdl,Hdr);
>> axis([0,1,-0.1,1.1]); title(’Frequency Samples: M=20’)
>> xlabel(’frequency in pi units’); ylabel(’Hr(k)’)
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