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Frequency Sampling Design Techniques 351
Frequency Samples: M=40,T1=0.39
Impulse Response
1
0.2
Hr(k)
0.39
h(n)
0.1
0
0
0 0.2 0.3 1
frequency in π units
−0.1
0 10 20 30 40
n
Amplitude Response
Magnitude Response
1
0
Hr(w)
0.39
Decibels
43
0
0 0.2 0.3 1
frequency in π units
0 0.2 0.3 1
frequency in π units
FIGURE 7.28 Optimum frequency design method: T 1 =0.39
□ EXAMPLE 7.16 Let us revisit our lowpass filter design in Example 7.14. We will solve it using two
samples in the transition band so that we can get a better stopband attenuation.
Solution
Let us choose M =60sothat there are two samples in the transition band. Let
the values of these transition band samples be T 1 and T 2. Then H r (ω) isgiven
by
H (ω) =[1,...,1 ,T
} {{ } 1,T 2, 0,...,0 ,T
} {{ } 2,T 1, 1,...,1 ]
} {{ }
7 ones
43 zeros
6 ones
From tables in [22, Appendix B] T 1 =0.5925 and T 2 =0.1099. Using these
values, we use MATLAB to compute h (n).
>> M = 60; alpha = (M-1)/2; l = 0:M-1; wl = (2*pi/M)*l;
>> Hrs = [ones(1,7),0.5925,0.1099,zeros(1,43),0.1099,0.5925,ones(1,6)];
>> Hdr = [1,1,0,0]; wdl = [0,0.2,0.3,1];
>> 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);
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