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Frequency Sampling Design Techniques 355
Highpass: M=33,T1=0.1095,T2=0.598
1
0.4
0.2
Impulse Response
Hr(k)
0.59
h(n)
0
0.109
0
1
0 .6 .8 1
frequency in π units
Amplitude Response
−0.2
−0.4
0 10 20 30
n
Magnitude Response
0
Hr(w)
0.59
Decibels
50
0.109
0
0 .6 .8 1
frequency in π units
0 .6 .8 1
frequency in π units
FIGURE 7.31 Highpass filter design plots in Example 7.18
MATLAB script:
>> M = 33; alpha = (M-1)/2; Dw = 2*pi/M; l = 0:M-1; wl = Dw*l;
>> k1 = 0:floor((M-1)/2); k2 = floor((M-1)/2)+1:M-1;
>> Hrs = [j*Dw*k1,-j*Dw*(M-k2)];
>> angH = [-alpha*Dw*k1, alpha*Dw*(M-k2)];
>> H = Hrs.*exp(j*angH); h = real(ifft(H,M)); [Hr,ww,a,P]=Hr_Type3(h);
The time- and the frequency-domain plots are shown in Figure 7.32. We observe
that the differentiator is not a full-band differentiator.
□
□ EXAMPLE 7.20 Design a 51-point digital Hilbert transformer based on the ideal Hilbert transformer
of (7.32).
Solution
From (7.32) the samples of the (imaginary-valued) amplitude response are
given by
⎧
⌊ ⌋ M − 1
−j, k =1,...,
⎪⎨
2
jH r (k) = 0, k =0
⌊ ⌋
⎪⎩ M − 1
+j, k = +1,...,M − 1
2
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