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Window Design Techniques 337
Solution There are two transition bands, namely, ∆ω 1
△
= ω1p −ω 1s and ∆ω 2
△
= ω2s −ω 2p.
These two bandwidths must be the same in the window design; that is, there is
no independent control over ∆ω 1 and ∆ω 2. Hence ∆ω 1 =∆ω 2 =∆ω. For this
design we can use either the Kaiser window or the Blackman window. Let us
use the Blackman Window. We will also need the ideal bandpass filter impulse
response h d (n). Note that this impulse response can be obtained from two ideal
lowpass magnitude responses, provided they have the same phase response. This
is shown in Figure 7.19. Therefore the MATLAB routine ideal lp(wc,M) is
sufficient to determine the impulse response of an ideal bandpass filter. The
design steps are given in the following MATLAB script.
>> ws1 = 0.2*pi; wp1 = 0.35*pi; wp2 = 0.65*pi; ws2 = 0.8*pi; As = 60;
>> tr_width = min((wp1-ws1),(ws2-wp2)); M = ceil(11*pi/tr_width) + 1
M = 75
>> n=[0:1:M-1]; wc1 = (ws1+wp1)/2; wc2 = (wp2+ws2)/2;
>> hd = ideal_lp(wc2,M) - ideal_lp(wc1,M);
>> w_bla = (blackman(M))’; h = hd .* w_bla;
>> [db,mag,pha,grd,w] = freqz_m(h,[1]); delta_w = 2*pi/1000;
>> Rp = -min(db(wp1/delta_w+1:1:wp2/delta_w)) % Actua; Passband Ripple
Rp = 0.0030
>> As = -round(max(db(ws2/delta_w+1:1:501))) % Min Stopband Attenuation
As = 75
%Plots
>> subplot(2,2,1); stem(n,hd); title(’Ideal Impulse Response’)
>> axis([0 M-1 -0.4 0.5]); xlabel(’n’); ylabel(’hd(n)’)
>> subplot(2,2,2); stem(n,w_bla);title(’Blackman Window’)
>> axis([0 M-1 0 1.1]); xlabel(’n’); ylabel(’w(n)’)
>> subplot(2,2,3); stem(n,h);title(’Actual Impulse Response’)
>> axis([0 M-1 -0.4 0.5]); xlabel(’n’); ylabel(’h(n)’)
>> subplot(2,2,4);plot(w/pi,db);axis([0 1 -150 10]);
>> title(’Magnitude Response in dB’);grid;
>> xlabel(’frequency in pi units’); ylabel(’Decibels’)
0
ω c2
π
+
−
0 ω c1
0 ω c1 π
ω c2
π
FIGURE 7.19
Ideal bandpass filter from two lowpass filters
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