《DSP using MATLAB》示例Example7.17

代码:
M = 40; alpha = (M-1)/2; l = 0:M-1; wl = (2*pi/M)*l; 
T1 = 0.109021; T2 = 0.59417456;
Hrs = [zeros(1, 5),T1,T2, ones(1, 7),T2,T1, zeros(1, 9),T1,T2, ones(1,7), T2,T1,zeros(1,4)];    % Ideal Amp Res sampled
Hdr = [0, 0, 1, 1, 0, 0]; wdl = [0, 0.2, 0.35, 0.65, 0.8, 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);
%Plot 
figure('NumberTitle', 'off', 'Name', 'Exameple 7.17a')
set(gcf,'Color','white'); 
subplot(2,2,1); plot(wl(1:21)/pi, Hrs(1:21), 'o', wdl, Hdr); axis([0, 1, -0.1, 1.1]);
xlabel('frequency in \pi nuits'); ylabel('Hr(k)'); title('Bandpass: M=40, T1=0.5941, T2=0.109');
grid on;
subplot(2,2,2); stem(l, h); axis([-1, M, -0.4, 0.4]); grid on;
xlabel('n'); ylabel('h(n)'); title('Impulse Response');
subplot(2,2,3); plot(ww/pi, Hr, wl(1:21)/pi, Hrs(1:21), 'o'); axis([0, 1, -0.2, 1.2]); grid on;
xlabel('frequency in \pi units'); ylabel('Hr(w)'); title('Amplitude Response');
subplot(2,2,4); plot(w/pi, db); axis([0, 1, -100, 10]); grid on;
xlabel('frequency in \pi units'); ylabel('Decibels'); title('Magnitude Response');
运行结果:

从幅度谱上看出,最小阻带衰减As大概60dB,满足设计要求。
    牢记:
1、如果你决定做某事,那就动手去做;不要受任何人、任何事的干扰。2、这个世界并不完美,但依然值得我们去为之奋斗。
 
                    
                
 
                
            
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浙公网安备 33010602011771号