数字信号处理实验五(3)

ylabel('Phase'); xlabel('Frequency (rad)');

Blackman:

N=64; k=0:N-1;

w=0.42-0.5*cos(2*pi*k/(N-1))+0.08*cos(4*pi*k/(N-1)); x=cos(2*pi/15*k).*w+0.75*cos(2.3*pi/15*k).*w; X=fft(x,N); subplot(2,1,1);

stem(k,abs(fftshift(X)));

ylabel('Magnitude'); xlabel('Frequency (rad)'); subplot(2,1,2);

stem(k,angle(fftshift(X)));

ylabel('Phase'); xlabel('Frequency (rad)');

??(0.1k)?2,6. 已知序列x[k]??e??0,2k?50 others利用FFT分析下列信号的幅频特性，频率范围为 Ω?[?π,π) ，N=500点。

(1) y[k]?x[2k]； (2) g[k]?x[4k]；

(3) 若将上述x[k]乘以cos(pk/2) ，重做(1)和(2)。 解答：(1)%y[k]?x[2k]

N=500;

K=-50:2:50;

w=-(0.1*K).^2/2;x=exp(w); X=fft(x,N); subplot(2,1,1);

stem(abs(fftshift(X)));

ylabel('Magnitude'); xlabel('Frequency (rad)'); subplot(2,1,2);

stem(angle(fftshift(X)));

ylabel('Phase'); xlabel('Frequency (rad)

%g[k]?x[4k]

N=500; K=-50:4:50; w=-(0.1*K).^2/2; x=exp(w); X=fft(x,N); subplot(2,1,1);

stem(abs(fftshift(X)));

ylabel('Magnitude'); xlabel('Frequency (rad)'); subplot(2,1,2);

stem(angle(fftshift(X)));

ylabel('Phase'); xlabel('Frequency (rad)');

(2) 将上述x[k]乘以cos(pk/2) %y[k]?x[2k]

N=500; K=-50:2:50; w=-(0.1*K).^2/2;

x=exp(w).*cos(pi*K/2); X=fft(x,N); subplot(2,1,1);

stem(abs(fftshift(X)));

ylabel('Magnitude'); xlabel('Frequency (rad)'); subplot(2,1,2);

stem(angle(fftshift(X)));

ylabel('Phase'); xlabel('Frequency (rad)');

%g[k]?x[4k]

N=500; K=-50:4:50; w=-(0.1*K).^2/2;

x=exp(w).*cos(pi*K/2); X=fft(x,N); subplot(2,1,1);

stem(abs(fftshift(X)));

ylabel('Magnitude'); xlabel('Frequency (rad)'); subplot(2,1,2);

stem(angle(fftshift(X)));

ylabel('Phase'); xlabel('Frequency (rad)');