求差分方程幅频和相频响应

clc;clear all;
Fs = 200;
b = [1 0 0 0 0 0 -2 0 0 0 0 0 1];
a = [1 -2 1];
N = 2048;
[h,w] = freqz(b,a,'whole',N);
figure;
subplot(2,1,1);
plot(w(1:N/2)/(2*pi)*Fs,20*log10(abs(h(1:N/2))))
xlabel(' Frequency (Hz)')
ylabel('Magnitude (dB)')

subplot(2,1,2);
plot(w(1:N/2)/(2*pi)*Fs,angle(h(1:N/2)))
xlabel(' Frequency (Hz)')
ylabel('Normalized Frequency (\times\pi rad/sample)')