How to plot the phase response of dft using fft in matlab

Question

SSG_newbiecoder 2018-7-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/408344-how-to-plot-the-phase-response-of-dft-using-fft-in-matlab

回答： David Goodmanson 2018-7-2

Hello, I have a doubt, I was trying to plot the phase response of a sin wave using fft, but I got something that is kind of like noise. The double sided spectrum is symmetric about fs/2. Can anybody help me out? Isn't phase response of a sine wave typically like a slanted line passing through the origin?

Fs=3000;
t=0:1/Fs:(0.5-1/Fs);
S=sin(2*pi*50*t);
figure(1),plot(S);
L=length(S);
f =Fs*(0:(L/2))/L;
Y = fft(S,L);
P2 = abs(Y/L);
f2=Fs*(0:(L-1))/L;
figure(2),stem(f2,fftshift(P2));
phase=angle(Y);
figure(3), stem(f2,phase)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

David Goodmanson 2018-7-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/408344-how-to-plot-the-phase-response-of-dft-using-fft-in-matlab#answer_327173

在 MATLAB Online 中打开

Hi SSG,

The angles look like noise because they are noise. The fft of the sine wave consists of peaks at +-50 Hz and is essentially zero at every other frequency, which is perfectly correct. (There are ways to mess up the fft and not get that exact result, but you avoided the pitfalls). Fftshift is already there in the code so you can see all this with

f2 = Fs*(-L/2:(L/2-1))/L;
figure(2),semilogy(f2,abs(fftshift(P2)));

At frequencies other than +-50 Hz, the amplitude is so small that the angle computed by the code is meaningless. It's noise.

What you may be thinking about for phase response is the following. Suppose you have a real symmetric function g(t)in the time domain. Its transform is a real symmetric function h(f) in the frequency domain. For the time-shifted function g(t-t0), its transform is h(f)*exp(-2*pi*i*f*t0). Using unwrap(angle(... on that function gives -2*pi*f*t0 which is a slanted line. With real data you can do a linear fit to find t0.