八度：FFT相位谱不正确

Question

A small program that I wrote in octave does not yield desired phase spectrum. The magnitude plot is perfect though.

 f = 200;
 fs = 1000;
 phase = pi/3; 
 t = 0: 1/fs: 1;
 sig = sin((2*pi*f*t) + phase);
 sig_fft = fft(sig);

 sig_fft_phase = angle(sig_fft) * 180/pi;

 sig_fft_phase(201)

sig_fft_phase(201) returns 5.998 (6 degrees) rather than 60 degrees. What am I doing wrong? Is my expectation incorrect?

Answer 1

In your example, if you generate the frequency axis: (sorry, I don't have Octave here, so Python will have to do—I'm sure it's the same in Octave):

faxis = (np.arange(0, t.size) / t.size) * fs

you'll see that faxis[200] (Python is 0-indexed, equivalent to Octave's 201 index) is 199.80019980019981 . You think you're asking for the phase at 200 Hz but you're not, you're asking for the phase of 199.8 Hz.

(This happens because your t vector includes 1.0—that one extra sample slightly decreases the spectral spacing! I don't think the link @Sardar_Usama posted in their comment is correct—it has nothing to do with the fact that the sinusoid doesn't end on a complete cycle, since this approach should work with incomplete cycles.)

The solution: zero-pad the 1001-long sig vector to 2000 samples. Then, with a new faxis frequency vector, faxis[400] (Octave's 401st index) corresponds to exactly 200 Hz:

In [54]: sig_fft = fft.fft(sig, 2000);

In [55]: faxis = np.arange(0, sig_fft.size) / sig_fft.size * fs

In [56]: faxis[400]
Out[56]: 200.0

In [57]: np.angle(sig_fft[400]) * 180 / np.pi
Out[57]: -29.950454729683386

But oh no, what happened? This says the angle is -30°?

Well, recall that Euler's formula says that sin(x) = (exp(i * x) - exp(-i * x)) / 2i . That i in the denominator means that the phase recovered by the FFT won't be 60°, even though the input sine wave has phase of 60°. Instead, the FFT bin's phase will be 60 - 90 degrees, since -90° = angle(1/i) = angle(-i) . So this is actually the right answer! To recover the sine wave's phase, you'd need to add 90° to the phase of the FFT bin.

So to summarize, you need to fix two things:

1. make sure you're looking at the right frequency bin. For an N -point FFT (and no fftshift ), the bins are [0 : N - 1] / N * fs . Above, we just used a N=2000 point FFT to ensure that 200 Hz be represented.
2. Understand that, although you have a sine wave, as far as the FFT is concerned, it gets two complex exponentials, at +200 and -200 Hz, and with amplitudes 1/(2i) and -1/(2i). That imaginary value in the denominators shifts the phase you expect by -90° and +90° respectively.
   • If you happened to have used cos , a cosine wave, for sig , you wouldn't have run into this mathematical obstacle 😆, so pay attention to the difference between sin and cos in the future 💪!

Answer 2

change to t=0:1/fs:1-1/fs; then

 sig_fft_phase(201)
 ans = -30.000