Trouble with understanding frequency bins in MATLAB's documentation (FFT Phase plots)

Hello everyone,

I'm working on plotting phase diagrams after performing an FFT, and although the plots appear to be correct, I'm getting what I see (rightly or wrongly) as 'extra' information.

I found some documentation that might illustrate what I mean: http://www.mathworks.co.uk/help/techdoc/math/brentm1-1.html

Scrolling down to the bit on plotting that signal's phase (unwrapped), the phase plot shows frequencies from -50 to 50 Hz, although the signal's highest frequency component should be 40 Hz. Thinking it might be the random noise that the code uses, I took that out, leaving only the 15 and 40 Hz components; however, the phase plot still shows phase over -50 to 50 Hz. I have a hunch this corresponds to the nyquist frequency, since the sampling frequency is 100 Hz; however, I don't quite understand why this happens or what it implies. Isn't the x-axis in this case showing frequency bins, and wouldn't this mean that according to the plot the phase is still changing at frequencies where the signal's amplitude is zero?

If anyone could shed some light on this, I would be very grateful!

Thanks,

David

Answer 1

[Wayne King]

Hi David, the sampling frequency is 100 Hz, so the frequency axis should run from -50 to 50 Hz.

The signal contains random noise:

fs = 100;                          % Sample frequency (Hz)
t = 0:1/fs:10-1/fs;                % 10 sec sample
x = (1.3)*sin(2*pi*15*t) ...       % 15 Hz component
  + (1.7)*sin(2*pi*40*(t-2)) ...   % 40 Hz component
  + (2.5)*randn(size(t));

The additive white Gaussian noise means that energy will be spread over the entire frequency interval from -50 to 50 Hz.

So it is not correct to say that the signal has no energy at frequencies other than 15 and 40 Hz.

[David]

Hi Wayne,

Thanks for your answer. As I said, I've also tried running the program without white noise (ie x = (1.3)*sin(2*pi*15*t) + (1.7)*sin(2*pi*40*(t-2)) ) to ensure that energy is only present at the 15 and 40 Hz bins. I understand that the FFT plot of the signal with white noise will clearly reflect the presence of noise. What I don't get is why the FFT phase plot of the signal without noise will also show phase changes at frequencies where there should be no energy (particularly when the FFT magnitude plot shows no energy in bin other than at 15 and 40 Hz).

[Dr. Seis]

The amplitudes at the frequencies other than 15 and 40 Hz are non-zero complex numbers. They have non-zero real and non-zero imaginary parts with amplitudes around the same order of magnitude, which means the angle between them (or phase) will also be non-zero. But if you look at the many, many orders of magnitude difference between amplitudes at 15 or 40 Hz and all other frequencies, you would know that the phase information outside of 15 and 40 Hz is pretty much garbage. This isn't recommended for all cases, but for this case try this out:

y = fftshift(fft(x));

subplot(3,1,1);plot(abs(y)); title('ABS');

subplot(3,1,2);plot(angle(y)); title('Phase ANGLE #1');

subplot(3,1,3);plot(angle(y).*(abs(y)>1e-6)); title('Phase ANGLE #2');

The last plot is basically multiplying the phase angle information by a 1 if the abs amplitude is greater that some tolerance (i.e., 1e-6) or 0 if the abs amplitude is less than that tolerance.

[David]

That shed a lot of light on the issue, thank you!

Answer 2

[Wayne King]

Hi David, Due to the fact that the fft() is a numerical implementation of the DFT (and not the DFT), you may find that even in the case of sinusoidal inputs, other DFT bins may not be identically zero, but will often be extremly small nonzero numbers. For example:

Fs = 100;
t = 0:1/Fs:10-1/Fs;  
x = (1.3)*sin(2*pi*15*t)+1.7*sin(2*pi*40*(t-2));
xdft = fft(x);
% now create a copy ydft and remove the DFT coefficients at 15
% and 40 Hz
bins = [151 401 601 851];
ydft = xdft;
ydft(bins) = [];
max(abs(ydft))

You see that the largest magnitude is only on the order of 10^{-11}.

Accordingly, you should not attach any physical intepretation to an argument (phase) based on such a small magnitude.

[David]

I see. My understanding of how the FFT function in matlab works is getting clearer, thank you very much.

So in the end, in the case of the signal we've been examining, what is the significance of examining the phase at a point where the signal's spectrum is of virtually zero energy? Am I right in assuming, in light of the answers that have been given, that all phase information in bins with negligible energy can be ignored?

[Wayne King]

yes