Finding Dominant Frequency and Phase using FFT

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Matt Gaidica
Matt Gaidica 2020-10-3
评论： Bjorn Gustavsson 2020-10-4
This is stemming from a forecasting problem, but my fundamental assumption about FFT appears to be incorrect. Below, I'm generating a pure sinusoid at 5Hz for 5 seconds (top). The amplitude of the FFT (padded, middle) confirms the dominant frequency. However, I'm unclear why phase at 5Hz would not equal 0 using angle() (bottom)? The ultimate goal is to find the dominant frequency and phase of a noisy signal using FFT, then forecast using a pure, phase-shifted sinusoid.
Fs = 125;
plotS = 5;
t = linspace(0,plotS,plotS*Fs);
fmod = 5;
pdelay = 0;
X = sin((2*pi*fmod*t) + pdelay);
close all
ff(800,600);
subplot(311);
plot(t,X);
hold on;
xlabel('Time (s)');
ylabel('Amplitude');
L = numel(t);
nPad = 5;
n = (2^nextpow2(L)) * nPad;
Y = fft(X,n);
f = Fs*(0:(n/2))/n;
P = abs(Y/n).^2;
A = angle(Y);
grid
subplot(312);
plot(f,P(1:n/2+1))
xlabel('Frequency (f)')
ylabel('|P(f)|^2')
xlim([0 10]);
grid
subplot(313);
plot(f,A(1:n/2+1));
xlabel('Frequency (f)')
ylabel('Phase (rad)');
xlim([fmod-2 fmod+2]);
grid

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采纳的回答

Bjorn Gustavsson
Bjorn Gustavsson 2020-10-4
Change your function to cos((2*pi*fmod*t) + pdelay) and re-run your code-snippet.
Also check what happens when you reduce your time-axis by one sample.
HTH
  2 个评论
Matt Gaidica
Matt Gaidica 2020-10-4
cos() does the trick! Thanks, Bjorn. I didn't reduce the time-axis, what is your thought behind that? Do you mean something like this right after t is declared?
t = t(1:end-1);
Bjorn Gustavsson
Bjorn Gustavsson 2020-10-4
Yup. I forgot you zero-padded your time-series, if you remove that you will see the difference between a signal that is perfectly periodic over your intervall and one that is not.

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