How to interpret the frequencies on a symmteric FFT plot.

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John Bach 2023-11-12

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评论： Star Strider 2023-11-12

Hi there,

I've produced an FFT plot to determine the frequency of vortex shedding from a cylinder. I've done some research and now understand why my plot exhbits perfect symmetry about a central line (as seen in the diagram below), however I'm still unsure of how to interpret this mathematically.

It seems logical to me that this would simply represent a second harmonic, although I'm not sure whether this is true.

Thank you in advance for your help with this.

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Answer 1

Star Strider 2023-11-12

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You need to use the fftshift function to put the zero frequency (D-C or constant offset) value in the centre. Then, depending on whether the fft vector has even or odd length, the ‘Fv’ (frequency vector) would be:

Fv = linspace(-Fs/2, Fs/2-Fs/length(s), length(s)); % EVEN 'length(s)' (Asymmetric)

Fv = linspace(-Fs/2, Fs/2, length(s)); % ODD 'length(s)' (Symmetric)

where ‘Fs’ is the signal sampling frequency and ‘s’ is the fft vector. You can then determine the frequencies from that. (I always calculate a one-sided fft. It’s easier to interpret.)

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Star Strider 2023-11-12

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I’m not certain what you’re doing (and it may not be an area of my expertise if I was), so I can’t comment on that.

It’s difficult to determine the frequency of the highest peak, although it may be about 2 (Hz?). The much smaller peak after it may be about 4 (Hz?), so it would classify as a harmonic. Other than that, I’m not certain how to interpret any of them. It might be worthwhile to significnatly lengthen the fft (use the nextpow2 function, and then increase that by powers-of-2 as required as the ‘NFFT’ argument) to improve the frequency resolution. That could tell you more about the frequency content of your signal without requiring more data.

I would do a one-sided Fourier transform, since that’s easier to interpret. A simple version of it and an example using it are included at the end.

Use the findpeaks function to accurately estimate the peak amplitudes and the frequency indices (or frequencies themselves, depending on how you choose the function arguments). That can make it easier to understand the fft data.

Example —

Fs = 1000;

L = 60;

t = linspace(0, Fs*L, Fs*L+1)/Fs;

s = sum(sin([1 50:50:450].'*2*pi*t));

[FTs1,Fv] = FFT1(s,t);

[pks,locs] = findpeaks(abs(FTs1)*2, 'MinPeakProminence',0.25);

PeaksData = table(Fv(locs).', pks, 'VariableNames',{'Frequency','Peak Magnitude'})

PeaksData = 10×2 table

Frequency Peak Magnitude _________ ______________ 1.0071 0.88891 50.003 0.97856 100.01 0.91645 149.99 0.91645 200 0.97856 250 1 300 0.97856 350.01 0.91645 399.99 0.91645 450 0.97856

figure

plot(Fv, abs(FTs1)*2)

hold on

plot(Fv(locs), pks, '^r')

hold off

grid

axis('padded')

function [FTs1,Fv] = FFT1(s,t)

s = s(:);

t = t(:);

L = numel(t);

Fs = 1/mean(diff(t));

Fn = Fs/2;

NFFT = 2^nextpow2(L);

FTs = fft((s - mean(s)).*hann(L), NFFT)/sum(hann(L));

Fv = linspace(0, 1, NFFT/2+1)*Fn;

Iv = 1:numel(Fv);

FTs1 = FTs(Iv);

end

Make appropriate changes to work with your data.

.

John Bach 2023-11-12

Thank you for these suggestions, that's extremely useful.

Really apprecaite everyone's help here.

Star Strider 2023-11-12

As always, my pleasure!

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Answer 2

Paul 2023-11-12

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Hi John,

The very short, and loose, answer to your question is as follows. Suppose we have a signal, like x[n] = cos(2*pi/N*k*n) (k,n and N are integers with k<N, 2*pi/N*k is the "frequency" in rad/sample). x[n] can be expressed as a sum of complex exponentials:

x[n] = cos(2*pi/N*k*n) = exp(1j*2*pi/N*k*n)/2 + exp(1j*(-2*pi/N*k)*n)/2

This is bascially what the FFT does. It determines the coefficients (to within a scale factor), in this case 1/2 and 1/2, that can be used to recononstruct x[n] as a sum of scaled complex exponentials. For a real signal x[n], we'll always have symmetry between the amplitude of the coefficients of the pairs of exponentials that correspond to +2*pi*k/N and -2*pi*k/N. Basically, the original cos has been "split" into two complex exponentials of mirror frequencies.

Because the complex exponential is perodic with period 2*pi, we can add an integer multiple of 2*pi to the second term

x[n] = exp(1j*2*pi/N*k*n)/2 + exp(1j*(2*pi/N*(-k)*n + 2*pi*n))/2

x[n] = exp(1j*2*pi/N*k*n)/2 + exp(1j*(2*pi*(-k/N+1)*n))/2

x[n] = exp(1j*2*pi/N*k*n)/2 + exp(1j*(2*pi/N*(-k+N)*n))/2

Looking at the second term on the right, we see that an equivalent frequency to -2*pi*k/N is 2*pi*(-k+N)/N, which is postive because N > k by assumption. That's the effect you're seeing in the "symmetric" FFT plot.