How to interpret the frequencies on a symmteric FFT plot.

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.

Answer 1

[Star Strider]

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.)

[Paul]

The expression for Fv in the ODD case is incorrect. We can see from a simple example

F0 = 10;
N  = 101;  % odd
Ts = F0/N/5;
Fs = 1/Ts;
t = (0:N-1)*Ts;
x = sin(2*pi*F0*t);
X = fftshift(fft(x));
Fv = linspace(-Fs/2,Fs/2,N);
figure
stem(Fv,abs(X))
xline(10,'r')
xlim([5 15])

The peak is slightly offset from +10 (and -10 on the other side). The correct expression is (one form of it)

Fv = linspace(-Fs/2,Fs/2,N)*(N-1)/N;
figure
stem(Fv,abs(X))
xline(10,'r')
xlim([5 15])

[Star Strider]

Noted.

[John Bach]

Thank you both for your help here, that's excellent!

Can I just confirm that the resulting frequencies would then just be the harmonics of the cylinder's vortex shedding?

[Star Strider]

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]

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

Really apprecaite everyone's help here.

[Star Strider]

As always, my pleasure!

Answer 2

[Paul]

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.

[John Bach]

Hi Paul,

Thanks for your repsonse, this is very helpful in understanding why this is the way it is!