How to find FFT of a non-uniformly sampled data?

I have a vibration data in time domain and want to convert it to frequency domain.However my data is not uniformly sampled, so I am having trouble applying fast fourier transform to it. I have attached my excel data file.Can you guys please help

Answer 1

[Star Strider]

Your data are very close to being uniformly sampled. The mean difference in sampling times (sampling interval) is 976.5623e-006, and the standard deviation is 496.2100e-009. You can interpolate them with the Signal Processing Toolbox resample function to be entirely uniformly sampled, then do the Fourier transform:

[D,S,R] = xlsread('vibration.xls');
t = D(:,1);
v = D(:,2);

tstats = [mean(diff(t))  std(diff(t))]                      % Information Only
tr = linspace(min(t), max(t), length(t));                   % Uniformly-Sampled Time Vector
vr = resample(v, tr);                                       % Resampled Signal Vector

L = length(tr);                                             % Signal Length
Ts = mean(diff(tr));                                        % Sampling Interval
Fs = 1/Ts;                                                  % Sampling Frequency
Fn = Fs/2;                                                  % Nyquist Frequency
FTvr = fft(vr)/L;                                           % Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                         % Frequency Vector
Iv = 1:length(Fv);                                          % Index Vector

figure(1)
plot(Fv, abs(FTvr(Iv))*2)
grid

You can use the File Exchange contribution NUFFT, NFFT, USFFT (link) if you want, but you probably don’t need it here.

[Vasav Ayush]

Thanks for your help, but I have a doubt......why did u multiply abs(FTvr(Iv ))by 2 in

plot(Fv, abs(FTvr(Iv))*2)

[Star Strider]

My pleasure.

The fft function calculates the two-sided Fourier transform, and the energy of the signal is divided evenly between the two halves. Here we are looking at the one-sided Fourier transform, so in order to represent the amplitudes correctly, it is necessary to multiply the amplitudes by 2.

[Pedro Andres Molina Cabrera]

// just a comment to those wondering the same:

The resampling step as written above is not necessary since both time vectors have the same size. Plotting vr and v shows the same data.

Perhaps an interpolation

y = resample(v, t, 1/mean(diff(t)) );

is more appropriate?

Cheers,

[Joseph Fox-Rabinovitz]

Rather than

tr = linspace(...
vr = resample(v, tr);

you probably want to use

[vr, tr] = resample(v, t);