FFT on low sample count signal

3 vues (au cours des 30 derniers jours)
Jona Gladines
Jona Gladines le 8 Jan 2019
Commenté : Star Strider le 14 Jan 2019
I have sampled data of a slowly variating signal, about 1.85e-4 Hz . However since the variation is so slow I only have about 1.5 periods of data sampled at a rate about 28 times higher, so every 188 seconds I take a sample of the signal for a total of 9600 seconds. I want to do a fft analysis on this data to find the most important spectral component.
This gives a good approximation of the data I have:
Fs = 0.0053; % Sampling frequency
T = 1/Fs; % Sampling period
L = 9800; % Length of signal
t=0:T:L; % Time Vector
freq=1/5400;
X = (0.06*sin(2*pi*freq*t)+15.3)+(0.08*(rand(size(t))-0.5));
figure;
plot(t,X)
title('Signal')
xlabel('t (seconds)')
ylabel('X(t)')
I tried an FFT on this, but could nog get any good results. The highest spectral component is in the same order of magnitute as the original signal, but not exactly what I hoped for. I'm only interested in the frequency of the component.
Y = fft(X);
f=linspace(0, fs/2, N/2);
figure;
plot(f,abs(Y(1:N/2)))
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
xlim([0 0.001]);
ylim([0 5]);
is there any way to improve without increasing the number of effective signal samples??

Réponse acceptée

Star Strider
Star Strider le 8 Jan 2019
Without seeing your signal, it is not possible to suggest a specif approach. One option may be doiing a nonlinear regression on your signal, using the approach in Curve fitting to a sinusoidal function (link).
  4 commentaires
Jona Gladines
Jona Gladines le 14 Jan 2019
Thanks,
This is indeed what I was looking for. I didn't think about resampling the data.
Jona
Star Strider
Star Strider le 14 Jan 2019
As always, my pleasure.
Resampling to uniform sampling intervals is necessary in order to get reliable results from the fft.

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