problem with frequency domain using fft on raw data

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omar Benamiar Messari
omar Benamiar Messari el 21 de Ag. de 2018
Editada: dpb el 22 de Ag. de 2018
I have a problem in my MATLAB program. I'm trying to find a cutoff frequency to create a low pass filter for compass data. I'm trying to go from the time domain to the frequency domain and find an Fc, so I used the FFT but it seems that's it's nor working.
This is what i have done:
if true
fs=1500;
dataset=xlsread('data.xlsx','Feuil1','A1:A363');
t=1:length(dataset);
L=length(dataset);
z=fft(dataset);
P2 = abs(z/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = fs*(0:(L/2))/L;
subplot(2,2,4)
plot(f,P1)
title('frequenciel')
xlabel('f(hz)');
ylabel('p1(f)');
end
but i have found something weird. wish is not what i was expencting
  3 comentarios
omar Benamiar Messari
omar Benamiar Messari el 22 de Ag. de 2018
hello, i'm expecting to see a big peak somewere on my frequency axes that i can shoose my fc . i also so that my bandwidth of my sensor is 10 khz which mean that normally i should have a peak on 10 khz. am I wrong ? sorry for my english.
dpb
dpb el 22 de Ag. de 2018
Editada: dpb el 22 de Ag. de 2018
You only sampled at 1500 Hz (fs=1500) according to your code so Nyquist frequency is 750 Hz.
The response of the sensor will depend on the frequency of the energy content of the input to it; "bandwidth" only means it is capable of responding up to that frequency (probably with some documented attenuation value) but you would have to sample at least 2X that frequency to see it and actually by 3-4X that to really reproduce the input well owing to Nyquist.
The FFT "knows nuthink!" about what the actual sampling frequency is; it is immaterial to the algorithm; the frequency for analysis/presentation is totally dependent upon what you say your sampling frequency was and there's absolutely no way to determine that from just the sampled data itself.

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Respuestas (1)

Dimitris Kalogiros
Dimitris Kalogiros el 21 de Ag. de 2018
Instead of
plot(f, P1)
i suggest to use
plot(f, 20*log10(P1));
  1 comentario
omar Benamiar Messari
omar Benamiar Messari el 22 de Ag. de 2018
this is really not what i'm expecting normally i should see a peak on 10000 khz because the bandwidth of the sensor is at 10khz normally using matlab i should see this on the frequency domain.i don't know i'm just confused.

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