How to chose FFT parameter ?
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Omar thamer
on 12 Oct 2013
Commented: Omar thamer
on 15 Oct 2013
Hello I have a discrete signal run for 512 Sec with one sample per sec. I chose the fft parameter as follow: Fs=1; nfft=1024; f=(0:nfft/2-1)*fs/nfft;
Is that correct ?
Whenever I do an fft for my signal i can seep the peak on zero only. Also there are some peaks on 1024, 512, 256,... Appreciate your help
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Accepted Answer
Image Analyst
on 15 Oct 2013
Edited: Image Analyst
on 15 Oct 2013
Your original question said 1 per second. Now you say 1 per day. Which is it?
You say you get a big spike at zero. That is because your signal does not have zero mean. It is shifted upwards - it has a mean of around 1 or 1.5 times ten to the (something) power. That means you have a lot of energy in the DC (zero frequency) component. I don't see the other spikes at the other frequencies - looks basically like random noise to me. Though because your signal is multiplied by a rect (basically a window the entire size of your signal) that means that your Fourier signal will be convolved with a sinc function, though a very narrow one. Sometimes this gives little ringing to the spikes. And of course if your signal is not the same at the left and right on the x axis, you will have a discontinuity which will also introduce higher frequencies. They way to reduce that is to window your signal. Look up Hanning or Hamming Window for more info.
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3 Comments
Image Analyst
on 15 Oct 2013
If you don't want the big spike at zero, you can subtract the mean from your signal so that the new mean is zero
newSignal = signal - mean(signal);
More Answers (1)
Youssef Khmou
on 12 Oct 2013
Fs should be at leats twice the maximum frequency in the signal , and the number NFFT increases resolution only , example :
Fs=80;
t=0:1/Fs:2-1/Fs;
y=sin(2*pi*t*35);
N=1024;
fy=fft(y,N);
freq=(0:N-1)*Fs/N;
figure, plot(freq(1:end/2),abs(fy(1:end/2)))
3 Comments
Youssef Khmou
on 15 Oct 2013
Edited: Youssef Khmou
on 15 Oct 2013
thats sound normal, the signal contains near zero frequency :
try :
figure, plot(abs(fft(signal)));
same spectrum?
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