How to chose FFT parameter ?

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

Answer 1

[Image Analyst]

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.

Helpful links:

http://www.mathworks.com/matlabcentral/fileexchange/28810-fourier-transform-demonstration/content/fourier_demo/html/fourier_demo.html

http://blogs.mathworks.com/steve/category/fourier-transforms/

http://www.cs.unm.edu/~williams/cs530/symmetry.pdf

[Omar thamer]

the data are for 512 days and sampled each day. You are right the data has no Zero mean and your assumption the data is multiplied by rect is true. Is it possible to detrend ? Is there anyway other than hanning or hamming to shift back the data to zero mean ? Thank you

[Image Analyst]

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);

[Omar thamer]

Thank you it works. Last question about sampling frequency To be precise my data for 512 working days and sampled each day. So length is 512. I put fs=1 n=1024 i still can see peaks on the fold on 2 base (i.e 1024, 512, 256, ...) which does not make sense. the bin obtained by these parameter is very high. (1/1024) Any suggestions ?

Answer 2

[Youssef Khmou]

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)))

[Omar thamer]

Thank you Youssef my data represents the volume change each day and for 512 days. I would say the max freq = 0.5 (1/day). So I put fs=1. Please have a lot into the signal and its fft. I dont know why am getting alwayse a peak on the zero frequency and other peaks on 1024, 512, 256,..... Any advice ?

<<

>>

[Youssef Khmou]

thats sound normal, the signal contains near zero frequency :

 try :

 figure, plot(abs(fft(signal)));

same spectrum?

[Omar thamer]

The fft plot is already for absolute values. I cant see the DC component you are talking about. the signal is varying very clearly. I agree with Image analyst that Zero component is due to shifted signal and no zero mean.