How to chose FFT parameter ?
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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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Image Analyst
2013-10-15
编辑:Image Analyst
2013-10-15
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:
3 个评论
Image Analyst
2013-10-15
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);
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Youssef Khmou
2013-10-12
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 个评论
Youssef Khmou
2013-10-15
编辑:Youssef Khmou
2013-10-15
thats sound normal, the signal contains near zero frequency :
try :
figure, plot(abs(fft(signal)));
same spectrum?
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