Using fft (Fast Fourier Transform)

Question

Using fft (Fast Fourier Transform)

CusterJess (Computer)

(OP)

26 Apr 10 11:15

Hi.

In the example on how to use fft, once they have the signal, they do this:

NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;

Meaning that they get the next power of 2 from length of y (the signal), and then they calculate the NFFT-points fft, to then divide it by L (length of y).
¿Can you tell me why they do that?

What I'm trying to do is capture a sound signal and then with the fft get the frequency of that signal.

Thank you!

GregLocock (Automotive) · Answer 1 · 2010-04-26 22:13:38

It is called zero padding. It is fast and has the advantage that usually small differences in the length of the signal don't affect the resolution.

Personally I don't like it and don't do it.

Cheers

Greg Locock

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electricpete (Electrical) · Answer 2 · 2010-04-26 23:09:00

The power of 2 FFT is more efficient. That would be the big driver. Some FFT algorithms like the one in excel won't except any input lenght that is not a power of 2.

The zero padding as Greg says doesn't hurt, in fact gives finer "frequency resolution"... sort of.

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CusterJess (Computer) · Answer 3 · 2010-04-27 06:03:27

Thank you very much for both of your answers!

The thing is that I capture the sound of a guitar, let's say an 'A', and after doing the fft and plot the result, I can see a peak where the main frequency should be, but it shows a number that is not supposed to be the one for 'A' (let's say I get a peak on 29, and the frequency's supposed to be 110), so I was looking for the answer on that business with the n-point fft...

¿You reckon what am I doing wrong?

Thanks again!

CusterJess (Computer) · Answer 4 · 2010-04-27 07:35:08

To be more specific:

in the example, after calculatin the fft with Y=fft(y,NFFT)/L , to show the graph (and get the frequency from there), they do:

f = Fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y(1:NFFT/2+1);

so maybe I should do something like that (not exactly that though, I guess, cos it gives me that wrong frequency I told you about), but I'm not really sure why they do the 2*abs(Y(1:NFFT/2+1). The only thing I kinda get is that they represent half of the signal only cos you have all the info you need there.

SomptingGuy (Automotive) · Answer 5 · 2010-04-27 08:33:16

The first bin is DC. The second bin is 1/T, where T is the length of your (possibly padded) signal. Other bins are multiples of this. Learning a formula for the frequency axis is (in my view) a waste of time. Derive it from first principles each time instead.

- Steve

GregLocock (Automotive) · Answer 6 · 2010-04-27 19:58:17

and as ever, posting your data files and scripts as an attachment is far more likely to get less cryptic replies.

Cheers

Greg Locock

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FeX32 (Mechanical) · Answer 7 · 2010-04-27 20:57:44

OP,
do a search. I recall vividly posting an example FFT script.

peace
Fe

CusterJess (Computer) · Answer 8 · 2010-04-28 05:22:10

Thanks a lot everyone.

I'll do a search around the forum to look for those fft examples, and if I'm still confused about it, I'll post the code I wrote so you can check it. :)

CusterJess (Computer) · Answer 9 · 2010-05-03 05:37:45

Ok, everything looks correct now :) Thanksalot everyone, you've been tremendly helpful.

Cheers!

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Using fft (Fast Fourier Transform)