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frequency vector in a fft

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SYML2nd
SYML2nd on 13 Jun 2019
Edited: Daniel Frisch on 31 Aug 2020
Hi all,
I am trying to do the fft of a signal (here attached cc.txt), unfortunately I do not understand which frequency I should associate at each value of the fft, in other word I did not understand how to create the frequency vector. I have tried to apply what I've read here https://it.mathworks.com/help/matlab/ref/fft.html and I have done this code, but unfortunately the frequency vector and the fft have different lenght, so an error results.
Can you help, what I am doing wrong in this code and why???
Thank you in advance
cc1=readtable('cc.txt')
cc=table2array(cc1(:,2:2))
L=length(cc)
ccf= fft(cc)
Tinc=0.001
Fs=1/Tinc
fn=Fs/2
L=length(cc)
f = Fs*(0:(L/2))/L;
plot(f,ccf)

Accepted Answer

Star Strider
Star Strider on 13 Jun 2019
Try this:
plot(f,ccf(1:numel(f))*2)
That should do what you want. (It plots the ‘positive half’ of the fft result, from D-C (0 Hz) to the Nyquist frequency, and the correct amplitude, multiplying the returned amplitude by 2.)

More Answers (3)

SYML2nd
SYML2nd on 13 Jun 2019
First of all thank you for your answer. But I have still doubts?
So was my frequency correct?
I didn't get why you multiplied the returned amplitude by 2
So, doing
ccf(1:numel(f))
am I am taking just part of the fft (from the first to the lenght of the f vector)?
Am I discarding a part of the fft because it exceeds the Nyquist limit?
Thank u in advance
  1 Comment
Star Strider
Star Strider on 13 Jun 2019
The Fourier transform output represents both the ‘positive’ and ‘negative’ frequencies (on either side of 0 Hz), so the energy is equally divided between the two sides. Multiplying by 2 corrects for this and approximates the amplitude of the time-domain signal.
Am I am taking just part of the fft (from the first to the lenght of the f vector)?
Am I discarding a part of the fft because it exceeds the Nyquist limit?
Exactly. It actually does not exceed the Nyquist frequency (although that is the maximum frequency you can uniquely represent in a sampled signal). It instead eliminates the mirror-image negative half of the fft output. You can see this by constructing an appropriate frequency vector, then using fftshift on both the frequency vector and the fft output. If ‘Fn’ is the Nyquist frequency, the result of the fftshift function will plot the fft output from -Fn to +Fn.

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Daniel Frisch
Daniel Frisch on 31 Aug 2020
Edited: Daniel Frisch on 31 Aug 2020
You can use my little function called easyFFT that does exactly that for you: it calculates & returns a frequency vector along with the spectrum.

Georges Theodosiou
Georges Theodosiou on 13 Oct 2020
Mr. Francesco Saverio Ciani,
Please lety me help you a bit. Function findpeaks() helps in your problem.
With regards and friendship, Georges Theodosiou

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