fft definition in matlab doc puzzling

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françois anquez
françois anquez le 16 Fév 2019
Commenté : françois anquez le 22 Fév 2019
Hi all,
I'm a bit puzzled with the definition of fft in the documentation :
The definition at the very end in "more about" states the fft as a projection on positive frequency (from 0 to Fs/N) while the examples suggest a "two sided" spectrum ie. with positive frequencies and negative ones (-Fs/2 to Fs/2).
To make my question more precise :
the definition in the doc is :
Given that the fft results in both positive and negative frequencies, I woumld have guess a defintion like :
Am I missing something ?
thanks in advance for your help,
Francois.
  2 commentaires
Walter Roberson
Walter Roberson le 17 Fév 2019
?? I do not see the word "projection" at all in that documentation ??
françois anquez
françois anquez le 19 Fév 2019
Dear Walter,
the word projection is not written. However suming on all is a projection on harmonic k.
all the best,
Francois.

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Réponse acceptée

Matt J
Matt J le 22 Fév 2019
Modifié(e) : Matt J le 22 Fév 2019
Given that the fft results in both positive and negative frequencies, I woumld have guess a defintion like :
Perhaps a better way to clear up the confusion is to point out that for any given sinuoid at some "positive" frequency , the corresponding negative frequency sinusoid is indentical, due to n-periodicity, to , or equivalently to , where .
Bust since and both lie in the interval 2...n, we can see that summing from k=1...n in the FFT is sufficient to capture all the same pairs of positive and negative frequencies as summing from -n/2 to +n/2.
  1 commentaire
françois anquez
françois anquez le 22 Fév 2019
thanks Matt this was the argument I was seeking for.
all the best,
francois.

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Plus de réponses (1)

Matt J
Matt J le 19 Fév 2019
Modifié(e) : Matt J le 19 Fév 2019
Your second definition only makes sense if you define the indexing in Y(k) to be modulo-n. Otherwise, it is not clear what it means for a vector Y to be indexed at negative k.
If you are defining Y(k) to be modulo n, then the two IFFT definitions are equivalent because Y(k) and W_n(j-1)(k-1) are then both n-periodic with respect to k. Therefore any sum over n successive indices k gives the same result.
  4 commentaires
françois anquez
françois anquez le 21 Fév 2019
function [coef1,coef2]=essaiFFT(X,k)
[N,~]=size(X);
N=N-1;
WN=exp(-i*2*pi/N);
coef1=0;
for jj=1:(N+1)
coef1=coef1+X(jj,1)*power(WN,(jj-1)*(k-1));
end % for jj
coef2=0;
for jj=(-(N/2)):(N/2)
coef2=coef2+X(jj+N/2+1,1)*power(WN,(jj-1)*(k-1));
end % for jj
end % function
Matt J
Matt J le 21 Fév 2019
Modifié(e) : Matt J le 22 Fév 2019
The test code should be as below. The main problem with your implementation is that in the first version of the coefficient calculation, you are putting the time origin at X(1) whereas in the second version, you are putting the time origin at X(N/2). Thus, the second version is really the transform of a phase-shifted version of X.
function [coef1,coef2]=essaiFFT_Matt(X,k)
N=numel(X);
WN=exp(-1i*2*pi/N);
coef1=0;
for j=1:N
coef1=coef1+X(j)*power(WN,(j-1)*(k-1));
end
coef2=0;
for j=(1:N) - floor(N/2) %floor() handles odd N
index=mod(j-1,N)+1; %modulo N transform for 1-based indexing
coef2=coef2 + X(index)*power(WN,(j-1)*(k-1));
end % for jj
end % function

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