What is different between FFT and SHIFTFFT

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University
University 2023-10-30
评论： University 2023-11-2
Please can someone explain why this code :
fftshift(fft((x - mean(x)).*hann(L), NFFT)/sum(hann(L)))
is better than
fftshift(fft(x))?
Why not fftshift(fft(x))?
Why subtracting mean(x) from fftshift(fft(x))?
From matlab documentation Hann returns an L-point symmetric Hann window. Please does this mean?
Why sum(hann(L)) ? Why not hann(L)?
  2 个评论
Walter Roberson
Walter Roberson 2023-10-30
Subtracting the mean gets you an fft() in which the first bin is 0 . But it is not clear that is productive since hann() starts with a 0.
sum(hann(L)) appears to be (L-1)/2 for L > 2 -- for L = 2 exactly the "0 at the end" rule takes precidence. For L = 1 exactly hann() is 1 .

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Chunru
Chunru 2023-10-31
Ran in:
First, it is not correct to say which is better unless you define what is "better". fft(x) has better resolution but higher sidelobe levels than fft(x.*hann(L)).
The substraction of mean(x) from x is to remove the DC component so that the spectrum has 0 (approximately) at 0 frequency.
fftshift is to re-arrange the frequency axis. fft alone has frequency from 0-fs. fftshift has frequency range of (-fs/2, fs/2).
Division by sum(hann(L)) is a normalization factor. With this normalization a unit amplitude signal exp(2i*pi*f) will have a unit spectrum value:
f0 = 1/4;
L = 512;
w = hann(L);
x = exp(2i*pi*f0*(0:L-1)'); % unit amplitude
y = fft(x.*w./sum(w)); % unit spectrum at f0
plot(abs(y))
To recap:
Why not fftshift(fft(x))? ==> You can use it if you don't mind the side lobe level.
Why subtracting mean(x) from fftshift(fft(x))? ==> To remove DC from the signal and spectrum.
From matlab documentation Hann returns an L-point symmetric Hann window. ==> Normalization factor
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Chunru
Chunru 2023-10-31
编辑：Chunru 2023-10-31
Ran in:
It seems there is a confusion here. Even the hanning window has 0 for t=0 (assuming L>>1). Multiply x(t) with h(t) will make a modified signal s(t)=x(t)*w(t) and s(t=0)=0. But this will not make F(f=0)=0 in general.
However, subtracting mean, i.e. x(t)-mean(x) will ensure its Fourier Transform has 0 (up to round-off error) amplitude for DC (ie first bin). So it DOES matter to substract mean if one wants to remove DC, whether h(0)=0 or not. See the illustration below (paying attention to DC component).
==> combination fft(x) .* hann(L) then the result is going to be the same (to within round-off error) as fft(x-mean(x)) .* hann(L)
Note that we are concerned with time domain window, x.*h, instead of frequency domain window fft(x).*h.
fftshift(fft( (x - mean(x)) .*hann(L), NFFT)/sum(hann(L)))
The above is a time domain window instead of freq domain window.
f0 = 1/4;
L = 512;
w = hann(L);
x = exp(2i*pi*f0*(0:L-1)') + 1; % unit amplitude
y = fft(x.*w./sum(w)); % unit spectrum at f0
subplot(121)
plot(abs(y)); title('With DC')
y1= fft((x-mean(x)).*w./sum(w));
subplot(122)
plot(abs(y1)); title('DC Removed')
University
University 2023-11-2
Thnak you so much. Your explanation really helped

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