Re: A question on fftshift

[Mark P. Esplin]

        I would say that the fftshift doc needs to be changed.  I use a Fourier 
transforms spectrometer.  If I put the input values that I use into your 
formula I get the right spectral frequencies.  If I use the formula in the 
fftshift docs I don't.

                                 -Mark Esplin

On Thursday 18 July 2002 12:46 pm, Steve M. Robbins wrote:
> Hi,
>
> I hope someone can tell me if the bug lies in the fftshift doc or
> my comprehension.
>
> My understanding of the discrete Fourier transform, confirmed by
> Numerical Recipes
>
>   http://www.ulib.org/webRoot/Books/Numerical_Recipes/bookcpdf/c12-1.pdf
>
> is that the frequency interval for the FFT output is [-fc,fc] where fc
> is the critical (Nyquist) frequency.
>
> For samples spaced Dt apart, fc = 1/(2*Dt), so I expect the output
> range to be [-1/(2*Dt),1/(2*Dt)].  Or, rather, the standard "wrapped"
> sequence of this.  Thus, I would expect that after "fftshift", the
> frequency values for the sample points should be
>
>   linspace( -1/(2*Dt), 1/(2*Dt) - 1/(N*Dt), N )
>
> for an input of N samples.
>
> The documentation for fftshift says otherwise, however.
>
>
>  - Function File:  fftshift (V)
>      Perform a shift of the vector V, for use with the `fft' and `ifft'
>      functions, in order the move the frequency 0 to the center of the
>      vector or matrix.
>
>      If V is a vector of E elements corresponding to E time samples
>      spaced of Dt each, then `fftshift (fft (V))' corresponds to
>      frequencies
>
>           f = linspace (-E/(4*Dt), (E/2-1)/(2*Dt), E)
>
>
> Who is correct?
>
> Thanks,
> -Steve
>
>
>
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