Is there a fast Fourier transform in Mathematica? Although looking in the help I could not find one.

I am looking to implement the equivalent of fft in MATLAB.

  • $\begingroup$ There is definitely something different going on with Fourier[] than the fft function in Matlab. If you take the fft(eye(n)) in Matlab you get the FourierMatrix of n. In Mathematica you do not. FourierMatrix[n] does exist, but the method of obtaining it via Fourier[IdentityMatrix[n]] does not work in Mathematica, so the fft and Fourier functions are different somehow. $\endgroup$
    – David
    Apr 14, 2018 at 22:30
  • $\begingroup$ I have put some notes on how Mathematica implements a Fourier transform here. $\endgroup$
    – Hugh
    Nov 4, 2021 at 6:47

2 Answers 2


Fourier[list] computes the discrete Fourier transform of list. I assume it uses the FFT when it can.

  • 9
    $\begingroup$ Note that the normalization used by Mathematica is quite different from conventional (e.g. physics, signal processing) normalizations. Check the definition you are using, and set the option FourierParameters accordingly. $\endgroup$ Feb 2, 2012 at 14:15
  • $\begingroup$ Is there a way to strip it out ? In Matalab syntax I need to do the following : function fs = MyFFT(fx) y = fftshift(fx); tmp = fft(y)/sqrt(length(fx)); fs = fftshift(tmp); $\endgroup$
    – 500
    Feb 2, 2012 at 14:27
  • $\begingroup$ fftshift[vec_?VectorQ] := RotateRight[vec, Quotient[Length[vec], 2]] $\endgroup$ Feb 2, 2012 at 14:39
  • 11
    $\begingroup$ For reference: MATLAB's fft(stuff) is equivalent to Mathematica's Fourier[stuff, FourierParameters -> {1, -1}]. To do fft(y)/sqrt(length(y)), one doesn't need to do an explicit division, as the adjustment of FourierParameters is all that's needed: Fourier[y, FourierParameters -> {0, -1}]. $\endgroup$ Feb 2, 2012 at 15:00
  • 1
    $\begingroup$ @A.Vieira, under the fifth and sixth bullets of "Details and Options" for Fourier[] and fourth and fifth bullets of "Details and Options" for FourierTransform[], are explanations on how the FourierParameters option affects them. There should be similar notations in the other Fourier-related functions of Mathematica that take the FourierParameters setting. $\endgroup$ Sep 8, 2017 at 14:29

Fourier uses FFT when possible


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