I apologize if this is a basic question, but here goes: The Discrete Fourier transform of a sequence of N complex numbers is uniquely defined by eq.1 in this wiki page,


There are no parameters in this definition of the Fourier transform. So why does the Fourier command in Mathematica need two parameters? I've looked at the Mathematica documentation, but can't find an explanation. Thanks for any help. Please pardon my lack of insight here.

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    $\begingroup$ Different fields use different definitions of the DFT. Look at the help file for FourierParameters: It says: A typical setting is FourierParameters->{a,b}. Some common choices for {a,b} are {0,1} (default), {-1,1} (data analysis), {1,-1} (signal processing). I guess the "signal processing" people wrote the wikipedia page. $\endgroup$ – bill s Sep 1 '20 at 23:43
  • $\begingroup$ @Bill Thanks bill. I was wondering if you could direct me to an alternative definition of DFT? What different definition would "data analysis" use. Thanks much for your help. $\endgroup$ – Chris Sep 2 '20 at 0:34
  • $\begingroup$ That's what the two parameters a and b define. Look in the "Details and Options" part of the help file for Fourier. $\endgroup$ – bill s Sep 2 '20 at 1:05
  • $\begingroup$ @bill Got it. Thanks. I should have gone to the details section earlier. I still don't understand the reason for these parameters, but at least now I have the mathematical definition where the parameters are defined. $\endgroup$ – Chris Sep 2 '20 at 1:27
  • $\begingroup$ @Chris, a parameter is just a normalization, b = -1/+1 is normal DFT usually computed with FFT. For some complex b Fourier[] is a generalization of DFT and it can't be computed directly with FFT, see this article for definition and some applications. $\endgroup$ – I.M. Sep 2 '20 at 2:57

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