2

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,

https://en.wikipedia.org/wiki/Discrete_Fourier_transform.

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.

Chris
  • 447
  • 2
  • 12
  • 4
    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. – bill s Sep 01 '20 at 23:43
  • @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. – Chris Sep 02 '20 at 00:34
  • That's what the two parameters a and b define. Look in the "Details and Options" part of the help file for Fourier. – bill s Sep 02 '20 at 01:05
  • @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. – Chris Sep 02 '20 at 01:27
  • @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. – I.M. Sep 02 '20 at 02:57
  • Some basic details on using Fourier can be found here although it looks like you are familiar with the basics. – Hugh Sep 02 '20 at 08:53
  • @I.M. The article you linked seems to explain the motivation for the two Fourier parameters in Mathematica. The article says "An impetus for studying the fractional Fourier transform is the existence of an algorithm .... that is significantly more efficient." I haven't read the whole paper, but I'm going to assume that this efficiency comes at a cost: You will lose some information. And now, I might understand the Mathematica documentation. Tell me if this makes sense: In data analysis, such as in a medical field (my situation), you general – Chris Sep 02 '20 at 15:51
  • comment above truncated; it continued: ", you generally have smaller data sets (such as blood pressure measurements versus time), and it is practical to do the full Fourier analysis. Therefore, these two parameters become irrelevant. By contrast, in signal analysis you will certainly have huge data sets, and you must be willing to lose some information to make the Fourier Transform practical. I could be wrong on this. I'm using some leaps in logic." – Mr.Wizard Oct 04 '20 at 18:56

0 Answers0