MathGroup Archive 2003

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Discrete Fourier Transform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg44893] Re: Discrete Fourier Transform
  • From: AES/newspost <siegman at stanford.edu>
  • Date: Sat, 6 Dec 2003 04:45:31 -0500 (EST)
  • References: <bm0juq$prl$1@smc.vnet.net> <bqpn51$91v$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <bqpn51$91v$1 at smc.vnet.net>, David Wood <me at floyd.attbi.com> 
wrote:

> > I am attempting to perform a discrete Fourier transform on a time series
> > using Mathematica 4.0.  (I'd like to determine the period of the signal.)
> > Unfortunately, I noticed that I cannot control the frequency resolution of
> > this transform.  I also have no idea what units the frequency is in.
> > Do you know of any way to get around the problem?
> 

I'd suggest that you need to write down on paper whatever definition of 
the continuous Fourier transform (CFT) pair you like (f and t, or omega 
and t, or whatever), making sure you've scaled them so they really are a 
self-transforming pair; and then write down the formal definition of the 
discrete Fourier transform (DFT) as given in The Mathematica Book in terms of r, 
s and n (or change this to i, j and np (number of points), or n, m and 
np, or whatever indices you like.

Then discretize the continuous transform by writing f = f_i deltaF, t = 
t_j deltaT, or whatever indices you like, and then make the adjustments 
you have to make to match up your discretized continuous Fourier 
transform (DCFT) to the Mathematica DFT for a given np value.  Note in 
particular that both of the DFT series repeat periodically, such that 
the highest frequency in the DFT is not np deltaF, but np deltaF/2.

You may find yourself still confused after doing this -- at least I 
still get myself messed up occasionally, after decades of working with 
transforms and using the DFT -- but at least you'll see where the 
conversion comes from.

[And as a side point, note that despite the suggestion above, the DFT is 
really *not* simply a discretized approximation of the CFT.  The DFT and 
CFT are separate free-standing totally independent exact transforms, in 
their separate discrete and continuous domains, though you can use the 
DFT to get a discrete approximation to the FFT if you use large enough 
np.]


  • Prev by Date: RE: Problem with Mathematica Cell Settings
  • Next by Date: Re: condition and pattern matching
  • Previous by thread: Re: Discrete Fourier Transform
  • Next by thread: Returning function with limitation on input