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MathGroup Archive 2002

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Re: Numerical Differentiation using Fourier Transform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32951] Re: [mg32941] Numerical Differentiation using Fourier Transform
  • From: Sseziwa Mukasa <mukasa at jeol.com>
  • Date: Thu, 21 Feb 2002 02:07:05 -0500 (EST)
  • Organization: JEOL (USA) Ltd.
  • References: <200202200626.BAA27697@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Mike wrote:

> Hi
>
> I need to differentiate a function that i only know numerically and
> for various reasons I have been asked to do this via a Fourier
> Transform Method.  Before doing it on the horrible function I am
> dealing with, I thought I would get to grips with the method using
> much easier functions.
>
> So say I want to differentiate
>
> E^(-x^2)
>
> w.r.t x using Fourier Transforms : analytically i could proceed as
> follows
>
> FourierTransform[E^(-x^2), x, w]] = 1/(Sqrt[2]*E^(w^2/4))
>
> multiply this by I*w and taking the inverse Transform yields
>
> (-2*x)/E^x^2
>
> as expected so i am along the right lines here.  Now I assume that I
> only know this function numerically and see if i can reproduce this
> result.  First I made a table of values :
>
> test = Table[E^(-x^2), {x, 0.01, 20, 0.01}];
>
> and find it's DFT using
>
> Fourier[test]
>
> now in order to find the derivative I need to multiply this by I*w and
> take the inverse transform.  My problem is how to do this?  what are
> my values of w?  The DFT looks very different to the one I found
> analytically and to be honest I don't have a clue of whats going on.
> Any help would be appreciated - Thanks
>

The values of w associated with the DFT of a vector of length n and
sampling period t are: if n is even w in [0, (n/2-1)/(n*t)] are in
positions [1,n/2] and [-1/(2*t) , -1/(n*t)] are in [n/2+1,n].  If n is
odd the first n/2+1 points contain [0,1/(2*t)], since
F(1/(2*t))=F(-1/(2*t)) for the DFT F of f DFT's of odd length are usually
computed such that the frequency range is [-(n-1)/(2*t),1/(2*t)] or some
other range instead, Mathematica does not support this however.
Furthermore there is no unique definition of the DFT, in particular there
is a sign convention: whether the basis functions are
Exp[-2*pi*I*j*k/(n-1)] or Exp[2*pi*I*j*k/(n-1)] and a normalization
convention: usually multiplication by a constant factor of 1/n when
applying the inverse transform but sometimes using 1/sqrt(n) for both the
transform and its inverse.  Mathematica supports different conventions
through the FourierParameters option to Fourier see the Help Browser for
more information.

Regards,

Sseziwa



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