Re: FFT in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg93663] Re: FFT in Mathematica*From*: "sjoerd.c.devries at gmail.com" <sjoerd.c.devries at gmail.com>*Date*: Fri, 21 Nov 2008 05:32:02 -0500 (EST)*References*: <gg3c81$k40$1@smc.vnet.net>

Hi Oliver, An FFT is meant for list of data points. Fourier[ ], the Mathematica function to perform an FFT, takes a list as argument and returns its FFT. However, since s(t) is a continuous function you will probably need a fourier transform, for which Mathematica has provided you with the conveniently named function FourierTransform[ ]. You can plot s(f) for instance like this (use Abs[ ] to show the amplitude and Arg[ ] for phase): Plot[FourierTransform[(1/(4*(Pi*1*t)^(3/2)))*Exp[-(3)/(4*1*t)], t, f] // Abs, {f, 0, 5}]. Cheers -- Sjoerd On Nov 20, 11:56 am, Oliver <sch_oliver2... at yahoo.de> wrote: > Hallo, > > i wanted to plot the following function: > > s(t) = (1/(4*(Pi*1*t)^(3/2))) * Exp[-(x^2 + y^2 + z^2)/(4*1*t)] > when x=1, y=1, z=1 > > so i wrote: > > eq = (1/(4*(Pi*1*t)^(3/2))) * Exp[-(x^2 + y^2 + z^2)/(4*1*t)] > > Plot[Evaluate[eq /. {x -> 1, y -> 1, z -> 1}], {t, 0, 3}] > > My Question is, how can i plot s(f), which is the function depending on t= he frequency. In other word, how to find the FFT of s(t) in Mathematica? > thanks.