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Re: FFT in Mathematica

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

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> 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.

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