       Re: FFT in Mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg93739] Re: FFT in Mathematica
• From: "Nasser Abbasi" <nma at 12000.org>
• Date: Mon, 24 Nov 2008 04:15:20 -0500 (EST)
• References: <gg8pas\$jtn\$1@smc.vnet.net>
• Reply-to: "Nasser Abbasi" <nma at 12000.org>

```"Oliver" <sch_oliver2000 at yahoo.de> wrote in message
news:gg8pas\$jtn\$1 at smc.vnet.net...
> Thanks a lot for ur very useful suggestions.
> well sorry for saying "FFT" but i actually meant the fourier Transform in
> analytical way and not in nummeric way.
> i plotted the FourierTransform of the function according to this:
>
> Plot[FourierTransform[(1/(4*(Pi*1*t)^(3/2)))*Exp[-(3)/(4*1*t)], t,
>   f] // Abs, {f, 0, 100}]
>
> but i actually got a very weird plot which does not make sense for the
> frequency.It produces a function in exponential behave.
> where should be the mistake??!
> maybe in the solution of the FourierTransform i got?  :
>
> ((-1 + Sign[f]) Sinh[
>  Sqrt[Abs[f]] Root[9 + #1^4 &, 4]])/(2 Sqrt \[Pi]^(3/2))
>
>

What is that "1/(4*(Pi*1*t)" that you have up there? Is this supposed to be
"1/(4*(Pi*I*t)" ? i.e. is it supposed to be "I" not a "1" ?

Why else would you multiply by "1" in there?

In Mathematica, sqrt(-1) is called "I". But if it is one, then this is the
FourierTransform you do get.

In:= g = (1/(4*(Pi*1*t)^(3/2)))*Exp[(-1*3)/(4*1*t)];
h = Assuming[Element[f, Reals],Integrate[g*Exp[(-I)*2*Pi*f*t],{t, -Infinity,
Infinity}]]

Out= Cosh[(1 + I)*Sqrt[f]*Sqrt[3*Pi]]/(2*Sqrt*Pi)

Nasser

```

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