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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[6] \[Pi]^(3/2))
>
> Thanks in Advance
>
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[27]:= 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[28]= Cosh[(1 + I)*Sqrt[f]*Sqrt[3*Pi]]/(2*Sqrt[3]*Pi)
Nasser
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