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