Mathematica 9 is now available
Student Support Forum
-----
Student Support Forum: 'Fourier Transform of a Lorentz function' topicStudent Support Forum > General > "Fourier Transform of a Lorentz function"

< Previous CommentHelp | Reply To Comment | Reply To Topic
Author Comment/Response
Pierre
09/10/12 1:33pm

In Response To 'Re: Fourier Transform of a Lorentz function'
---------
Thanks
You right, i was expecting a result with {0,2Pi} parameters for signal processing.
It works fine now for the 'complex' Lorentzian function: Boltz = 1/(1 + (2*(x-xo)/Dx)^2)
TransfBoltz gives: (Pi/2)*Dx*Exp[-Pi*Dx*Abs[X]]
And that is what i was expecting.

But still, even with {0,2Pi} parameters, the simplest Lorentzian function:
Boltz = 1/(1 + x^2)
TransBoltz =
FourierTransform[Boltz, x, X, FourierParameters -> {0, 2 Pi}]
Gives:
Pi*Exp[-2Pi*Abs[X]]

When theorical Fourier Transform (i've checked it on internet but i'm not 100% sure) should be:

Pi*Exp[-Abs[X]]

As you see, the Exp slope is very different...
And this result cannot be obtained with any parameters...

URL: ,

Subject (listing for 'Fourier Transform of a Lorentz function')
Author Date Posted
Fourier Transform of a Lorentz function Pierre 09/09/12 4:25pm
Re: Fourier Transform of a Lorentz function Bill Simpson 09/09/12 9:22pm
Re: Re: Fourier Transform of a Lorentz function Pierre 09/10/12 1:33pm
< Previous CommentHelp | Reply To Comment | Reply To Topic