       RE: Fourier Transform and convolution.

• To: mathgroup at smc.vnet.net
• Subject: [mg26617] RE: [mg26583] Fourier Transform and convolution.
• Date: Sat, 13 Jan 2001 22:36:01 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Hi,

I was doing a Fourier transform using the Excel Tool Pack and I have not
Now with Mathematica, I have tried your codes and it seems to work very
well.

Cheers,

-----Original Message-----
From: BobHanlon at aol.com [mailto:BobHanlon at aol.com]
To: mathgroup at smc.vnet.net
Subject: [mg26617] Re: [mg26583] Fourier Transform and convolution.

Needs["Statistics`NormalDistribution`"];

dist = NormalDistribution[mu, sigma];

p = PDF[dist, x]

1/(E^((x - mu)^2/(2*sigma^2))*(Sqrt[2*Pi]*sigma))

f = FourierTransform[p, x, w, FourierParameters -> {1, 1}]//PowerExpand

E^(I*mu*w - (sigma^2*w^2)/2)

f == CharacteristicFunction[dist, w]

True

InverseFourierTransform[f^2, w, x,
FourierParameters -> {1, 1}]//PowerExpand

1/(E^((x - 2*mu)^2/(4*sigma^2))*(2*Sqrt[Pi]*sigma))

This is N(2*m, Sqrt * sigma)

% == PDF[NormalDistribution[2*mu, Sqrt*sigma], x]

True

Bob Hanlon

In a message dated 2001/1/9 2:17:00 AM, Adil.Reghai at dresdnerkb.com writes:

>We wanted to perform the distribution of a sum of two independent variables
>X and Y which are both N(0,1).
>The way We wanted to do it is to use the convolution property. In order
>to
>compute this convolution we perform
>a Fourier Transform of the pdf n(0,1) and then square each point. After
>that
>we performed an inverse Fourier transform.
>Instead of obtaining a N(0,sqrt(2)) we obtain a distribution which is
>shifted in the form of a U.
>

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