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Re: FFT of gaussian [message #74390 is a reply to message #74389]

Wed, 12 January 2011 06:28

Kenneth P. Bowman
Messages: 585
Registered: May 2000

Senior Member

In article
<e8f1db77-67a5-491f-9671-fbd85c50a7b6@a10g2000vby.googlegroups.com>,
Gray <graylikethecolor@gmail.com> wrote:

> Hi all,
>
> Here's something I don't really understand. The Fourier transform of
> a Gaussian function is another Gaussian... so why if I create a
> Gaussian and run FFT do I not get a Gaussian? Is it because my
> Gaussian vector is discrete? How can I fix this?

You get the discrete transform of a discrete approximation to
a Gaussian, which is also a discrete approximation to a Gaussian.

Try this:

IDL> x = -5.0d0 + 0.1d0*dindgen(101)
IDL> y = exp(-(x^2))
IDL> plot, x, y
IDL> yt = fft(y)
IDL> plot, shift(abs(yt),50)

Don't forget, the IDL always does a full complex FFT.

Ken Bowman

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[Message index]

		Re: FFT of gaussian By: Mark Piper on Wed, 12 January 2011 12:20
		Re: FFT of gaussian By: Gray on Wed, 12 January 2011 09:42
		Re: FFT of gaussian By: Kenneth P. Bowman on Wed, 12 January 2011 06:28

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