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Re: FFT of gaussian [message #74389 is a reply to message #74382]

Wed, 12 January 2011 09:42

Gray
Messages: 253
Registered: February 2010

Senior Member

On Jan 12, 9:28 am, "Kenneth P. Bowman" <k-bow...@null.edu> wrote:
> In article
> < e8f1db77-67a5-491f-9671-fbd85c50a...@a10g2000vby.googlegroup s.com >,
>
> Gray <grayliketheco...@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

Ah, the shift(abs()) helped very much! Thanks :)

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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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