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I have read in many texts that the Fourier Transform of a Gaussian is yet another Gaussian, however how does the mean and standard deviation change?

Also if we convolve a Gaussian with itself then we get a wider Gaussian, this is equivalent to the product with the Fourier Transform of the Gaussian with itself. Will this still be a wider Gaussian?

Thanks

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    $\begingroup$ This is in every elementary textbook (AND you can do the computation yourself) so I would not call this a research level question. Voting to close. $\endgroup$
    – Igor Rivin
    Commented Mar 29, 2012 at 15:23

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The formula for transforming a 0 mean Gaussian says $F_x\[e^{-ax^2}\](k)=\sqrt{\frac{\pi}{a}}e^{-\pi^2k^2/a}$ so the standard deviation certainly changes. Indeed the inverse proportionality is an example of the Heisenberg phenomenon.

Changing the mean of the input by translating its graph will multiply the output by a phase factor.

A widening self-convolution in one domain corresponds to a narrowing self-multiplication in the other domain.

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