My Math Forum Convolution of two Gaussian Functions

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 January 28th, 2009, 07:45 AM #1 Newbie   Joined: Jan 2009 Posts: 2 Thanks: 0 Convolution of two Gaussian Functions Hi, i have to convolve this two functions $g(x)=e^{-\pi(\frac{x}{a})^2}$ $h(x)=e^{-\pi(\frac{x}{b})^2}$ i now the solution, it's: $\frac{a\cdot b}{\sqrt{a^2+b^2}}\cdot e^{-\frac{\pi u^2}{a^2+b^2}$ so i've no idea which steps i have to go to get this solution, hope anybody can help me
January 28th, 2009, 03:52 PM   #2
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Re: Convolution of two Gaussian Functions

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 Originally Posted by scratch Hi, i have to convolve this two functions $g(x)=e^{-\pi(\frac{x}{a})^2}$ $h(x)=e^{-\pi(\frac{x}{b})^2}$ i now the solution, it's: $\frac{a\cdot b}{\sqrt{a^2+b^2}}\cdot e^{-\frac{\pi u^2}{a^2+b^2}$ so i've no idea which steps i have to go to get this solution, hope anybody can help me
There are two ways.

The most direct way is write down the convolution integrand and play around with the exponent.

An indirect, but simpler way, is to take the Fourier transforms, which is very easy for these functions. The transforms will look very similar to the functions themselves. Then multiply the transforms together and take the back transform.

 January 29th, 2009, 04:22 AM #3 Newbie   Joined: Jan 2009 Posts: 2 Thanks: 0 Re: Convolution of two Gaussian Functions thank u very much, i got it
 October 9th, 2015, 03:29 AM #4 Newbie   Joined: Oct 2015 From: Austria Posts: 2 Thanks: 0 If your Gaussians are univariate, a whole solution is available here: http://www.tina-vision.net/docs/memos/2003-003.pdf
 October 13th, 2015, 07:13 AM #5 Newbie   Joined: Sep 2015 From: New york Posts: 17 Thanks: 4 This is equivalent to the statement that if X1 and X2 are normally distributed with mean zero and variances s1^2 and s2^2, then their sum is normally distributed with zero mean and variance s1^2 + s2^2

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### convolution of two gaussians

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