From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: Help with convolution theorem
Date: 16 Oct 1999 08:51:49 -0500
Newsgroups: sci.physics,sci.math
Keywords: convolution and more general integration kernels

In article <7u8odn$sug@r02n01.cac.psu.edu>, chiz <smm22@psu.edu> wrote:
>Hi All

>I have a question.  I need to know in words what convolution is and why it
>is important for spectroscopists.  I understand all the math, I just cant
>get that physical picture.

This problem is far more general than convolution.  The general idea
is that there is an input, which for simplicity will be assumed to be
a density f.  But what is observed instead is g, where 

*	g(x) = \int f(y) K(x,y) dy.

In the case where K(x,y) = h(x-y), this is a convolution.

However, what is observed is typically not g, but g plus an error.

>When spectroscopists say they are deconvoluting some peaks in the
>frequency domain, they break frequency domain data into many smaller (say)
>gaussians so when all the gaussians are added together, you get the freq
>domain data back.  But, by looking at the math, it looks like a convolution
>means a multiplication in the frequency domainnot an addition.  Yes?

The problem is to come up with an approximate solution for f;
it must be approximate because of observational and numerical
errors, as well as the model not being exactly correct.  

When it comes to getting numerical answers, use what works, even
if it seems mysterious.  Applying Fourier transforms to *, the
equation becomes g^(t) = f^(t)*h^(t).  This is easier to solve
than *, and one can figure out how to handle the errors.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558