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From: cjwein@watcgl.waterloo.edu (Chris J. Wein)
Newsgroups: comp.dsp
Subject: Re: Why Don't My FFT Deconvolutions Work!!
Message-ID: <1991Feb15.011915.29463@watcgl.waterloo.edu>
Date: 15 Feb 91 01:19:15 GMT
References: <1991Feb9.034956.7993@bilver.uucp> <1991Feb14.042835.5631@lynx.CS.ORST.EDU>
Distribution: na
Organization: University of Waterloo
Lines: 28

In article <1991Feb14.042835.5631@lynx.CS.ORST.EDU> cosgrok@jacobs.CS.ORST.EDU (Kevin Cosgrove) writes:
>In article <1991Feb9.034956.7993@bilver.uucp> alex@bilver.uucp (Alex Matulich) writes:
>[stuff deleted]
>>You will recall that the deconvolution of a function f with a response
>>function g is
>>
>>F(f decon g) = F(f) / F(g)
>>
>>where F() is the Fourier transform.  See the problem?  You have to DIVIDE
>>by the Fourier transform of g.  In all my experiments, for any response
>>function I try, F(g) is zero somewhere (or near zero) which completely
>>messes up the result.
>
>Don't get too frustrated.  We don't want to see you end up in a hospital
>or anything -- geese that's it!  Could l'Hopital's rule help evaluate
>the function when the denominator goes to zero?
>

L'Hopital's Rule only works when both the numerator and the denominator
become zero.  The best example is finding the value of (sin x)/x at x=0.
A singularity is a singularity and using l'Hopital's rule doesn't make it
go away magically.

-- 
==============================================================================
 Chris Wein                           | cjwein@watcgl.waterloo.edu 
 Computer Graphics Lab, CS Dept.      | cjwein@watcgl.uwaterloo.ca
 University of Waterloo               | (519) 888-4548