From: Bill Stewart <stewarts@ix.netcom.com>
Date: Wed, 31 Jul 1996 07:27:13 +0800
To: alexf@iss.net
Subject: Re: Digital Watermarks for copy protection in recent Billbo
Message-ID: <199607301939.MAA08382@toad.com>
MIME-Version: 1.0
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>Now, would you mind doing a little translation (for the laymen), 
>since I didn't understand?

We did Fourier transforms in third--or-fourth semester calculus in college,
but then I _was_ an engineer; electrical engineers would go on to do
lots more of this stuff, since frequencies and waveforms are their territory.
Essentially, you can look at "most" continuous functions in normal time-space,
or you can represent them in a frequency space instead,
and you can reproduce the original function by transforming from
the frequency space back to the time space.  The "Lebesgue" bit
is a precise definition of "most".  

(For most of the math I did in college, "Lebesgue" was a phrase meaning 
"/* you are not expected to understand this */",
and it and Measure Theory got trotted out to clarify rigorously
when functions are well-behaved enough for the stuff we were learning to apply.
Most functions you use are Lebegue integrable, unless you use stuff like
"f(x) = 0 if x is rational and 1 if x is irrational".)

Discrete Fourier Transforms are a related analysis technique that work
on sets of numbers such as equally-spaced samples from a continuous function.
The Fast Fourier Transform is a particularly efficient way to do DFTs,
which was a breakthrough that made them practical to do on computers,
and Jim was reminding the previous poster that for the problem at hand,
determining the frequency spectrum of whatever-it-was, that DFTs aren't
what you need; you need the regular continuous Fourier transform.

At 01:04 PM 7/29/96 +0000, you wrote:
>> Jim Choate <ravage@einstein.ssz.com> writes:
>
>> You want a continuous Fourier transform, not a discrete one, to
>> determine the frequency spectrum of the waveform being sampled.
>> The FFT is simply an algorithm for computing the DFT without
>> redundant computation.  In general, any Lebesgue integrable
>> complex function will have a Fourier transform, even one with a
>> finite number of discontinuities. The reverse transform will
>> faithfully reproduce the function, modulo the usual caveats about
>> function spaces and sets of measure zero.
#			Thanks;  Bill
# Bill Stewart, +1-415-442-2215 stewarts@ix.netcom.com
# <A HREF="http://idiom.com/~wcs"> 
#			Dispel Authority!