[music-dsp] Gaussian FIR filter

Dave Gamble signalzerodb at yahoo.co.uk
Fri Sep 3 06:44:43 EDT 2004

On 2 Sep 2004, at 20:05, Pierre Richemond wrote:

> Hi all,
> Just passing as well...
I'm having lots of trouble understanding this.

>> It kinda feels like there might be some reccurence based on 
>> integration
>> by parts; but it's not obvious...
>> So... how does one prove it? :)
> It's a lot easier to use contour integration in the complex plane :p
> ( http://mathworld.wolfram.com/ContourIntegration.html )
> Since the function z->exp(-z^2) is holomorphic everywhere, its 
> integral on
> any closed path is zero.
Hmm... not convinced I believe you, but I've never looked at e(-z^2) 
over the complex plane.
I could believe that it is holomorphic everywhere...

> Given l, and a variable R, the result still holds by picking a 
> rectangular
> contour with edges R, R+i.l, -R+i.l, -R.
Agreed. This is forming a rectangle with one edge from -R to R with 
width l.
Accepting that exp(-z^2) is holomorphic, we know that the sum of these 
integrals totals zero.

Still, we need exp(-z^2)*exp(-i*z*f)... I guess the 'far' edge (from 
-R+i.l to R+i.l) is following
exp(-(z+il)^2) = exp(-(z^2+2il-l^2)) = exp(l^2) exp(-z^2 + 2il).

I can imagine that you get some cancellation to end up with a nice 
clean result... is that the case?

> Simplify the integral on each of the four segments, and let R go to
> infinity to get exp(-l^2). sqrt(Pi) = lim(R->infinity)
> Int(exp(-t^2-2i.t.l),t,-+infinity)
> which is pretty much the result !
I think I maybe understand... how far off am I?

>> With that proof in hand, the uncertainty principle must follow quite
>> easily :)
> Yes, it gives the special case when the time-frequency product is 
> maximal,
> yet the general inequality is derived with Cauchy-Schwarz...
Mmmmmm. Wavelets...


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