[PD] Cross correlation using FFT

[Mathieu Bouchard]

On Fri, 7 Apr 2006, Jamie Bullock wrote:

> I'm still trying to work this out and getting into more knots. I thought
> I'd strip it back to the basic maths.
> 
> Am I right in saying that the basic operation FFT(x) * conj(FFT(y))
> works like this:
> 
> (x + ix')(y - iy') = (xy + x'y') + 0i ?

no, the case when imaginaries cancel happens only when you multiply a 
number by its own conjugate, and then it gives the square of the 
magnitude:

(x + ix')(y - iy') =  xy + xiy' + ix'y - ix'iy'
                   = (xy + x'y') + (xy' + x'y)i

so in the case when x=y and x'=y',

(x + ix')(x - ix') = (xx + x'x') + (xx' + x'x)i
                   = (x^2 + x'^2) + 0i

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| Mathieu Bouchard - tél:+1.514.383.3801 - http://artengine.ca/matju
| Freelance Digital Arts Engineer, Montréal QC Canada