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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