[PD] help with rfft~
Mathieu Bouchard
matju at artengine.ca
Wed Nov 30 00:44:52 CET 2005
On Tue, 29 Nov 2005, thewade wrote:
> So which is better for convolution: polar or rectangular? Some processing must
> happen for convolution, because just multiplying the real outputs of two rfft~
> objects and the two complex outputs makes for very poor convolution. Do most
> convolution tools (PD or otherwise) convert to polar, multiply the
> frequencies, divide by niquist (I have no idea here), then use one of the the
> signal phases and convert back to rectangular for rifft~, or is there some
> oter way?
excuse me? multiplying the frequencies? i think you mean, for each
frequency, multiply the amplitudes together.
multiplication of two complexes x+y*i and x'+y'*i in cartesian
representation is:
x'' = x*x' - y*y'
y'' = x*y' + y*x'
Then with a polar representation:
x = r*cos(a)
y = r*sin(a)
It becomes as easy as:
a'' = a+a'
r'' = r*r'
Or even easier in complex log representation:
s+a*i = log(x+y*i)
s = log r
a'' = a+a'
s'' = s+s'
However, it only makes sense if you are going to make many multiplications
between polar forms. If not, then the sin,cos conversions and their
inverses r=sqrt(x*x+y*y),a=atan2(y,x) become more of a burden than a
shortcut.
____________________________________________________________________
Mathieu Bouchard - tél:+1.514.383.3801 - http://artengine.ca/matju
Freelance Digital Arts Engineer, Montréal QC Canada
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