[PD] plotting freq response of [cpole~] (or "vcf")

[forrest curo]

This may help... [online draft of Puckette's book]:

http://msp.ucsd.edu/techniques/latest/book-html/node105.html

"The main reason we use complex numbers in electronic music is because they
magically automate trigonometric calculations. We frequently have to add
angles together in order to talk about the changing phase of an audio
signal as time progresses (or as it is shifted in time, as in this
chapter). It turns out that, if you multiply two complex numbers, the
argument of the product is the sum of the arguments of the two factors.
[etc]"


On Tue, Aug 12, 2014 at 5:34 PM, Alexandre Torres Porres <porres at gmail.com>
wrote:

> > Frequency response is normally computed in terms of
>
> > magnitude and phase--because the result of applying
>
> > a filter is to multiply the magnitudes and shift (add) the
>
> > phases.
>
>  That seems clear for me. I know how to get both mag/phase but my patch
> is simplified to get the magnitude only. I also know how to get mag & phase
> with real/Imaginary parts too. Where I get stuck is the z transform deal.
> More precisely, adapting the patch to a complex version.
>
> For instance, it works on plotting the freq response of a real pole with
> an input of the filter coefficient. But I’d like to plot the freq response
> of complex pole, from the real and imaginary part of the coefficient.
>
>
>
> > To put it in terms of 'f' in Hz relative to the sampling frequency, use
> > w=(2*pi/Fs) * f,  with Fs=sampling frequency in Hz
>
>
>
> Yeah, the patch already calculates frequency in rad/sample. More over, it
> uses complex frequencies, which are the cosine and sine of the freq in
> rad/sample.
>
>
>
> Now, as I said before, I know the transfer function of [cpole~] is is
> H(Z) = 1/(1 - aZ^-1) – just like the [rpole~] by the way – but that is not
> clear on how to deal with a complex coefficient.
>
>
> > The next problem: you get a complex number in the denominator.
>
>
>
> I guess you mean what I just said :)
>
>
> > Multiply numerator and denominator by the conjugate and split into
>
> > real and imaginary parts before applying the magnitude and phase
>
> > calculations to get your spectrum.  Your coefficient 'a'  is a complex
>
> > number, so work carefully with the conjugate math to separate the
>
> > real and imaginary parts.
>
> well, if this is the solution to my problem, I don’t think I could follow
> what you meant.
>
>
>
> Anyway, I’m attaching a much more objective and simpler version of the
> patch I’ve sent before. It also has a descriptive text that explains the
> patch and the issue. I think I’m really close to nailing this. I just need
> a tiny hand with the math.
>
>
>
> Thanks
>
> Alex
>
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