[PD] plotting freq response of [cpole~] (or "vcf")
Alexandre Torres Porres
porres at gmail.com
Wed Aug 13 02:34:40 CEST 2014
> 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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