> 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