one doubt emerges really soon anyway. Since they are complex (there are two coordinate numbers for each pole and zero) how do I get only one number by, for example, summing or multiplying one pole to the other? as in:

b1 = -(P0 + P1)
b2 = (P0*P1)

cheers!


2013/9/24 Alexandre Torres Porres <porres@gmail.com>
damn, it says it cant load it here :P

but this seems like a simple formula to try out, from what you copied here. If that's all, and if I got what it means, I can see a patch coming right now :) let's see!

thanks


2013/9/24 Funs Seelen <funsseelen@gmail.com>
Hi Alexandre,

This is the online tool: http://kmt.hku.nl/~pieter/cgi-bin/resp/nph-PZT.cgi.

It starts with an example and every time you refresh the page it gives you a new one. If you scroll down there's a link that tells you how the coefficients were calculated, e.g.:

2 zeros give 3 coefficients:

a0 = G
a1 = -G(Z0 + Z1)
a2 = G(Z0*Z1)

2 poles give 3 coefficients:

b0 = 1
b1 = -(P0 + P1)
b2 = (P0*P1)

The linear difference equation is derived from these as you can see.

Regards,
--Funs


On Tue, Sep 24, 2013 at 7:36 AM, Alexandre Torres Porres <porres@gmail.com> wrote:
for what i see, it's not some sort of straight formula, right? seems a bit more complicated than that. 

cheers


2013/9/23 Funs Seelen <funsseelen@gmail.com>
On Mon, Sep 23, 2013 at 5:35 PM, Alexandre Torres Porres <porres@gmail.com> wrote:

Cool.
 

> For extra inspiration you could have a look at PoZeTools

It sure does look like what I need. Thanks. But extracting what I need to know about the math of converting from coordinates to coefficients was just over my head :P unfortunately, sorry.

I was hoping for something simpler, like just the operations needed. If the info is in code, I need it to more explicit. I'd really appreciate if anyone knows how to read from this and just points it out for me so I can put it in a patch. 

I'm assuming it's rather simple math

I remember I once learned how to do this but never repeated the practice. If I find time to do that I would gladly try to figure it out again, but if someone more experienced feels the urge to chime in before that time I would be very happy too :).