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
This is the online tool:
http://kmt.hku.nl/~pieter/cgi-bin/resp/nph-PZT.cgi.
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:
thanks, here's a pic of what I have so far
https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/11212_101518729960466...
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 :).