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
Hey Alexandre,
This blog - EarLevel
Engineeringhttp://www.earlevel.com/main/2003/02/28/biquads/ -
really helped with my understanding of poles/zeros and biquads. Hope it's
useful!
Cheers,
Joe
On 24 September 2013 06:36, Alexandre Torres Porres por...@gmail.comwrote:
for what i see, it's not
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
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
hey joe, this blog is awesome, I stumbled upon it too, they even have an
applet that does the job I want, but no code or formulas around :P it's the
closest thing I found on the subject in the internet...
weird how I can't seem to find these formulas on google and all...
cheers
2013/9/24 Joe
On Tue, Sep 24, 2013 at 2:36 PM, Alexandre Torres Porres
por...@gmail.comwrote:
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
It doesn't load here either. Perhaps the server is too busy since I put
this link here and
On Tue, Sep 24, 2013 at 2:50 PM, Alexandre Torres Porres
por...@gmail.comwrote:
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
well, not sure what you mean, again way over my head, but I was giving it a
hard shot in the dark and it seemed to have worked out :)
I just summed both parts of Z0, for instance, and tried the given math,
numbers came out!
now to make more tests and see if this is consistent, then finish the
On Tue, Sep 24, 2013 at 3:08 PM, Funs Seelen funssee...@gmail.com wrote:
On Tue, Sep 24, 2013 at 2:50 PM, Alexandre Torres Porres por...@gmail.com
wrote:
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
so you're basically saying all i need to use is use only the real part,
right?
my frankenstein was working and alive for several times until i tried some
bandpass coeff, let's se if i fix this now :)
2013/9/24 Funs Seelen funssee...@gmail.com
On Tue, Sep 24, 2013 at 3:08 PM, Funs Seelen
hey, starting to see what you mean much more clear, cool, really excited.
Thanks a lot!
2013/9/24 Alexandre Torres Porres por...@gmail.com
so you're basically saying all i need to use is use only the real part,
right?
my frankenstein was working and alive for several times until i tried
On Tue, Sep 24, 2013 at 3:35 PM, Alexandre Torres Porres
por...@gmail.comwrote:
so you're basically saying all i need to use is use only the real part,
right?
No, I meant that I have the idea that the imaginary part in the calculated
coefficients will disappear automatically if you add
...@gmail.com
Subject: Re: [PD] from poles/zeros to biquad coefficients - how to?
(something like max's z-plane)
Date: September 24, 2013 12:36:27 AM CDT
To: Funs Seelen funssee...@gmail.com
Cc: pd-lista puredata pd-list@iem.at
for what i see, it's not some sort of straight formula, right
...@gmail.com
*Subject: **Re: [PD] from poles/zeros to biquad coefficients - how to?
(something like max's z-plane)*
*Date: *September 24, 2013 12:36:27 AM CDT
*To: *Funs Seelen funssee...@gmail.com
*Cc: *pd-lista puredata pd-list@iem.at
for what i see, it's not some sort of straight formula, right
On 24/09/13 21:46, Funs Seelen wrote:
On Tue, Sep 24, 2013 at 3:35 PM, Alexandre Torres Porres
por...@gmail.comwrote:
so you're basically saying all i need to use is use only the real part,
right?
No, I meant that I have the idea that the imaginary part in the calculated
coefficients will
On Tue, Sep 24, 2013 at 6:32 PM, Simon Wise simonzw...@gmail.com wrote:
On 24/09/13 21:46, Funs Seelen wrote:
On Tue, Sep 24, 2013 at 3:35 PM, Alexandre Torres Porres
por...@gmail.comwrote:
so you're basically saying all i need to use is use only the real part,
right?
No, I meant that
On Tue, Sep 24, 2013 at 6:13 PM, Alexandre Torres Porres
por...@gmail.comwrote:
after some shots in the dark, adjustments and stuff, I was able to make it
work really well... thanks a lot again, will put this out hopefully soon
after I clean it up and include some features. Cheers
Great!
Hi Alexandre,
Great that you're doing this! For extra inspiration you could have a look
at PoZeTools (http://kmt.hku.nl/~pieter/SOFT/RESP/html/PoZeTools.html).
It's great software by Pieter Suurmond. He's the one who taught me filter
design and he probably has some info about Z-transform on his
On Mon, Sep 23, 2013 at 5:35 PM, Alexandre Torres Porres
por...@gmail.comwrote:
thanks, here's a pic of what I have so far
https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/11212_10151872996046683_1825736206_n.jpg
Cool.
For extra inspiration you could have a look at PoZeTools
Great that you're doing this!
thanks, here's a pic of what I have so far
https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/11212_10151872996046683_1825736206_n.jpg
For extra inspiration you could have a look at PoZeTools
It sure does look like what I need. Thanks. But extracting what I
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 funssee...@gmail.com
On Mon, Sep 23, 2013 at 5:35 PM, Alexandre Torres Porres por...@gmail.com
wrote:
thanks, here's a pic of what I have so far
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