Sure, that wouldnt' be too hard with Tcl/Tk. Data structures would
probably not work so well. Ever looked at Kzrysztof's [Scope~]?
It's in cyclone.
.hc
On Sep 15, 2008, at 7:21 PM, Martin Peach wrote:
Any chance of an interactive implementation like zplane in max/msp?
Looks like it
Charles Henry wrote:
There is no zero at z=0. I'm not sure about this one--but it seems as
though it's impossible to have a zero at z=0?
[shrug] i don't even know what a 'zero' is. and that's after reading
several different webpages that attempted to explain biquad filtering to
me. i just
Charles Henry wrote:
z^-1 is the unit delay operator
so,
Y(z)=z*X(z) means y(n) = x(n+1)
while this is not impossible... it's non-causal. For real-time
filtering, you can't already know the sample that comes next.
I've been thinking about it for a little while now. I hope we can
Damian Stewart wrote:
[shrug] i don't even know what a 'zero' is.
a (probably a little too simple) explanation:
a zero is a valley. its depth is 0 on its deepest
point. a pole is a very (very,very,...) high
mountain. both things exist on the complex plane
(just a plain, where all points have 2
On 21/09/2008, at 16.02, Charles Henry wrote:
For real-time filtering, you can't already know the sample that
comes next.
Isn't that what windows (=(?) blocks in Pd-lingo) is for?
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Charles Henry wrote:
There's two problems with your patch.
There's a sign error in one of the expr which calculates the imaginary value.
The second problem (in the same expr) was a * vs / error ( you put /
2*$f1 instead of / 2 / $f1). Does that make a difference?
Anyway, I jiggled with it
On Sat, Sep 20, 2008 at 7:09 AM, Damian Stewart [EMAIL PROTECTED] wrote:
however, there's some weird gain thing going on. check out the attached:
notice how the left channel output (biquad~ output) drops gain, while the
right channel output (handmade) does not.
any ideas how to fix this?
Hallo Chuck,
thank you soo much! All is clear now. It seems I already was close or
at least on the right path. The check for real roots is very nice as
well, I had a bad feeling ignoring them completely and just crossing
fingers in my patch.
Now it isn't difficult anymore to replace the message
Hallo,
martin brinkmann hat gesagt: // martin brinkmann wrote:
Claude Heiland-Allen wrote:
cpole~ czero~ rpole~ rzero~ have signal inlets for filter control, so i
assume sufficiently smooth changes will not cause clicks
that is true, but it looks like no one has made a (usual
Frank Barknecht wrote:
Hallo,
martin brinkmann hat gesagt: // martin brinkmann wrote:
Claude Heiland-Allen wrote:
cpole~ czero~ rpole~ rzero~ have signal inlets for filter control, so i
assume sufficiently smooth changes will not cause clicks
that is true, but it looks like no one has
Claude Heiland-Allen wrote:
Frank Barknecht wrote:
Hallo,
martin brinkmann hat gesagt: // martin brinkmann wrote:
Claude Heiland-Allen wrote:
cpole~ czero~ rpole~ rzero~ have signal inlets for filter control, so i
assume sufficiently smooth changes will not cause clicks
that is true, but
Claude Heiland-Allen wrote:
http://ccrma.stanford.edu/~jos/filters/BiQuad_Section.html
you need to munge the biquad coefficients from the form expected by Pd's
biquad~ to the form of the difference equation at the bottom - should
just be some scaling and sign stuff.
then you can
Frank Barknecht wrote:
that is true, but it looks like no one has made a (usual lp,hp,etc.)
filter with these objects until now.
Except Miller. [1]
you are right, there are some usual filters made with
pole/zero objects, the shelving and peaking ones, a nonresonant lp
and a bp, iirr. and the
Charles Henry wrote:
wow - this is a bit above my head at the moment. i can go from equations in
C to pd, but only if i have the equations themselves to look at...
It's easy. I've done it half a dozen times. For example,
http://lists.puredata.info/pipermail/pd-list/2007-01/046315.html
On Mon, Sep 15, 2008 at 7:49 AM, Damian Stewart [EMAIL PROTECTED] wrote:
dude, that's not 'easy'. i can barely remember how the quadratic formula
works with pen and paper, let alone in C, and let alone to the point where i
could confidently transfer from C to Pd. i'd have no idea where the bugs
Hallo,
Charles Henry hat gesagt: // Charles Henry wrote:
On Mon, Sep 15, 2008 at 7:49 AM, Damian Stewart [EMAIL PROTECTED] wrote:
dude, that's not 'easy'. i can barely remember how the quadratic formula
works with pen and paper, let alone in C, and let alone to the point where i
could
Hallo,
Frank Barknecht hat gesagt: // Frank Barknecht wrote:
1*z^2 - fb1*z^1 - fb2 = 0
zb = (-fb1 +- sqrt(fb1*fb1 - 4*fb2)) / 2
Sorry, the one above is wrong, of course. Given the minus-signs in the
original transfer formula it should be:
zb = (fb1 +- sqrt(fb1*fb1 + 4*fb2)) / 2
if I'm
Frank Barknecht wrote:
Hallo,
Frank Barknecht hat gesagt: // Frank Barknecht wrote:
1*z^2 - fb1*z^1 - fb2 = 0
zb = (-fb1 +- sqrt(fb1*fb1 - 4*fb2)) / 2
Sorry, the one above is wrong, of course. Given the minus-signs in the
original transfer formula it should be:
zb = (fb1 +-
Any chance of an interactive implementation like zplane in max/msp? Looks
like it could be done with data structures or gem...
Martin
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On Mon, Sep 15, 2008 at 9:39 AM, Damian Stewart [EMAIL PROTECTED] wrote:
Frank Barknecht wrote:
Hallo,
Frank Barknecht hat gesagt: // Frank Barknecht wrote:
1*z^2 - fb1*z^1 - fb2 = 0
zb = (-fb1 +- sqrt(fb1*fb1 - 4*fb2)) / 2
Sorry, the one above is wrong, of course. Given the minus-signs in
Hallo,
Damian Stewart hat gesagt: // Damian Stewart wrote:
Frank Barknecht wrote:
Hallo,
Frank Barknecht hat gesagt: // Frank Barknecht wrote:
1*z^2 - fb1*z^1 - fb2 = 0
zb = (-fb1 +- sqrt(fb1*fb1 - 4*fb2)) / 2
Sorry, the one above is wrong, of course. Given the minus-signs in the
There's two problems with your patch.
There's a sign error in one of the expr which calculates the imaginary value.
The second problem (in the same expr) was a * vs / error ( you put /
2*$f1 instead of / 2 / $f1). Does that make a difference?
Anyway, I jiggled with it enough to get it right.
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