
kyro
@chaosprint said:
For example, I would like to trigger a sound playback multiple times
it layers the playback at that particular rateYou mean, like a polyphonic sampler? If so I'd load the sample in a table and have an instance of tabread~/tabread4~ for each "layer".

kyro
@jameslo said:
If you follow the algorithm in http://cgm.cs.mcgill.ca/~godfried/publications/banff.pdf, E(5,13) = 1001010010100, But if you use the much simpler one from @Stutter you get 1001001010010.
If you flip stutter's result three digits to the left you get banff's. It's a matter of a [ 3] somewhere.

kyro
If I had to buy something today I'd go for Intech EN16. Fighting my impulse buy a while back, I dug out my old dusty bcr2000, found out it had a software editor and could be set in absolute mode. And I don't need anything more.
With encoders set to absolute, no need for twoway communication, no more jumps in values, everything stays in the patch. I have a set of WIP abstractions that steal the focus of the shared unique controller, map and scale CC increments/decrements to the relevant patch parameters and deals with store/restore presets.
I just wish there were something to faders like what encoders are to knobs.

kyro
http://www.pdtutorial.com/english/ch03s08.html was a great help for me back in time.


kyro
@ddw_music very nice! With your downward edge detection I can now divide a phasor~ by an arbitrary number. There's a thing I cannot wrap my head around though: how to reset the counter to zero every divided downward edge. So it can safely be run indefinitely.
I don't understand why feeding the rpole~ argument with ((1(divided downward edge detection)) delayed by one sample) only resets it once in a while and not every cycle.


kyro
so I totally gave up on solving my maths problems in pd. I used FAUST instead and run my tonestack~.dsp in [faustgen~].
you can access the 3 parameters with [low/med/top $1( from 0 to 1.
first time with FAUST, looks awesome. 
kyro
Wew lads! What a complicated topic. Finding real roots for the denominator, finding either 3 real or 1 real and 2 conjugated complex roots for the numerator... The current filter is 100% highly unstable, I hope I'll get it working eventually.

kyro
so far my best guess would be https://en.wikipedia.org/wiki/Cubic_equation#Trigonometric_solution_for_three_real_roots