[sdiy] Polymoog resonator question

Donald Tillman don at till.com
Sat Oct 6 21:01:21 CEST 2018


> On Oct 6, 2018, at 11:43 AM, Richie Burnett <rburnett at richieburnett.co.uk> wrote:
> 
> To state it again, it's the *gain* that matters, and implementing it like shown in the schematic means that you can use a common "log taper" pot to get an approximately exponential frequency scaling for the cutoff frequency control.  If you had instead chosen to make the R of the integrators variable (like in the polymoog resonator schematic,) it would require an "anti-log" taper pot to get the right feel for the frequency control which is harder to get.

The pot taper issue gets interesting.

An audio taper pot isn't really an exponential curve; it's two linear curves spliced together at the center.

And the Polymoog resonator uses linear pots in series with the input resistor, and the gain is proportional to the inverse of the resistance:

TFintegrator = 1 / (2400 + (1-x)10000) sC

Where x is the pot travel, 0.0 to 1.0.

And you can plot that, and it's log, on the Grapher app on a Mac, or any plotting facility, and check it out.

So the Polymoog style curve turns out to be reasonably close to an exponential curve, closer than an audio pot.

  -- Don
--
Donald Tillman, Palo Alto, California
http://www.till.com




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