[sdiy] Polymoog resonator question

David G Dixon dixon at mail.ubc.ca
Sat Oct 6 22:59:13 CEST 2018

And the 2164 is exponential, or "volts/octave" which, to me, makes more
sense than the linear pots in the Moog unit.  In any case, the original
resonator is "set it and forget it", so I don't really see why the taper or
feel of the pot (or slider) would matter all that much.

> -----Original Message-----
> From: Synth-diy [mailto:synth-diy-bounces at synth-diy.org] On 
> Behalf Of Donald Tillman
> Sent: Saturday, October 06, 2018 12:01 PM
> To: Richie Burnett
> Cc: synth-diy at synth-diy.org
> Subject: Re: [sdiy] Polymoog resonator question
> > 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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