[sdiy] Talking of Q

[Olivier Gillet]

> Isn't resonance/oscillation regarded as "infinite Q"? Or am I totally wrong about that?

Yes ; a non infinite Q means that the self oscillation is decaying over time.

> If so, most synth filters go from zero to infinity. I suppose the usable part of that range is probably much less - which might be where 0-500 comes in.

With Q < 0.5 you can have a severe gain loss in the pass-band too.
This can be seen by looking at the transfer function (assuming an
overall unity gain):

H(s) = 1 / (s^2 / (w0)^2 + 1 / Q . (s / w0) + 1)

If s = jw with w << w0 ; the term in s^2 can be ignored ; but 1 / Q (s
/ w0) can get high when Q gets closer to 0.

> I always pictured it as something that looks like x^2 - a graph curving off upwards exponentially towards y-axis infinity. Maybe that's not how it is...

On a 2-poles SVF design, Q is proportional to the reciprocal of the
gain of the feedback/damping cell ; so if you have an exponential VCA
there, it follows an upward exponential curve. The curve won't reach
infinite Q / zero damping because the VCA can't be exactly silenced.
By forcing extra feedback through an additional resistor (this was
discussed here some time ago), your response becomes something like:

Q = 1/2 max(gain on damping feedback path - extra feedback amount, 0.0) ^ -1

So

Q = 1/2 max(10^-(Q control CV x scaling) - extra feedback amount, 0.0) ^ -1

With an exponential VCA in the feedback path.

Here's what the theory gives for the Shruthi-1 dual SVF (blue curve):

http://img805.imageshack.us/img805/6472/screenshot20110617at218.png

Green curve is the Q progression using a linear attenuator in the
resonance feedback (you see the slow build up - which is typical for
example of many 4-pole LPF designs where resonance is done with a
linear VCA or a linear pot). Red curve is the Q curve when using the
distortion from an OTA on the Q CV, then a linear VCA [a trick I've
seen here: http://www.emulatorarchive.com/assets/PDF/SSM2033%20SSM2044%20Circuits.pdf]

Olivier