[sdiy] How does resonance work?

[David G Dixon]

> Recently I've started building the Jupiter 8 filter using the 
> IR3109 and also with discrete OTAs. I couldn't get the 
> resonance to work, so I looked at the other implementations 
> of the 3109 in filter circuits. I started wondering:
> 
> Is resonance as simple as a VCA in the feedback path of the 
> gain cells?
> 
> I know that resonance is an increase in emphasis of the 
> cutoff knee of the filter; the highest frequency you can hear 
> from the LPF. So, although the fundamental frequency of the 
> signal passing through it might be quite low you hear the 
> whooshy self-oscillation as a higher freq.
> 
> Please correct/educate me.

To understand the resonance of any filter circuit, you really need to have
the transfer function for that circuit.

Resonance emphasizes the cutoff frequency.  This is not the highest
frequency you can hear from the LPF.  It is the frequency which is 6dB down
at zero resonance.

Resonance works by decreasing the denominator of the transfer function.  At
a certain amount of loop gain, the denominator goes to zero.  At this point,
the filter self-oscillates.

As an example, here is the transfer function for a 24dB cascaded-stage
(Roland-style) filter:

V_out/V_in = 1/[(s+1)^4+G] 

           = 1/[s^4 + 4s^3 + 6s^2 + 4s + 1 + G] 

           = 1/[(w^4 - 6w^2 + 1 + G) + j(4w - 4w^3)]

where s = the Laplace variable, G = loop gain, j = the imaginary number
(root -1), and w (omega) = f/f_Cutoff.

Note that, at the cutoff frequency, w = 1.  Hence, at a loop gain of G = 4,
the transfer function becomes:

V_out/V_in = 1/[(1 - 6 + 1 + 4) + j(4 - 4)] = 1/0 = infinity.

This means that a "pole" (an infinity) is at the cutoff frequency along the
imaginary axis, which is when a filter self-oscillates.

Other filters have other transfer functions, and other conditions for
self-oscillation.