# [sdiy] SVFs with different gains in the integrators

Tom Wiltshire tom at electricdruid.net
Tue Dec 15 17:36:04 CET 2020

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> On 15 Dec 2020, at 13:24, Guy McCusker <guy.mccusker at gmail.com> wrote:
>
>> This simply implies that the denominator of a S-V filter changes from a normalized s^2 +(1/Q)s + 1  to  s^2 + (A1/Q)s + A1A2 where the integrator gains change from 1 to A1 and A2 respectively, -1/Q is feedback VB to input, and the feedback from VL to input is -1.  Just a modified 2nd-order, and the quadratic equation yields the exact new poles (hence the exact SHAPE of the frequency response) and corresponding (re-interpreted?) performance parameters.  [For example, Omega-Zero moves from 1 to sqrt(A1A2) as someone here offered.]   Everything should follow easily, obviously, and correctly.  No magic in this - right?
>
>
> I guess the interesting -- though admittedly not magical -- element of
> this to me is that the Q of the transfer function you have written is
> not given by the parameter called Q but by Q *  sqrt(A2/A1), so the Q
> of the filter is influenced by the ratio of the integrators' unity
> gain frequencies as well as by the feedback from the bandpass. Andrew
> Simper pointed out that this is also relevant to standard
> implementations because presumably the two integrators are not
> perfectly matched, though I suppose the influence of their mismatch is
> not that great.

The square root helps makes sure it’s not that significant.

Imagine A1 and A2 are out by up to 10% so the worst case is either

1.1/0.9 = 1.22, sqrt(1.22) = 1.105

0.9/1.1 = 0.81, sqrt(0.81) = 0.90

So a 10% variation in the integrator frequencies (both of them) gives a 10% variation in Q. That’s not that huge in Q terms, and we might hope that we get closer matching between our integrators than 10%.

Tom

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