[sdiy] Ladder filters and gain drop, that old chestnut

[Andrew Simper]

On 30 August 2015 at 20:40, Richie Burnett <rburnett at richieburnett.co.uk> wrote:
>> The gain needed for an
>> n stage cascade is 1/(cos(pi/n)^n), so the gain for self oscillation
>> is as follows:
>>
>> n=3 gain=8
>> n=4 gain=4
>> n=5 gain=2.9
>> n=6 gain=2.4
>> n=7 gain=2.1
>> n=8 gain=1.9
>
> Thanks for posting the figures Andy. I was too lazy to work them all out!

You're welcome!

For anyone wanting the gory detail you "just" have do the following.
First take the laplace transform of a one pole low pass:

g / (g + s), where g = 1/RC

Then solve for when the phase shift is pi/n, when s = -sqrt(-1) = -i

g / (g + i) = g^2/(1+g) - i g/(1+g)

then take the arctan of this and solve for when g is pi/n:

arctan (g^2/(1+g), g/(1+g)) == pi/n

g = cot(pi/n)

then evaluate the gain of the filter at this point:

sqrt (g/(g+i) * g/(g-i))^n = (g/sqrt(1+g^2))^n, where g = cot(pi/n)

then you need to take the 1 on this to get the makeup gain you need:

self osc gain = 1 / (cot(pi/n) / sqrt (1+cot^2(pi/n))^n) = 1 / (cos(pi/n)^n)

I would not like to try and derive the same thing for the unbuffered
(e.g. 303) case.



>> The TB-303 diode cascade is a 4 pole filter, but the stages are not
>> buffered from each other, so the gain at the cutoff isn't -3dB, it's
>> lower...
>
> I know what you mean, but we need to be careful with terminology here. Most Electrical Engineer
> types will say >the gain *IS* -3dB at the cutoff frequency, because the -3dB amplitude point is the
> definition of "cutoff >frequency". It's just convention really.
>
> -Richie,

Great, thanks for pointing out the correct terminology, that makes sense.

Cheers,

Andy