Hi Stu,

> thanks tim, that's really helpful. of course, like all the best answers, it raises more questions :)
>
> like this one: what's so special about a cut off of 1kz and a slope of 30db that the filtered signal stays in phase with the original up the cut off frequency

Indeed, that is an interesting observation! And the short answer is I have no idea. However, having run a few more simulations, at frequencies above and below 1kHz, and stepping the resonance from min to max, it appears to be something special about the fifth stage, and doesn't appear to be particularly frequency related, nor related too much to the resonance (though it is more pronounced at higher res settings): for the stage below (24dB), below the cut-off it peaks above zero, and for above (36dB), it falls away from zero before reaching the cut-off.

I'm guessing it is just some funny facet of the algebra concerned. The (normalized) transfer function for 'stage n', n=1 to 8 (so 30dB is n=5), appears to be

H(s)=1/((s+1)^n+k*(s+1)^(2n-8))

where k is the gain in the feedback loop. If one were to be bothered to work out the phase from this, tan^-1 im(H(wj))/re(H(wj)), I dare say it might be possible to see what it is about n=5 that keeps it nearly zero for w<1 (for 'w', read 'omega'). I have to say I'm not even remotely tempted, far too busy doing other things!

Tim