[sdiy] Pole Dancing

[Donald Tillman]

> On Sep 18, 2018, at 12:34 PM, Richie Burnett <rburnett at richieburnett.co.uk> wrote:
> 
> That's cool Don.  Great programming, and thanks for sharing.  I've had this on my "to do" list for sometime, but can strike it off now someone else has done it well.
> 
> I always found it fascinating how the Moog filter's poles move as global feedback is applied around the ladder.  Particularly how poles that are slightly spread out on the real axis due to component tolerances "pair up" with their nearest neighbour and then split again to form *two resonant (complex) pairs*.

You bring up an interesting point.  What's the effect of the Moog poles being mistuned a bit?  The answer is, not much.  Worst case, you may need a little more feedback to get the job done.

Even if the tuning is substantially off, the two right poles will pair up, and the two left poles will pair up, and they'll split off into a very similar X pattern.

>  It is interesting to me because the 4-pole moog filter with resonance applied actually behaves like two cascaded 2-pole filters with different cutoff and resonance values.  As the poles split in the classic X formation, the two that make a dash for the jw axis are highly resonant and dominate the frequency response, and the two that head leftwards are still slightly resonant but are at a higher natural frequency and very highly damped.


Indeed.  And you can also check out the individual contributions of the poles.

Given a Moog Ladder quad pole at X=-1 and feedback of -2.5.

You can show the rightmost poles with a double pole around -0.1 with -0.8 feedback, and see the resonance and 12dB/octave slope.  

And separately show the leftmost poles with a double pole around -2.0 with -0.9 feedback, and see that contribution is a 12dB/octave slope only, and not significant until about 3x the tuned frequency.

So the rightmost poles really do predominate at medium and higher resonances.

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Also note, a Moog Ladder pole pattern is nothing like a Butterworth filter, where all the poles contribute roughly equally.

  -- Don
--
Donald Tillman, Palo Alto, California
http://www.till.com