[sdiy] Poles of a diode ladder filter

[Don Tillman]

   > Date: Sun, 26 Mar 2006 23:35:48 -0500 (EST)
   > From: Aaron Lanterman <lanterma at ece.gatech.edu>
   > 
   > So, tomorrow's lecture is on the Moog ladder, which I think I can
   > fake my way through OK.

Hey Aaron,

Are you going to talk about the ladder pole positions, and the pattern
they make when feedback is applied?  Excellent.

   > I wanted to talk about the EMS/TB-303 style ladder that uses
   > diodes (or diode connected transistors) instead of transistors -
   > in those cases, it's like an _unbuffered_ RC ladder.
   > 
   > So, I wondered what the transfer function looked like and where
   > the poles go.
   > 
   > Ugh.
   > 
   > I set up my ladder, and then did repeated Thevinin
   > equivalents. (I'm praying there's an easier way but I didn't see
   > one).
   > 
   > After an ungodly amount of algebra, I found (for a normalized
   > RC=1 filter)
   > 
   > s^4 + 7 s^3 + 15 s^2 + 10 s + 1
   > 
   > Has anyone else ever tried to slog this out?

Someone would have to be nuts to do that... Oh yeah, I did that and
posted the result to SDIY ten years ago.  I've included a copy of the
message below.

Yeah, you got it right (or you made the same mistakes I did).

  -- Don

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


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>From till Tue May 28 16:13:24 -0700 1996
To: synth-diy at horus.sara.nl
In-reply-to: <31A1111F at MSMGATE.M30X.NBG.SCN.DE> (message from Haible_Juergen#Tel2743 on Mon, 20 May 96 17:39:00 PDT)
Subject: Re: butter and pole locations
Reply-to: don at till.com
From: Don Tillman <don at till.com>
Organization: Don's house, Palo Alto, California

   From: Haible_Juergen#Tel2743
   Date: Mon, 20 May 96 17:39:00 PDT

   (BTW, has anybody yet calculated the exact locations / pole spread
   for the diode ladders?)
   The Roland cascaded SV 4p-filters might be different ... anybody
   knows if *they* start on an arc (as in a butterworth)?

Derriving the transfer function isn't all that difficult, but it's a
very mistake-prone process.

The generalized version (any values for the Rs and Cs) is too long for
me to conveniently write out, but if you set all the Rs equal and all
the Cs equal, the transfer function becomes:

  Vout              1
  ---- = ----------------------------
   Vin   s^4 + 7s^3 + 15s^2 + 10s + 1

(I *think* this is correct.  If anybody cares they can do it out
independently and we'll compare notes.)

Which means the poles are at (normalized to 1Hz for clarity):
   2.35Hz, 3.53Hz, 1.00 Hz, 0.12 Hz

As expected we have four real poles.

Adding a feedback factor simply increments the 0th coefficient by that
amount, soooo, with a feedback gain of 1 the poles are:
   2.62Hz, 3.41Hz, 0.59 Hz,  0.38 Hz

At a feedback gain of about 1.07, the lower two poles meet.

Feedback gain of 2:
   3.00Hz,  3.15Hz, double pole at 0.56 Hz (Q= 0.66)

Not too interesting. 

Feedback gain of 3:
   3.14 Hz (Q= 0.50), 0.64 Hz (Q= 0.84)

(Note the higher two poles have met.)
	
At a feedback gain of 4:
   3.20 Hz (Q= 0.51), 0.70 Hz (Q= 1.04)

The lower pole pair is starting to show some gain.

At a feedback gain of 5:
   3.25 Hz (Q= 0.51), 0.75 Hz (Q= 1.25)

At a feedback gain of 10:
   3.45 Hz (Q= 0.52), 0.96 Hz (Q= 2.98)

At a feedback gain of 15:
   3.60 Hz (Q= 0.52), 1.11 Hz (Q= 9.57)

Here's some serious resonance.  And it starts oscillating with a
feedback gain of about 18.3.  The higher pole pair never really gets
off the ground

Executive summary: not too exciting.

  -- Don

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