[Martin Taylor 2005.03.04.20.19]

[From Bill Powers (2005.02.26,1201 MST)]

The question remains, how would reversing the sign of feedback at one level have the effect of reversing the sign of feedback at a higher level? I'm sure there are cases where this would happen, but I can't think of any just now. If a lower-level system runs away, it would do so by making the perceptual signal avoid the reference setting, but that could go in either direction; p < r and p getting smaller, or p > r and p getting larger. One of those cases would make p change the right way for the next system up to retain negative feedback.

Actually, I don't think this is correctly analyzed. Rick's simulation seems to show what actually happens. It's a kind of _intrinsic_ positive feedback that we haven't discussed before and that I never thought through until now. I hope I've thought it through properly now.

Positive feedback means that any small error is increased through the action of the output affecting the perceptual signal so as to move the perceptual signal further from its reference value. We have hitherto talked about a simple loop, in which the output has a defined effect on the perceptual signal.

Now we are talking about a loop in which one element is itself a feedback loop. To talk about it, I need some symbols. Call the higher level perception, reference, error and output P2,R2, E2, and O2 respectively. Let O2 be the reference signal for several lower level loops, so O2 -> R11 = R21 ... = R2n. Then we have the obvious O11 ... O1n, P11 ... P1n, and E11 ... E1n as the remaining required symbols.

If Loop 1 is a negative feedback loop, then to a first approximation P11=R11. P2 = f(P11, ... P1n), and is directly influenced by O2.

If Loop 11 is in positive feedback, there's a problem. Start with Loop 11 being in equilibrium (E11 = 0) and (we assume) no disturbance. Until R11 changes, P11 will maintain its value, and P2 will, as well (assuming no disturbances affect P12 ... P1n).

Now let there be a small disturbance affecting P1k. This will affect P2, and change O2, changing R11 ... R1n. If Loop 11 is in positive feeback, P1n will then go into an exponential runaway (without limit in a linear system), which is likely to affect P2 in an exponentially increasing degree. O2 will need to change to compensate for this runaway.

The only way that O2 could halt the runaway in Loop 11 (and thereby halt its own runaway) would be if it could at some point adjust R11 to match the accelerating value of P11 while at the same time bringing its own error to zero through a fortuitous combination of values of P12 ... P1n.

So without computing a loop gain, it seems that reasonable kinds of function P2 = f(P11, ... P1n) are likely to lead Loop 2 into a runaway if P11 goes nto a runaway state.

What happens now if we reverse the sign of the output O2? In the ordinary case, if one sign of the output function leads to positive feedback, the other sign should lead to negative feedback. But it doesn't happen in this case. The runaway happens for either sign of the output function. Whatever fluctuation there is in the output creates an equivalent fluctuation in the reference signal at R11, and causes the perceptual signal P11 to diverge exponentially. If the function P2 = f(P11 ... P1n) allows it, P2 will go into exponential runaway, increasing the error E2, increasing the output O2, usually enhancing the virulence of the exponential runaway of P11.

That's what Rick found to happen in the spreadsheet, I think. He was surprised that changing the sign of the output didn't change the positive feedback to negative. Maybe the above is an appropriate explanation.

Now back to the dog-work!

Martin