Positive feedback

[Bruce_Abbott]

[From Bruce Abbott (950113.1125 EST)]

Rick Marken (960112.2330) --

Me:

you say that Killeen's model is a positive feedback system that
is in equilibrium (which means a loop gain < 1.0).

Bruce Abbott (960112.2015)

To be in equilibrium the loop gain has to be 1.0

This is not consistent with Bill's analysis of P-systems (p. 143
in LCS). Bill claims that a positive feedback loop is stable only
when loop gain is _less than_ 1.0; as soon as the loop gain becomes
1.0 the system becomes unstable.

I think we are talking about different circumstances. Bill is talking about
a variable that can take on both positive and negative values. In that
case, when g < 1.0 the following will be true:

   X(n) < g*X(n-1), X > 0

   X(n) = g*X(n-1), X = 0

   X(n) > g*X(n-1), X < 0

Or, to put it simply, the deviation of X from zero will be reduced with each
iteration, regardless of whether X is above or below zero. Once X reaches
zero, it will stay there. An example would be a ball rolling down the
curved sides of a bowl. Move it up a side and release it, and it will
always return to the bottom.

In this system, if the gain is 1.0, X(n) = X(n) regardless of the value of
X. Such a system is neither stable nor unstable; it is astable. An example
would be a ball resting on a flat, horizontal surface. Wherever you put it,
it stays there when you release it.

If the gain is > 1.0, then any small deviation of X from zero leads to
runaway, because the absolute value of X(n) will be larger than the absolute
value of X(n-1) on each cycle. Such a system is unstable. This is the ball
resting at the top of a hill: any slight push sends it accelerating down.

The kind of system I was describing has a variable that ranges from zero on
up, and the gain is positive. If the gain is > 1 then

    X(n) > g*X(n-1)

and the system runs away to infinity unless it encounters an upper limit, in
which case the limit forces X(n) = g*X(n-1). Now, if this is true, then

    g = X(n)/X(n-1) = 1.0 at the limit.

If the gain is positive and < 1 then

    X(n) < g*X(n-1)

and the system runs away to zero, at which point it encounters its lower
limit. At this point X(n) = X(n-1) = 0.0, and once again the gain is
effectively 1.0 since X remains unchanged over cycles. [Mathematically, the
gain is undefined, since g = X(n)/X(n-1) = 0.0/0.0 is undefined.]

When g = 1.0 then

   X(n) = g*X(n-1) and the system remaines in its current state over iterations.

Now, imagine that g is a negative function of X such that g is always
positive throughout the range of X. Assume the following:

   When X = a, g = 1.0;
   When X < a, g > 1.0;
   When X > a, g < 1.0;

When X = a, then g = 1.0 and X(n) = g*X(n-1). The system will remain at a.

When X < a, then g > 1.0 and X(n) > g*X(n-1). X will approach a from below
as a negative exponential.

When X > a, then g < 1.0 and X(n) < g*X(n-1). X will approach a from above
as a negative exponential. By linking g to X in this way, we have converted
this bistable, positive feedback system into a stable, negative feedback
system whose stable value is a.

I suspect that Bill P. can provide a simpler description of this system, but
I believe that it works as I suggest.

In order to make the loop dynamically stable I had to use a _leaky_
integrator in the output function, so that:

o := o + (k.2i-o)

If the output function is a simple integral (o := o + k.2i) then the
loop variables (o and i) go to infinity no matter what the loop gain
(unless the loop gain is 0.0).

I can stabilize the loop with a loop gain of 1.0 or higher by making the
integrator "leak" more than it increments:

o := o + (k.2i-k.3o)

By making k.3 large enough, you can stabilize the loop even with very
high gain.

What do you suppose your leaky integrator is doing to the loop gain on each
iteration? When you find the right answer to that question, you will know
why this positive feedback loop stabilizes.

Regards,

Bruce

[_Martin_Taylor]

[Martin Taylor 960115 12:00]

Rick Marken (960112.2330)

Rick, What did you actually do?

1:

The system I tested was defined by the following
system of equations:

i = k.1o + d
o = k.2i

The loop gain is k.1*k.2, which is the dimensionless product of all
signal and variable amplifications around the loop.

or 2:

In order to make the loop dynamically stable I had to use a _leaky_
integrator in the output function, so that:

o := o + (k.2i-o)

or 3:

If the output function is a simple integral (o := o + k.2i) then the
loop variables (o and i) go to infinity no matter what the loop gain
(unless the loop gain is 0.0).

An integrator has no "gain" as such. The "gain" is a function of frequency,
doubling with every halving of frequency. So the pure integrator will have
a gain greater than any specified limit at some sufficiently low frequency,
no matter what the other linear factors are around the loop.

If the integrator is leaky, its low-frequency gain has an upper bound, and
the maximum gain at any frequency can be limited by linear factors elsewhere
in the loop.

If you used form (1), as you said, then you do have a gain that is flat
across frequency, but it is computationally unstable because there is
significant gain at frequencies above the Nyquist limit of half the
computational sampling rate at any one point, and because of the
computationally induced pseudo-delay around the loop.

For computational stability, you have to ensure that there is sufficient
low-pass filtering in the signal path that two things are true: (1) the
sampling rate is fast enough that there is effectively no energy at
frequencies higher than fc = 1/2dt where dt is the sampling interval, and
(2) the loop delay of the loop being simulated is long compared with the
delay introduced by the fact that you can't determine the value at the
output of a stage on computational step N until you have found the value
of its input during computational step N-1. Around a loop with possibly
varying reference and disturbance, you need computation at at least four
places: output function, comparator, perceptual function, CEV (i.e. where
the output combines with the disturbance). You can combine some of these
into a single computational stage under appropriate circumstances (as in
your equations (1)), but you can't get away without at least two
computational stages (again as in your equations (1)) to determine the effect
of an output change on itself. If the loop you are simulating has a loop
delay that is not substantially longer than that, you can't believe in the
results of your simulation.

There's a complication in combining requirements (1) and (2). Any low-pass
filter introduces its own delay, on the order of deltaT = 1/2W, where
W is its bandwidth. So by low-pass filtering to avoid the Nyquist sampling
error, you introduce extra pseudo-delay around the loop. The only way to
avoid this is to increase substantially the sampling rate in comparison
to the loop delay of the loop being simulated. If you don't do this, you
can't believe that your simulations tell you ANYTHING about the analogue
loop you are simulating.

An integrator (leaky or not) is a simple low-pass filter that reduces
high-frequency energy in a signal by half for every doubling of frequency,
which is why integrator output stages ensure comupational stability if
the computing sample period is long enough, and why the "slowing factor"
works.

4:

I can stabilize the loop with a loop gain of 1.0 or higher by making the
integrator "leak" more than it increments:

o := o + (k.2i-k.3o)

By making k.3 large enough, you can stabilize the loop even with very
high gain.

Do you now see what this does?

All of which has no bearing on whether Bruce is right or not.

It's not clear to me what the effect would be of a system whose gain is
disturbed, and for which the gain is the input to a perceptual function
of another system controlling to keep that gain at 1.0. Also, it's not
clear what the meaning of the whole verbal description would be if the
system whose gain is controlled has an integrator in its output function.

Martin

[Bruce_Buchanan1]

From Bruce Buchanan (950204.1245 EST)

Bill Powers (950203.0930 MST)] writes:

Negative feedback is goal-seeking feedback. Positive feedback is goal-
avoiding feedback and generally creates a destructive runaway condition
or extreme dynamic instability, also destructive.

So positive feedback is not a healthy condition for any organism or
organization. In an organism it is not usually pro-survival. In an
organization, it can be partly offset by the negative feedback efforts
of the members of the organization, but still has bad effects. . . .

I understand that this might be the point of view of the control systems
engineer. From the point of view of the biologist and living systems,
however, this perspective is incomplete.

The S-shaped curve of natural growth involves processes of positive
feedback in the early stages, when the active elements are few and the
resources in surplus supply. At that stage growth may be exponential.
Sooner or later, however, space and material resources - nutrients, blood
supply - become relatively sparse, and further growth brings about
negative feedback.

Indeed the effect of all the positive feedback processes I can think of -
high explosives, avalanches, atmospheric updrafts that produce hurricanes,
earthquakes, etc. - may release immense energies quickly, but always come
up against larger limits within which new equilibria will be established,
i.e. negative feedback will govern. Similarly in economics and politics and
science, the power of a new product or idea takes fire for a while, perhaps
to reshape the landscape. Speaking of physical systems, it might also be
said that solar energies which provide for all life originate in atomic
reaction processes in positive feedback, maintained in balance by forces of
gravity. So while positive feedback may be destructive in some ways, it
seems to be a necessary aspect of larger constructive processes.

If managers remain unable to
recognize this positive feedback effect and just let it go on, the only
end-point will be an economic disaster.

Perhaps managers should recognize this effect and seek to understand and
take advantage of it, allowing product lines to expand in accordance with
demand but not to outrun the capacity of the business to maintain other and
future services. It is not positive feedback that is the hazard as much as
pride, ignorance and overweening ambition - a cybernetic interpretation of
an old story!

Cheers!

Bruce B.

[Bill_Leach1]

<[Bill Leach 950209.22:40 EST(EDT)]

Bruce Buchanan (950204.1245 EST)

The S-shaped curve of natural growth involves processes of positive
feedback in the early stages, when the active elements are few and the
resources in surplus supply. At that stage growth may be exponential.
Sooner or later, however, space and material resources - nutrients,
blood supply - become relatively sparse, and further growth brings
about negative feedback.

The abundance of resources is not positive feedback... it may well be
considered to be a "positive condition" but any growth reduces the level
of these resources which reduces the capability for growth. The feedback
is negative if the term applies at all: As long as sufficient resources
exist, growth continues. The problem there (as I understand what you are
saying) in applying the term in reference to the resources is that the
resources are an independent variable. There is a correlation between
resource quantity and growth but resources don't CAUSE growth but rather
only make growth possible.

Indeed the effect of all the positive feedback processes I can think of
- high explosives, avalanches, atmospheric updrafts that produce
hurricanes, earthquakes, etc. - may release immense energies ...

Of these mentioned, I only see one that typically involves a positive
feedback process (high explosives). Fire often involves positive
feedback in terms of drafting (thermal gradient increases air flow which
increases oxidation which increases thermal gradient, etc.) and I have to
admit that my knowledge of atomospherics is too limited to know if
positive feedback processes are involved or not.

Fusion and fission both involve positive feedback but as you mention,
negative feedback is always ultimately limiting (and controlling where
sustained reactions occur).

-bill