# S-R for I/O relations of functions

[From Bruce Nevin (980303.1324)]

Martin Taylor 980303 11:20--

using S-R ideas when dealing with input-output relations of
functions, and control loop ideas when dealing with the behaviour of
control loops.

If the function is within a control loop (input function, output function)
how is it useful to talk about its output in S-R terms? Its output is a
contributor to its input, with lag that effectively does not matter for
reasons discussed. R is part of S. S-R analysis is inappropriate even when

Bruce Nevin

[From Oded Maler (980303)

[From Bruce Nevin (980303.1324)]

Martin Taylor 980303 11:20--
>using S-R ideas when dealing with input-output relations of
>functions, and control loop ideas when dealing with the behaviour of
>control loops.

If the function is within a control loop (input function, output function)
how is it useful to talk about its output in S-R terms? Its output is a
contributor to its input, with lag that effectively does not matter for
reasons discussed. R is part of S. S-R analysis is inappropriate even when

Excuse me, if, for example, f1=x-x* and f2=x*-x (x is state and x*
reference), their closed-loop behavior will be rather different,
isn't it?

--Oded

[From Bruce Nevin (980303.1554 EST)]

Oded Maler (980303)

[...] if, for example, f1=x-x* and f2=x*-x (x is state and x*
reference), their closed-loop behavior will be rather different,
isn't it?

See B:CP p. 62. (There's a typo: the discussion of a subtractor is in
Chapter 3, p. 28.) Bill assumed that the perceptual signal is inhibitory
and the reference signal excitatory for reasons that I don't recall. If you
reverse the signs as you suggest, you still get the delta. The sign would
have to be reversed at effectors as well. We assume comparators are
organized the same way throughout the hierarchy. It's conceivable that
they're not (with the sense of error output also reversing appropriately),
but unlikely.

Bruce Nevin

[From Bill Powers (980303.1421 MST)]

Bruce Nevin (980303.1554 EST)--

See B:CP p. 62. (There's a typo: the discussion of a subtractor is in
Chapter 3, p. 28.) Bill assumed that the perceptual signal is inhibitory
and the reference signal excitatory for reasons that I don't recall.

Strictly a convention, to be overridden by any actual data. There are data
to suggest that the second level of control (in the brainstem) contains the
reverse connections. The sensory feedback going into the motor nuclei is
excitatory, and the reference signals coming from above (from the
cerebellum, for instance) are inhibitory. Obviously the comparators are
part of the motor nuclei, and in fact they are bunched together in one
place at the input to the motor nuclei.

If you
reverse the signs as you suggest, you still get the delta. The sign would
have to be reversed at effectors as well. We assume comparators are
organized the same way throughout the hierarchy. It's conceivable that
they're not (with the sense of error output also reversing appropriately),
but unlikely.

It's possible. I once speculated that the signs reversed with every
successive level (on the basis of observing a sample of one), but couldn't
think of any reason to support that idea.

Best,

Bill P.

[From Oded Maler (980304)

Bruce Nevin (980303.1554 EST)]

Oded Maler (980303)

>[...] if, for example, f1=x-x* and f2=x*-x (x is state and x*
>reference), their closed-loop behavior will be rather different,
>isn't it?

See B:CP p. 62. (There's a typo: the discussion of a subtractor is in
Chapter 3, p. 28.) Bill assumed that the perceptual signal is inhibitory
and the reference signal excitatory for reasons that I don't recall. If you
reverse the signs as you suggest, you still get the delta. The sign would
have to be reversed at effectors as well. We assume comparators are
organized the same way throughout the hierarchy. It's conceivable that
they're not (with the sense of error output also reversing appropriately),
but unlikely.

I'm not sure I was understood. I have two "open-loop" functions f1(x) and
f2(x). One of them, when connected in a closed loop will lead to negative
feed-back, the other will lead to a positive feed-back. This was supposed
to be a counter-example to your previous suggestion that it is not useful
to talk about the I/O properties of a function.

--Oded

[From Bruce Nevin (980304.0916)]

Oded Maler (980304)--

I have two "open-loop" functions f1(x) and
f2(x). One of them, when connected in a closed loop will lead to negative
feed-back, the other will lead to a positive feed-back. This was supposed
to be a counter-example to your previous suggestion that it is not useful
to talk about the I/O properties of a function.

It is not a counter-example. It is a departure from the universe of discourse.

The subject here is negative feedback control. If you reverse the sense of
the inputs to a comparator without simultaneously reversing the effect of
error output (at effectors or at reference inputs at a lower level) you no
longer have negative feedback control. I don't know why you want to talk
about positive feedback, which is incompatible with control, but it is not
what we are talking about here.

What I said:

If the function is within a control loop (input function, output function)
how is it useful to talk about its output in S-R terms? Its output is a
contributor to its input, with lag that effectively does not matter for
reasons discussed. R is part of S. S-R analysis is inappropriate even when