Perseptions we don't control

[Bj_rn_Simonsen]

[From Bjorn Simonsen (2007.01.13,10:50 EUST)]

Will anybody comment the way I use the
negative feedback model below?

The negative feedback model explains how
living organisms control their perceptions.

Changes in the environment tend to disturb
the current perception by pushing it away from its reference standard. The
error signal result in feedback effects that oppose the disturbance and the
perceptual signal is brought back to its reference standard.

Most of what we perceive is not involved in
control.

Is it possible to use the negative feedback
control model when we describe perceptions we don’t control and say the
following: “ All perceptual signals have the reference value (very near). Therefore
the error is (very near) zero. No actions and no feedback effects. Our
perceptions are now our representation in the brain of what is going on outside
us”.

bjorn

···

[Richard_Marken2]

[From Rick Marken (2007.01.13.1130)]

Bjorn Simonsen (2007.01.13,10:50 EUST)--

Will anybody comment the way I use the negative feedback model below?

The negative feedback model explains how living organisms control their perceptions.

Changes in the environment tend to disturb the current perception by pushing it away from its reference standard. The error signal result in feedback effects that oppose the disturbance and the perceptual signal is brought back to its reference standard.

I'd say the error signal results in output variations that act on the controlled perception via the feedback path through the environment to counter the effects of disturbances to that perception.

Most of what we perceive is not involved in control.

Who knows?

Is it possible to use the negative feedback control model when we describe perceptions we don�t control and say the following: � All perceptual signals have the reference value (very near). Therefore the error is (very near) zero. No actions and no feedback effects. Our perceptions are now our representation in the brain of what is going on outside us�.

No It is not necessarily true that all perceptual signals have a reference value. (I don't know what "very near" refers to). Also, zero error does not necessarily mean no action and no feedback effect.

Best

Rick

Richard S. Marken Consulting
marken@mindreadings.com
Home 310 474-0313
Cell 310 729-1400

[Bj_rn_Simonsen]

[From Bjorn Simonsen
(2007.01.13,23:45 EUST)]

From Rick Marken (2007.01.13.1130)

No It is not necessarily true that all perceptual
signals have a

reference value. (I don’t know what “very
near” refers to). Also, zero

error does not necessarily mean no action and no
feedback effect.

If there is no reference value, I think it is zero
(very near). Then the error becomes negative and that doesn’t work. Therefore
the error is zero (very near).

I think that if the error is very near (not exact) zero
there need not be any action and feedback effect. But I can’t explain how there
is an action and feedback effect if the error is zero (exact zero).

You didn’t answer if we can use the negative feedback
loop model if we don’t control a perception. I think we can and we should, but
I may be wrong.

bjorn

···

[Martin_Taylor2]

Re: Perseptions we don’t
control
[Martin Taylor 2006.01.13.17.54]

[From Bjorn Simonsen (2007.01.13,23:45
EUST)]
From Rick Marken (2007.01.13.1130)

No It is not necessarily true that all perceptual
signals have a
reference value. (I don’t know what “very
near” refers to). Also, zero
error does not necessarily mean no action and no
feedback effect.

If there is no reference value, I think it is zero
(very near). Then the error becomes negative and that doesn’t work.
Therefore the error is zero (very near).
I think that if the error is very near (not exact)
zero there need not be any action and feedback effect. But I can’t
explain how there is an action and feedback effect if the error is
zero (exact zero).

Think of the simplest control system, the one that is often used
as an example. It has an integrator as its output function. If the
error is zero, it measn that the output doesn’t change. It doesn’t
mean there’s no output.

If the disturbance value doesn’t change, and the reference value
doesn’t change, and the control system has had time to bring the
output to a level that nearly compensates for the disturbance, you
wouldn’t want the control system suddenly to stop its output just
because the error has now reached nearly zero, would you?

You didn’t answer if we can use the negative
feedback loop model if we don’t control a perception. I think we can
and we should, but I may be wrong.

Our
sensory input channels have many millions of degrees of freedom per
second. Our output channels (muscles and chemical) have at most low
hundreds. How could we possibly be controlling all our perceptions at
any one moment? it makes absolutley no sense to suggest that we
might.

Of all the
perceptions we have at any one moment, we cannot control anywhere near
1%. Perhaps .01%, but maybe not eve that. The question you should be
asking is how and under what conditions we transfer control from one
we are controlling to one we had not been controlling but now need to
control. (My own view is that this isone of the two main domains of
conscious perception, but let’s not get into that again).

Martin

[Bj_rn_Simonsen]

[From Bjorn Simonsen (2007.01.14,23:05 EUST)]

[Martin Taylor 2006.01.13.17.54]

Think of the simplest control system, the
one that is

often used as an example. It has an integrator as its

output function. If the error is zero, it means that the

output doesn’t change. It doesn’t mean there’s no output.

OK

If the disturbance value doesn’t change,
and the reference

value doesn’t change, and the control system has had

time to bring the output to a level that nearly compensates

for the disturbance, you wouldn’t want the control system

suddenly to stop its output just because the error has

now reached nearly zero, would you?

Let us say d = 5 and r = 5 and the output function
has an integrator, then o = 5. This lead to p = 0, r = 5, e = 5 and the e is not still zero. What is wrong
here.

Let us say you are correct and I don’t quite
understand you. Then this must be a special example. Normally d is changing and
also r. Therefore your example normally exists for just a short, very short
time. – Where am I wrong.

You didn’t answer if we can
use the negative feedback loop model if we >>don’t control a perception. I
think we can and we should, but I may be wrong.

Our sensory input channels have many millions of
degrees

of freedom per second. Our output channels (muscles and

chemical) have at most low hundreds. How could we possibly

be controlling all our perceptions at any one moment? it

makes absolutley no sense to suggest that we might.

Did you misunderstand me here. Of course we can’t
control all our perceptions at any moment.

My question was if we can use Bill’s Negative Feedback
System for controlling Perceptions if what we perceive isn’t involved in
control. Watching the moon rise e.g. Rick confirmed later that we can.

bjorn

[Martin_Taylor2]

Re: Perseptions we don’t
control
[Martin Taylor 2007.01.14.19.58]

[From Bjorn Simonsen (2007.01.14,23:05
EUST)]
[Martin Taylor 2006.01.13.17.54]

If the disturbance value doesn’t change, and the
reference

value doesn’t change, and the control system has had

time to bring the output to a level that nearly compensates

for the disturbance, you wouldn’t want the control system

suddenly to stop its output just because the error has

now reached nearly zero, would you?

Let us say d = 5 and r = 5 and the output function has
an integrator, then o = 5.

I suppose such a condition could occur, but it would be rather
unlikely, and very transient. Why would the output be 5, unless
previous history had brought about this strnage condition (e.g. the
disturbance had been zero, and in the last microsecond had jumped to
5).

This lead to p = 0, r = 5, e = 5 and the e is
not still zero. What is wrong here.

Figure out what will happen now, with such a large error.
The output will change until (asymptotically) the error approaches
zero. At this point, the output will be zero, won’t it, if the
disturbance is exactly what is needed to bring the perception to
its reference value.

Let us say you are correct and I don’t quite
understand you. Then this must be a special
example.

It’s your example, not mine.

Normally d is changing and also r. Therefore
your example normally exists for just a short, very short time. -
Where am I wrong.

Again, the example is not mine. All I said was the standard
“the output tends toward a value that brings the perception
toward its reference value.” That’s no special case.

You didn’t answer if we can use the negative
feedback loop model if we >>don’t control a perception. I think
we can and we should, but I may be wrong.

Our sensory input channels have many millions of
degrees

of freedom per second. Our output channels (muscles and

chemical) have at most low hundreds. How could we possibly

be controlling all our perceptions at any one moment? it

makes absolutley no sense to suggest that we
might.

Did you misunderstand me here. Of course we can’t
control all our perceptions at any moment.
My question was if we can use Bill’s Negative
Feedback System for controlling Perceptions if what we perceive
isn’t involved in control.

No, I don’t understand you. If there’s no feedback loop, you
aren’t controlling. That’s clear. If what we perceived isn’t involved
in control, it isn’t involved in control, so it’s not part of the
negative feedback loop of any control system. Obviously I don’t
understand your question, which sounds like “If there isn’t any
X, is there any X?”

Watching the moon rise e.g. Rick confirmed later that
we can.

Would you be watching the moon
rise if you didn’t want to?

Martin

[_Erling_Jorgensen]

[From Erling Jorgensen (2007.01.14 22:00 EST)]

Bjorn Simonsen (2007.01.14,23:05 EUST)

Martin Taylor 2006.01.13.17.54

Hi Bjorn,

you wouldn't want the control system
suddenly to stop its output just because the error has
now reached nearly zero, would you?

Let us say d = 5 and r = 5 and the output function has an
integrator, then o = 5. This lead to p = 0, r = 5, e = 5
and the e is not still zero. What is wrong here.

If the reference is 5, & the disturbance is 5 to a not-yet-
acted-upon perception, then the perception is already meeting
its reference, & there would be no _change_ in output. It
certainly wouldn't start climbing/integrating toward o = 5.

It's hard to think of examples where a disturbance constitutes
all that is needed to make a perception match its reference,
but maybe the following situation fits. When one is sledding
down a hill, ordinarily the disturbance provided by gravity
is enough to satisfy the desired perception of motion.

At some of the ski resorts near where I live in New Hampshire,
they have added the capability to go "tubing," (& my daughter
is quite eager to have me take her!) The idea is to rent an
inner tube, ride the lift up the slope, & lie face forward on
the tube as you hurtle down the slope.

The initial push-off at the top (i.e., the starting output
for the motion-sensing control system hooked to the exhilaration-
sensing control system), is not in itself the full degree of
desired downward motion. But from that point on, gravity
provides the necessary disturbance to get the tube up to an
exhilarating speed.

I am told, by other parents who have taken their children,
that the top of the slope can be a little intimidating to some
children, & it is wise for the parent (hooked to a second
inner tube behind the one the child is on) to drag their feet
on the way down, at least on the first few runs. In other
words, the disturbance provided by gravity can easily accelerate
the motion perception beyond its reference, at which point the
foot-dragging output acts to adjust the speed, back from a
frightening level to only an exciting one. But soon the kids
get the hang of it, or get into the thrill of it, & often want
gravity to manage the whole speed again.

I think this example shows a disturbance doing much of the work
of getting the perception to its desired level, during which
time little corrective output is needed; (i.e., low or no error
means little change in output). It also shows the braking output
kicking in as gravity-acceleration gets too high. There's no
good way to accelerate the tube beyond what gravity provides, so
once the run is underway this seems to be a uni-directional
control system that can only modulate speed downward. I suppose
the control systems for even greater speed are to sneak the tube
up to "Suicide Run" or whatever the top of the big-kids' hill may
be called!

I realize this does not address the main thrust of what you are
raising with this thread --

My question was if we can use Bill�s Negative Feedback System
for controlling Perceptions if what we perceive isn�t involved
in control.

When Martin raised the issue of how many more perceptual degrees
of freedom we have compared to output degrees of freedom (several
orders of magnitude higher, by his approximations), I heard him
giving an indirect reply to your query. Because of such a
difference in degrees of freedom, he concludes that the key
question is how we switch from not controlling to controlling,
& back again. As a corollary, the answer would be yes, we should
use to same theoretical approach to analyze such systems, in their
various ways of operating.

In fact, in B:CP, chapter 15, Bill uses the same organizational
model (diagrammed in figure 15.3, p.221), to analyze the modes of
a) controlling, b) passive observation without controlling, c)
automatic controlling without awareness or memory, & d) imagination,
which in different configurations can generate remembering,
imagining, planning, thinking, sleeping/dreaming, closure, &
hallucinating.

That's a pretty impressive list, emerging from the same basic
model. It's certainly enough to convince me of the heuristic
worth of trying to apply this model to a wide variety of phenomema.
It leaves unanswered (as does B:CP) the question of how the
perceptual switch & the memory switch of the model get "flipped."
(And that is essentially the question Martin was revisiting,
with the hint of a trail marker at his preferred solution.)
But it does provide a single unifying schema for approaching
questions of control as well as not controlling.

My own interests as a clinician are often focused on how people
arrive at useful references, for getting things _under_ control.
Sometimes that involves modeling potential behaviors, as in skill
acquisition groups. Sometimes it means looking at the uncontrolled
aspects themselves, as in what is called motivational interviewing.
Sometimes it means empathic listening & circular questioning, akin
to the method of levels. Sometimes it means interrupting the
runaway feedback from meta-levels, as when panic becomes anxiety
over having anxiety. Sometimes it means stepping back & looking at
acceptance strategies -- cp. "...serenity to accept the things I
cannot change..." -- where the error-gap is reduced by making what
one wants match what one is getting.

With many of these questions, the basic structure of a negative
feedback control loop, together with its modular arrangement in
hierarchical form, & a small set of postulates for memory &
reorganization, proves very helpful in keeping one's balance
amidst the complexity, & in generating fruitful perspectives to
explore.

So I think I join you & Martin & Rick & others, in answering yes
to your question, that this is a very robust model for examining
a wide range of human functioning.

All the best,
Erling

[Bill_Powers1]

Let us say d = 5 and r = 5 and
the output function has an integrator, then o = 5. This lead to p = 0, r
= 5, e = 5 and the e is not still zero. What is wrong
here.
[From Bill Powers (2007.01.16.0658 EST)]

Bjorn Simonsen
(2007.01.14,23:05 EUST) –

The calculations are wrong. If d = 5 and o is initially zero, then p = 5.
If p = 5 and r = 5 then e = 0. And if e = 0,. the integral of e remains
zero, and o = 0, which means the system is in equilibrium.

On the other hand, if d = 0
and r = 5, and the output is initially zero, then the error is 5 and the
output begins to change at the rate of 5Ko units per second, positive.
Ko is the factor in the output integrator – let’s say it’s 20. After
0.01 seconds, the output will be 1, the perception will be 1, and the
error will be 4 units. Now the output will be changing at the rate
of 420 or 80 units per second, which in 0.01 second will change
the output by 0.8 units. The output goes to 1.8 units and the error
reduces to 3.2 units.

Keep going, and you will see the error gradually reduce to zero, the
output increase to 5, and the perception also increase to 5, to match the
reference signal. If d is changing, the final state will include changing
values of the other variables, p, e, and o. How much the change will
depend on how fast d changes and how large it is.

Best,

Bill P.

[Bj_rn_Simonsen]

[From Bjorn Simonsen (2007.01.15,14:25 EUST)]

Martin Taylor
2007.01.14.19.58

Your example
with an integrator in the output function confuse me a little after reading “From Bill Powers (2007.01.16.0658
EST)”, but here below I put Bill away.

If the disturbance value
doesn’t change, and the reference

value doesn’t change, and the control system has had

time to bring the output to a level that nearly compensates

for the disturbance, you wouldn’t want the control system

suddenly to stop its output just because the error has

now reached nearly zero, would you?

Let us say d = 5 and r = 5 and
the output function has an

integrator, then o = 5.

I
suppose such a condition could occur, but it would be

rather unlikely, and very transient. Why would the output

be 5, unless previous history had brought about this

strnage condition (e.g. the disturbance had been zero, and

in the last microsecond had jumped to 5).

I agree. A
main point in my understanding PCT is that it is a dynamical system. Problems
come into being each time I stop the system and analyse the static picture. In
my example above I took as my starting point e.g. a situation where I perceived
what I wish to perceive, r=3 and d=3. Then something happened. (continue after
1 and 2)

( digression:
I will in detail describe two situations that I imagine describes our changing
life. I can’t imagine other ways describing how our perceptions change. I
appreciate if any comments the two ways and eventually tell me other ways)

I take a bath. The temperature
is 40 degrees C, d = 3 and I wish the temperature to be 40 degrees C, r = 3.
Everything is OK. Suddenly one person opens the warm water tap. And after 1 minute
the temperature is 55 degrees C. The d
changes from 3 to 5. The r is 3 and I try stop the warm water tap and to leave
the bath. In the course of 1 minute I change my wish and now I wish the
temperature to be 55 degrees C. I sit down again and adapt the new temperature.
When d=5 and r=5 (I have been in the bath for some minutes), then the r = 0.
But because the output function has an integrator o = 5. This value could be
different, but I looked at Bills Live Block Diagram and changed d up 0.2, looked
at p and e and o. Once more I changed d up 0.2 and looked at p, e and o. I
changed d this way up to 5 and looked at p, e and o. Bill’s Live Block has not
an integrator as output function, but I imagined a little. Therefore I put o = 5.

I take a bath. The temperature
is 40 degrees C, d = 3 and I wish the temperature to be 40 degrees C, r = 3.
Everything is OK. Suddenly I wish to raise the temperature to 55 degrees C. Now
r = 5, d = 3 and I open the warm water tap (action). The water gets warmer. r =
5 and d rises from 3 to 5. It takes about one minute. After the one minute d=5,
r=5 and e = 0. The output function has an integrator and I have set o = 5 (look
above).

My main point
with 1 and 2 is that our life is dynamic. Parts of our perceptions change because
things in our environment change or because we wish our environments to change.
I don’t know other initiators.

(stop
digression) (continue here)

After the
happening r = 5 and d = 5. This was my thinking to get the control system’s
output to a level that nearly compensates for the disturbance. Look below at
your statement.

If the disturbance value doesn’t
change, and the reference

value doesn’t change, and the control system has had

time to bring the output to a level that nearly compensates

for the disturbance, you wouldn’t want the control system

suddenly to stop its output just because the error has

now reached nearly zero, would
you?

I didn’t understand that the control system would stop without an
integrator. I thought the o became very near zero and no more actions and
feedback effect without an integrator. After 2 minutes the phone is ringing and
d changes and life continues.

This lead to p = 0, r
= 5, e = 5 and the e is not still zero. What is wrong here.

Figure out
what will happen now, with such a large error.

The output
will change until (asymptotically) the error

approaches
zero. At this point, the output will be zero,

won’t it, if the disturbance is exactly what is needed to

bring the perception to its reference value.

I don’t follow your thinking. It looks like
the integrator is gone away.

Let me start again.

Me:

Let us say d =
5 and r = 5 and the output function has an integrator, then o = 5. This lead to
p = 0, r = 5, e = 5 and the e is not
still zero. What is wrong here.

Let us say you
are correct and I don’t quite understand you. Then this must be a special
example. Normally d is changing and also r. Therefore your example normally
exists for just a short, very short time. – Where am I wrong.

In my two
first sentences I tried to exemplify your statement:

Martin:

If the
disturbance value doesn’t change, and the reference

value doesn’t change, and the control system has had

time to bring the output to a level that nearly compensates

for the disturbance,

Me
exemplifying Martin

Let us say d =
5 and r = 5 and the output function has an integrator, then o = 5.

If the picture
is so, then p = 0, r = 5. In the output function will the integrated input
value compensate the disturbance. Therefore p = 0.

If p = 0 and r
= 5, then e=5. It isn’t me that make it “so great”, it is the integrator.

I must be
wrong, because with an error like 5 the output value will grow and grow. Therefore
I asked “Where am I wrong”.

you
wouldn’t want the control system

suddenly to stop its output just because the error has

now reached nearly zero, would you?

I didn’t think
the control system would stop because the error reaches zero. I think I
sometimes have error like zero myself. Then I perceive what I wish to perceive
and that is wonderful. But then the phone starts ringing.

Again, the
example is not mine.

Yes, but the
integrator is your and it disturbed my thinking.

All I said
was the standard "the output tends toward

a value that brings the perception toward its reference

value." That’s no special case.

I understand
your statement if d = 0. But if there
is a d =5 and an integrator in the output function, then o = 5 and this
compensates the d = 5 and this doesn’t bring the perceptual signal toward its
reference value. And that is a special case.

I am afraid I
am a muddler.

bjorn

You didn’t answer if we can
use the negative feedback loop model if we >>don’t control a perception.
I think we can and we should, but I may be wrong.

Our sensory input channels have
many millions of degrees

of freedom per second. Our output channels (muscles and

chemical) have at most low hundreds. How could we possibly

be controlling all our perceptions at any one moment? it

makes absolutley no sense to suggest that we might.

Did you misunderstand me here. Of
course we can’t control all our perceptions at any moment.

My question was if we can use Bill’s
Negative Feedback System for controlling Perceptions if what we perceive isn’t
involved in control.

No, I don’t
understand you. If there’s no feedback loop, you aren’t controlling. That’s
clear. If what we perceived isn’t involved in control, it isn’t involved in
control, so it’s not part of the negative feedback loop of any control system.
Obviously I don’t understand your question, which sounds like “If there
isn’t any X, is there any X?”

Watching the moon rise e.g. Rick
confirmed later that we can.

Would you be watching the moon rise if you
didn’t want to?

I looked at
Bill’s Live Block. I put d = 0 and r =5. Changing r up 0.2 very quick to r =5,
the e changed. The e didn’t surpass 0.5 and ended at 0.05 and p =4.95.

I see my fault
now. It is a result of making the dynamical system static. e is still zero.

Your first
sentence; Yes. Now I am uncertain about your second sentence. How can the
output be zero when the output function has an integrator.

[Martin_Taylor2]

[Martin Taylor 2007.01.15.11.22]

[From Bjorn Simonsen (2007.01.15,14:25 EUST)]
>Martin Taylor 2007.01.14.19.58

Your example with an integrator in the output function confuse me a little after reading "From Bill Powers (2007.01.16.0658 EST)", but here below I put Bill away.

As far as I can see, Bill said what I said, but expanded on it to illustrate how the change happens smoothly. If you didn't understand what I said, work through what Bill said, and maybe then you will understand me better.

>If the disturbance value doesn't change, and the reference
>value doesn't change, and the control system has had
>time to bring the output to a level that nearly compensates
>for the disturbance, you wouldn't want the control system
>suddenly to stop its output just because the error has
>now reached nearly zero, would you?

>>Let us say d = 5 and r = 5 and the output function has an integrator, then >>o = 5.

>I suppose such a condition could occur, but it would be
>rather unlikely, and very transient. Why would the output
>be 5, unless previous history had brought about this
>strnage condition (e.g. the disturbance had been zero, and
>in the last microsecond had jumped to 5).

I agree. A main point in my understanding PCT is that it is a dynamical system. Problems come into being each time I stop the system and analyse the static picture. In my example above I took as my starting point e.g. a situation where I perceived what I wish to perceive, r=3 and d=3.

You didn't. With the numbers you give, you would peceive what you wish to perceive only if p = 3. For p to be 3 when d = 3, o would have to be zero since p = o + d. You had r = 5 and you set o = 5. Along with d = 5, this would make p = 10, a long way from what you want to perceive.

Then something happened. (continue after 1 and 2)

( digression: I will in detail describe two situations that I imagine describes our changing life. I can't imagine other ways describing how our perceptions change. I appreciate if any comments the two ways and eventually tell me other ways)

1. I take a bath. The temperature is 40 degrees C, d = 3 and I wish the temperature to be 40 degrees C, r = 3. Everything is OK. Suddenly one person opens the warm water tap. And after 1 minute the temperature is 55 degrees C. The d changes from 3 to 5. The r is 3 and I try stop the warm water tap and to leave the bath. In the course of 1 minute I change my wish and now I wish the temperature to be 55 degrees C. I sit down again and adapt the new temperature. When d=5 and r=5 (I have been in the bath for some minutes), then the r = 0.

What temperature would r = 0 correspond to? Assuming your numbers correspond linearly to temperature (3 = 40C and 5 = 55C, so a change of 2 represents 15C) r = 0 means you now want the temperature to be 17.5C. The 55C bath is now MUCH too hot! Maybe you mean o = 0, which would be correct.

But because the output function has an integrator o = 5.

Leaving aside the question of what the form of the output function has to do with your choice of starting situation, I must ask: What does o = 5 correspond to in this scenario? You have mentiond two kinds of action: turning the tap off, which doesn't immediately change your perception of being hot or cold, and getting out of the bath, which does. Neither of these seem to correspond obviously to a number that represents a point on a continuum of possibilities, though I suppose turning the tap to a particular flow rate would. I guess you could have added a possible action to turn the cold tap on, in which case giving the output a numeric value would make sense (it could represent the flow rate of hot tap as positive and of cold tap as negative).

The output function in any _particular_ control system is specific to that control system. The canonical control system used as a type example has an integrator as its output function. An integrator has a smooth, continuous, output. Your choice of action, either getting out of the bath or not getting out, is discrete. Clearly that cannot be the output from a simple integrator. There has to be at least a threshold element or a switch somewhere in the output part of the feedback loop.

2. I take a bath. The temperature is 40 degrees C, d = 3 and I wish the temperature to be 40 degrees C, r = 3. Everything is OK. Suddenly I wish to raise the temperature to 55 degrees C. Now r = 5, d = 3 and I open the warm water tap (action). The water gets warmer. r = 5 and d rises from 3 to 5. It takes about one minute. After the one minute d=5, r=5 and e = 0. The output function has an integrator and I have set o = 5 (look above).

Again, what does o = 5 represent in this scenario? Is it now the flow rate of the hot tap? If it is, I expect you would turn the tap off (making o be zero) when the bath reaches the desired temperature.

My main point with 1 and 2 is that our life is dynamic. Parts of our perceptions change because things in our environment change or because we wish our environments to change. I don't know other initiators.

It's an open question whether we can make our perceptions change without acting to influence our environment. One can, in imagination, and since imagined perceptions can and do combine with data derived from the senses, I suppose even strictly classical HPCT allows that perceptions can change because we wish our environments to change. I think I've talked myself into agreeing with you here.

After the happening r = 5 and d = 5. This was my thinking to get the control system's output to a level that nearly compensates for the disturbance.

So, at this point, the output is near zero, which would exactly compensate for the disturbance and make the perception equal the reference.

Look below at your statement.
>If the disturbance value doesn't change, and the reference

value doesn't change, and the control system has had
time to bring the output to a level that nearly compensates
for the disturbance, you wouldn't want the control system
suddenly to stop its output just because the error has
now reached nearly zero, would you?

I didn't understand that the control system would stop without an integrator. I thought the o became very near zero and no more actions and feedback effect without an integrator.

The output should indeed become near zero if the disturbance happened by chance to bring the environmental situation to be what you wanted to perceive. That's rather a special case, though. It is more probable that some continuing action will be needed if the perception is to stay near its reference value. If r = 3 and d = 2, you would need a persistent output o = 1 to keep p at 3, whatever the output function. That's nothing to do with whether the output function is an integrator.

After 2 minutes the phone is ringing and d changes and life continues.
>> This lead to p = 0, r = 5, e = 5 and the e is not still zero. What is wrong here.

>Figure out what will happen now, with such a large error.
>The output will change until (asymptotically) the error
>approaches zero. At this point, the output will be zero,

won't it, if the disturbance is exactly what is needed to bring the perception to its reference value.

I don't follow your thinking. It looks like the integrator is gone away.

Read Bill's message that you cited. He says this in more detail. The integrator doesn't change. It's always there, and so is the feedback, even when the output value happens for the moment to be zero.

Let me start again.

Me:
Let us say d = 5 and r = 5 and the output function has an integrator, then o = 5. This lead to p = 0, r = 5, e = 5 and the e is not still zero. What is wrong here.

Let us say you are correct and I don't quite understand you. Then this must be a special example. Normally d is changing and also r. Therefore your example normally exists for just a short, very short time. - Where am I wrong.

In my two first sentences I tried to exemplify your statement:
Martin:
>If the disturbance value doesn't change, and the reference

value doesn't change, and the control system has had
time to bring the output to a level that nearly compensates
for the disturbance,

Me exemplifying Martin
Let us say d = 5 and r = 5 and the output function has an integrator, then o = 5.
If the picture is so, then p = 0, r = 5.

No. If d = 5 and o = 5, p = 10. What does "and the output function has an integrator" hade to do with this?

If p = 0 and r = 5, then e=5. It isn't me that make it "so great", it is the integrator.

No. It's you who set the output to be this value. An integrator could conceivably arrive at those values, but only if something drastic had changed in the past history of the system. The numbers you provide are very far from the equilibrium state of the control system. The various values (or at least o and p) will change rapidly away from this state.

I must be wrong, because with an error like 5 the output value will grow and grow. Therefore I asked "Where am I wrong".

Well, for one thing, you aren't adding o and d to get p. For another, you've specified a positive feedback loop instead of a negative feedback loop (or maybe these are both the same thing?).

>All I said was the standard "the output tends toward

a value that brings the perception toward its reference
value." That's no special case.

I understand your statement if d = 0.

It has no connection with the value of d. It's the nature of a control system to produce output that works with the effects of a disturbance so as to bring the perception near its reference value and keep it there.

But if there is a d =5 and an integrator in the output function, then o = 5 and this compensates the d = 5

No. o = 5 would make things even worse. If d = 5, then o has to approach -5 if r = 0. And that has nothing to do with the form of the output function.

How can the output be zero when the output function has an integrator.

What's the problem? If the error has been negative as much as it has been positive, the integral will be zero, won't it?

Don't worry too much about the form of the output function. I introduced the integrator into the discussion because that's often used as a kind of easy idealization that makes it easy to calculate the time course of the waveforms in a control loop. Real output functions may be very different. One could be an on-off switch, for example. Getting out of the bath or not getting out would be the overt result of an output funvtion that contained a switch.

But that doesn't change the fact that good control means that the output works with the disturbance so as to bring the perception near its reference value. Sometimes "works with" means acting in the same direction (e.g. r = 5, d = 3, in which case o = 2 brings p to its reference), sometimes it means acting in opposition (e.g. r = 3, d = 5, in whch case o = -2 brings p to its reference value). And feedback doesn't stop when o = 0!

Martin

[Bj_rn_Simonsen]

[From Bjorn Simonsen (2007.01.17,08:00 EUST)]

Martin Taylor 2007.01.15.11.22

As far as I can see, Bill said what I said, but
expanded on it to

illustrate how the change happens smoothly. If you
didn’t understand

what I said, work through what Bill said, and
maybe then you will

understand me better.

Yes, I will. But Bill doesn’t have an integrator as
its output function. And it is the integrator effect I will learn about.

I had problems with the integrator, but I think I got
it. I just comment the beginning of your mail that made things clear to me.
Thank you.

Let us say d = 5 and r = 5 and the output
function has an

integrator, then o = 5.

I suppose such a condition could occur,
but it would be

rather unlikely, and very transient. Why
would the output

be 5, unless previous history had brought
about this

strange condition (e.g. the disturbance
had been zero, and

in the last microsecond had jumped to 5).

I agree. A main point in my understanding PCT
is that it is a

dynamical system. Problems come into being
each time I stop the

system and analyse the static picture. In my
example above I took as

my starting point e.g. a situation where I
perceived what I wish to

perceive, r=3 and d=3.

You didn’t. With the numbers you give, you would
perceive what you

wish to perceive only if p = 3. For p to be 3 when
d = 3, o would

have to be zero since p = o + d. You had r = 5 and
you set o = 5.

Along with d = 5, this would make p = 10, a long way
from what you

want to perceive.

Your two last sentences. Yes I see that o = 5 is a too
great value. But o will get a value. Can we say o = 1 or o =0.7 (output
function is an integrator) Then p will be 6 or
p = 5.7. Less than above, but p
will be greater than the value I wish to perceive.

I think I got the integrator. o will increase, but
from the moment p = o+ d is greater than r, e will be negative which it can’t
be. It is therefore zero and o will not increase over the value that gives p =
r.

Am I correct?

bjorn

[_Erling_Jorgensen]

[From Erling Jorgensen (2007.01.17 0915 EST)]

Bjorn Simonsen (2007.01.17,08:00 EUST)

Hi Bjorn,

I think I follow where you are running into difficulty.

...from the moment p = o+ d
is greater than r, e will be negative which it can�t be. It is
therefore zero and o will not increase over the value that gives p = r.

To get a bidirectional control system in neural tissue which cannot
register negative numbers, the reference signal can be split into two
signals, one of which activates an inhibitory neuron (is that a glial
cell? I'm not sure), before feeding into the comparator. The same is
done with a perceptual signal, feeding one copy through an inhibitory
neuron.

So then, it becomes two parallel control systems. In one, the comparator
gets an excitatory reference & an inhibitory perceptual signal, so that
subtraction can occur. With the other, the comparator gets an inhibitory
reference & an excitatory perception, again to get the subtraction.
Both systems are working with positive signals, but operating in opposite
directions in terms of correcting error.

Naturally, this is just the scematic wiring of the model, which simplifies
a process likely involving many, many actual synapses.

All the best,
Erling

[Martin_Taylor2]

Re: Perseptions we don’t
control
[Martin Taylor 2007.01.17.10.10]

[From Bjorn
Simonsen (2007.01.17,08:00 EUST)]
Martin Taylor
2007.01.15.11.22

As far as I can
see, Bill said what I said, but expanded on it to
illustrate how
the change happens smoothly. If you didn’t
understand
what I said,
work through what Bill said, and maybe then you
will
understand me
better.
Yes, I will. But
Bill doesn’t have an integrator as its output function. And it is
the integrator effect I will learn about.

He did in the message you referenced [From Bill Powers
(2007.01.16.0658 EST)].

“On the other hand,
if d = 0 and r = 5, and the output is initially zero, then the error
is 5 and the output begins to change at the rate of 5*Ko units per
second, positive. Ko is the factor in the output integrator
-”

I had problems with
the integrator, but I think I got it. I just comment the beginning of
your mail that made things clear to me. Thank you.

Let us
say d = 5 and r = 5 and the output function has an
integrator, then o =

I
suppose such a condition could occur, but it would
be
rather
unlikely, and very transient. Why would the output
be 5,
unless previous history had brought about this
strange
condition (e.g. the disturbance had been zero, and
in the
last microsecond had jumped to 5).

I agree. A
main point in my understanding PCT is that it is a
dynamical
system. Problems come into being each time I stop
the
system and
analyse the static picture. In my example above I took
as
my starting
point e.g. a situation where I perceived what I wish
to
perceive,
r=3 and d=3.

You didn’t.
With the numbers you give, you would perceive what
you
wish to
perceive only if p = 3. For p to be 3 when d = 3, o
would
have to be zero
since p = o + d. You had r = 5 and you set o = 5.
Along with d =
5, this would make p = 10, a long way from what
you
want to
perceive.

Your two last
sentences. Yes I see that o = 5 is a too great value. But o will get a
value. Can we say o = 1 or o =0.7 (output function is an integrator)
Then p will be 6 or p = 5.7. Less than above, but p will be greater
than the value I wish to perceive.

You simply CAN’T choose p, o, and d all independently. p = o + d,
always, whatever the form of the output function. If you know p = 3
and d = 3, you KNOW that o = 0. If o = 5 and d = 5, you KNOW p = 10 at
this moment.

All right, you now want to arbitrarily give o a value of 1 (then
p = 6) or of 0.7, making p = 5.7. Why?

I think I got the
integrator. o will increase,

decrease, actually, under the conditions you specify.

but from the
moment p = o+ d is greater than r, e will be negative which it can’t
be.

Why not? The error is what it is. It is the amount by which the
perceptual value differs from the reference value. You can’t
arbitrarily declare that the external world is always going to provide
perceptions above (or always below) the ever-changing reference
value.

Are you mixing up possible implementation details with the
functional operation of a control system? Can you imagine any kind of
effective control if the only possble operation was to increase the
output? If the implementation mechanism happens to be one way, any
effective control system would have to have two of them operating in
opposition and adding their effects, wouldn’t it? Bill has often
pointed this out when you have seemed to talk of unidirectional
comparators.

Heer are two rather bad ASCII diagrams, one showing how the
comparator acts in the control system, the other showing how it might
be implemented using one-way subtraction elements. You have to view
these pictures using a mono-space font.

ref |ref

                                 >

v----<—>-----v

per— -
O-----error–
perc-- + O----v

     >   
   >________ | _______

     >   
            
   v

     >   

•   |
  

     >   

V______________ -__O--------+

^
V
^ V

error

Functional Same comparator

representation implemented
using one-way

of
comparator
subtraction elements

Does this help? The same thing applies to
one-way output. If the chosen implementation is one-way, use two of
them, one working in each direction (like our arm muscles, for
example). Any real-world control system would,
so our theoretical control systems should, too, if they want to
simulate what might happen with a real world control
system.

In the drawings above, I combined the two
one-way error signals to send a functionally two-way error signal to
the output function. You could do it differently. For example, you
could send each error signal to a one-way output function and combine
the outputs to make a functioning control system (of course, you would
have to have some kind of cross-connection to ensure that the combined
tensions (thinking opposed muscles) didn’t keep building up. But
that’s an implementation detail.

The one thing to note is that the
comparator is a function, not an implementation. If a proposal for an
implementation does not prform a function that experiment or
observation suggests is going on, then you may question either the
implementation proposal or the function. What you may not do is to
attempt to analyze a theoretical structure and impose on it an
implementation that does not perform the functions specified in the
theory.

Martin

[Bill_Powers1]

Yes, I will. But Bill doesn’t
have an integrator as its output function. And it is the integrator
effect I will learn about.
[From Bill Powers (2007.01.17.1710 MST)]

Bjorn Simonsen
(2007.01.17,08:00 EUST) –

I am getting a strong feeling that you and I do not mean the same thing
by the word “integrator.” My model did have an integrating
output function, as Martin Taylor attests.

Martin is also quite correct in saying that you can’t just abitrarily set
the output to some number. The output is produced by the error signal and
the output function. In fact, the only two variables you can adjust while
the system is running are the disturbance and the reference signal – all
the other variables in the loop are dependent on those two
variables.

Best,

Bill P.

[Bj_rn_Simonsen]

[From Bjorn Simonsen (2007.01.18,11:45 EUST)]

[From Bill Powers (2007.01.17.1710 MST)]

I am getting a strong feeling that you and I do not

mean the same thing by the word “integrator.” My

model did have an integrating output function, as

Martin Taylor attests.

Maybe this is the root for my
misunderstanding. I understand am not familiar enough with the function of an
integrator. I look upon an integrator
as a formula or quality that presents a totality.

I look upon an integrator as a formula or
quality that gets a value from another formula and remember it. The next value
from the other formula is added up and so on. If the other formula calculate a new
value 60 times a second to be 0.21, 0.23, 0.38, 0.17, 0, 0, 0, ………, then the
integrator show the value 0.21, 0.44, 0.82, 0.99. 0.99, 0.99, 0.99 ……… .

But it isn’t there my false knowledge has it’s
root.

The root of my misunderstanding has been the Disturbance.
I have thought upon the disturbance as more or less the only environment, the
real, real World. If I saw a folded newspaper on the table, d =5. Then I wished
to perceive what was laying under the newspaper, r = 3. I lifted the newspaper
away (action) and the disturbance changed to represent a pair of glasses on the
table, d = 3.

If I did that on your LiveBlock, first r = 3,
then the o increased to 2.97, Then I changed d to be 3. Now the o was reduced
to 0.00. The p was 3. I perceived what I wished to perceive.

Therefore I said that your LiveBlock had no integrator as output function.

I now see that if d = 3 the action that
lifted the newspaper away is to be find in the effect of the feed back and the
d continues to be 3 ( of course d may change, but that is another case).

Am I nearer now. If you answer this, I may be
able to excuse my error towards Martin and Erling later.

Martin is also quite correct in saying that you can’t just arbitrarily set
the >output to some number. The output is produced by the error signal and
the >output function. In fact, the only two variables you can adjust while
the >system is running are the disturbance and the reference signal – all
the other >variables in the loop are dependent on those two variables.

I will excuse my error towards Martin when I
am sure in my understanding.

As I mentioned above I adjusted both r and d.

bjorn

[Bill_Powers1]

If the other formula
calculate a new value 60 times a second to be 0.21, 0.23, 0.38, 0.17, 0,
0, 0, ………, then the integrator show the value 0.21, 0.44, 0.82, 0.99.
0.99, 0.99, 0.99 ……… .
[From Bill Powers (2007.01.19.1025 MST)]

Bjorn Simonsen
(2007.01.18,11:45 EUST) –

Yes, that is my idea, too. I think we were talking about different
examples. I thought you were referring to my post in which I described an
integrating output function. But in this post now, you refer to the Live
Block Diagram, which contains a “leaky” integrator, not a pure
integrator.

The time constant in the output function corresponds to the slowing
factor (SF) in the output function of other models we use, as
follows:

o := o + (G*e - o)/(SF)*dt

By expanding this expression we can see how it relates to a pure
integrator:

o := o + G*e/(SF)*dt - o/(SF)*dt

The factor dt is the duration of one iteration of the simulation
program. SF is then in units of seconds (time constant =
SF).

In a pure integrator, the last term on the right is missing , so the
integrator gain is just G/(SF). With the second term present, the output
loses a fraction of its value every second, that fraction being 1/SF.
When there is leakage, the output comes to equilibrium (given a constant
value of e) when the last two terms on the right are equal:

G*e/(SF) = o / (SF), or

o = G*e

So the output is simply proportional to the error signal when the error
signal varies slowly in comparison to the time constant. For rapid
changes in e, the output looks like the time integral of e.

Then I wished to perceive what
was laying under the newspaper, r = 3. I lifted the newspaper away
(action) and the disturbance changed to represent a pair of glasses on
the table, d = 3.

If I did that on your LiveBlock, first r = 3, then the o increased to
2.97

Yes, and the input quantity and the perceptual signal both became 2.97,
for an error signal of 0.03. If the output function had been a pure
integrator, the output would have become 3.00, the input quantity and the
perceptual signal would have been 3.00, and the error signal would have
been 0.0.

Then I changed d to be
3. Now the o was reduced to 0.00. The p was 3. I perceived what I wished
to perceive.

Yes, now the disturbance was holding the input quantity and the
perceptual signal at the same value as the reference signal, so the error
was zero, and the output was also zero. If the output function had been a
pure integrator, this same result would have been seen, exactly. Setting
the disturbance to 3 would have increased the perceptual signal so it
became greater than the reference signal. The result would be a negative
error signal, which would cause the output integrator to integrate toward
zero. The negative error would have become smaller, and the output would
have fallen more slowly, until finally the error signal reached exactly
zero and the output also became exactly zero.

As Martin Taylor explained, this same kind of behavior could be produced
by combining two one-way control systems, one acting for p >= r and
the other for p <= r. If the two systems had identical parameters,
except for signs, the result would be the same as for the single
bidirectional system in the Live Block Diagram.

Therefore I said that your
LiveBlock had no integrator as output function.

The difference between a pure integrator and a leaky integrator in the
output function is very small when the steady-state gain of the leaky
integrator is high (it is set by default to 100 in the LBD). It is the
difference between a steady-state mismatch of 1 per cent between p and r,
and a mismatch of 0.

I don’t understand what your idea of the disturbance is. In the LBD, it
is a physical variable (called Disturbance) separate from the controlled
variable that the system perceives. The system does not perceive the
disturbance itself. The reference signal has nothing to do with the
setting of the disturbance. Perhaps this would be clearer if I used a
small circle to indicate the location of the disturbing variable below
the input quantity.

Best.

Bill P.

[Bj_rn_Simonsen]

[From Bjorn Simonsen (2007.01.25,13:45 EUST)]

From Rick Marken (2007.01.13.1130)

From Erling
Jorgensen (2007.01.17 0915 EST)

Martin Taylor 2007.01.17.10.10

From
Bill Powers (2007.01.19.1025 MST)]

I am
sorry I needed so long time for my answering.

I have worked through this thread once more. I have
re-read statements I have expressed and questions I have asked on the basis of
what you have told me is a wrong understanding of how the negative feedback
loop works.

I have lived in a world where (confer Bill’s
LiveBlock) I thought that the effect of the feedback quantity affected the
disturbance to change toward the reference value when I perceived what I wished
to perceive. When I wished to perceive r = 3, the negative feedback effect lead
to a perceptual signal, p = 3. If the disturbance, d =2, I said to myself: “Now
you perceive what you wish to perceive, therefore d = 3”. I changed the d to be
3 and saw that o became zero (the system itself didn’t change d, but I did).
And that was OK because o = 0 lead to no actions, and no actions are necessary
because I perceive what I wish to perceive.

I equalized my p and d and Fred Nichols distinct
statement told me what I did myself and what I at the same time denied that I
did;

From Fred Nickols (2007.01.18.1531
EST)

It

seems to me that I have been equating
“perception” with “sensory input” and

that, I think would be a mistake
(except at lower levels).

That was of course WRONG. All of you have told me that
I was wrong and B:CP expresses the same.

Now I say when I wish to perceive r = 3, the negative
feedback lead to a perceptual signal, p = 3. If the disturbance, d = 2, I NOW
say to myself that negative feedback effect AND the disturbance affect the
Input Quantity that lead the perceptual signal to be 3. (r = 3, o = 1 and d= 2
lead to p = 3).

Martin, Erling, Rick, Fred and Bill and other have given comments for years that
should have lead me to correct understanding, but … .

I am sorry for molesting you for my slowness. Now I am
better, I think (and hope).

Let me refer to your last mails and confirm central
points.

[From Rick Marken
(2007.01.13.1130)]

I’d say the error signal results in output
variations that act on the

controlled perception via the feedback path
through the environment to

counter the effects of disturbances to that
perception.

Yes, if the disturbances push a perception away from
its reference value, the feedback effect will bring the perceptual signal back
to the reference signal (normally).

Thank you, Rick.

Martin Taylor
2007.01.14.19.58

This lead to p = 0, r = 5, e
= 5 and the e is not still zero. What is wrong here.

Figure out
what will happen now, with such a large error.

The output will change until (asymptotically) the error

approaches zero. At this point, the output will be zero,

won’t it, if the disturbance is exactly what is needed to

bring the perception to its reference value.

Of course,
Martin. Thank you.

Let
us say d = 5 and r = 5 and the output function has an

integrator,
then o = 5. This lead to p = 0, r = 5, e = 5

and
the e is not still zero. What is wrong here.

If
the reference is 5, & the disturbance is 5 to a not-yet-

acted-upon
perception, then the perception is already meeting

its
reference, & there would be no change in output. It

certainly
wouldn’t start climbing/integrating toward o = 5.

Of course,
Erling. Thank you. I learned from your “tubing” trips.

Because of such a difference in degrees of
freedom, he

concludes that the key question is how we switch from

not controlling to controlling, & back again.

As a corollary, the answer would be yes, we should

use to same theoretical approach to analyze such
systems, in their

various ways of operating.

OK.

In fact, in B:CP, chapter 15, Bill uses the same
organizational

model (diagrammed in figure 15.3, p.221), to
analyze the modes of

a) controlling, b) passive observation without
controlling, c)

automatic controlling without awareness or memory,
& d) imagination,

which in different configurations can generate
remembering,

imagining, planning, thinking, sleeping/dreaming, closure,
&

hallucinating.

I will always remember figure 15.3 p 221 (p 223 2.
ED.) And the passive mode described very well what happens when perceive and
not control.

My own interests as a clinician are often focused
on how people

arrive at useful references, for getting things
under control.

Sometimes that involves modeling potential
behaviors, as in skill

acquisition groups. Sometimes it means looking at the uncontrolled

aspects themselves, as in what is called
motivational interviewing.

Sometimes it means empathic listening &
circular questioning, akin

to the method of levels. Sometimes it means interrupting the

runaway feedback from meta-levels, as when panic
becomes anxiety

over having anxiety. Sometimes it means stepping back & looking at

acceptance strategies – cp. "…serenity to
accept the things I

cannot change…" – where the error-gap is
reduced by making what

one wants match what one is getting.

Yes I think clinicians have challenging work to do. I
didn’t understand “cp” in your last sentence.

But I appreciated that clinicians
(you) focus on how people arrive at useful references, for getting things
under control. I look upon clinicians as disturbances. Some times people
don’t control any perceptions where variables in the clinician is involved
(passive o.m… Sometimes they try to control a perception, but they experience
conflict. Sometimes they reorganize and sometimes they control perceptions on a
higher level.

I think the negative feedback loop demonstrate very
well how people arrive at useful references, for getting things under
control. If the disturbance is the starter, three things can happen; passive
observation mode, control and conflict. If the reference is the starter, two
things can happen; control and conflict. Am I right?

From Erling
Jorgensen (2007.01.17 0915 EST)

To get a bidirectional control system in neural
tissue which cannot

register negative numbers, the reference signal
can be split into two

signals, one of which activates an inhibitory
neuron (is that a glial

cell? I’m not sure), before feeding into the
comparator. The same is

done with a perceptual signal, feeding one copy
through an inhibitory

neuron.

Yes, normally I have learned that the perceptual
signal enters in the inhibitory sense ( minus sign) and the reference signal
enters in the excitatory sense (positive sign), and if both signals are
reversed, the effect is the same.

If p>r; e<0 and negative numbers cannot be
registered as you say.

If p<r; e>0 and a pos. error may lead to an
output quantity.

If p=r; e=0 and this may lead the output quantity to
maintain its value if the output function has an integrator.

In Neurology, I think they refer to this as
neuromodulation. And I think this is fundamental for developing different
psychiatric medicines. Is that correct?

Martin Taylor 2007.01.15.11.22

You didn’t. With the numbers you give, you would perceive
what you

wish to perceive only if p = 3. For p to be 3 when
d = 3, o would

have to be zero since p = o + d. You had r = 5 and
you set o = 5.

Along with d = 5, this would make p = 10, a long
way from what you

want to perceive.

Yes, I see it.

I take a bath. The temperature is 40 degrees C, d = 3 and I

wish the temperature to be 40 degrees C, r =

3. Everything is OK.

Suddenly one person opens the warm water tap.
And after 1 minute the

temperature is 55 degrees C. The d changes
from 3 to 5. The r is 3

and I try stop the warm water tap and to leave
the bath. In the

course of 1 minute I change my wish and now I
wish the temperature

to be 55 degrees C. I sit down again and adapt
the new temperature.

When d=5 and r=5 (I have been in the bath for
some minutes), then

the r = 0.

What temperature would r = 0 correspond to?
Assuming your numbers

correspond linearly to temperature (3 = 40C and 5
= 55C, so a change

of 2 represents 15C) r = 0 means you now want the
temperature to be

17.5C. The 55C bath is now MUCH too hot! Maybe you
mean o = 0, which

would be correct.

Yes, it was a write fault. It should be e= 0 because
d=5 and r=5.

Leaving aside the question of what the form of the
output function

has to do with your choice of starting situation,
I must ask: What

does o = 5 correspond to in this scenario? You
have mentioned two

kinds of action: turning the tap off, which
doesn’t immediately

change your perception of being hot or cold, and
getting out of the

bath, which does. Neither of these seem to
correspond obviously to a

number that represents a point on a continuum of
possibilities,

though I suppose turning the tap to a particular
flow rate would. I

guess you could have added a possible action to
turn the cold tap on,

in which case giving the output a numeric value
would make sense (it

could represent the flow rate of hot tap as
positive and of cold tap

as
negative).

The way I thought was; I was sitting in the bath,
wishing the temperature to be 40 degrees C when the water became more and more
hot, from 40 degrees toward 55 degrees C. Then I changed my wish of bath temperature
to be 55degrees C. I think this is a possible happening. As long as the
temperature rose, I did nothing. The o increased because of the integrator. When
the temperature was 55 degrees the o had reached the output quantity 5. This
value of o lead my muscles to stop the hot water tap. Now I perceived the
temperature to be 55 degrees C, p = 5 as I wished, r = 5. Now e = zero and o
continues to have the value 5 because of the integrator (5+0=5, ….5+0=5). I see
the problem that o=5 would lead my muscles to stop the water tap when the water
tap already was stopped. You asked me once earlier; “you wouldn’t want the control system
suddenly to stop its output just because the error has now reached nearly zero,
would you?”

If it
didn’t have an integrator as an output function, the smaller (smaller than 5)
output quantities would stimulate the muscles to switch off the warm water tab
until o became zero. And then the muscles would stop switching off the tab.

But that doesn’t change the fact that good control
means that the

output works with the disturbance so as to bring
the perception near

its reference value. Sometimes “works
with” means acting in the same

direction (e.g. r = 5, d = 3, in which case o = 2
brings p to its

reference), sometimes it means acting in
opposition (e.g. r = 3, d =

5, in which case o = -2 brings p to its reference
value). And feedback

doesn’t stop when o = 0!

This is quite clear. But I am not sure if I understand
your last sentence. The way I understand it is that the feedback function is
working with insert value o = 0 (?).

From Bill Powers (2007.01.19.1025 MST)]

I don’t understand what your idea of the
disturbance is. In the LBD, it is a physical >variable (called Disturbance)
separate from the controlled variable that the system >perceives. The system
does not perceive the disturbance itself. The reference signal >has nothing
to do with the setting of the disturbance.

Yes, of course.

bjorn

···

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[Martin_Taylor2]

Re: Perseptions we don’t
control
[Martin Taylor 2007.01.25.12.33]

[From Bjorn Simonsen (2007.01.25,13:45
EUST)]

We seem now to be clear on a lot that was causing a problem
before. But apparently we still are not out of the woods
completely.

But that doesn’t change the fact that good
control means that the
output works with the disturbance so as to bring
the perception near
its reference value. Sometimes “works with”
means acting in the same
direction (e.g. r = 5, d = 3, in which case o = 2
brings p to its
reference), sometimes it means acting in
opposition (e.g. r = 3, d =
5, in which case o = -2 brings p to its reference
value). And feedback
doesn’t stop when o = 0!

This is quite clear. But I am not sure if I understand
your last sentence. The way I understand it is that the feedback
function is working with insert value o = 0 (?).

The
feedback exists because the feeback loop is connected. The value
of the feedback signal at any moment is whatever it might be. It can
be 1, 100, -20, 0, or anything else. The signal is there and has a
value. Zero is a value, like any other.

If the
feedback circuit is disconnected, there is no feedback signal.
Numerically, of course, it has the same effect as having a feedback
signal that is for the moment zero. But not having a signal is
functionally quite different from having a signal, whatever the value
of that signal.

The way I thought was; I was sitting in the bath,
wishing the temperature to be 40 degrees C when the water became more
and more hot, from 40 degrees toward 55 degrees C. Then I changed my
wish of bath temperature to be 55degrees C. I think this is a possible
happening. As long as the temperature rose, I did nothing. The o
increased because of the integrator. When the temperature was 55
degrees the o had reached the output quantity 5. This value of o lead
my muscles to stop the hot water tap. Now I perceived the temperature
to be 55 degrees C, p = 5 as I wished, r = 5. Now e = zero and o
continues to have the value 5 because of the integrator (5+0=5,
Š.5+0=5). I see the problem that o=5 would lead my muscles to stop the
water tap when the water tap already was stopped. You asked me once
earlier; “you wouldn’t want the control system suddenly to
stop its output just because the error has now reached nearly zero,
would you?”
If it didn’t have an integrator as an output
function, the smaller (smaller than 5) output quantities would
stimulate the muscles to switch off the warm water tab until o became
zero. And then the muscles would stop switching off the
tab.

You don’t have to worry about the possibility of an output
integrator in this situation, because the physics of the bath tap
constrain the effects of your output on your perception more than does
any plausible structure of the output function. The bathwater itself
serves as a leaky integrator (no reflection on the solidity of your
bathtub). If you leave it alone, it will slowly cool toward room
temperature. If you put in a stream of hot water, it will slowly raise
its temperature to that of the input stream (assuming you have an
infinitely big bathtub and its cooling rate is infinitely slow).

But you have a tap that emits water at some temperature below
100C and at a finite rate that would probably take minutes to fill a
reasonably sized bathtub. It doesn’t matter what your muscles do, you
can’t heat the bathwater moe quickly than the temperature and stream
rate of the hot water allows, and unless you use the cold tap, you
can’t affect the cooling rate of the water at all significantly (yes,
you can splash around and expose more water surface to the cool air,
but that’s not going to be a signification effect). The net result is
that your perception of temperature is affected by a leaky integrator
fed by an on-off switch (hot tap fully on or fully off). So it doesn’t
matter what kind of output function you have. All it needs to do it
turn the switch. You won’t get quicker heating by putting more force
into your turnng of the tap, and you won’t get quicker cooling by
turning it off more forcefully.

So forget about what your muscles are doing, and consider the
physical limitations of the situation.

When I said: “you wouldn’t want the control system suddenly
to stop its output just because the error has now reached nearly zero,
would you?” I could have been referring either to a situation in
which a continued push against a disturbance was required to keep the
perception near its reference or to the fact that you would want to
disconnect the feedback circuit because for the moment the error was
near zero. I don’t remember which.

Martin

[Bj_rn_Simonsen]

[From Bjorn
Simonsen (2007.01.25,21:15 EUST)]

[Martin Taylor
2007.01.25.12.33]

Martin:

And feedback doesn’t stop when o = 0!

Bjorn:

The way I understand it is that the feedback
function is

working with insert value o = 0 (?).

Martin:

The feedback exists because the feedback loop is
connected.

The value of the feedback signal at any moment is whatever

it might be. It can be 1, 100, -20, 0, or anything else. The

signal is there and has a value. Zero is a value, like any other.

Your first sentence is OK. The two other sentences
tell me that the feedback signal is dependent on the insert value, o. It
happens that the feedback signal is zero (it has no effect on the Input
quantity, and that is also an effect). When o = 0, the feedback effect is also
zero. But it depends on which feedback
function we use (it could be a function with “memory”, even though I can’t
describe any.)

If the feedback circuit is disconnected, there is
no feedback

signal. Numerically, of course, it has the same effect as

having a feedback signal that is for the moment zero. But

not having a signal is functionally quite different from

having a signal, whatever the value of that signal.

Yes, that’s OK. But let us talk about the value zero.

The idea that zero is a value as good as any value.

If we leave out the value zero, we have to disconnect
the actual function.

If we allow for negative numbers, zero is an important
transition from positive to negative values. But in the system (not in the
environment) there are no negative values. There are inhibiting values, but
they are positive.

The feedback is functioning in the environment and I
can imagine negative feedback values.

You don’t
have to worry about the possibility of an

output integrator in this situation, because the physics

of the bath tap constrain the effects of your output on

your perception more than does any plausible structure

of the output function.

I understand
that you say that the integrated value, o = 5 may lead to stop the hot water
tap. But when it is if it cannot be more off. So it doesn’t matter.

The
bathwater itself serves as a leaky integrator (no

reflection on the solidity of your bathtub). If you leave

it alone, it will slowly cool toward room temperature.

If you put in a stream of hot water, it will slowly raise

its temperature to that of the input stream (assuming

you have an infinitely big bathtub and its cooling rate

is infinitely slow).

Yes, that’s
OK.

bjorn

···

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[Martin_Taylor2]

Re: Perseptions we don’t
control
[Martin Taylor 2007.01.25.17.15]

[From Bjorn Simonsen (2007.01.25,21:15
EUST)]
[Martin Taylor 2007.01.25.12.33]

Your first sentence is OK. The two other sentences
tell me that the feedback signal is dependent on the insert value, o.
It happens that the feedback signal is zero (it has no effect on the
Input quantity, and that is also an effect). When o = 0, the feedback
effect is also zero. But it depends on which feedback function we use
(it could be a function with “memory”, even though I can’t
describe any.)

It doesn’t matter how the feedback value that adds to the
disturbance value gets to be what it is. If it comes from a system
with memory (as almost all physical systems do), so be it. At THIS
nanosecond, it has THAT value.

Incidentally, what is an “insert signal”? You have used
the term more than once. What is the source of this signal and where
is it inserted? The two “insert signals” I understand are
the reference signal and the disturbance signal (though Tom Bourbon
did experiment with another, that controlled the output gain). Do you
mean either or both of these?

But in the system (not in the environment) there
are no negative values. There are inhibiting values, but they are
positive.

You are still mixing up implementation details with functional
analysis. Functionally, if the physics of the situation admit negative
values, then so must the function of the control system. Left-right
placement of a table-mat admits negative values of left-ness from a
central position, whereas the brightness of a light does not. It would
be pretty silly if someone controlling for having the table-mat
centrally placed couldn’t move it because it happened to be the wrong
side of centre!

The feedback is functioning in the environment and I
can imagine negative feedback values.

You had better, if your control systems are going to compensate
for disturbances that would by themselves bring controlled perceptions
to values above their reference values!

Internal to a control system, the mechanisms may be anything at
all. They become important if you are trying to see whether you can
trace the influences through the physiology (if the control system is
in an animal). If you are trying to figure out the parameters of
control, mechanisms may be important for setting limits (e.g. I can’t
lift 200 kg, though some people apparently can; it takes a few msec
for optical signals to travel from retina to brain, so you have to
allow for some transport lag when simulating a control system that
includes vision). But if you know that the control system must be able
to compensate for errors of both signs, there is really no point in
making arguments that errors can only be corrected if they are
positive.

Think function, not mechanism, unless the mechanism necessarily
affects the function in some way important to you.

Martin