# Knot on dot \[Was Perceived Present \[was Shared references\]\]

**URL:** <http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095>\
**Category:** CSG2004\
**Created:** [August 5, 2004, 4:56pm UTC](http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095 "2004-08-05T16:56:34Z")\
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**Author:** ![Bill\_Powers1](http://discourse.iapct.org/letter_avatar_proxy/v4/letter/b/8e7dd6/32.png) [@Bill\_Powers1](http://discourse.iapct.org/u/Bill_Powers1)\
**Post date:** [August 5, 2004, 4:56pm UTC](http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095/1 "2004-08-05T16:56:34Z")

</div>

> > > When I control “knot on the dot” and the “knot is  
> > > on the dot” I don’t affect the
> 
> > > environment by any actions. The reference is a1, the perceptual  
> > > signal is a1, the
> 
> > > error is zero, the feedback signal is zero (I think) and the  
> > > disturbance "knot on
> 
> > > the dot" results in a perceptual signal a1. This is  
> > > continued without any affects
> 
> > > till the “knot is not on the dot”
> 
> .
> 
> [Martin]\>Huh? There’s no perceptual signal if the knot is on the  
> dot?
> 
> Did I say that?  
> [From Bill Powers (2004.08.05.0845 MDT)]  
> Bjorn Simonsen (2004.08.05,12:18  
> EuST)–
> 
> > [From Bjorn Simonsen (2004.08.04,09:30 EuST)]  
> > p=  
> > ki\*ke\*ko\*e +kd\*d. I meant to  
> > say that when „knot is on the dot“, e=0. Then p = kd\*d.  
> > I say “huh?” too. What happened to the effect of the output on  
> > qi? The knot is in the environment, and qo has to affect it  
> > continuously to counteract the continuing effect of d. If d is a  
> > steady pull on one rubber band, qo has to be a steady pull in the other  
> > direction on the other rubber band. The force produced by the output is  
> > certainly an effect on the environment, isn’t it?  
> > Let’s analyze the rubber band experiment. To make the analysis easy,  
> > let’s assume that the rubber bands are identical, so the knot will always  
> > be halfway between their movable ends. I’ll also ignore the resting  
> > length of the rubber bands, saying it is zero. Notice that we’re defining  
> > all the environmental variables as positions, so their values can be  
> > added algebraically. This means that we have a much simplified dependence  
> > of the knot position, qi, on the disturbance position d and the output  
> > position qo.

1. qi = (qo + d)/2  
That’s the environment equation.  
The controller equations are  
p = qi  
qo = Ko\*(r - p), which reduces to
2. qo = Ko\*(r - qi)  
We’ll solve for qo by subsituting the right side of equation 1 for  
qi:  
qo = Ko\*[r - (qo+d)/2]  
Expanding:  
qo = Ko_r - Ko_qo/2 - Ko_d/2  
Collecting terms in qo on the left:  
qo + qo_Ko/2 = Ko_r - Ko_d/2  
Extracting the common term qo on the left:  
qo\*(1 + Ko/2) = Ko_r - Ko_d/2  
Dividing both sides by (1 + Ko/2):  
Ko_r - Ko_d/2  
qo = --------------- .  
1 + Ko/2  
For an exact solution we can subsitute specific values for d and Ko and  
see what qo will be. First, however, a value for r is needed. The  
reference condition for qi is “knot over dot.” If we place the  
dot conveniently at the zero position of our measurement scale, The  
desired position of qi is zero, which is where the dot is, so the  
reference signal is equal to zero: that makes the solution even  
simpler.  
Let’s say r = 0, Ko = 100, and d = -20 cm (20 centimeters in the negative  
direction)  
0 cm \* 100 + 20 cm \* 100/2  
qo = --------------------------  
(1 +  
100/2)  
qo = 1000/51 cm  
qo = 19.61 cm  
The position of the knot, qi, will be (by equation 1) the average of qo  
and d, or  
qi = (19.61 - 20)/2 or  
qi = -0.195 cm  
So this proportional control system with an output gain of 100 can keep  
the knot within 2 mm of the center of the dot when the disturbance is 20  
cm, and even closer when the disturbance is smaller.  
Exercise 1: Redo the calculation with an output gain of 2000. What are  
the values of qo and qi?  
Exercise 2: Redo the calculation with the reference condition set to  
“knot 5 cm to the right of the dot” (r = 5.00).  
Exercise 3: How big is the error when Ko becomes infinite?  
Exercise 4: How will this result be affected if the rubber bands are  
identical but their elastic constants are nonlinear (i.e., change with  
the amount of stretch)?  
Exercise 5: Derive the control system equations for the case when the  
rubber bands have different resting lengths and/or different elastic  
constants Kd and Ke. To do this exercise you have to do some physical  
analysis of the environment and the equations will look considerably more  
complex. The results, however, will be the same if you express the output  
as output position minus resting length, and the disturbance as the  
disturbance position minus resting length. Assume that the disturbance  
position is never less than the resting length of the associated rubber  
band.

> **···**
>
> * * *
> 
> It was this view of a proportional control system that led William Ashby  
> to say that a compensating system was in principle a better control  
> system than a negative feedback control system. The compensating system  
> can be adjusted to make the effect of a disturbance on a controlled  
> variable zero, whereas a proportional negative feedback control system  
> could not – not quite.  
> This shows how limited Ashby’s understanding of control systems was. In  
> the first place, as exercise 3 shows, the error can be made as small as  
> desired in a negative feedback control system (if it remains stable)  
> simply by raising the gain. And second of all, the system will remain  
> stable if, for any gain, a slowing factor is inserted that reduces the  
> amount of error correction that takes place per unit time. Such a slowing  
> factor is needed when there is a transport lag in the system.  
> Finally, in the limiting case of infinite gain, the required slowing  
> factor turns the output function into a pure integrator with a suitable  
> gain factor. And when the output function is a pure integrator, the error  
> that remains when the system reaches equilibrium will be _exactly  
> zero_ – just as for the ideal compensating system.
> 
> However, when the error is zero in a system with an integrating output  
> function, the output is not zero, unlike the situation in a proportional  
> control system. It is some steady value, just enough to cancel exactly  
> the effect of a constant disturbance on the input quantity. If a positive  
> error develops, the output will integrate upward, increasing until the  
> error is again exactly zero. A negative error will produce the opposite  
> action, thye output integrating downward until there is again exactly  
> zero error in the steady state.
> 
> Of course when the disturbance changes, it will take a little time for  
> the output to cancel it exactly once again. And if the disturbance  
> changes continuously, the output will never quite catch up. But remember  
> why we need the integrator in the first place: because there is a  
> transport lag in the system. Without the transport lag, or with a very  
> very small transport lag, we could crank Ko up toward infinity and get  
> essentially zero error (there is zero transport lag in the algebraic  
> equations above: all variables in a system of algebraic equations change  
> at literally the same time: there is no time). By adjusting the slowing  
> factor, we can make the error correction as fast as possible for any  
> given amount of transport lag.
> 
> Obviously, a compensating system with a transport lag would also have to  
> allow error while the disturbance was changing, so it cannot act  
> perfectly in those circumstances any more than the negative feedback  
> control system can. If one can postulate a compensating system with no  
> internal tranport lag, then one can postulate a negative feedback control  
> system with no internal tranport lag. However you slice it, the result  
> comes out that the compensating system is no faster than the negative  
> feedback control system, and keeps the error in the controlled quantity  
> no smaller, even in principle.
> 
> There are many aspects of control not yet considered, and when they are  
> considered they make negative feedback control both more accurate and  
> simpler than compensating control.
> 
> This, clearly, is going to be in the new book. So thank you, Bjorn,  
> for your very productive mistake.
> 
> Best,
> 
> Bill P.

---

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**Author:** ![BJORN\_SIMONSEN](http://discourse.iapct.org/letter_avatar_proxy/v4/letter/b/8baadc/32.png) [@BJORN\_SIMONSEN](http://discourse.iapct.org/u/BJORN_SIMONSEN)\
**Post date:** [August 6, 2004, 11:46am UTC](http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095/2 "2004-08-06T11:46:16Z")

</div>

[From  
Bjorn Simonsen (2004.08.06,13:45 EuST)]

> [From  
> Bill Powers (2004.08.05.0845 MDT)]

> > Bjorn Simonsen (2004.08.05,12:18 EuST)–

> > > [From Bjorn Simonsen (2004.08.04,09:30 EuST)]

> > > When I control “knot on the dot” and the “knot is on  
> > > the dot” I don’t affect the

> > > environment by any actions. The reference is a1, the perceptual  
> > > signal is a1, the

> > > error is zero, the feedback signal is zero (I think) and the  
> > > disturbance "knot on

> > > the dot" results in a perceptual signal a1. This is continued  
> > > without any affects

> > > till the “knot is not on the dot”

.

[Martin]\>Huh? There’s no perceptual signal if the knot is on the dot?

> > Did I say that? p= ki\*ke\*ko\*e +kd\*d.  
> > I meant to say that when „knot is on the dot“, e=0. Then p = kd\*d.

> I say “huh?” too. What happened to the effect of the output on  
> qi? The knot is in

> the environment, and qo has to affect it continuously to counteract  
> the continuing

> effect of d. If d is a steady pull on one rubber band, qo has to be a  
> steady pull in the

> other direction on the other rubber band. The force produced by the output  
> is

> certainly an effect on the environment, isn’t it?

What I  
said above is “When I control “knot on the dot” and the “knot is  
on the dot” I don’t affect the environment by any actions.” Paraphrased; “When  
I wish to hold the “knot on the dot” there is a reference like a1. When I keep “the knot is on the  
dot”, the perceptual signal is also like a1. Now, e=zero. I don’t exercise any actions except for the  
tension to keep the “knot is not  
on the dot”.

That  
should give p = kd\*d.

I don’t  
understand why you ask “What happened to the effect of the output on qi?” The effect of the output is qo = Ko\*(r - p).

I  
thought that if the “knot is not on the dot” and I wish to keep the “knot  
is not on the dot”, then r=p and r-p =0. Now if you go back to \>\>  
(above) p must be p = kd\*d.

I have  
read what you have written below and I will end this mail with a question and  
an example. I have also redone your examples as you asked for.

> Let’s analyze the rubber band  
> experiment. To make the analysis easy, let’s assume

> that the rubber bands are identical, so the knot will always be halfway  
> between

> their movable ends. I’ll also ignore the resting length of the rubber  
> bands, saying

> it is zero. Notice that we’re defining all the environmental variables as  
> positions,

> so their values can be added algebraically. This means that we have a much

> simplified dependence of the knot position, qi, on the disturbance position  
> d

> and the output position qo.

OK

> 1. qi = (qo + d)/2

That’s the environment equation.

The controller equations are

p = qi

qo = Ko\*(r - p), which reduces to

1. qo = Ko\*(r - qi)

We’ll solve for qo by subsituting the right side of equation 1 for qi:

qo = Ko\*[r - (qo+d)/2]

Expanding:

qo = Ko_r - Ko_qo/2 - Ko\*d/2

Collecting terms in qo on the left:

qo + qo_Ko/2 = Ko_r - Ko\*d/2

Extracting the common term qo on the left:

qo\*(1 + Ko/2) = Ko_r - Ko_d/2

Dividing both sides by (1 + Ko/2):

```
  Ko*r - Ko*d/2

```

qo = --------------- .

```
   1 + Ko/2

```

OK

.

For an exact solution we can subsitute specific values for d and Ko and see

what qo will be. First, however, a value for r is needed. The reference

condition for qi is “knot over dot.” If we place the dot conveniently  
at the

zero position of our measurement scale, The desired position of qi is zero,

which is where the dot is, so the reference signal is equal to zero: that

makes the solution even simpler.

Let’s say r = 0, Ko = 100, and d = -20 cm (20 centimeters in the negative  
direction)

```
 0 cm * 100 + 20 cm * 100/2

```

qo = --------------------------

```
        (1 + 100/2)

```

qo = 1000/51 cm

qo = 19.61 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,  
or

qi = (19.61 - 20)/2 or

qi = -0.195 cm

So this proportional control system with an output gain of 100 can keep the  
knot within 2 mm of the center of the dot when the disturbance is 20 cm, and  
even closer when the disturbance is smaller.

OK

Exercise 1: Redo the calculation with an output gain of 2000. What are the  
values of qo and qi?

Let’s  
say r = 0, Ko = 2000, and d = -20 cm (20 centimeters in the negative direction)

```
 0 cm * 2000 + 20 cm * 2000/2

```

qo = --------------------------

```
        (1 + 2000/2)

```

qo = 2000/1001 cm

qo = 19.98 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,  
or

qi = (19.98 - 20)/2 or

qi = -0.0099 cm

\*\*\*The higher gain, the greater the output per  
unit error. \*\*\*

Exercise 2: Redo the calculation with the reference condition set to “knot  
5 cm to the right of the dot” (r = 5.00).

Let’s  
say r = 5, Ko = 100, and d = -20 cm (20 centimeters in the negative direction)

```
 5 cm * 100 + 20 cm * 100/2

```

qo = --------------------------

```
        (1 + 100/2)

```

qo = 1500/51 cm

qo = 29,41 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,  
or

qi = (29,41 - 20)/2 or

qi = 4.7 cm

_**If the reference increases (I wish more  
strongly to “keep the knot on the dot”), then qi also increases.**_

Exercise 3: How big is the error when Ko becomes infinite?

Let’s say r = 0, Ko = 1000,000, and d =  
-20 cm (20 centimeters in the negative direction)

```
 5 cm * 1000000 + 20 cm * 1000000/2

```

qo = --------------------------

```
        (1 + 1000000/2)

```

qo = 10,000,000/500,001 cm

qo = 19.99 cm

The position of the knot, qi, will be (by equation 1) the average of qo and d,  
or

qi = (19.99 - 20)/2 or

qi = 0 cm

_**If the gain becomes /(almost) infinite,  
there are no qi.**_

> Exercise 4: How will this result be affected if the rubber bands are  
> identical

> but their elastic constants are nonlinear (i.e., change with the amount of  
> stretch)?

_**Still the perceptual signal will be kept equal to the reference signal,  
provided that the loop gain never drops below some minimum value **_. _** qi will no longer numerical be equal the  
perceptual signal except at maximum (the normalizing point) (B:CP 276-279).**_

> Exercise 5: Derive the control system equations for the case when the  
> rubber

> bands have different resting lengths and/or different elastic constants Kd  
> and Ke.

> To do this exercise you have to do some physical analysis of the  
> environment

> and the equations will look considerably more complex. The results,  
> however,

> will be the same if you express the output as output position minus resting  
> length,

> and the disturbance as the disturbance position minus resting length.  
> Assume

> that the disturbance position is never less than the resting length of the  
> associated

> rubber band.

I will  
do this and come back later. I have noted me your comments.

> **···**
>
> * * *
> 
> > This shows how limited Ashby’s understanding of control systems was. In
> 
> > the first place, as exercise 3 shows, the error can be made as small as
> 
> > desired in a negative feedback control system (if it remains stable) simply
> 
> > by raising the gain. And second of all, the system will remain stable if,  
> > for
> 
> > any gain, a slowing factor is inserted that reduces the amount of error
> 
> > correction that takes place per unit time. Such a slowing factor is needed
> 
> > when there is a transport lag in the system.
> 
> May I  
> ask a question about Gain and the Slowing factor?
> 
> As I  
> wrote to Martin, I define Gain as a value we put in a Simulating model to get  
> precise control. The higher value in Gain, the greater output per unit error.  
> (I think I have this from Rick). High Gain means precise control, but systems  
> with high Gain respond more slowly. The value of Gain and the Slowing factor  
> must be inversely related.
> 
> Different  
> people wish different wishes in a different degree. Let us have an example where  
> Mr. A wishes to educate himself to a certain profession. He has a family with  
> small kids. He looks at his future profession in an economical way, a family  
> way, from his interests and from the viewpoint of “the best for the state”.  
> When he controls his perception of future profession, he has activated a most  
> extensive control system.
> 
> Mr. B  
> wishes to educate to the same profession just because he must have a future  
> job. When he controls his perception of future profession, he has activated a smaller  
> control system.
> 
> And  
> now to my question. Is the (p – r) value for Mr. A influenced of a greater Gain  
> than the same value for Mr. B?
> 
> Can we  
> say that because Mr. A controls a greater control system, it takes longer time before  
> his p = r?
> 
> Here  
> is my example.
> 
> Let’s  
> say r = 0, Ko = 100, and d = 0 cm (0 centimeters in the negative direction)
> 
> ```
> 0 cm * 100 + 0 cm * 100/2
> 
> ```
> 
> qo = --------------------------
> 
> ```
> (1 + 100/2)
> 
> ```
> 
> qo = 0/51 cm
> 
> qo = 0 cm
> 
> The position of the knot, qi, will be (by equation 1) the average of qo and d,  
> or
> 
> qi = (0 - 0)/2 or
> 
> qi = 0 cm
> 
> If you  
> go to the top where I wish to keep “the knot on the dot” and if I in a moment  
> perceive “the knot on the dot”, then p = kd\*d.
> 
> P = 0  
> = r
> 
> Let’s  
> say r = 5, Ko = 100, and d = 0 cm (0 centimeters in the negative direction)
> 
> ```
> 5 cm * 100 + 0 cm * 100/2
> 
> ```
> 
> qo = --------------------------
> 
> ```
> (1 + 100/2)
> 
> ```
> 
> qo = 5/51 cm
> 
> qo = 0.098 cm
> 
> The position of the knot, qi, will be (by equation 1) the average of qo and d,  
> or
> 
> qi = (00098 - 0)/2 or
> 
> qi = 0 cm
> 
> If you  
> go to the top where I wish to keep “the knot on the dot” more strongly and if I  
> in a moment perceive “the knot on the dot”, then p = kd\*d.
> 
> P = 0  
> = r
> 
> bjorn

---

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**Author:** ![Bill\_Powers1](http://discourse.iapct.org/letter_avatar_proxy/v4/letter/b/8e7dd6/32.png) [@Bill\_Powers1](http://discourse.iapct.org/u/Bill_Powers1)\
**Post date:** [August 6, 2004, 5:36pm UTC](http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095/3 "2004-08-06T17:36:47Z")

</div>

> I don’t understand why you ask  
> “What happened to the effect of the output on qi?” The effect of  
> the output is qo = Ko\*(r - p).  
> [From Bill Powers (2004.08.06.0944 MDT)]

Bjorn Simonsen  
(2004.08.06,13:45 EuST)–

What I said  
above is “When I control “knot on the dot” and the “knot  
is on the dot” I don’t affect the environment by any actions.”  
Paraphrased; “When I wish to hold the “knot on the dot”  
there is a reference like a1. When I keep “the knot is on the  
dot”, the perceptual signal is also like a1. Now, e=zero. I  
don’t exercise any actions except for the tension to keep the “knot  
is not on the dot”.

OK, I can see the problem now. You are thinking of action as being only a  
change of output. You think of a steady output, like holding the end of a  
rubber band against a disturbing force, as “no action” if the  
hand is not moving. Do I have that right?

In control theory, qo is a measure of the amount of output from the  
control system, whether it is constant or changing. When I say  
“action” I mean simply the amount of output at any given  
instant, however we are measuring it. I do not mean “change of  
output.”

That is the output, not the effect of the output. The output is the  
position of your hand, the effect of the output is a force applied to the  
knot. Notice that if the error is exactly zero in the equations you are  
using, the output will be zero. If the output is zero there is nothing to  
resist the disturbance because there is no pull opposed to it.

> Exercise  
> 1: Redo the calculation with an output gain of 2000. What are the values  
> of qo and qi?
> 
> Let’s say r =  
> 0, Ko = 2000, and d = -20 cm (20 centimeters in the negative direction)
> 
> ```
> 0 cm * 2000 + 20 cm * 2000/2
> 
> ```
> 
> qo = --------------------------
> 
> ```
> (1 +
> 
> ```
> 
> 2000/2)
> 
> qo = 2000/1001 cm

This is wrong: 20\* 2000/2 = 20,000, not 2000

> qo  
> = 19.98 cm

This is right. A more exact number is 19.98002. You’ll see why I say  
that, shortly.

> The  
> position of the knot, qi, will be (by equation 1) the average of qo and  
> d, or  
> qi = (19.98 - 20)/2 or  
> qi = -0.0099 cm
> 
> _**The  
> higher gain, the greater the output per unit  
> error.**_

(A more exact number is qi = -0.0999001)

That is true. Notice that the error is also smaller when the output gain  
is higher – but it is not zero. In a proportional control system, the  
error is never zero except when the disturbance alone can keep the  
perceptual signal equal to the reference signal. Since p = qi, we have p  
= 0.0990001.The reference signal is r = 0, so the error (r - p) is (0 -  
(-0.0999001)) or

e = 0.0999001. Multiply that error by the gain of 2000 to calculate the  
output quantity, and we get

qo = 2000 \* 0.0999001 = 19,98002.

Remarkable, no? The amount of error that remains is exactly enough to  
produce exactly the amount of output needed to cancel just enough of the  
effect of the disturbance to leave exactly that amount of error. So you  
see that in a proportional control system, the error can’t go to zero,  
with a single exception: when the disturbance has just the right  
magnitude. I leave it as an exercise to figure out what that amount of  
disturbance is.

> Exercise  
> 2: Redo the calculation with the reference condition set to “knot 5  
> cm to the right of the dot” (r = 5.00).
> 
> Let’s say r = 5, Ko = 100, and d = -20 cm (20 centimeters in the negative  
> direction)  
> 5 cm \* 100 + 20 cm \* 100/2  
> qo = --------------------------  
> (1 +  
> 100/2)  
> qo = 1500/51 cm  
> qo = 29,41 cm  
> The position of the knot, qi, will be (by equation 1) the average of qo  
> and d, or  
> qi = (29,41 - 20)/2 or  
> qi = 4.7 cm
> 
> _**If the reference increases (I wish more strongly to “keep the knot  
> on the dot”), then qi also  
> increases.**_

This is not the correct interpretation. If the reference is not zero,  
then the control system will try to keep the knot at a position away from  
the dot. The dot is still at position zero, so when the center of the  
knot is exactly on the center of the dot, the knot’s position is also  
zero. Here we have the center of the knot being at 4.7 cm, which means  
4.7 cm to the right of the dot (with positive measured to the right).  
Note that the reference signal was set to 5.0 cm. The knot is 0.29411 cm  
to the left of the reference position, and 4.70589 cm away from the  
dot.

The reference does NOT determine “how strongly you want to keep the  
knot on the dot.” It determines what state of the perception you  
want to maintain: on the dot (r = 0) or in this case, 5 cm to the right  
of the dot (r = 5). How strongly you react to errors depends on the loop  
gain, not on the reference signal.

> Exercise 3: How big is the error  
> when Ko becomes infinite?  
> Let’s say r = 0, Ko = 1000,000, and d = -20  
> cm (20 centimeters in the negative direction)  
> 5 cm \* 1000000 + 20 cm \* 1000000/2  
> qo = --------------------------  
> (1 +  
> 1000000/2)  
> qo = 10,000,000/500,001 cm  
> qo = 19.99 cm  
> The position of the knot, qi, will be (by equation 1) the average of qo  
> and d, or  
> qi = (19.99 - 20)/2 or  
> qi = 0 cm
> 
> _**If the gain becomes /(almost) infinite, there are no  
> qi.**_

You didn’t answer the question: qi is not the error. The error is the  
difference between qi and the position implied by the reference signal,  
which in this case is zero, the same position as the dot. Since the dot  
is at zero and the reference position is at zero, there is no error and  
the error is zero.

Saying “there is no qi” does not mean the same thing as saying  
qi = 0.

> > Exercise  
> > 4: How will this result be affected if the rubber bands are identical  
> > but their elastic constants are nonlinear (i.e., change with the  
> > amount of stretch)?
> 
> _**Still  
> the perceptual signal will be kept equal to the reference signal,  
> provided that the loop gain never drops below some minimum  
> value **_. _** qi will no longer numerical be equal  
> the perceptual signal except at maximum (the normalizing point) (B:CP  
> 276-279).**_

This is both correct and not correct. You are correct in saying that the  
behavior will be the same as with linear rubber bands. If the two rubber  
bands are identical, the knot will always be exactly midway between the  
free ends even if the rubber bands are nonlinear. So the behavior of the  
whole system will be exactly the same.  
But it is not correct to say that qi will no longer be equal to the  
perceptual signal, because p = qi, always, and that is not changed by  
changing the properties of the rubber bands.  
It is also not correct to say that the perceptual signal will be kept  
_equal_ to the reference signal. In Example 1 above, I showed why  
there must be an error, exzctly the _right_ error. If the error were  
zero, the output position would be zero and the disturbance from the  
other rubber band would cause an error.

DO NOT THINK OF DISTURBANCES, INPUTS, ERRORS, PERCEPTIONS, AND OUTPUTS AS  
EVENTS OR CHANGES. THINK OF THEM AS STEADY QUANTITIES OR  
MAGNITUDES.

You need differential equations to deal with rates of change as  
variables.

> > Exercise 5: Derive the control  
> > system equations for the case when the rubber
> 
> > bands have different resting lengths and/or different elastic  
> > constants Kd and Ke.
> 
> > To do this exercise you have to do some physical analysis of the  
> > environment
> 
> > and the equations will look considerably more complex. The results,  
> > however,
> 
> > will be the same if you express the output as output position minus  
> > resting length,
> 
> > and the disturbance as the disturbance position minus resting length.  
> > Assume
> 
> > that the disturbance position is never less than the resting length  
> > of the associated
> 
> > rubber band.
> 
> I will do this  
> and come back later. I have noted me your comments.
> 
> * * *
> 
> As I wrote to  
> Martin, I define Gain as a value we put in a Simulating model to get  
> precise control. The higher value in Gain, the greater output per unit  
> error. (I think I have this from Rick). High Gain means precise control,  
> but systems with high Gain respond more slowly.

They don’t respond more slowly just because the gain is higher. You have  
to MAKE them respond more slowly by changing the slowing factor, if you  
want them to remain stable. High gain means precise control, but it also  
means that instability is likely. If instability does result from  
increasing the gain, you can correct that problem by making the action  
slower.

But in algebraic analyses of control systems, there are no slowing  
factors because we are not using differential equations. There are no  
time delays.

> The  
> value of Gain and the Slowing factor must be inversely  
> related.

Yes, if you want the highest precision possible. You can always use a  
lower gain and leave the slowing factor unchanged, if it doesn’t matter  
that the system will correct errors more slowly. Usually you want to use  
a slowing factor just sufficient to prevent instability. You don’ usually  
want control to be any slower than necessary.

> Different  
> people wish different wishes in a different degree. Let us have an  
> example where Mr. A wishes to educate himself to a certain profession. He  
> has a family with small kids. He looks at his future profession in an  
> economical way, a family way, from his interests and from the viewpoint  
> of “the best for the state”. When he controls his perception of future  
> profession, he has activated a most extensive control  
> system.

You seem to be picturing this as a single control system. I prefer to say  
that one control system controls only one variable. You are speaking of  
controlling more than one variable, so more than one control system is  
needed. Mr. A is controlling many variables at the same time, so there  
are many control systems in operation at the same time, each one still  
controlling only one variable.

> Mr.  
> B wishes to educate to the same profession just because he must have a  
> future job. When he controls his perception of future profession, he has  
> activated a smaller control system.

Mr.B is controlling fewer variables – only one, in fact. There is only  
one control system (that you are considering).

> And  
> now to my question. Is the (p – r) value for Mr. A influenced of a  
> greater Gain than the same value for Mr. B?
> 
> Can we say that because Mr. A controls a greater control system, it takes  
> longer time before his p = r?

No. All these variables in Mr. A are at about the same level or  
organization, I would guess. Each control system has its own gain, which  
determines how much the output of that system will change for a unit  
change in error signal in that system. Mr. A is controlling several  
different perceptions, p1, p2, p3 … relative to several different  
reference signals, r1, r2, r3 … Each perception is controlled by a  
different control system, independent of the others at this level. The  
speed of control in each system is determined by the gain and slowing  
factor in each system, which is independent of the others. Since all  
these systems operate in parallel, there is nothing about having more  
than one system in action that would slow them down, unless they conflict  
with each other.

When you talk about different degrees of wishing, I think you’re  
confusing wishing for different amounts of something and acting with  
different amounts of effort to correct a unit of error. If you want to  
earn $100,000 per year, you are wishing for more income than if you want  
to earn $10,000 per year. You aren’t wishing any harder, you’re just  
wishing for a larger amount.But if the amount you actually earn is $1000  
less than that amount you wish to earn, how hard are you willing to work  
to bring the amount closer to the amount you want? The man who  
earns $99,000 per year will probably not increase his efforts as much as  
the man who earns $9000 per year, so in fact the man with the lower  
wished-for income will try harder than the other man to correct the same  
error (can you think of a way to contradict this?). This is a matter of  
the loop gain, not the setting of the reference signal. The reference  
signal specifies only how much of the perception you will try to  
maintain. It doesn’t specify how hard you will try to maintain it when  
the perception falls a little short.

I think what people mean by wishing harder is imagining a larger effort  
to attain the same wish. You know, you clench your fists and squint your  
eyes and squeeze your breath and tense your muscles when you wish harder,  
but that has nothing to do with the wishing part – it has to do with  
making more effort.

> Here  
> is my example.
> 
> Let’s say r = 0, Ko = 100, and d = 0 cm (0 centimeters in the negative  
> direction)
> 
> ```
> 0 cm * 100 + 0 cm * 100/2
> 
> ```
> 
> qo = --------------------------
> 
> ```
> (1 +
> 
> ```
> 
> 100/2)
> 
> qo = 0/51 cm
> 
> qo = 0 cm
> 
> The position of the knot, qi, will be (by equation 1) the average of qo  
> and d, or
> 
> qi = (0 - 0)/2 or
> 
> qi = 0 cm
> 
> If you go to the top where I wish to keep “the knot on the dot” and if I  
> in a moment perceive “the knot on the dot”, then p =  
> kd\*d.
> 
> P = 0 = r

Yes,in this case the reference position is zero, there is no disturbance,  
so no output effort is needed to keep p = r. I don’t know yet what this  
is an example of, but I’m trying to follow you.

> Let’s  
> say r = 5, Ko = 100, and d = 0 cm (0 centimeters in the negative  
> direction)
> 
> ```
> 5 cm * 100 + 0 cm * 100/2
> 
> ```
> 
> qo = --------------------------
> 
> ```
> (1 +
> 
> ```
> 
> 100/2)
> 
> qo = 5/51 cm

Mistake here. Should be

qo = 5\*100/51 = 500/51 = 9.8 cm.

This means that the next result is wrong, too

> qo  
> = 0.098 cm
> 
> The position of the knot, qi, will be (by equation 1) the average of qo  
> and d, or
> 
> qi = (00098 - 0)/2 or
> 
> qi = 0 cm

This is wrong even without the above mistake: (0.098 - 0)/ 2 is 0.049,  
not 0. But since the 0.098 should be 9.8, the answer should be

qi = (9.8 - 0)/2 = 4.6.

Since p = qi, p = 4.6. The reference value was 5.0, so the error (r-p) is  
-0.4 cm. In the second case the error has increased when the reference  
position is moved away from the dot. By “reference position” I  
mean the position where the knot would be IF the error were zero. But  
notice that the knot is now 4.6 cm away from the dot.

Keep trying!

Best,

Bill P.

---

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**Author:** ![BJORN\_SIMONSEN](http://discourse.iapct.org/letter_avatar_proxy/v4/letter/b/8baadc/32.png) [@BJORN\_SIMONSEN](http://discourse.iapct.org/u/BJORN_SIMONSEN)\
**Post date:** [August 8, 2004, 7:10am UTC](http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095/4 "2004-08-08T07:10:58Z")

</div>

[From  
Bjorn Simonsen (2004.08.08,08:36 EuST)]

> [From  
> Bill Powers (2004.08.06.0944 MDT)]

> > Bjorn Simonsen (2004.08.06,13:45 EuST)–

> > What  
> > I said above is “When I control “knot on the dot” and the “knot  
> > is on the dot” I don’t

> > affect the environment by any actions.” Paraphrased; “When I wish to  
> > hold the "knot on the

> > dot" there is a reference like a1. When I keep “the  
> > knot is on the dot”, the perceptual signal

> > is also like a1. Now, e=zero. I  
> > don’t exercise any actions except for the tension to keep the

> > “knot is not on the dot”.

> OK, I can see the problem now. You are thinking of action as being only a  
> change of output.

> You think of a steady output, like holding the end of a rubber band against  
> a disturbing force,

> as “no action” if the hand is not moving. Do I have that right?

Yes and No. We must look at my  
Statement in correct relationship. Rohan Luland raised “Rohan Luland  
about 2004.08.03, 15:42 EuSt” the question:

> it  
> also raises the question for me, which may be very naive, " do we have to  
> be able to affect

> the environment or the perception in the environment for it to be in a  
> control loop" are there

> some where with either choose to control it or if can not we choose quickly  
> to control

> something else when error is experienced (binary sort of system). it may  
> not look like a

> control loop but still is

I marked the two first lines in  
his section, because I am taken up with the idea _**“that we  
control all control system we can control and we perceive all control system  
we “just” can perceive/(notice) \_ ALL THE TIME. But we are not conscious/aware  
of all these perceptions.”**_ And I answered “From Bjorn Simonsen (2004.08.04,09:30  
EuST)]”

When  
I control “knot on the dot” and the “knot is on the dot” I don’t affect the  
environment

by any actions. The reference is a1,  
the perceptual signal is a1, the error is zero, the feedback

signal is zero (I think) and the disturbance “knot on the dot” results in a  
perceptual signal a1.

This is continued without any affects till the “knot is not on the dot”

This was an example when we  
" do we do not have to be able to affect the environment or the perception  
in the environment for it to be in a control loop". Just that moment. Here  
I don’t think about action as being only a change of output. But I think of a steady output, like holding  
the end of a rubber band against a disturbing force, as “no action”  
if the knot is not moving away from the.

> In control theory, qo is a measure of the amount of output from the control  
> system,

> whether it is constant or changing. When I say “action” I mean  
> simply the amount

> of output at any given instant, however we are measuring it. I do not mean  
> “change of output.”

May I meet this with a comment? An action or behavior is in  
response to the presence of an error. The emitted action is transmitted  
purposefully, with the intention of changing the state of the world so that my  
perception can be made a desired state of goal, which reduces the error signal  
to (about) zero. Before my perception can be made a desired state of goal qo  
has a value different from zero. At the same time an action is exercised. This  
action changes the state of the world and my perception. When my perception  
changes, the error and qo changes. When my perception is made a desired state  
of goal, qo _is_ (about) zero. At  
this time _no_ action is  
exercised. When there is no action, the  
state of the world doesn’t change, and my perception doesn’t change. Neither does  
the error or qo. At this moment when my perceptions have made the desired state  
of goal (qo = zero and no actions), I just notice my perceptions, I don’t  
control them.

I think there is coherence  
between action and “change of output”.

> > I  
> > don’t understand why you ask “What happened to the effect of the output on  
> > qi?”

> > The effect of the output is qo = Ko\*(r - p).

> That is the output, not the effect of the output. The output is the position  
> of your

> hand, the effect of the output is a force applied to the knot. Notice that  
> if the error

> is exactly zero in the equations you are using, the output will be zero. If  
> the output

> is zero there is nothing to resist the disturbance because there is no pull  
> opposed to it.

Yes you are correct, but I  
meant to answer your question a ”What happened to the effect of the output on qi?”  
with the answer: “The effect of the output is zero because qo = Ko\*(r - p) AND  
(r-p) = zero.

NOW READ THE NEXT SECTION TWICE.

In an ordinary rubber band game  
it happens that the knot is on the dot, but using many decimals, qo is never  
ZERO.

If you and I play a SPECIAL  
rubber band game where we agree upon the dot, put the knot over the dot and  
keep it so. Then we say “if you stretch the rubber band I shall try to keep  
“knot over the dot”. The time _before_  
you stretch the rubber band the knot is over the dot, I stretch with just the  
same power as you in opposite direction (linear output function). NOW p=r, e=0,  
qo =0, p=0. Controlling means\*\*\*: A is said to control B IF for every disturbing influence acting  
on B, A generates an action that tends strongly to counteract the effect of the  
disturbing influence on B.\*\*\*

NOW we have an example where a  
perception is just noticed and not controlled. Have I learned? If I have learned,  
my comment to Rohan Luland was wrong. Because in my example, no control took  
place.

> > Exercise  
> > 2: Redo the calculation with the reference condition set to “knot 5 cm to  
> > the right of the dot” (r = 5.00).

Let’s say r = 5, Ko = 100, and d = -20 cm (20 centimeters in the negative  
direction)  
5 cm \* 100 + 20 cm \* 100/2  
qo = --------------------------  
(1 + 100/2)  
qo = 1500/51 cm  
qo = 29,41 cm  
The position of the knot, qi, will be (by equation 1) the average of qo and d,  
or  
qi = (29,41 - 20)/2 or  
qi = 4.7 cm

> > _**If the reference  
> > increases (I wish more strongly to “keep the knot on the dot”), then qi  
> > also increases.**_

> This is not the correct interpretation. If the reference is not zero, then  
> the control system

> will try to keep the knot at a position away from the dot. The dot is still  
> at position zero,

> so when the center of the knot is exactly on the center of the dot, the  
> knot’s position is

> also zero. Here we have the center of the knot being at 4.7 cm, which means  
> 4.7 cm to the

> right of the dot (with positive measured to the right). Note that the  
> reference signal was

> set to 5.0 cm. The knot is 0.29411 cm to the left of the reference  
> position, and 4.70589 cm

> away from the dot.

> The reference does NOT determine “how strongly you want to keep the  
> knot on the dot.”

> It determines what state of the  
> perception you want to maintain: on the dot (r = 0) or in this

> case, 5 cm to the right of the dot (r = 5). How strongly you react to  
> errors depends on the

> loop gain, not on the reference signal.

Here I am learning aspects of  
PCT. The way I understand your comments is that you think upon a linear input  
function where the knot is moving along a “horizontal” line through the dot.  
The game doesn’t tolerate movements orthogonal this line ((r=5) is 5 cm to the  
right of the dot). When orthogonal movements are actual, we talk about a  
nonlinear input function.

> > Exercise  
> > 3: How big is the error when Ko becomes infinite?  
> > Let’s  
> > say r = 0, Ko = 1000,000, and d = -20 cm (20 centimeters in the negative  
> > direction)  
> > 0 cm \* 1000000 + 20 cm \* 1000000/2  
> > qo = --------------------------  
> > (1 +  
> > 1000000/2)  
> > qo = 10,000,000/500,001 cm  
> > qo = 19.99 cm  
> > The position of the knot, qi, will be (by equation 1) the average of qo and d,  
> > or  
> > qi = (19.99 - 20)/2 or  
> > qi = 0 cm

> > _**If the gain  
> > becomes /(almost) infinite, there are no qi.**_

> You didn’t answer the question: qi is not the error. The error is the  
> difference between

> qi and the position implied by the reference signal, which in this case is  
> zero, the same

> position as the dot. Since the dot is at zero and the reference position is  
> at zero, there

> is no error and the error is zero.

> Saying “there is no qi” does not mean the same thing as saying qi  
> = 0.

Let me answer the question  
after your comment. See above.

```
            e= qi – r

            qo = 

```

10,000,000/500,001 = 19.99996

```
            qi = (19.99996 –20) = - 0.00004

            e = -0.00004 – 0 = -0.00004

           ***If the gain becomes

```

/(almost) infinite, the error becomes still smaller than when the gain is  
great.\*\*\*

> ………………. The reference signal specifies only how much of the perception you

> will try to maintain. It doesn’t specify how hard you will try to maintain  
> it when the

> perception falls a little short.

Thank you.

You have got an insight in my  
way of thinking when you wrote:

> DO  
> NOT THINK OF DISTURBANCES, INPUTS, ERRORS, PERCEPTIONS, AND

> OUTPUTS AS EVENTS OR CHANGES. THINK OF THEM AS STEADY

> QUANTITIES OR MAGNITUDES

I will.

bjorn

---

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**Post date:** [August 8, 2004, 1:38pm UTC](http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095/5 "2004-08-08T13:38:49Z")

</div>

Re: Knot on dot [Was Perceived Present [was Shared  
referen  
[Martin Taylor 2004.08.08.0927]

I’m away for about 3 weeks starting in a few hours, so this has  
to be quick.

> [From Bjorn Simonsen (2004.08.08,08:36  
> EuST)]
> 
> > [From Bill Powers (2004.08.06.0944  
> > MDT)]
> 
> > > Bjorn Simonsen (2004.08.06,13:45  
> > > EuST)–
> 
> > OK, I can see  
> > the problem now. You are thinking of action as being only a change of  
> > output.
> 
> > You think of a steady output, like holding the end of a rubber  
> > band against a disturbing force,
> 
> > as “no action” if the hand is not moving. Do I have that  
> > right?
> 
> Yes and No. We must look at my Statement in correct  
> relationship. Rohan Luland raised “Rohan Luland about 2004.08.03,  
> 15:42 EuSt” the question:
> 
> > it also raises the question for me, which may be  
> > very naive, " do we have to be able to affect
> 
> > the environment or the perception in the environment for it to be  
> > in a control loop" are there
> 
> > some where with either choose to control it or if can not we  
> > choose quickly to control
> 
> > something else when error is experienced (binary sort of system).  
> > it may not look like a
> 
> > control loop but still is
> 
> I marked the two first lines in his  
> section,

As you should, becasue what it asks is “Is there any way of  
affecting our perception other than by having an influence on the  
environment?” It involves the notion of controlling in  
imagination, and whether our imagination affects what we perceive from  
the environment.

But nothing of that shows up in the rest of your message.

> I don’t think about action as being only a  
> change of output. But I think of a steady output, like holding the end  
> of a rubber band against a disturbing force, as “no action”  
> if the knot is not moving away from the.

You should be VERY careful not to treat a variable that has a  
value of zero as the same as the absence of the variable. “x=0”  
is a very different statement from “there is no such thng as  
‘x’”.

> > In control theory, qo is a measure of the amount of output from  
> > the control system,
> 
> > whether it is constant or changing. When I say “action”  
> > I mean simply the amount
> 
> > of output at any given instant, however we are measuring it. I do  
> > not mean “change of output.”  
> > May I meet this with a comment? An action or behavior  
> > is in response to the presence of an error.

No. Same mistake. It is a function of the magnitude of the error  
and of the history of the error. In a control loop “the presence  
of an error” is guaranteed by the “wiring” of the  
control loop. A comparator exists, and has values at its inputs.  
Therefore it has a value at its output, which is labelled “the  
error”.

-----------------a different  
problem------------------------------

> NOW READ THE NEXT SECTION TWICE.  
> In an ordinary rubber band game it happens that the  
> knot is on the dot, but using many decimals, qo is never  
> ZERO.  
> If you and I play a SPECIAL rubber band game where we  
> agree upon the dot, put the knot over the dot and keep it so. Then we  
> say "if you stretch the rubber band I shall try to keep “knot over  
> the dot”. The time _before_ you stretch the rubber band the  
> knot is over the dot, I stretch with just the same power as you in  
> opposite direction (linear output function). NOW p=r, e=0, qo =0, p=0.  
> Controlling means\*\*\*: A is said to control B IF for every  
> disturbing influence acting on B, A generates an action that tends  
> strongly to counteract the effect of the disturbing influence on  
> B.\*\*\*  
> NOW we have an example where a perception is just  
> noticed and not controlled. Have I learned? If I have learned, my  
> comment to Rohan Luland was wrong. Because in my example, no control  
> took place.

You are still controlling a perception, but it is a perception  
that your action matches what the other is doing, not of the position  
of the knot. The position of the knot now is a side-effect of your  
control for matching the other’s actions.

----------------problem 3--------------------

> Here I am learning aspects of PCT. The way I  
> understand your comments is that you think upon a linear input  
> function where the knot is moving along a “horizontal” line  
> through the dot. The game doesn’t tolerate movements orthogonal this  
> line ((r=5) is 5 cm to the right of the dot). When orthogonal  
> movements are actual, we talk about a nonlinear input  
> function.

No we don’t. It’s  
still linear, but two perceptions are being controlled independently  
by two different control systems, one for position aling the line of  
the rubber band, one for position orthogonal to that line.  
“Non-linear” means something quite different, and  
potentially could apply separately to each of those control systems.  
“Non-linear” shows up in the equations for the control  
system itself. For example, instead of q0 = ke, you might have q0 = k

- log(e) or q0 = k \* e^2. In human perception, you often have  
something along the lines of p = log(s) or p = s^p.

Martin

---

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**Post date:** [August 8, 2004, 3:06pm UTC](http://discourse.iapct.org/t/knot-on-dot-was-perceived-present-was-shared-references/6095/6 "2004-08-08T15:06:42Z")

</div>

> Yes and No. We must look at my  
> Statement in correct relationship. Rohan Luland raised “Rohan  
> Luland about 2004.08.03, 15:42 EuSt” the question:  
> When I control “knot on the dot”  
> and the “knot is on the dot” I don’t affect the environment
> 
> by any actions. The reference is a1, the perceptual signal is a1,  
> the error is zero, the feedback
> 
> signal is zero (I think) and the disturbance “knot on the dot” results in  
> a perceptual signal a1.
> 
> This is continued without any affects till the “knot is not on the  
> dot”  
> [From Bill Powers (2004.08.07.0642 MDT)]

Bjorn Simonsen  
(2004.08.08,08:36 EuST)–

> This  
> was an example when we " do we do not have to be able to affect the  
> environment or the perception in the environment for it to be in a  
> control loop". Just that moment. Here I don’t think about action as  
> being only a change of output. But I think of a steady output, like  
> holding the end of a rubber band against a disturbing force, as “no  
> action” if the knot is not moving away from  
> the.

I don’t know of any case where we do not have to be able to affect the  
environment or the perception for it to be a control loop. When you say  
“be able to affect the environment”, I understand that this  
means having an output device (an arm and a hand) connected to the  
environment (the finger of the hand hooked through the rubber band) so  
that the output device can affect the controlled quantity, the position  
of the knot. If I removed the rubber band from your finger, you would no  
longer be able to control the position of the knot, although you could  
still see it. I don’t see how you could continue to control the position  
of the knot if you could no longer affect its position. Please  
explain.

> When  
> my perception changes, the error and qo changes. When my perception is  
> made a desired state of goal, qo _is_ (about) zero. At this time  
> _no_ action is exercised. When there is no action, the state  
> of the world doesn’t change, and my perception doesn’t change. Neither  
> does the error or qo. At this moment when my perceptions have made the  
> desired state of goal (qo = zero and no actions), I just notice my  
> perceptions, I don’t control them. think there is coherence between  
> action and “change of output”.

You are defining control in too limited a way, so that one particular  
state of an active control system is defined as “not  
controlling.” But what I am trying to get across is that that in  
most cases, that state of small error cannot be maintained against a  
steady disturbance unless the control system is producing a continuous  
and NON-ZERO output. You still seem to be assuming that if there is no  
_change_ in the output, there is no output. That is not  
true.

> > > I  
> > > don’t understand why you ask “What happened to the effect of the output  
> > > on qi?”
> 
> > > The effect of the output is qo = Ko\*(r - p).
> 
> > That is the output, not the effect of the output. The output is the  
> > position of your
> 
> > hand, the effect of the output is a force applied to the knot. Notice  
> > that if the error
> 
> > is exactly zero in the equations you are using, the output will be  
> > zero. If the output
> 
> > is zero there is nothing to resist the disturbance because there is  
> > no pull opposed to it.

> Yes  
> you are correct, but I meant to answer your question  
> a ”What happened to the effect of the output  
> on qi?” with the answer: “The effect of the output is zero because  
> qo = Ko\*(r - p) AND (r-p) = zero.

No, (r-p) is NOT zero. It is a small amount, which is then amplified by  
the gain in the output function (Ko) to put the hand in the right  
position to counteract the effect of the disturbance on the knot. If the  
error became larger, the hand would move farther from the zero position.  
That would move the knot off the dot even further, causing the error to  
increase, so the hand would move back toward the dot again. Except that  
this sequential description is wrong: the hand simply moves to the  
position where the knot is just far enough off the dot (not very far) for  
the resulting error to keep the hand where it is. Control continues  
whether the hand and knot are moving or stationary.

> NOW  
> READ THE NEXT SECTION TWICE.
> 
> In an ordinary rubber band game it happens that the knot is on the dot,  
> but using many decimals, qo is never ZERO.  
> NOW we have an example  
> where a perception is just noticed and not controlled. Have I learned? If  
> I have learned, my comment to Rohan Luland was wrong. Because in my  
> example, no control took place.

I read that twice and it is still misleading. You don’t need “many  
decimals” to show that qo is not zero when the knot is on the dot.  
You made some arithmetic mistakes that made this seem to be true, but in  
your example, the output was actually something like 9 units, not zero,  
even omitting the decimal places.  
Let’s diagram the rubber band experiment (this will give some hints about  
how to analyze the more general case in one of the Exercises). Let’s  
start with the rubber bands at their resting lengths, the knot over the  
dot, and both the experimenter’s hand (He) and the controller’s hand (Hc)  
holding the ends of the rubber bands. Under these conditions we can that  
He = 0, Hc = 0, and qi = 0, where qi is the position of the knot. The dot  
always remains at zero.

5…4…3…2…1…0…1…2…3…4…5

He=============8==============Hc  
-5…-4…-3…-2…-1…-0  
knot  
0…1…2…3…4…5

x

dot  
The scale on the top applies to the knot (the figure 8) and dot (the x)  
only  
Now consider what happens when the disturbance, Hd, changes from 0 to -3  
units. First, let’s see what would happen to the knot if the gain of the  
control system was zero.

-5…-4…-3…-2…-1…-0…1…2…3…4…5

```
 (d) Hd===================8===================Hc

```

(qo)  
-5…-4…-3…-2…-1…-0  
knot(qi)  
0…1…2…3…4…5

x

dot  
We have a disturbance d of -3 units and an output quantity Hc of 0 units,  
so the position of the knot is (He + Hc)/2 = (-3 + 0)/2 = -1.5 units. The  
8, which is the knot, is at the position -1.5 on its scale. The reference  
signal is set to zero and p = qi (qi is the knot position), so the error  
signal will have a value of (r - p) = [0 - (-1.5)] = +1.5  
units.  
If the gain Ko is greater than zero, the error signal will produce  
output, and Hc will be farther to the right. Suppose the gain is 1. In  
that case the output will be  
qo = Ko_r - (Ko_d/2)/(1 + Ko/2) or  
qo = 0 - (1\*(-3)/2)/(1+1/2) or  
qo = 1  
The diagram will look like this:

-5…-4…-3…-2…-1…-0…1…2…3…4…5

Hd=====================8======================Hc  
-5…-4…-3…-2…-1…-0  
knot  
0…1…2…3…4…5

x

dot  
The theoretical knot position is -1, which is what we see.  
Now if we let the gain be 100, the error will become so small that it  
looks like zero in this diagram. The output quantity will be further  
still to the right, and we will see this:

-5…-4…-3…-2…-1…-0…1…2…3…4…5

Hd=========================8==========================Hc  
-5…-4…-3…-2…-1…-0  
knot  
0…1…2…3…4…5

x

dot  
Now the knot is over the dot as nearly as we can see on this scale. The  
actual value of qo will be 2.941176471 units (to the right) and the knot  
position will be -0.02941164 units (to the left). The error signal will  
be positive by the same amount since p = qi and e = r - p. The reference  
signal r is zero, Ko is 100, so the output quantity Hc (otherwise known  
as qo) will be 2.941164 units as nearly as my calculator can show. That’s  
the same number we started with so we know the calculation is right. The  
rounding error in this result is 0.00000071, which is just because my  
calculator displays only 12 characters and rounds the result.

If you and I  
play a SPECIAL rubber band game where we agree upon the dot, put the knot  
over the dot and keep it so. Then we say “if you stretch the rubber band  
I shall try to keep “knot over the dot”. The time _before_ you  
stretch the rubber band the knot is over the dot, I stretch with just the  
same power as you in opposite direction (linear output function). NOW  
p=r, e=0, qo =0, p=0. Controlling means\*\*\*: A is said to control B IF  
for every disturbing influence acting on B, A generates an action that  
tends strongly to counteract the effect of the disturbing influence on  
B.\*\*\*

Sorry, no. You are controlling even if the output is constant, and even  
if it is zero. You would not be controlling if the loop were broken,  
which is a different matter. Suppose that with the rubber bands just  
relaxed at zero, I quietly cut your end of the rubber bands while you’re  
not looking, and arrange the cut ends so they still look joined. Now,  
even if you think you have control, you do not, because the environment  
equation has changed. It is no longer qi = (d + qo)/2, but qi = d (using  
the diagram above). Now you have no control. Your output no longer has  
any effect on qi. The effect of qo on qi in our equation is zero, so it  
doesn’t matter how large or small qo is. It still has no effect, and  
there is no control.

> > > Exercise  
> > > 2: Redo the calculation with the reference condition set to “knot 5  
> > > cm to the right of the dot” (r = 5.00).
> 
> > This is not the  
> > correct interpretation. If the reference is not zero, then the control  
> > system
> 
> > will try to keep the knot at a position away from the dot. The dot is  
> > still at position zero,
> 
> > so when the center of the knot is exactly on the center of the dot,  
> > the knot’s position is
> 
> > also zero. Here we have the center of the knot being at 4.7 cm, which  
> > means 4.7 cm to the
> 
> > right of the dot (with positive measured to the right). Note that the  
> > reference signal was
> 
> > set to 5.0 cm. The knot is 0.29411 cm to the left of the reference  
> > position, and 4.70589 cm
> 
> > away from the dot.
> 
> > The reference does NOT determine “how strongly you want to keep  
> > the knot on the dot.”
> 
> > It determines what state of the perception you want to  
> > maintain: on the dot (r = 0) or in this
> 
> > case, 5 cm to the right of the dot (r = 5). How strongly you react to  
> > errors depends on the
> 
> > loop gain, not on the reference signal.
> 
> Here I am learning aspects of PCT. The way I understand your comments is  
> that you think upon a linear input function where the knot is moving  
> along a “horizontal” line through the dot. The game doesn’t tolerate  
> movements orthogonal this line ((r=5) is 5 cm to the right of the dot).  
> When orthogonal movements are actual, we talk about a nonlinear input  
> function.

No, we just add a second control system that controls the knot position  
in the y direction, independently of the first one. The new qi is  
controlled by a new qo: a position of the end of the rubber band at right  
angles to the line of the rubber bands.

But you didn’t comment on my demonstration that your calculation was  
wrong. Increasing the reference signal does not make the system try  
harder; it changes the location where the knot is to be maintained,  
moving it to the right.

> > Exercise 3: How big is  
> > the error when Ko becomes infinite?

> > > _**If  
> > > the gain becomes /(almost) infinite, there are no qi.**_
> 
> > You didn’t answer the question: qi is not the error. The error is the  
> > difference between
> 
> > qi and the position implied by the reference signal, which in this  
> > case is zero, the same
> 
> > position as the dot. Since the dot is at zero and the reference  
> > position is at zero, there
> 
> > is no error and the error is zero.
> 
> > Saying “there is no qi” does not mean the same thing as  
> > saying qi = 0.
> 
> Let me answer the question after your comment. See  
> above.

I think you missed my meaning: the problem may be one of translation into  
English. When you say “There is no qi” this is taken to mean  
“There is no variable that is being controlled.” or “there  
is no knot.” That is not the same as saying “There is a  
variable that is being controlled, and value at which it is being  
controlled is zero.”

I hope that this is helping.

Best,

Bill P.