When I control “knot on the dot” and the “knot is

on the dot” I don’t affect theenvironment by any actions. The reference is a1, the perceptual

signal is a1, theerror is zero, the feedback signal is zero (I think) and the

disturbance "knot onthe dot" results in a perceptual signal a1. This is

continued without any affectstill 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=

k_{i}*k_{e}*k_{o}*e +k_{d}*d. I meant to

say that when knot is on the dot, e=0. Then p = k_{d}*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.

- qi = (qo + d)/2

That’s the environment equation.

The controller equations are

p = qi

qo = Ko*(r - p), which reduces to - 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*Ko/2 = Ko

Collecting terms in qo on the left:

qo + qo*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.