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
Did I say that?
[From Bill Powers (2004.08.05.0845 MDT)]
Bjorn Simonsen (2004.08.05,12:18
[From Bjorn Simonsen (2004.08.04,09:30 EuST)]
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
- 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
qo = Ko*[r - (qo+d)/2]
qo = Kor - Koqo/2 - Kod/2
Collecting terms in qo on the left:
qo + qoKo/2 = Kor - Kod/2
Extracting the common term qo on the left:
qo*(1 + Ko/2) = Kor - Kod/2
Dividing both sides by (1 + Ko/2):
Kor - Kod/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
Let’s say r = 0, Ko = 100, and d = -20 cm (20 centimeters in the negative
0 cm * 100 + 20 cm * 100/2
qo = --------------------------
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
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.