Re: No Action Required?
[From Fred Nickols (2004.12.11.1435 EST)]
Many thanks for the reply, Martin, and
many more for the animated drawing. I sat and watched it for several
minutes. (How did you do that?)
Before I burden you with any more
questions, let’s see if I have the symbols correct (I know I don’t
have one of them right):
R = reference signal
P = perceptual signal
E = error the difference between R & P
The little blue circle where R and P are
inputs is the comparator.
What do you call the red triangle? I
would call it an effector.
O = output
D = disturbance
Output and disturbance are inputs to a
little green circle. That little green circle is what I would think of as
S = ??? Does it refer to signal of
some kind? My current understanding of such matters is that I would think
of it as the net effect of my actions (output) plus whatever effect the disturbance
is having on the controlled variable. However, that seems to be incorrect
in light of what you’re saying.
So, what is S and what is the yellow
One other question: You show arrows going
off at angles from O and P. What do those represent?
More questions embedded below…
[Martin Taylor 2004.12.11.13.14]
[From Fred Nickols (2004.12.11.1145 EST)]
I thought “disturbances” were
factors that interfered with my control of a variable.
This isn’t actually wrong, but it’s
misleading, and it is probably not a good idea to think of disturbances like
Why is it misleading? Because
"interfering with my control of a variable’ seems to imply that something
has affected your ability to influence the variable. In other words,
thinking of the canonical diagram of an
it would be altering the bottom feedback path. A disturbance doesn’t do
[From Fred: What are you designating
the “bottom feedback path”? From O to little green circle or
on through to the yellow rectangle? But a disturbance would alter the
perceptual signal, right?]
Why is it not actually wrong? What a disturbance does, if you do
nothing to oppose it, is alter the value of the input to the perceptual
function, and therefore the value of the perceptual signal. So its immediate
effect is to create error. The output function inputs the error and outputs an
effect on the disturbed variable, reducing the error. So, if you think
“interfering with my control” is synonymous with “increasing the
error in the controlled variable”, then a disturbance does that – without
interfering with the ECU’s ability to control.
[From Fred: Would I be correct in
assuming that “the perceptual function” is the yellow rectangle and
that S is the input to it? If so, and if the little green circle
represents the controlled variable “out there”, then output and
disturbance are both affecting it and through it, the input (S) to the
perceptual function. If I’ve got this wrong feel free to whack me
alongside the head.]
A disturbance isn’t the only thing that can create momentary error. The
same happens if the reference value changes. Disturbance and Reference are the
only two signals that enter the loop from outside. So, even if there is nothing
in the environment affecting the controlled variable except the output of the
control system, transient errors may still occur. But you wouldn’t say that a
reference signal interferes with your control of a variable, would you? So I
think it’s not a good idea to think of a disturbance interfering with your
[From Fred: Well, actually, I would
say that changing a reference value does (or can) interfere with my ability to
control a variable. If the new value represents a degree of control that
greatly exceeds anything I’ve accomplished before it seems to me that it
might be beyond my current ability to control.]
So, if the value of the disturbance is zero, and the error is zero, the
output must be zero, and your “No action required” suggestion would
[From Fred: What you say directly
above is what I thought I was saying first time around. I obviously
didn’t say it very well.]
But if there is a steady external influence on the environmental
variable, then to maintain zero error in the perceptual signal, the output must
be equal, opposite, and continuing.
[From Fred: I take the sentence
immediately above to mean that if disturbances are present that I must act to
keep the controlled variable in whatever state I want it to be.]
[From Fred Nickols (2004.12.10.1825 EST)]
Further, “keep doing what you’re doing” implies ongoing
action. What about start up or initiation?
Does the foregoing answer that? Start-up implies a change of reference
signal, doesn’t it?
of the way a simple integrator works is that, first of all, it has more
one input. It was also a dynamic device. If its inputs go
to zero, its
outputs do too.
I don’t know what you called an integrator in your electro-mechanical
analogue computer days, but what I meant was a function that implemented the
mathematical operation of integration. I can’t easily use math notation here,
but in words:
x(t1) = x(t0) + (integral from time t0 to t1 of y(t) dy)
Here, y(t) is the input waveform to the integrator, x(t) its output.
if y(t) has been zero between t0 and t1, x(t1) = x(t0), not zero.