[From Bill Powers (960117.1010 MST)]
Bill Leach 960115.22:51 U.S. Eastern Time Zone --
You're really got me puzzled:
In a "bidirectional" or effectively bidriectional control system, a
"reference of zero" result in for a particular controlled
perception means that a loop has a non-zero reference (ie: The
zero reference is inverted to maximum).
Is that really what you mean? If the perceptual signal can go both
positive and negative, and the output likewise, wouldn't a reference
signal of zero bring the input to zero, in the middle of the range?
And this takes a little interpretation:
In a simple unidirectional control loop the reference of zero means
that the perception is not controlled and may assume any (real)
value.
If the perceptual signal, error signal, and output function are limited
to positive values only, and the perceptual signal is subtracted from
the reference signal, what you say is right. The control range is only
between zero and the setting of the reference signal.
···
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Gary Cziko (960116) --
You asked me privately about a graphics "BGI" error when using Demo1 and
Demo2 on a PS2 computer. I told you that the ".bgi" files were missing,
forgetting that these demos have the graphics interface integrated into
the program.
We had a problem like this a few years ago, and someone on the net
solved it. It seems there is a setup procedure for the PS2 that affects
the graphics mode so it's not standard for other PC-compatibles, or for
my programs. I don't recall who it was who solved the problem, but
perhaps that person will now repeat the fix for you.
Bill
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Rick Marken (950116.2100) --
I fixed up the simulation, added the slowing factor so that
o := o + slow * gain * i
and, sure enough, the loop only ran away when gain (loop gain)
was >= 1.0.
This still isn't right. You have the slowing factor simply multiplying
the gain. Your system will be stable if slow*gain is less negative than
-1, but not if it is greater. That's because you just have a system with
a gain that is slow*gain. What you really need is this:
o := o + slow*(gain*i - o)
"Slow" has to be a number less than 1. If all other gain factors in the
loop are 1, the minimum value for the slowing factor is 1/(1 + gain).
That will correct the error in one iteration. For smaller values of
"slow" the approach to steady state will be slower.
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Best to all,
Bill P
P.S. We just had our first real snow of 1995-96. Eight inches of heavy
wet stuff, and more to come in a couple of days. The ski resort
operators are weeping with joy.