Feedback: Positive and Negative

[From Rick Marken (2015.02.08.1145)]
RM: Negative feedback exists when the product of all the multipliers of the loop variable are negative. Consider the simultaneous equatoins that define a basic control loop:
o = r - p (1)
p = k.o*o + k.d*d (2)
RM: In these equations the only variables in the loop are p and o. Both r and d are independent influences on the loop so the sign of these variables doesn't count in determining whether these equations define a positive or negative feedback loop. So the polarity (positive or negative) of the loop defined by equations 1 and 2 is determined by the product of the signs of the effect of p on o and that of o on p. From (1) the sign of the effect of p on o is minus and from (2) the sign of the effect of o on p is plus. So the product of the signs is minus, indicating that these equations define a negative feedback loop.
RM: This procedure can be used to determine whether the equations that define a mass spring system describe a positive or negative feedback loop. The equivalent of equations 1 and 2 above for a mass spring system considered as a control loop are:
o = s * (r - Pos) (3)

Pos = - o + Force (4)
RM: If r is 0, as assumed by equilibrium theorists, then the polarity of this loop is positive since Pos has a negative effect on o and o has a negative effect on Pos. Since a mass-spring system does not behave like a positive feedback system (with exponentially increasing output) the loop can only be made stable by eliminating r so that
o = s*Pos (3a)
which is basically Hooke's law. So it is incorrect to imagine that an equilibrium system has a reference specification for the variable that is returned to its resting state when the Force that causes a displacement from this state is removed.

RM: When we solve the simultaneous equations for a control loop (equations 1 & 2) for p (the controlled variable) we get:
p = k.o/(1+k.o)*r + k.d/(1+k.o)*d
which, assuming k.o, the output gain, is very large, simplifies to:
p = r + (1/k.o)*d (5)
RM: When we do the same thing for the simultaneous equations for an equilibrium system (equations 3a and 4) we get
Pos = Force / (1+s) (6)

RM: Since Pos is the equivalent of p and Force is the equivalent of d equation 6 can be written as
p = d/(1+s) (7)
RM: Comparing equation 5 to equation 7 reveals the important difference between an equilibrium and a control system. Both are formally negative feedback loops. But equation 5 shows that, in a control system, the value of the input variable, p, is determined mainly by the reference specification,r. Equation 7 shows that, in an equilibrium system, the "input" variable (the position of the mass or of the pendulum bob) is determined completely by the disturbing force. There is no disturbance resistance at all.
RM: This is a very interesting discovery, I think. It shows that a system can formally be a negative feedback system (not technically an open loop system) and still act exactly like an open loop system. I think this explains why equilibrium systems -- negative feedback systems that act exactly like open loop systems -- could fool people into thinking that they are similar to control systems. In fact, equilibrium systems are not anything like control systems; they don't resist disturbance at all and they certainly don't bring variables to reference states -- they have no reference specifications. They are open-loop causal system systems masquerading as control-like systems by being analyzable as negative feedback systems. Equilibrium systems are very much l like Lewis Carroll's Boojum; they look like a Snark and so if you're hunting Snark you can be fooled into thinking you have a Snark when it's really a Boojum. Of course, the difference between Boojums and equilibrium systems is that when you find one, thinking you've found something like a control system, you don't softly and suddenly vanish away, as you can see.
Best
Rick

···

--
Richard S. Marken, Ph.D.
Author of <http://www.amazon.com/Doing-Research-Purpose-Experimental-Psychology/dp/0944337554/ref=sr_1_1?ie=UTF8&qid=1407342866&sr=8-1&keywords=doing+research+on+purpose>Doing Research on Purpose.
Now available from Amazon or Barnes & Noble

Dear CSG,

Is there any difference between calculating the output using o = r - p vs o = old value + change in value?

kind regards,

Philip

···

[philip.2.8.2015.12:09 pm]

[Martin Taylor 2015.02 08,23,00]

Apart from the fact that they are unrelated, what difference are you

thinking of? And how does your question relate to PCT? “Output” in a control system isn’t “r-p”. If you intended to use the
usual symbols, you would write e = r-p, but you used “o”, so your
must be thinking of something else. And how does a difference
equation for “o” relate to anything?
Martin

···

[philip.2.8.2015.12:09 pm]

Dear CSG,

    Is there any difference between calculating the output

using o = r - p vs o = old value + change in value?

kind regards,

Philip

[From Rick Marken (2015.02.08.2130)]

···

Is there any difference between calculating the output using o = r - p vs o = old value + change in value?

RM: Yes. o = r-p is the equation for the output of a control system in a “quasi-static” analysis. Quasi-static means that the analysis ignores the fact that variables must change over time. So when we write a control system as the simultaneous equations

philip.2.8.2015.12:09 pm]

o = r-p

p = k*o + d

it implies that the value of o is caused by p at the same time as the value of p is caused by o. Of course, this can’t happen in reality but the quasi-static analysis is useful let’s us see how the variables in a control system are related using simple algebra. But when we want to write digital simulations of control system behavior we have to take into account the fact hat the variables do not change simultaneously. So we let the variables change a little during each step of the simulation loop. This can be done by having o change only a fraction, k, of the total amount that it would change based on r-p during each step of the loop, as in

o := k*(r-p)

but sometimes we want the output to be proportional to the integral of the error over time. So we can write something like:

o := o + (k(r-p)-o)/slow

Note that both of these are computer statements and that the := is a replacement operator and not an equal sign. So in that statement the effect of error is accumulated (integrated) into memory location o.

Best

Rick


Richard S. Marken, Ph.D.
Author of Doing Research on Purpose.
Now available from Amazon or Barnes & Noble