Hi all,
I assume that a process with quadratic behaviour can not be controlled with a simple PI-controller (if I do not use a gain scheduling
controller). But how do I prove this mathematically?
pt
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It *can* be controlled perfectly well with a PI controller as long as you keep away from the gain reversal. Before you embark on your maths, you need to specify exactly what it is that you want to prove.
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On Fri, 25 Nov 2011 13:30:50 +0800, Bruce Varley wrote:

Well, as long as you don't have to control across the gain reversal, and as long as your bandwidth is high enough at the low gain end of the operating range, and as long as your damping is low enough at the high gain end of the operating range.
After that point, consider that gain scheduling can make you a hero, but it can also make you look like a buffoon. (Nothing, but nothing, equals the impression you can make by spending an hour in the conference room explaining why your approach is superior, then walking out into the lab or the factory floor and blowing something up).
--
Tim Wescott
Control system and signal processing consulting
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Yep, anyone who works in the process industries has had more experience with this than they may realise. Heaps of flow controllers utilise equal percentage control valves, whose installed characteristic is often just as, or more bendy than a quadratic. The limited operating range of the plant usually protects you from the nonlinearity, as long as you don't make the mistake of doing the tuning at a low operating point.
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I meant the problem with the gain reversal. What is the way to prove mathematically that this is a problem?
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Your problem is still a semantic one. If you want to prove that conditions can exist where positive feedback will cause the loop response to be unbounded, then that's trivial. If you want more than this, then what do you mean by 'problem'?
Sorry to appear difficult, but I can assure you that many real world control projects are derailed by failure to clarify situations like this.
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Bruce Varley schrieb:

Lets start here with the trivial. Under which conditions is the loop response unbound?

No problem. As you may assume, I found "the problem" in a complex control loop. Don't know where the instability results from. So I try to divide the system or the problem into smaller pieces. The quadratic behaviour is one of those, as I see that the derivate of the process function changes its sign. Meaning, the process function is not monotonous (?). I hope to get a better understanding if I can put it into mathmatics and would be happy if someone can give me a hint how to setup the mathematical prove.
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On Fri, 25 Nov 2011 07:16:06 -0800, pt wrote:

That may not be so trivial after all. But if you have an element in the system whose instantaneous gain is changing from positive to negative, then it is very doubtful that a system whose linearized model is stable for one case will be stable for the other, unless you have a very strange (and very nonlinear) controller.

You mean monotonic (you should be able to look up "monotonic function" in Wikipedia). A monotonous process function would be boring: malfunctioning equipment is, unfortunately, not often boring.

The most direct way that I could see to do this would be to take my system model and do a root locus plot around the changing gain of the part, over the expected gain range of the part. If you've got just one pole at the origin (from the controller integrator) then the system will either be stable for some values of positive gain and not at all for negative, or the converse.
In fact, the only way that I could see that you might have a system with overall stability in the face of an element whose gain changes sign would be if that portion of the system were loaded down somehow with an element (or elements) that were so very dissipative that they would swamp out the quadratic element. I can't even begin to think of a concrete example of this outside of electronic circuits, and within that discipline the "swamping out" is generally pretty wasteful in terms of energy and often space.
--
Tim Wescott
Control system and signal processing consulting
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Actually, I have no clue how the transfer function with quadratic gain would look like.
pt
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wrote:

Only one unstable scenario is needed for a proof, for a given configuration. What's wrong with just simulating your chosen system?
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I would like to simulate the system after I have understood what's happening at zero gain. For this I prefer Tim's approach.
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This sort of thing does happen occasionally. My experience is that you won't get expected results :) Since real world systems have limits (say saturation) they can occasionally be made to work; sometimes in odd ways. Why not explicitly explicitly describe the system and the part that you think has quadratic part and gain reversal? Give the physical reason you believe that this is happening. There are plenty of people here who are quite knowledgeable about a wide range of real world applications. I don't think this is a problem that can be treated in the abstract. BTW: Scilab is a free program that has a graphical interface allowing a variety of non-linear blocks and allowing the execution the block diagram. A physical description and a accurate executable block diagram would probably lead to really effective answers. In engineering mathematical answers typically require an approach from the bottom up. Understanding the physical system and identifying a mathematical models that describe it.
Ray
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On Sat, 26 Nov 2011 14:14:36 -0800, pt wrote:

If you mean "transfer function" in the sense of a Laplace transfer function or a z-domain transfer function, that's because the whole concept of a transfer function with quadratic gain is entirely meaningless: transfer functions are a way of modeling system behavior in the Laplace or z domain, and modeling in those domains only makes sense for linear, time invariant (or shift-invariant, for z domain) systems.
If you want to use transfer function analysis then you need to linearize the system around an operating point, then find the transfer function of the resulting linear model. Doing so and finding that the resulting linearized system is unstable means that your overall system is unstable around that operating point. In a wide sense this may or may not mean that you're in trouble: some systems work perfectly well under these circumstances, but usually only if the nonlinearity is well behaved and makes them oscillate around the desired set point. In your case, there's a good change that getting on the wrong side of that quadratic gain curve will mean that the system will go berserk.
If the system, once you cross over the zero-gain part of the quadratic characteristic, tends to go further in that direction (instead of breaking into a sinusoidal oscillation, for instance), then you can be pretty sure that it'll go berserk. Even if it does something more benign but still out of the correct operating range, then you have a problem.
Why can't you just restrict the system from driving that part into the negative gain region?
--
Tim Wescott
Control system and signal processing consulting
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Hi Tim

Yes. I was on the wrong track.

I have started with a simple example. With the transfer function of a PI controller G_C(s)=K_C*(1+T_C*s)/T_C*s and the transfer function of a PT1 process G_P(s)=K_P/(1+T_P*s), I build the closed loop and calculate the characteristic equation
1+(K_P*K_C*(1+T_C*s))/((1+T_P*s)*T_C*s)=0
Now, I take the part between "1+" and "=0", name it G_CP and figure out the zeros and poles. These should be
s_z1 = -1/T_C (zero 1)
s_p1= 0 (pole 1)
s_p2=-1/T_P (pole 2).
Finally the curve parameter K_0=K_P*K_C/T_P.
With this I could play with the time constants and gains of process and controller. As you suggested, I can try to minimize K_P like it would be in the area of the gain reversal. I can also define a process with negative gain and examine the stability on the other side of the gain reversal. I have the feeling that both, the gain reversal and the small gain cause instability. I think you wrote about the small gain problem before but I wasn't sure what you mean.
I chose some values for the time constants and gains and used a root locus java applet on some web site. My first question is, can you suggest a web page with root locus applet? My second question is, how do I evaluate the stability with the root locus plot? I have literature about this but after reading I still have no clue.
Example values: T_P=T_C=500ms, K_C=K_P=1 Applet Page: http://users.ece.gatech.edu/bonnie/book/OnlineDemos/InteractiveRootLocus/applet.html numerator setting: "0.0 0.5 1.0" denominator setting: "0.3 0.5 0.0" gain: 0 to 5

Some stupid designed it in the middle of the operating range. A design change is my first choice. But still I would like to understand what is happening in that region.
Best regards,
pt
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On Wed, 30 Nov 2011 13:55:17 -0800, pt wrote:

Get a copy of Scilab, and use the "Evans" function (because the _full_ name of the root locus plot is the _Evans_ root locus plot).

The root locus shows the paths (the loci) of the roots as the gain changes. If you have a root that falls outside of the stability region (for the Laplace domain that's for any root with a non-negative real part) then that root is unstable. Roots close to the stability boundary indicate a system that will take a long time to settle, and that may not be robust to changes in system parameters.

InteractiveRootLocus/applet.html
Oh Joy!!

Depending on the plant you may not be able to stabilize it at all without going to wacky extremes -- sign changes in your gains are Very Bad Things.
But like everything, it depends.
--
Tim Wescott
Control system and signal processing consulting
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I have severe problems with my scilab installation. For some reason I can't get the scilab plot working. There seems to be a problem with the interface java to openGL which stopped functioning after a kernel security update. This may take a while. But I'm eager to try out your proposal.
pt
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On Tue, 20 Dec 2011 13:39:45 -0800, pt wrote:

Ouch! I know nothing!
Except that you could try scicoslab. There used to be one Scilab group, and one Scilab. Then they (apparently) underwent miosis, and now there's two. Switching from one to another is -- in the extreme -- a way to dodge bugs.
--
Tim Wescott
Control system and signal processing consulting
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Tim,

[...]
After a long time I found a bugfix for scilab 5.3 on ubuntu 11.10 here (http://www.equalis.com/forums/posts.asp?topic=321201 ). The solution proposed by R. Rivi�re on 11/22/2011 worked fine.
And I plotted the root locus for the linearized quadratic function. We discussed that the two problems are the small gain left and right of the gain reversal and the gain reversal itself. The root locus plots I saved at http://imageshack.us/photo/my-images/687/gainreversal.png/. As far as I understood the roots for gains less the -0.1 move towards the non-negative area and would be unstable.
pt