Friday's PID controller design quiz

Hello guys,

I suggest to play a little for Friday night quiz. Objective is to design PI controller that will provide "good" response of the system that is given in attached picture. Plant is 1/s(Tms+1), where Tm time constant is equal 0.052.

I tried to design PI controller in form Kp(1+sTp)/s i.e. to determine Kp and Tp for system to be critcially dumped. I failed and I know why. Now I wonder if it is really possible to design PI controller to make system's response critically dumped. Can you design controller to prevent overshoot? What would be formal math procedure?

http://www.plctalk.net/qanda/uploads/Quiz.jpg

Pandiani said:

Hello guys,

I suggest to play a little for Friday night quiz. Objective is to design PI controller that will provide "good" response of the system that is given in attached picture. Plant is 1/s(Tms+1), where Tm time constant is equal 0.052.

I tried to design PI controller in form Kp(1+sTp)/s i.e. to determine Kp and Tp for system to be critcially dumped. I failed and I know why. Now I wonder if it is really possible to design PI controller to make system's response critically dumped. Can you design controller to prevent overshoot? What would be formal math procedure?

http://www.plctalk.net/qanda/uploads/Quiz.jpg

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I'm spending my Friday night emptying beer cans

curlyandshemp said:

I'm spending my Friday night emptying beer cans

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I can empty many beer cans and still provide the answer(s). The correct ones too! I have the critically damped solution memorized. The over damped solution I have worked out somewhere in one many Mathcad worksheets. I will hold back and let the others think on this.

Pandiani, did you work this out for your class project? The solution is just a simpler form of the factoring problem a few weeks back. I am sure the same people will find answers. The difference is that this requires a symbolic solution.

The PI controller doesn't look right. What is Tp?
Should be Kc*(1/(Ti*s)+1).

Peter Nachtwey said:

Pandiani, did you work this out for your class project?

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I found that example in my old notebook from undergraduate studies.

Peter Nachtwey said:

The PI controller doesn't look right. What is Tp?
Should be Kc*(1/(Ti*s)+1).

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It does, just it is non standard form. Kp*(1+sTp)/s is also PI conroller, just Kp and Tp are not proportional gain and integral time constant, that's all.

Pandiani said:

It does, just it is non standard form. Kp*(1+sTp)/s is also PI conroller, just Kp and Tp are not proportional gain and integral time constant, that's all.

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This is the kind of **** that academics do to torture poor students. No real engineer would express a PI controller in this way. And they say the CIA and Bush are bad.

The controller
Gc(s)=Ki/s + Kp
The motor
Gm(s)= K*a/(s*(s+a))
where K is the motor gain and 'a' is the corner frequency or the pole at -a.

The closed loop transfer function is:

T(s)=Gc(s)*Gm(s)/(1+Gc(s)*Gm(s))

I change labels from Kp to K to avoid form confusion. I knew Peter would object...
Here's is through analysis

Hint

1. Calculate the PID gains symbolically in terms of the plant parameters and the desired closed loop pole locations. Can Matlab calculate the PID gains symbolically?

2. Set the Kd to 0. This will tell you the relationship between the desired closed loop pole locations. You will find that there is one spot where the system will be critically damped. The are many solutions where the system will be over damped. That is where all the poles are on the negative real axis but as one poles is moved away from the break away spot toward 0, the other poles will move away from the break away spot towards -infinity. Since the pole closest to 0 dominates and slows down the system its response gets slower or more over damped.

I had to solve this problem when I made the auto tuner for the RMC100. The user moves a slider bar to chose between over damped and critically damped response. What the user doesn't know is that he is really just moving the desired location for the closed loop poles on the negative real axis. When the slider bar is in the conservative tuning range the response is normally over damped and the Kd gain is 0. As the slider bar is moved to the more aggressive direction the poles converge at the break away point. This is the point where all three poles are equal and Kd still equals 0. As the user moves the slider bar even more into the aggressive zone the tuning stays critically damped and all three poles are moved together in the negative direction. This increases the gains and the bandwidth.

One more hint. If the open loop pole location is at -pole then the breakaway location will be at -pole/3.

I'm surprised that nobody tried to confirm or reject thesis that for this particular structure, system will be unstable if Tp <Tm.
Peter,
My first approach was to find K and Tp with simple pole placement method we discussed earlier. In attached file you can see that.

What is Kd? I'm not using any derivative term in this task.

Isn't the system transfer function in #6 only from reference to output?

Original problem and solution illustrated below. Perhaps I'm not alone in refreshing my Matlab skills?

Tm = 0.052; Tp = 0.468; K = 13.7;
sys = tf([Tp*K K],[Tm 1 K*Tp K]);
step(sys,0:0.01:2),shg

spaderkung, object is to find K and Tp so the response doesn't have overshoot...
I failed to find it, and want to induce some general conclusions fro this particular example...

Very good guys

However, I wonder if the right solution would have been found without hints. It looks like Matlab makes it easy but it is just one solution of many. Matlab found the critically damped solution and not the more generic and flexible over damped solutions.

You both have the right answer for a critically damped response. So why does it over shoot? There is a simple answer. I can make mine so it doesn't over shoot. Hint, what is there in the closed loop transfer function that has been completely ignored?
These are the kind of things you learn to answer very quickly when making motion controller because customer don't like over shoot.

Edit. Make a Bode plot and a pole zero plot. Maybe that will give you a clue. I know it will you just have to know when you see it.

Peter Nachtwey said:

However, I wonder if the right solution would have been found without hints.

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I did this procedure before I decided to start a thread. I don't consider this a solution. My last .pdf is basically what i had before. Procedure that we discussed few weeks ago with choosing critically damped will work only if transfer function has no zeros.
Just one thing: do you considres response critically damped if it has real poles: p1=p2=...=pn? Can you say critically damped when you have overshoot?
I'm interested in terminology.

What about Tp<Tm? Can anyone confirm what I concluded from couple of examples I've done before.

Edit:
I've just spoken with Peter, it seems there are some things that can be unclear to readers. Structure of system is given, you cannot change that. Also, input to the system is Step function. Objective is to try to find K and Tp that will provide response without overshoot. I said in my first post that I have failed and I know why. In other words, solution is to find K and T without overshoot. I didn't use correct terminology. I said I failed to obtain K and Tp for critically damped system, but in fact I failed to find K and Tp without overshoot.
At first I thought that critically damped system would not provide overshoot. I stand corrected.

My answer

ftp://ftp.deltacompsys.com/public/NG/Mathcad%20-%20t1p1%20pi%20od%20Pandiani.pdf

It is the zero that causes the overshoot. The pole locations calcuated by Pandiani and Spaderkung are correct. The zero is added by the PI controller. To fix the problem, the zero must be removed. This can be done by putting the proportional term in the feed back path only. This way the porportional gain only works on the PV or actual position in this case. Putting the proportional gain in the feed back path only make the controller an I-P controller. You can see the changes are minimal. Look at the Bode plot and the different transfer functions. You can see the controller in the forward path has only the integrator term.

Sorry i'm late. I had a pretty busy weekend.

As I read this over and looked at Pandinai's original plant two things came to mind. The first was that this was a type 1 system. In a 'perfect' type 1 system (which is what the transfer function suggests; no losses) you can't get away from overshoot if you have an integrator in the controller. By the way, I also held the misconception that Pandiani did about critically damped systems and overshoot. But with a little nudge I see how they are separate things.

The second thing that came to mind was that you will end up with more poles than you have controller gains. You don't have enough gains to influence all the poles with a PI controller. This thought turned out to be incorrect, as you guys have shown. I keep thinking of the integrating nature of a type 1 system as increasing the order of the system. For the purposes of design and evaluation I don't know that I can do that.

So if I understand the I-P controller correctly, from a physical standpoint, the proportional term working on feedback only will ultimately counteract the integral sum so you have a zero output when you reach the setpoint. Interesting implementation.

And now for a question. I understand how pole locations influence a systems response and have a reasonable feel for the physical items in a system that will end up becoming poles in a transfer function. However, I have no feel for what in a physical system will represent a zero. I can appreciate from a mathematical standpoint what a zero is. But I don't have a physical feel for one.

Keith

I know what you mean.

kamenges said:

However, I have no feel for what in a physical system will represent a zero. I can appreciate from a mathematical standpoint what a zero is. But I don't have a physical feel for one.

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In this case the zero isn't caused by a physical object. It is just some math in the controller. Planes and boats all have zeros in the right half of the s plane. That is why they go in the wrong direction first. This is called non-minimum phase. In both case the energy isn't stored or transfered. It uses the forward momentum or kinetic energy to convert to sideway motion. The best I can think of is that zeros tap energy from some other source.