PD(PI) Math Model (e.g.Tank Level)

JCH · 2007-05-21T15:34:52+00:00

Anything could be thought in mA and minutes. level moves v1' = 15 mA/minProcess transfer function F1(s)A1 * v1' = v2 = 0.3mA (Valve step)A1 = 0.3mA/(15mA/min) = 0.02minF2(s) Feedforward controller PD(PI)---------------------------------------Error e = u - v1v2 = u + B1 * u' + ^^^^^^^^^^^ Recognize PD1! (=PD) + K * (e + 30 * Integrator(e)) + z1 ^^^^^^^^^^^^^^^^^^^^^^ ^^ PI!K = 1Integration[0...t] e dt + C, C=0Forced disturbance z1z1=0.1 at t=0.3F1(s) Process transfer function-------------------------------A1*v1' + v1 + z2 = v2 ^^Forced disturbance z2z2=-0.1 at t=0.5F3(s) Target filter (1st order)-------------------------------C1*u' + u = w (w = scheme)Data used:A1 = 0.02 minB1 = A1C1 = A1 (for full compensation)v1 ist NOT missing because you can integrate v1'.v1 = Integration[0...t] v1' dt + CNote:A tank is a machanical integrator with delta flow input (v2) and level v1 output.

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Peter Nachtwey 19 years ago

Obviously there are systems like that. However, you have redefined your valves once again so that reality matches the math. You have also added another output to the controller so your math matches reality. None of this was in your original post. If you would have mentioned all of this up front I would have thought it was OK. I still wouldn't let you off the hook though. I would still point out the high or infinite gain controller will not work. Why do you ignore this? Even the TankLevel.sce control starts to look rough when I truncate the PV to match a feedback resolution of 0.001m.

See ftp://ftp.deltacompsys.com/public/NG/TankLevelWithQuantizing.gif ftp://ftp.deltacompsys.com/public/NG/tanklevel.sce ftp://ftp.deltacompsys.com/public/NG/TankLevel.gif

Compare with my previous TankLevel.gif. You can see how just a little reality can ruin a theoretically perfect control scheme. My gains aren't near as high as yours. I haven't even included noise or ripple caused by the in and out flow. Tim has been giving me/us a pass so far. Your high gain system would be very hard on the valves if it work at all. The control output would probably just go from one extreme to the other.

Quantizing/feedback resolution can't be ignored. This is one of the biggest factors that limits gains. BTW, what is your sample time?

Peter Nachtwey

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JCH 19 years ago

I published my actual solution on May/21 2007 (today: May/28 2007)

Read carefully first and last sentence:

k->infinity applies to feedback control. Examples: k=infinity x = u/(1+1/infinity) = u Steady state error 0% k = 100 x = u/(1+1/100)=u/1.01=0.99*u Steady state error of 1% k = 1000 ... I guess that 95% P-controller are used in level control. One does not care about error. If you care then PI-controller and PID-controller are used. Amplifiers and transmitters (e.g. temperature, 4...20mA output) are pure P-controllers with internally high gain k. k=10V/1µV=10^6, error=0.00001=0.001% If you can't utilize k=10^6 for some reason you must reduce k for having stable conditions.

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JCH 19 years ago

"Peter Nachtwey" schrieb im Newsbeitrag news:vJ6dnfpyYcuNLsTbnZ2dnUVZ snipped-for-privacy@comcast.com...

I specified all that could be specified. You find this in

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1, 2, 3, 4

Look at page 4 for higher target change and keeping precision.

Note: I have made 1 error, found it myself and corrected it on the same day (May/21 2007, introduction of split range control). I defined PD(PI) Math Model (e.g.Tank Level) and gave you 'all' data (page 2). It was not intented to simulate just PID control with fixed set point. All (note: all) what I have written so far is correct.

All diagrams show: BENCHMARK SCHEME

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Peter Nachtwey 19 years ago

I wouldn't say that. A simple proportional band controller will do the trick as long as the proportional band doesn't let the tank over flow. This way the gains can be very low and the ripple in the tank and quantizing of the feedback will not be noticeable.

In reality I wouldn't spend the time on it. The only reason why I am bothering with this lost cause is that you will corrupt other rookies that are just lurking and trying to learn.

You shouldn't be teaching anybody and you refuse to learn. Why do you bother? You seem think you know it all already so why bother with us?

I don't understand why you are so stubborn about this. Can't you see that just a little noise or quantizing greatly affects my controller and my open loop gains are much lower than your gains. If have said this so many times and then you claim I am just saying that you are wrong without reason.

I don't see why you refuse to show your calculations like I have shown mine.

You refused to answer many of my questions.

Believe me, your precious calculations are not a top secret if if they were right.

I don't see why you refuse to answer my questions or look at my Tanklevel.sce and ask why I did things the way I did.

YOU WILL ALWAYS BE WRONG AS LONG AS YOU INSIST THE OPEN LOOP GAIN IS K BECAUSE NOTHING HAS AN INFINITE BANDWIDTH. DON'T FEEL HURT AND CLAIM I AM ACTING LIKE THE POPE. I AM JUST TELLING IT LIKE IT IS.

It is time for a new horse. It wouldn't drink. I tried shoving it into the water but it wouldn't budge. This horse is ready for the glue factory.

Peter Nachtwey

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JCH 19 years ago

"Peter Nachtwey" schrieb im Newsbeitrag news:T4KdncPr3tKz48bbnZ2dnUVZ snipped-for-privacy@comcast.com...

[...]

Topic: PD(PI) Math Model(e.g.Tank Level) with BENCHMARK SCHEME ^^^^^ Anything is defined and solved in 4 pages:

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is 100% calculated (formulae in page 2).

Your response: ^^^^^^^^^^^^^ Normal PID level control without BENCHMARK SCHEME test conditions.

Please, EQUAL CONDITIONS! Then I can compare.

Peter warns students that what I am writing is wrong, completely wrong. But he can't tell what is wrong. He just knows that's wrong. The pope can argue this way. But not engineers.

You still ignore the BENCHMARK SCHEME in your simulation. Then your control is much (note: much) easier than using BENCHMARK SCHEME.

Please, be fair and use the BENCHMARK SCHEME! All what I have done is based on it.

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