Minor Loop Control questions

I designed a lead-lag series compensator controller for my plant that looks like
Gc = 10(1+s/58.4)(1+s/100)/(1+s/8)(1+s/700)
my plant is Gp = 100/s(1+s/100)(1+s/1000).
I successfully simlutate and prove a stable design with the goals of velocity constant = 1000 phase margin = 50 deg.
Now, I am converting this controller to a minor loop controller and have successfully suimulated it and verified stability of the minor loop and overall plant---but I have some questions that I don't understand:
1) by basing the design on the previously determined Gc, do I need to worry about velocity constant? (ie will the minor loop feedback have to be re-adjusted to guarantee kv00)?
2) my original gaincrossover was calculated to be about 140 rad/s. Unexpectedly, the Bode plots for the minor loop compensator move the gain crossover from 140 to about 60. Is this to be expected? if so, why?
My matlab/simulink seem to indicate all is OK--I'm just bugged by the fact of gain crossover frequency moving so far to the left and giving a more sluggish step-response than the supposed equivalent series lead-lag compensator?
My web searches have yields nothing useful on minor loop techniques/comparisons.
My approximated minor feedback block is:
0.0125s/(1+s/58.4) and is applied around the (1/(1+s/100)) term of the plant.
Comments?
THanks,
Bo
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Some comments: http://home.arcor.de/janch/janch/_news/20070416-bo / Page 1: Calculation Page 2: Block Diagram
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Regards/Gre
Jan C. Hoffmann http://home.arcor.de/janch/janch/menue.htm
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Changes: http://home.arcor.de/janch/janch/_news/20070416-bo /
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Regards/Gre Jan C. Hoffmann http://home.arcor.de/janch/janch/menue.htm
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Jan,
Thanks for taking the time to reply and for your posted site. I apologize, but I really didn't follow how the posted info relates to my questions though. I think I have safely concluded that my minor loop feedback block cannot cause the overall design to meet the kv requirement. ie. I must insert some series gain of 10 to get kv00 because no modification of H can effect kv. Is this an accurate statement? And it follows that by adding a series gain of 10 somewhere in the plant, that my crossever frequency would likely move back to 140 rad/s--or at least near it.
Thanks,
Bo
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(...)

(...)
Don't worry, me neither, so you are not alone :). [Jan, what the ... are you talking about?]
Kv=lim s->0 (OpenLoopTF) So, it depends on your TF's (transfer functions) if inner loop can influent on Kv. If they are zero class type or something A/sB (A,B 's' polynomials).
The structure is crucial. This the first thing we have to determine before any calculation. If you change structure, you change everything.
--
Mikolaj

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napisa(a):

Sorry, first I didn't recognize that it was homework. In real life I would have a solution for

similar to I discussed with Peter's process transfer function.
I couldn't recognize a minor loop!
I'm talking about 100% time-compensation using PD^3 controller with loop gain as high as possible avoiding or minimizing controller integral function.
Using Peter's process transfer function he posted:
See revised version: http://home.arcor.de/janch/janch/_news/20070424-mikolaj /
Page 1: Math model: Fig. 1 Loop model: Fig. 2 The behavoir would then be like an amplifier, Fig. 3.
Page 2: If setting a benchmark test w the result could be like: Page 2
That's what I had done if it were not homework but real work to be done.
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(...)

What's it all to do with the Bo's question about Kv and inner loop? And what is PD^3, where is this terminology from? And where do you get process model with this all derivatives? You show us some theory while we are talking about something completely different. You should start new subject for what you are trying to show and please do it if you want discuss it.

Yes? Are you sure that calculating 3'rd derivative in 8bit MCU is what we are commonly do in practise??
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Mikolaj

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napisa(a):

<cited> Sorry, first I didn't recognize that it was homework. In real life I would have a solution for ... similar to I discussed with Peter's process transfer function. </cited>

A 3'rd derivative could be implemented easily with an industrial PC.
Examples http://www.msc-tuttlingen.de/deutsch/ipc/industrie-pc/index.html http://www.automatas.org/xycom/pc_industrial.htm ...
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(...)

Of course they can! Especially when build a satelite or something.
My second order controller will cost 1$, how much will cost your industrial PC? And mind that your PC not always will be usefull. So much for reality.
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Mikolaj

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napisa(a):

If quality does not count make a good 2nd order least-square approximation with the data from 3rd order. Or just prove what happens if dropping the 3rd coefficient.
Example for dropping http://home.arcor.de/janch/janch/_news/20070424-mikolaj2 /
Least-square approximation for 2nd order would be much better.
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What are you talking about this time? I have no idea how to talk to you to be understand. And once again, please, start new subject and tell me or us, what are you trying to show or say.
I give up, somebody, give me a hand, please. Maybe someone should talk to him in his language?
I remind you what we were we talking about in last post. I asked you where are 3rd order derivatives from, (how you get them from Bo's transfer function) not about 3rd order minimalization to 2nd order (there are far way better methods then LS, try to aproximate nonminimumphase object haha). So stop producing diagrams and start answering. And add white noise to your 'u' signal (whatever you are calculating there).
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Mikolaj

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Sorry about this Bo for this tangent but we need a reality check.
Jan, your method will not work in a real system. Look at your graphs. In each example I have seen the response is almost perfect and instantaneous. That just can happen in the real world. You are assuming that your controller has infinite power and infinite resolution and that you can measure the higher order derivatives of the feedback
Jan, stop using simulations by converting the s domain into the time domain. That just doesn't show the practical limitations. You must implement the controller and the system separately. The controller must have a +/- 10 volt or +/-100% control output is a limitation. Second, limit the feedback resolution to 10 microns for starters. Third, the actuator or plant must be digital too and have a zero order hold ( or worse, dead time ). Use the discrete state space or use the z transform tables. Look at my simulations.
I think instructors that do simulations by directly converting a closed loop transfer functions to a time response are committing a crime or at least misleading students. It may be easy but it isn't reality. Example. http://www.engin.umich.edu/group/ctm/PID/PID.html Look at the PID response at the bottom of the page. This doesn't take into account that the control output could be saturated and therefore impossible
I know that I haven't shown the effects of quantizing on my worksheets but believe me I do quantize all my feedback in my more serious simulations.
Peter Nachtwey

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Peter Nachtwey wrote:
Newsbeitrag

-- snip --

-- snip --

It's interesting that you should mention that, because I'm teaching a course this spring.
I will be showing simulations that directly convert transfer functions to time responses -- but I'm also trying to be sure to keep the notion of a nonlinear system alive in the student's mind by mentioning it at least once a week. The problem that one has is that one wants to keep the subject accessible, which means sticking with linear control for the most part. OTOH, I don't want to lead them into thinking that purely linear system analysis is adequate for all purposes.
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Tim Wescott
Wescott Design Services
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You better do digital simulations. Look at my .pdf files to which I have posted links. You can see that I rely on digital simulation and all of those systems I have simulated are linear, at least in the s domain. Are you going to teach using Scilab? If so you should work out a general simulator that will take the plant transfer function in the s and convert it to the z domain. The controller should be be implemented separately. This way the student find out about the realities of quantizing and integrator wind up. Knowing how to implement a good PID that handles saturation well is important too. I feel as strongly about digital simulation as you do about poles and zeros. You can see there were two cases where systems with good looking Bode plots in the iterative tuning thread that failed to work when digitally simulated. I think Jan has been mislead by not digitally simulating his systems too.
The +/ 100% control output and quantizing problems CAN NOT BE IGNORED. It is easier to ignore the ZOH problem because the sample rates today can be so fast relative to the system. However, if I were controlling small motors with very small time constants I must take the ZOH into account. The +/- 100 control signal limitation is alway a problem with systems where the customer want to go very fast. The quantizing problem shows up in higher order systems where one must use the second derivative or error in acceleration.
BTW, I post from Google about half the time. I can't tell one post from another except my e-mail address changes. I think Google has fixed their news group posting feature so posts made from it look like any other.
Peter Nachtwey
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In real life you have nonlinearities, more or less. If more then you can measure and calculate the parameters at 20, 50 and 90% position (Z). Your control device will operate with B1(Z), B2(Z), and B3(Z). You add this feature also using least-square approximation methods. This will cover any state of the process. The accuracy could be about more then 95%.
If your equipment cannot bear the theoretical step changes then you have to limit actions using speed limiter, damper, etc.[1] Last not least you have to implement safety equipments, too.
My point is I had to do all this with try-and-error methods, and that could be implemented more sophistically using computer techniques.
[1] Note for Peter: Using filter T*u' + u = w http://home.arcor.de/janch/janch/_news/20070417-nachtwey /
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Please click again for the newest revision: http://home.arcor.de/janch/janch/_news/20070417-nachtwey /
It should be clearer now.
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Regards/Gre http://home.arcor.de/janch/janch/menue.htm
Jan C. Hoffmann eMail aktuell: snipped-for-privacy@nospam.arcornews.de
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: : Please click again for the newest revision: : http://home.arcor.de/janch/janch/_news/20070417-nachtwey / : : It should be clearer now. : I understand what you are trying to do. I have for the last few posts. I just couldn't believe you think that this would really work. Now show us a digital simulation where the controller generates an output. that must be limited to +/-10 volts or +/- 100%. Show us how the systm will react when the feedback reslution is 10 microns.
In your system the controller gain is K*FS2(s). The resulting gains will be extremely high. The resulting transfer function is always K/(K+1) because FS1(s) and FS2(s) cancel each other out and you seem to think you can keep increasing K without practical limit..
You have been mislead and you are misleading others. What bothers me is that one of Tim's students may read this and do direct s domain to time domain conversion and think what you are showing us is wonderful.
How many times must I say that you need to do your simulation in discrete time and implement the controller and plant separately. Do you know how to convert the s domain to the z domain? Doesn't a lingering doubt bother you? Just try it, pretty please. Your eyes will be opened.
Peter Nachtwey
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K/(K+1) = 1/(1+1/K) = 1 K => infinity or big number if exact time-compensated
That is theoretical the same as an amplifier or P-controller in closed loop feedback control with high gain.

What I am doing so far is pure mathematics with interpretations. E.g. I don't have a closed loop working, I calculate the closed loop (blue line): http://home.arcor.de/janch/janch/_news/20070417-nachtwey /
Example for time-comensation n=1 PD^n controller PD^n process: http://home.arcor.de/janch/janch/_news/20070418-nachtwey /
Anyone who understands the mathematics here will agree that n=3 will work, too. Peter, if you have doubts let's discuss this example (n=1). This is basically the same.
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: > : : > : Please click again for the newest revision: : > : http://home.arcor.de/janch/janch/_news/20070417-nachtwey / : > : : > : It should be clearer now. : > : : > I understand what you are trying to do. I have for the last few posts. I : > just couldn't believe you think that this would really work. Now show us : > a : > digital simulation where the controller generates an output. that must be : > limited to +/-10 volts or +/- 100%. Show us how the systm will react : > when : > the feedback reslution is 10 microns. : > : > In your system the controller gain is K*FS2(s). The resulting gains will : > be : > extremely high. The resulting transfer function is always K/(K+1) because : > FS1(s) and FS2(s) cancel each other out and you seem to think you can keep : > increasing K without practical limit.. : : K/(K+1) = 1/(1+1/K) = 1 : K => infinity or big number : if exact time-compensated
Doesn't it bother you that any step input would result in a step response. So how much energy does it take to instantly move an object between two points? How much energy does it take to instantly heat a system? : : That is theoretical the same as an amplifier or P-controller in closed loop : feedback control with high gain.
No, a simple proportional control is not the same as a matter transporter. It will not move an object instantly between two points. A simple proportional system cannot instantantly heat a system. That would take something nuclear. : : > You have been mislead and you are misleading others. What bothers me is : > that one of Tim's students may read this and do direct s domain to time : > domain conversion and think what you are showing us is wonderful. : > How many times must I say that you need to do your simulation in discrete : > time and implement the controller and plant separately. Do you know how : > to : > convert the s domain to the z domain? Doesn't a lingering doubt bother : > you? : > Just try it, pretty please. Your eyes will be opened.
I can see your eyes are still shut. : : What I am doing so far is pure mathematics with interpretations.
Math is just a tool to help us understand physics. It is the physics of energy transfer you need to understand. There is always a limit to the power that can be applied to heat something or make something move so it can't be done instantly.
E.g. I : don't have a closed loop working, I calculate the closed loop (blue line): : http://home.arcor.de/janch/janch/_news/20070417-nachtwey /
What you are doing is science fiction. : : Example for time-comensation n=1 : PD^n controller : PD^n process: : http://home.arcor.de/janch/janch/_news/20070418-nachtwey / : : Anyone who understands the mathematics here will agree that n=3 will work, : too. Peter, if you have doubts let's discuss this example (n=1). This is : basically the same.
No, your transfer function is always K/(K+1) which means there is no response time in you systems and that isn't realistic.
Peter Nachtwey
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If you tune power stations you know by experience that this type of steam generator and turbine can be power-incread 10%/minute. Your employer sold for millions of dollars the control sytem that garantees 10%/minutes. The time delays are minutes, not seconds.
u = power target x = power (PV)
On a first try, u is limited to 1%/minute. If all is going smoothly you go to 2%/minute, and so on.
The operator would never have a change to move the plant with a step signal. There will always be a target speed limiter inside the control system.
If finaly power change of 10%/minute is proved and accepted by the staff of the power station my job was done.
What I was basically trying to do you find in the diagram: http://home.arcor.de/janch/janch/_news/20070419-nachtwey / Picture on bottom
V(theoretical) = high (or whatever is possible) Time-compensated(theoretical) = 100% (or whatever is possible)
In the times I tuned power stations with try-and-error method I could reach 10%/minute 'without destroying' power stations.
That last diagram you must realy understand: x ~ u c can be 1
This gives you 'linearized' output.
V is nonlinear (valves, pumps, etc.). The integral part of PIDs is worsening time behavior of the loop. You must therefore minimize the integral impact. All these thoughts apply also to your system. In your system I've chosen a filter for example, not a speed limiter.
Peter, you have modified my idea to feedback the total process transfer function in a Matlab test. You should try using feedback the total process transfer function. I have the target signal damped by a filter with T=0.25.
If it works you have the advantage that you or anybody else can change from try-and-error methods to calulate the system in less time, having only one parameter to adjust (filter).
To make it clear: Your result you presented was ok for me. I was just trying to calculate a similar result, and if possible do it slightly better, have better influence to control behavior.
What I'm up to is to find also process transfer functions by good approximation methods and calculate the whole stuff. Calculation is better than just trying.
General note: If PT1 can be controlled by PD why not PT3 by PD3?
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