Advanced Control. Off topic poll.

This is why I need to find someone that can verify the calculations in the pdf I posted above. Salesmen lie whenever their mouth is moving. Especially in China. I can provide irrefutable proof but no one understands it.

Do you understand what is in my pdf? ALl the math is there. No steps are skipped.

Note, I didn't actually go so far as to check your math, but when I looked through it, there was something that stuck out to me while wearing my skeptical "all sales guys lie" hat.

The equations at the bottom of the first page for the "desired characteristic equation", where the alphas and betas (i think) are introduced. It says "This characteristic equation will have the best response and smallest closed loop time constant", but you might need a reference to back up that assertion.

The results in the multiple start/stop graph seem counter-intuitive to me, based on a high level understanding of P control (ramp up speed, then slow down as you get closer). Maybe that's the difference between the ideal closed loop case and the simple open loop one? Obviously a bit out of my depth here, but possibly representative of your audience, whether or not that's what you were looking for.

The math in general looked more or less familiar compared to what I had seen in school. HOWEVER, if the audience never had that background, I'm not sure if even the top of the first page would make sense.

:confused: not sure if that helps any?
 
Do you understand what is in my pdf? ALl the math is there. No steps are skipped.

You must take bigger steps than I do. I need to take a few more small steps to follow things through. Page 1 and 2 are correct depending on how you regard the notes below.

Note, you list the equation for the "desired characteristic equation", but it doesn't seem to be a "desired" equation, but rather the form for a desired equation. Page 2, you show you can get the characteristic equation in that form using the appropriate gain Kp as determined by the formula, however alpha and beta are fixed, as it is determined by lambda and ohmega n, of the given system.

On page 2, I agree with your first paragraph. On the second paragraph, you state there are two solutions both are valid, but it looks like you are showing the same solution twice. I think Beta should have the sign changed for the second equation (but it really doesn't change your result, since Beta is part of a complex conjugate). I agree with the rest of the equations on page 2.


I do not have time to verify the following pages, but I might be able to later on in the week.
 
The equations at the bottom of the first page for the "desired characteristic equation", where the alphas and betas (i think) are introduced. It says "This characteristic equation will have the best response and smallest closed loop time constant", but you might need a reference to back up that assertion.
This is an excellent observation.
There is no standard method of determining the best pole locations. Schools teach root locus but they don't say anything about what the best gain should be.

So I used some common sense. The closed loop transfer function has 3 poles. One is real, it could be placed at -λ. The other two are complex at -⍺+/-jβ.
For the fastest decay of error it is best to have the real part of the poles as far away from the j⍵ axis as possible. exp(-4*t) decay faster than exp(-2*t). The problem is that as the real pole is made more negative, the real parts of the complex poles become less negative so the error ( and oscillation ) will decay slower. The fastest response will be to have all the real parts of the pole be at the same distance from the j⍵ axis. So what I did is substitute ⍺ for λ so that all 3 real parts of these poles as as far from the j⍵ axis as possible. Now when I solve for ⍺ the result is 2*ζ*⍵/3 or the error will decay at exp((-2*ζ*⍵/3)*t).
What does this mean? Basically the PLC guys that are trying to tune a hydraulic system with P only control are limited by the hydraulic design. If the hydraulic designers design a system with a low natural frequency and damping factor and they expect the PLC control guy to compensate by tuning, it just isn't possible.

Notice that the time to reduce the error to within 1% of the set point ( 5 time constants ) is a good part of a second no matter how small the move is.


The results in the multiple start/stop graph seem counter-intuitive to me, based on a high level understanding of P control (ramp up speed, then slow down as you get closer).
Normally there would be a smooth motion profile but the Laplace transform for a step in the input ( set point ) is simple. I am assuming the step is small so the control signal to the valve does not saturate.

Maybe that's the difference between the ideal closed loop case and the simple open loop one? Obviously a bit out of my depth here, but possibly representative of your audience, whether or not that's what you were looking for.
If the control output saturates at 100% there isn't any difference between open loop and closed loop.

The math in general looked more or less familiar compared to what I had seen in school. HOWEVER, if the audience never had that background, I'm not sure if even the top of the first page would make sense.

:confused: not sure if that helps any?

You asked a very good question.
 
Note, you list the equation for the "desired characteristic equation", but it doesn't seem to be a "desired" equation, but rather the form for a desired equation.
The equation with the ⍺ and β is the desired characteristic equation and it has 3 poles just like the actual characteristic equation. The values of ⍺ and β can be chosen so the error decays as quickly as possible. There are 3 coefficients for the 3 lower powers of s that must be equated. Then it is possible to solve for the three unknowns, Kp, ⍺ and β in terms of the open loop gain K, natural frequency and damping factor.

Page 2, you show you can get the characteristic equation in that form using the appropriate gain Kp as determined by the formula, however alpha and beta are fixed, as it is determined by lambda and ohmega n, of the given system.
Yes!!! Another good observation. It is impossible to move the complex poles locations independently of the real pole. If you move the real pole to be more negative the complex poles will become less negative and slow the response. If you move the complex more negative the real pole will become less negative and slow the response. I explained above, the only place where all 3 poles are as far away from the j⍵ axis is when the real parts are all at -2*ζ*⍵/3. As a control guy you have no control over this except to make the response slower.

On page 2, I agree with your first paragraph. On the second paragraph, you state there are two solutions both are valid, but it looks like you are showing the same solution twice.
Yes, there are two solutions for β because a complex pole has a + and - imaginary part.

Good questions.
 

Similar Topics

This video shows the things I think about on page 1. I use Laplace transforms a lot. Laplace transforms allow one to express differential...
Replies
2
Views
1,793
Instead of this info getting lost in another thread I will start a new thread. For those that don't know you can search for Advanced Control on...
Replies
8
Views
2,619
sampling data. I will probably use this in a magazine article.
Replies
18
Views
8,286
This topic deserved to be a separate thread with and Advanced Control header. The first video was made from videos a couple of years back. You...
Replies
3
Views
1,491
Hello dear experts! I use WinCC Advanced Runtime V15 as a HMI for the process. I grab values from AB Micro 850 PLC via Modbus TCP. On the HMI...
Replies
2
Views
2,359
Back
Top Bottom