How to get PID Padameters for a "lowpass with integrator" process

Hello Peter,
for better understanding of the falling curve of the impulse-response, I modelled the motor in scicos. And now I understand the behaviour.
Responsible for the shape is the speed independent load torque. My modell is adjusten in a way, that I have the same behaviour as at a real run. If you are interested, I will send you the scicos modell.
Wolfgang
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I am glad you found out why the response looked the way it did. It looked to me like the response was exponential but decaying to an offset other than 0 as if you had another active load. You should also plot the control output to the motor.
I will look at it your Scicos file. I have never used Scicos, just Scilab, but I am willing to give it a try.
BTW, I agree with your comments about the documentation. I am looking into new tools like perhaps Python or Java augmented by a graphing package and math library. I have found the new release of Scilab to be so buggy that I just can't get behind it anymore.
I did buy software to do video captures of my screen and audio but I found out that my test system had been 'modified'. I need to get my system put back together. It did look like this.
ftp://ftp.deltacompsys.com/public/Pictures/Motor%20test%20system.JPG
It is just two DC 200 watt motors running in current mode as yours are. Then I will be able to play too and make some vodcasts.
Peter Nachtwey
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Wolfgang, I am about to do a first on this use group. I am posting a link to a vodcast about how we do auto tuning. Notice that the open loop control signal doesn't need to be a pulse or a step. I can use just about any control signal I want. The real trick is how the data in the graphs is processed.
The video is about 30 MB because I needed to use 1280x720 HD so the numbers on the screen can be seen. This is my first crude unscripted vodcast. Download the video first and then play. Otherwise the parts of the screen that are changing will be blurry when limited by internet speeds..
ftp://ftp.deltacompsys.com/public/videos/AutoTuneMotor.mp4
Notice that I still think in terms of hydraulic systems using terms like extend and retract. It is hard to talk intelligently for 15 second let alone 7 minutes. I will work on making scripts in the future.
Peter Nachtwey
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It has come to my attention that QuickTime is required for view the file.
ftp://ftp.deltacompsys.com/public/videos/AutoTuneMotor.divx Can be viewed using VLC. I am still experimenting with the best compression methods DivX seems to be the best so far.
Peter Nachtwey
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The closed loop transfer function of your approach is: TFCL = lambda^3/(s + lambda)^3 TLTF = lambda^3/[s^3 + 3*lambda*s^2 + 3* lambda^2 * s + lambda^3]
if I ignore the higher order of s I get the CLTF_approx
CLTF_approx = lamda^3/[3*lambda^2*s + lambda^3] = lamda^3/[lambda^3*((3/lambda)*s + 1] = 1/[(3/lambda)*s + 1]
So the time constant is 3/lambda.

Please read exact what you wrote above:
Part 1:"but the response will be very slow as it takes about 9 lambda to get within %1 of a set point in response to a step change" If you write this, then the time to reach setpoint is growing with lambda! t_within1% = 9* lambda !!!!
Part 2:"The larger the lambda the faster the response as the poles move farther to the left on the negative real axis"
Thats correct, but Part 1 is an antilogy to Part 1

Thank you for advice
So - now I have calculated the approach for my integrating process and for the low-pass-process.
At first : Integrating process with f(s) = k/{s*(1+s*T)]
I have scanned the approach:
http://www.loaditup.de/334115.html
With the right mouse button, you can save it.
To the result (at the bottom): 1)The system has a unique solution 2) If I want a control without D-Part, then lambda must be 1/(3*T) That means, with a PI controller, I cant vary lambda?!?!
At 2nd to the low-Pass-Process with f(s) ? k/(1+s*T) reachable at
http://www.loaditup.de/334114.html
To the result: 1) I get only 2 equations for the 3 unknowns (Kp, Ki, Kd) 2) I set K to any value and calculate with the 2 equations Ki and Kd
How do you think about it?
Best regards
Wolfgang
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Yes, because the CLTF is a third order system! I just take the desired settling time and divide by 9 to get the close loop transfer function time constant, tc. If lambda is then 1/tc.

Your solution is good.

Yes, IF you want all three pole at the same place.

Yes, but now your desired closed loop characteristic equation is (s +lambda)^2*(s+mu) As lambda gets bigger, mu will get smaller. All three poles will converge at 1/(3*T). When the three poles aren't at 1/(3*T) the system is over damped.
Now you have new home work. Find the PI gains that result in the closed loop characteristic equation being (s+lambda)^2*(s+mu) AND (s+lambda)*(s+mu)^2. You still have three equations, one for each of the powers of s except s^3. Now you solve for Ki, Ki, mu as a function of K, tau and lambda. In the case of (s +lambda)^2*(s+mu) you are only placing two poles and the third pole is calculated from the plant data and the lambda poles. In the second case, (s+lambda)*(s+mu)^2, you are only placing one pole
Ackermaan's equation assume all the poles are going to be placed. You can see that there is more flexibility solving the problem symbolically.

Solving for Ki and Kd doesn't make much sense, it might work but I think the obvious thing to do is to solve for Ki and Kp. Try that instead. You are getting the hang of it.
Peter Nachtwey
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