Fall 2012 ME 395 - GSI Josh Lacey

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 Proportional gain and K_cl

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twhalm



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Join date : 2012-09-13

PostSubject: Proportional gain and K_cl    Thu Nov 01, 2012 12:09 pm

In the lecture slides we are told that a proportional controller us simply u=K_cl*(r-w). Is K_cl related to the "P" that we changed as input in LabView?

Also, is "P" of any use when predicting our closed loop time constant?
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GSI Overlord
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PostSubject: Re: Proportional gain and K_cl    Sat Nov 03, 2012 1:35 pm

There is a bit of difference between the notation in your lecture slides (which corresponds to the standards of the controls world) and what you saw on the LabVIEW interface. The P that you were setting in lab is the proportional gain corresponding to Kp in the slides. And yes, you will need that P value, as that is your closed loop gain for the proportional controller (K_cl) that you need in the calculation of the closed loop time constant from the model.
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sinalex



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Join date : 2012-09-12

PostSubject: Re: Proportional gain and K_cl    Sun Nov 04, 2012 10:38 pm

I have a question about the SS error and it has to do with the K_cl.

For one of our P values, we inputted a very small percentage for 0% overshoot. When I was plugging it into the formulas on the lecture slides, I got an error of around 99% or something absurd like that. I evaluated the theoretical SS speed as well and it was way smaller than the RPM that we measured in the lab. From my understanding, the output voltage is determined by K_cl(r - omega), but with the small P that we inputted into the computer for each value of omega, the output voltage would be very small. I'm quite sure what I'm doing wrong.
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GSI Overlord
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PostSubject: Re: Proportional gain and K_cl    Sun Nov 04, 2012 11:40 pm

You will find for very small values of Kp, the proportional gain, there will be a very high steady state error. If you think of the formula, you can imagine as Kp->0, steady state error goes to unity.

You actually aren't asked to evaluate the theoretical steady state speed of the closed-loop controllers by the task letter (just the time constant and the steady state error), so you don't need to worry about that.
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tuesday ninja



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Join date : 2012-09-15

PostSubject: Re: Proportional gain and K_cl    Mon Nov 05, 2012 9:26 pm

GSI Overlord wrote:
You will find for very small values of Kp, the proportional gain, there will be a very high steady state error. If you think of the formula, you can imagine as Kp->0, steady state error goes to unity.

You actually aren't asked to evaluate the theoretical steady state speed of the closed-loop controllers by the task letter (just the time constant and the steady state error), so you don't need to worry about that.


Do you mean we aren't asked to calculate the theoretical steady state error? The equation for %steady state error approaches unity as K_cl approaches zero.

In the sticky "Important lab 6 information" you say, "Also, you will need another plot that has steady-state error plotted against proportional gain and again two curves for modeled and experimental." A bit confused.
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GSI Overlord
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PostSubject: Re: Proportional gain and K_cl    Tue Nov 06, 2012 2:34 am

You ARE asked to evaluate the steady state error by the task letter, but you do not have to use the model to evaluate the steady state speed. The plot you will need to show is the steady state error against the proportional gain for both the theoretical and experimental cases. You can calculate the theoretical steady state error using the formula in the slides ess = r/(1+K*Kcl), and the experimental steady state error will be calculated from your data, of course.
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