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Homework Help: P-Controller Design for Steady State Error

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data

    So this is a bit of a two-part question and I'm unsure which part I'm not doing right (or both!).
    i) Find the closed loop transfer function of the system shown
    ii) Design a proportional controller for the system to give a 10% steady state error

    Any help, hints, suggestions would be greatly appreciated.


    2. Relevant equations
    Go = 4.0
    α = 0.168
    β = 0.0047
    Z1 = 8.9
    G(sen2) = 5.33*10^-5
    The reference level is 40mm (it's a twin water tank system)

    3. The attempt at a solution

    Firstly, the CLTF

    And the P-controller

    Thanks for any help :D
  2. jcsd
  3. Jun 16, 2012 #2

    rude man

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    1. In your CLTF panel. what's with the last two equations? You didn't try to isolate the pole locations... not that there was any reason to ... but your 1st equation is correct. I'm too lazy to work thru the rest of that panel.

    2. As for the kp computation: Assuming your final T(s) is correct, just take
    lim s → 0 of T(s). Using your derivation of T(s) I make that to be
    T(0) = kpG0Gsens/(β + GsensCG0z1)
    which you set to 0.1 & solve for kp.
  4. Jun 16, 2012 #3
    Oh, ok. Thanks guys. Maybe it is correct, or at least fundamentally on track - I'm not so worried about the actual numbers as the method.

    Rude Boy, the final 2 lines of the CLTF are the characteristic equation - just the denominator of the transfer function.
    To isolate the pole locations, do I make set the numerator to equal zero and solve for S?
    Similarly, to isolate zero locations I set the denominator to zero and solve for S?

    Thanks! :)
    Last edited: Jun 16, 2012
  5. Jun 17, 2012 #4

    rude man

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    Well, there was really no need to find the poles (yes, they are found by setting the denominator to zero) since you weren't asked to find the time response, just the steady-state error. And that, for a unity step input, is just T(0).
    (Reason: step input transform is 1/s but the final-value theorem says lim s→ 0 of sT(s) so the s's cancel).

    If you had been asked to find the time response to a step input you would, after finding the poles, have to do a fractional expansion of the entire T(s), includng the numerator, and then done the inverse transform on each of the terms of that expansion.

    Or, if you got lucky, you might have found the entire T(s) in a table of transforms, then there would have been no need to find the poles.
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