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Controlled systems - Closed loops

  1. Mar 1, 2014 #1

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    1. The problem statement, all variables and given/known data[/b]
    http://snag.gy/m9Iq0.jpg

    2. Relevant equations



    3. The attempt at a solution

    I know that the sensistivity of a closed for at small change is given by the formula S = ∂ Ln T/∂ Ln G
    And T(s) = G_c(s)G(s)/1 + G_c(s)G(s)
    But since i don't have any G_c(s), i don't no how to solve it, and they get the result they get..
     
  2. jcsd
  3. Mar 1, 2014 #2

    rude man

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    You did not attach the figure. I'll assume the total forward gain is G Gc and the feedback gain is 1.

    So G Gc = 100/(τs+1).
    You don't need G and Gc separately. Call G Gc = P (open-loop or "loop" gain) and T = P/(1+P) = closed-loop gain.

    You can just compute T(s) from P(s), then get the new time constant τ', then a given dτ results in a corresponding dτ'.

    Express as fractional devialtions: dτ/τ → dτ'/τ'. You will find the proportionality dτ'/τ' / dτ/τ
    is a constant = K. So a 1% change in τ results in a K % change in τ'.

    You will also find that a change in τ does not affect the dc gain of either P (obvious) or T.
     
  4. Mar 1, 2014 #3

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    I am not quite sure..
    About the picture you are correct.. => http://snag.gy/sceeD.jpg
    But i don't understand, what about the (d/ds)G part.. since the sensistivity is a ratio of the T and d derivative.
     
  5. Mar 1, 2014 #4

    rude man

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    You don't have to use the complementary sensitivity formula you cited (S = dT/T / dP/P).
    I have never used it so I'd have to go thru it myself. Unless the instructor specifically wants you to use that formula I would proceed as I suggested. It's very simple. The question is just if τ changes by 1%, how much does the new time constant change.

    As soon as you compute T(s) you will see that the form of T is the same as the form of P = k/(τs+1). Except there is a new k' and a new τ'.
     
  6. Mar 1, 2014 #5

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    I am a bit confused my T(s) looks like this.
    I don't see how this should lead to the answer http://snag.gy/XGASC.jpg
     
  7. Mar 1, 2014 #6

    rude man

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    Carry ou the product terms in the denominator in your last expression. What do you get?

    (I suggest letting k = 100 and using k instead of the numerator until the end).
     
  8. Mar 1, 2014 #7

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    I get K /101 + st..
     
  9. Mar 1, 2014 #8

    rude man

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    \
    \
    Or, K/(K + 1 + sτ), right?

    So divide numerator and denominator by (K+1) & what do you get?

    BTW don't use t for the original time constant τ. t is always time, a variable.
     
  10. Mar 1, 2014 #9

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    Well.. the same? I am kinda confused on where you are heading...
     
    Last edited: Mar 2, 2014
  11. Mar 1, 2014 #10

    rude man

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    Try to make your T(s) look like a/(bs + c).
     
  12. Mar 4, 2014 #11

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  13. Mar 4, 2014 #12

    rude man

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    Disregard your last line.
    Go
    T(s) = k/(k+1+τs) from previous line;

    = k/(k+1) / (1 + τs/(k+1))

    Now you can let k' = k/(k+1)
    and τ' = τ/(k+1).

    What does your new expression for T(s) look like then?
     
  14. Mar 4, 2014 #13

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    T(s) = k'/1+ t'...

    I feel pretty confused right now.. I feel like a dummy, which just does what someone say..
    Could it be possible to explain how come to do, and how it all is connected?

    As far as i've understood i need to calculate, T(s) and then T'(s) ,and G'(s)
    And take S = (T'(S) /T(s))/(G'(s)/G(s)) is that correct?

    This is my solution for a)..
    where is my mistake??

    http://snag.gy/O0Z4e.jpg
     
    Last edited: Mar 4, 2014
  15. Mar 4, 2014 #14

    rude man

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    As I said, I don't have the time to learn about stability coefficient S right now.

    But: can't you see that your T(s) looks just like your G(s) except for the values of gain and time constant?
    So if you change the original time constant in G by say 1 msec., how much does that change the time constant in T? How much does that change the dc gain in T?
     
  16. Mar 4, 2014 #15

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    I just realised the answer is written on the back..

    I says http://snag.gy/tr5eD.jpg

    If I change the time, it's both the same.
     
  17. Mar 4, 2014 #16

    rude man

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    No, if you change τ by Δτ, τ' changes by Δτ/101. You can see that from τ' = τ/(k+1) = τ/101. And that's what your written answer says also.
     
  18. Mar 4, 2014 #17

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    But the answer isn't the same as the one written in my book..

    I just took a screenshot of the part of my book which is about the topic

    http://snag.gy/WDedY.jpg

    Here is the equation, i've been using.
     
  19. Mar 4, 2014 #18

    rude man

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    Your question had two parts.

    I only addressed the part that asked for the effect on T(s) for a change in τ. And the answer is in τ' = τ/(k+1). So dτ' = dτ/(k+1) and that is the answer they gave also. And the change in k' for a change in τ is zero. That defines the changes in T(s) for a change in τ completely.


    I'm not familiar with the S parameter, as I said. I don't have a physical feel for a derivative with respect to s, which is an operator (= d/dt) or complex frequency variable (= σ + jω). This is just because I was never exposed to this parameter, and I have to say I see no point in it, but that's just me. I have no physical feel for taking the derivative of any F(s) with respect to s.

    Anyway, to get S just compute S = ∂T/T / ∂G/G. That's what they want.
     
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