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Root locus question

  1. May 6, 2015 #1
    1. The problem statement, all variables and given/known data
    I've been trying for a week to understand root locus and how it works but what i got is a big confusion that's why i'm asking here for help. Why do we only look at open loop transfer function of the system when we are supposed to know the stability of the closed loop system? I don't even understand the use of this method at all, why don't we look at the poles of the closed loop system which are at 1 / K * Gd in the example below. I don't want you to solve me any problems but please give me the intuition behind this concept.
    Thank you so much!
     

    Attached Files:

  2. jcsd
  3. May 7, 2015 #2

    Hesch

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    Gold Member

    The short answer is: You don't.

    Say you have a feed forward transfer function, A(s), in the loop, and a feed backword transfer function, B(s). Mason's rule then says, that the transfer function for the closed loop will be:

    H(s) = A(s) / ( 1 + A(s) * B(s) ).

    In your attached file, A(s) = k / ( s3 - 3s2 - 10s ), B(s) = 1.

    Using Mason's rule you will find:

    H(s) = k / ( s3 - 3s2 - 10s + k ).

    The characteristic equation for the closed loop is: s3 - 3s2 - 10s + k = 0.

    By solving this equation with different values of k, you can plot the root locus.

    ( It seems to be a highly unstable loop. )
     
    Last edited: May 7, 2015
  4. May 7, 2015 #3

    NascentOxygen

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    Staff: Mentor

    Are you sure you're not thinking of the Nyquist Plot?
     
  5. May 7, 2015 #4

    LvW

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    Yes - Hesch is right. The denominator roots of the closed-loop transfer function are identical to the poles of this function.
    And the root locus, therefore, gives you the locus of the poles with the gain as parameter.
    Because of stability criteria, these poles (the roots of the denominator) must not move to the right half of the s-plane.
    Hence, you can see for which gain values the poles remain within the left half - indicating stability.
     
  6. May 7, 2015 #5

    Hesch

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    Two of the poles will never be in the left half if gain is positive.

    One of the poles will never be in the left half if gain is negative.
     
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