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    Drawing a root locus and finding the dominant roots

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    Please see my attached questions. I actually only need no. 24 and 25.

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    We can write this function as the ratio of two polynomials in , a numerator polynomial and a denominator polynomial , thus

    On examination we can equate

    In a negative feedback system the closed loop transfer function is given in general by

    As for the feedback function we have (for a unity feedback system) and thus we get a closed loop transfer function as given by {1} below


    Putting in values for the specific case and we obtain {2}


    The denominator of which we define as the "characteristic equation", call it , for . Thus we can write as {3}


    - Finding the "finite" Poles and zeros of the open loop transfer function

    Hence looking at these polynomials separately when either the numerator or denominator equate to zero to find the finite (when ) poles and zeros of the open loop transfer function.

    - From the numerator:

    has one finite zero at obtained by equating the numerator to zero, thus

    We thus denote the number of zeros by a parameter

    - From the denominator:

    has roots (finite poles) when (duplicate pole) and when

    We thus denote the number of zeros by a parameter

    Using Evans rules we have

    Asymptotes to the root loci and thus poles that go to infinity as

    Therefore for the particular case we have


    - The intersection of the asymptotes on the real axis in the plane can further be deduced from another of Evans rules, namely

    For the particular case the summation of poles is

    The summation of zeros is

    Thus we get an intersect point for asymptotes of

    - The angles that the asymptotes make with one another are determined from


    When we get

    When we get

    - We get the intersection of the root locus plot with the real axis in the plane when

    Using the quotient rule for differentiation and the notation ,

    Thus we can determine the intersect of the root locus with the real axis in the plane when we have the condition below

    Looking at out numerator and denominator polynomials again we have



    Thus on substitution we solve for


    (double root)

    We thus get an asymptotic intersection at

    - Root Loci plots

    The root loci plots are shown on the plot below. We have one root loci on the real axis to the left of an odd number of poles and zeros between the two poles at , this has zero extension as one cannot draw an extended line between 0 and 0 of zero length. We have one root locus on the real axis extending between and As defined by the rules the root locus must extend away from an odd ...

    Solution Summary

    The expert draws a root locus and finding the dominant roots/