A simple root loci question

Discussion in 'Homework Help' started by blazedaces, Dec 9, 2008.

  1. blazedaces

    Thread Starter Active Member

    Jul 24, 2008
    So, if I have a unity feedback controller with transfer function

    G_p(s) = \frac{K(s+1)}{s^2(s+9)}

    Is my root loci going to be of G_p(s) OR is it of

    T(s) = \frac{G(s)}{1+KG(s)} = \frac{s+1}{s^3+9s^2+Ks+K}


    That's my question. Now, it seems more likely it's the first one, since why would we be told to graph the root loci of a graph with unobtainable poles...

    And, if that is the case, does that mean to graph the root loci of a non-unity feedback controllers do I need to convert it to a form that resembles a unity feedback controller?

    Thanks a lot guys,
  2. Papabravo


    Feb 24, 2006
    Poles come from the polynomial in the denominator. Zeros come from the poynomial in the numerator. For a third order polynomial there will be at least one real root. The other two roots will be real or will be a complex conjugate pair. The parameter K in the denominator changes the character and location of the poles. The whole purpose of the exercise is to understand how the roots move as a function of K. In particular it is important to find values of K for which the system is unstable.

    Positive real roots disappear or the system does!
  3. vvkannan

    Active Member

    Aug 9, 2008
    Hi blazedaces,
    I think i can answer this question as i studied this last semester.
    Actually each branch of the root locus originates from an open-loop pole corresponding to k=0 and terminates at open loop zero.
    So in the characteristic equation if give you 0 for 'k' you get the open loop poles as the roots of the equation.
  4. blazedaces

    Thread Starter Active Member

    Jul 24, 2008
    I must have phrased the question incorrectly, but I found out the answer in the book. Thank you both for your help.