# A simple root loci question

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

1. ### blazedaces Thread Starter Active Member

Jul 24, 2008
130
0
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,
-blazed

2. ### Papabravo Expert

Feb 24, 2006
11,542
2,386
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
138
11
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
130
0
I must have phrased the question incorrectly, but I found out the answer in the book. Thank you both for your help.

-blazed