# The center and radius of root locus

With this opened-loop transfer function,

\begin{align*} L(s) & = \dfrac{K(s+2)}{s(s+1)} \end{align*}

I would like to plot the root locus of this system. Based on the textbook, I got the following:

\begin{align*} & 1 + L(s) = s^2 + (K+1)s + 2K = 0 \\\ & n = 2, m = 1, p_1 = 0, p_2 = -1, z_1 = -2 \end{align*}

\begin{align*} & θ = \dfrac{Nπ}{n - m} = \dfrac{Nπ}{2-1} = Nπ, \quad N = \pm1, \pm3,.. \end{align*}

Therefore, I learned that the locus is indeed a circle. Besides, the system with only two poles and only one zero has a center at its zero. In this case, the center is,

$$O(-2, 0)$$

However, I also learned from a textbook, it said that the center of gravity can be calculated via:

$$\sigma_c = \dfrac{1}{n-m}(\sum_{i=1}^np_i - \sum_{j=1}^mz_j) = \dfrac{1}{2-1}(0+(-1)-(-2)) = 1$$

This contradicts what I know, and I believe I must have mixed something different together, probably the branches. What is wrong with my idea?

• What you call 'center of gravity' is the abscissa of the intersection of the asymptotes of the root locus and the real axis. In this case the one asymptote is the real axis and the calculation doesn't make sense. reference.wolfram.com/language/ref/RootLocusPlot.html#249082789 Jul 28, 2023 at 20:25
• @SubaThomas Do you mean this is kind of a special case in which the calculation does not provide any useful meaning? Jul 29, 2023 at 2:14
• Yes. That is the case here. Jul 30, 2023 at 2:18