# root locus of $t(s)=\frac{K(s+2)}{(s+3)(s^2+2s+2)}$?

I calculated the breakaway points. They come out to be

Root 1: -2.481
Root 2: -0.689
Root 3: 1.17

because of root locus properties only the -2.481 will fall on root locus . But when I saw the solution there were no going towards breakaway points. I just want to know when to calculate the breakaway points.

A breakaway point can only occur between two adjacent real open-loop poles. A break-in point can only occur between two adjacent real open-loop zeroes.

A real solution to $\frac{dK}{ds}=0$ is not a sufficient condition to establish the existence of a breakaway or break-in point.

To find all the points with multiple roots you need to solve the two equations: $$c=0$$ and $$\frac{\partial c}{\partial s}=0$$ where c is the characteristic equation. Then choose those solutions whose $K$ values are in the domain of interest.

The break-away and break-in points are a subset of the above solutions, and are those whose s values are real.

So in your example, the two equations are: $$K (s+2)+(s+3) \left(s^2+2 s+2\right)=0$$ and $$K+s^2+2 s+(s+3) (2 s+2)+2=0$$ and the solutions are:

So you have no break-away or break-in points if $K>0$, and one such point if $K$ can also take the value of $-1.90665$.

• how you calculated $K+s^2+2 s+(s+3) (2 s+2)+2=0$ for K how can you solve it – Boris Oct 19 '15 at 14:39
• I solved the two equations simultaneously for K and s (using Mathematica). – Suba Thomas Oct 19 '15 at 14:46