Finding the region of stability of a system

Question

Suppose we have a closed loop system controlled by some microcontroller $K$

First we take the open loop gain which is $\frac{K}{s(s+6)}$.It has 1 pole at the origin and at $s=6$ and 0 zeros.

So we would except to have 2 regions when $Re(s)<0$ where our system is stable.Something like this:

But according to the diagram we have only 1 region always for $Re(s)<0$ where our system is stable(for $-6<s<0$)

So what gives?We need a 3rd virtual pole to split the 1 big region into 2 regions to get what we have predicted and the 3rd virtual pole will come from the expression of the closed loop gain

Now lets take the closed loop gain equation which is simply $\frac{K}{s(s+6)+K}$

Lets take the polynomial of the denominator $s^{2}+6s+k=0 $.

Lets apply Routh's algorithm and for our system to be stable we need $k>0$

Now lets solve for $k$ which gives $k = -s^{2}-6s$ and differentiate $\frac{dk}{ds} = -2s-6=0\rightarrow s = -3$ so our virtual pole will be at s = -3

Now if there was a critical frequency over which the system would become unstable from Routh's algorithm we would get 2 conditions for k but now we get only so the system is stable for $-3>s>-6$ and $0>s>-3$

Am I correct?

Answer 1

First of all I think this is a question more suited for EE.SE as this is purely a control theory question (not that this website doesn't have the expertise to answer such a question).

Second of all it doesn't have two poles at the origin it just has one (second-order so it can only have two poles overall).

Third of all, you have a closed loop system with unity feedback, a plant and a controller. Now your plant has stable poles and your controller is a constant gain $K$. In this case, you shouldn't really strive to find the range of $K>0$ which makes your system stable since it is already stable for all $K>0$ as follows:

The char eqn is $1+GK=0$.

$$1+\frac{K}{s(s+6)}=0$$ $$s^2 +6s + K = 0$$

Using the Routh Hurwitz test:

\begin{align*} s^2 & \quad 1 \quad K\\ s^1 & \quad 6 \quad 0\\ s^0 & \quad K\\ \end{align*}

As there should be no sign changes meaning every row should be greater than zero. As such

\begin{align*} 1 + K > & \; 0\\ K > & \; -1 \end{align*}

But the last row of the table already has a requirement that $K>0$. This is the case because your plant has no unstable poles. This means that any positive gain value for the controller $K$ will ensure stability in this closed-loop system. And if we look at the root locus we see how the roots traverse a path as we change $K$: