# How can a stable system have a RHP pole?

I am working with a control system which has an unstable pole in the process. So I have the following transfer functions for the controller and for the process:

G = 10/((s+10)*(s-1));
K1 = K*(s+1)/s;


where G is the process and K1 is the controller. In the controller I have to choose a gain K such that the system is stable, and using the Routh Criterion I have found that it is stable for $$K>1.125$$, so I have chosen has value 4 because it gives me good performances. Now, I have that if I do the root locus I have:

L1 = K1*G;
figure;
rlocus(L1)


so I have a pole in the RHP. But I have that with an appropriate value of the gain of the controller we have a stable system. But in this case the pole should be in the LHP, not in the RHP.

So, what I don't understand is how it is possible that the system is stable even if it is present a pole in the RHP?

Can soebody help me?

How can a stable system have a RHP pole?

It can't.

Now, I have that if I do the root locus I have: (image left out of quote) so I have a pole in the RHP.

This is where you misunderstand. The word "locus" means "The set of all points whose coordinates satisfy a given equation or condition". The traces in that root locus show all possible locations for the poles -- and there's a lot of possible locations for poles inside the stability region.

But I have that with an appropriate value of the gain of the controller we have a stable system. But in this case the pole should be in the LHP, not in the RHP.

In this case the poles are in the LHP. You would find the gain where all the poles are stable (my intuition tells me you want the real part of your complex poles is in the neighborhood of $$-2.5\mathrm{s^{-1}} < \mathcal{Re}(s) < -2.0\mathrm{s^{-1}}$$), and then your system will be stable.

• Might want to check again your inequality
– Ben
Dec 20 '19 at 20:57
• Thanks. Copy pasta error... Dec 20 '19 at 21:30