# Can an unstable plant be stabilized with a lead-lag controller?

I have the following plant transfer function

$$P(s) = 500\cdot\frac{s-5}{s^2 - 625}$$

this function is not stable. I want to know if the system can be stable with a controller of the form

$$C(s) = k\cdot\frac{s+a}{s+b}$$

I can see mathematically how it's not possible, since (non-unity) closed loop transfer function: $$\frac{1}{1 + P(s) \cdot C(s)}$$ is not stable, but I wanted to know if there is some other way of knowing that a lead-lag controller will not stabilize this system, maybe a deeper insight to what a lead-lag can or cannot do?

• Have you considered looking at its Bode plot? Jun 2, 2022 at 6:56
• Note: The CLTF expression you wrote is incorrect (assuming unity feedback). Jun 2, 2022 at 6:59
• @SPARSE I looked at the bode plot, but I did not manage to extract information from it that will help me analyze the effect of a lead-lag on the closed-loop system, I will try to look it up, EDIT: I understand now :) Jun 2, 2022 at 7:10
• Have you tried to look at the root locus?
– KBS
Jun 2, 2022 at 7:47
• The Routh-Hurwitz criterion is clear about it. Jun 2, 2022 at 8:45

That pole-zero pair on the right side of the $$s$$-plane is a problem. There's no amount of poles or zeros that we can add to the left side that will bring all closed-loop poles there for any gain value.
For completeness, a causal second-order system of the form $$K{s+a \over (s-b)(s+c)}$$ can stabilize the plant, but then the compensator itself is open-loop unstable.