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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?

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    $\begingroup$ Have you considered looking at its Bode plot? $\endgroup$
    – SPARSE
    Jun 2, 2022 at 6:56
  • $\begingroup$ Note: The CLTF expression you wrote is incorrect (assuming unity feedback). $\endgroup$
    – SPARSE
    Jun 2, 2022 at 6:59
  • $\begingroup$ @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 :) $\endgroup$ Jun 2, 2022 at 7:10
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    $\begingroup$ Have you tried to look at the root locus? $\endgroup$
    – KBS
    Jun 2, 2022 at 7:47
  • $\begingroup$ The Routh-Hurwitz criterion is clear about it. $\endgroup$
    – Cesareo
    Jun 2, 2022 at 8:45

1 Answer 1

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As KBS suggested, we can look at the system's root locus to gain a deeper insight into what's happening.

System's root locus

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.

Our only hope to stabilise this system with simple poles and zeros is to add a pole in between that right side pole-zero pair, so as to create a breakaway point from the real axis, and a suitable zero in the left side that will allow us to move the closed-loop poles to the left side for some gain value.

Compensated root locus

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.

Root locus, 2nd-order compensator

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