Suppose we have a closed-loop system $H(s)=\frac{A(s)}{1+A(s)f(s)}=\frac{A(s)}{1+T(s)}$. I've seen the stability of the system stated a couple ways:

  1. If $H(s)$ has any poles in the RHP, then it is unstable since the system response contains a term of the form $e^{s_{RHP}t}$, where $S_{RHP}$ has a positive real part and so will grow without bound over time.
  2. If $1+T(s)=0$ for any $s$ (or equivalently $T(s)=-1$ for some $s$) then the system is unstable since there exists some $s$ such that the transfer function blows up.

What confuses me is how these definitions are supposed to be equivalent. In particular, the second definition seems to say that if $H(s)$ has any poles, then the system is unstable. Why is this the case? Take for example a simple low-pass filter, $H(s)=\frac{1}{1+RCs}$, which has a pole at $s=-1/RC$. This pole is in the LHP and so by (1) is stable, however (2) seems to consider it unstable.

What is the relationship between (1) and (2)? In what situations, if any, are they equivalent/does one imply the other?

  • 1
    $\begingroup$ 2. is not right. Solving for1+T(s) = 0 will give you the poles. The system is unstable only if these poles lie on the RHP, as 1 says. What is the source for 2? $\endgroup$
    – orchi_d
    May 9 '21 at 9:42

Only the first criterion is correct when referring to input-output stability. The second criterion is just a way to compute the poles, as mentioned in a comment. All realizable continuous-time systems using lumped elements have poles, which doesn't contradict the stability requirement.

Poles in the right half-plane (RHP) make a causal system unstable. Note that if we let go of the causality requirement, RHP poles don't cause instability but they make the system non-causal.

Note that poles on the imaginary axis cause transients that don't die out, which is undesirable in almost all practical applications. So in practice you want all poles to lie on the open left half-plane, i.e., $\textrm{Re}\{s_{\infty,k}\}<0$, where $s_{\infty,k}$ is a pole of the system.

  • $\begingroup$ To pick nits to a finer degree: poles in the RHP make a causal system unstable -- but given a a system that is purely reverse-causal (i.e., all responses happen before the input), any LHP poles would make the system unstable. So if you're going to start playing with non-causal system theory, you need to parse out your statements with care. $\endgroup$
    – TimWescott
    May 9 '21 at 18:06
  • $\begingroup$ Thanks for the information. To clarify something, is the reason that (for causal systems) RHP poles cause instability while LHP ones don't is because, while both can cause the transfer function to go to infinity for certain inputs, the LHP response will continue even if the input goes away, while the RHP response will continue growing indefinitely? $\endgroup$
    – knzy
    May 9 '21 at 18:46
  • $\begingroup$ @knzy: As you wrote in your question, RHP poles correspond to modes of the system that grow exponentially with time, whereas LHP correspond to modes that decay over time. $\endgroup$
    – Matt L.
    May 9 '21 at 18:48
  • 1
    $\begingroup$ @knzy: Yes, we're talking about BIBO stability here. The relation between causality and stability is actually part of Laplace transform theory, especially the region of convergence (ROC) of the bilateral Laplace transform. In practical electronics you would of course only consider causal systems. Still it's good to know that in some cases it can be useful to talk about ideal, non-causal systems, and these systems can have poles in the RHP without becoming unstable. $\endgroup$
    – Matt L.
    May 10 '21 at 15:54
  • 1
    $\begingroup$ @TimWescott: I think it's good to point out once in a while that the statement "RHP poles cause instability" is true only for causal systems. Of course, causal systems are the ones we're usually interested in, but we regularly come across transfer functions of non-causal systems, e.g., in the classical design of analog filters, where the magnitude squared functions have poles in both half-planes, yet they're stable. $\endgroup$
    – Matt L.
    May 10 '21 at 15:58

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