# Linear system stability criteria

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?

• 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? Commented May 9, 2021 at 9:42

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.