I am trying to understand the whole topic of stability of a linear system. Some things are clear but others are blurry, I'd like to clarify it. I looked all over the internet but couldn't find an answer.
Things I'm sure about:
- If all the poles have negative real parts the system is stable.
- If one pole or more has a positive real part the system is unstable.
- If there are no poles with positive real part and there are simple poles with no real part, the system is partially stable.
The things I don't understand:
- What happens if there are no poles with positive real part and I have simple poles on the imaginary axis with multiplicity? I have seen an example that shows that a system can be stable in this case, how can I know?
This is the example function that I have seen:
$10\frac{1+\frac{s}{2}}{s^2(1+\frac{s}{16})}$
If you test stability with the Nyquist plot you get that it is stable.
- Following that, I don't really know what kind of stability I test with the Nyquist plot. As I understand it right now (as I wrote above) the stability I talked about is asymptotic. But here, because part of the poles have non-negative real parts it doesn't count as asymptotically stable. So what is the stability we are talking about in the Nyquist plot.
Thank you very much for reading. I hope you can help me here. By the way I have an examination in 2 days so I hope someone can help me soon (I know it's rude, but sorry it's the truth).
I'm assuming you are only familiar with transfer function model and not state space model. So I will try to explain the concepts only using transfer function.
The things you are sure about are true for the impulse response of the system (or zero input with nonzero initial conditions, but you don't see it in transfer function). By the way as a general rule of thumb I suggest you always ask the question "for which input?" when thinking about the concepts in linear control theory.
Disclaimer. All the explanations below are given for LTI systems in transfer function model.
Asymptotic stability: It is the case when the impulse response of the system goes to $0$. It happens when all the poles have negative real parts.
Exponential stability: It is the case when the impulse response can be bounded by an exponentially decaying function. In general ES implies AS, but in LTI case they are equivalent.
Marginal stability: It is the case when the impulse response does not go to $0$, but also does not blow up either. It happens when there are simple poles on imaginary axis (and not any poles at the RHP of course).
Lyapunov stability: It is the case when the impulse response is bounded, i.e. does not blow up. (Actually it is a very rich theory with too much applications). It happens when the system is marginally or asymptotically stable.
All these definitions are done for impulse response of a system. But another useful definition is BIBO (Bounded Input Bounded Output) stability, given for bounded inputs.
BIBO stability: It is the case when the output stays bounded for a bounded input. It happens when all the poles are at LHP.
By looking at the Nyquist plot you check the stability of the closed loop system by using the open loop system model under the assumption of unit feedback. So it is entirely possible that an open loop system is unstable but it becomes stable with feedback. It also gives the gain and phase margin of the feedback for which the closed loop system stays stable.
Edit. The idea behind Nyquist plot is the fact that closed loop system (under unit feedback) poles are equal to the complex numbers such that $1+G(s)=0$, which are different than open loop system poles. So, it must be $G(s)=-1$ (hence the encirculation conditions of $-1$). But this is still hard to calculate, so the idea is checking only the critical points, i.e. when $s=j\omega$ and deduce the number of stable/unstable/critical closed loop system poles without ever calculating the actual values and using only the open loop transfer function.