Difference between Lyapunov and linear stability criteria

[Alejandro11]

Dear all,
Consider the connection of two electrical circuits. Both circuits, Z1 and Z2, are stable and only one of them is non-passive. I.e., the eigenvalues are located in the LHP but Re{Z2(jw)}<0 in a frequency range.
For studying the closed-loop stability, you represent the linear system by its ODEs and find the solution of the characteristic equation det(A-λI)=0. In this specific case, you might obtain all the eigenvalues λ of the equations of motion in the LHP and thus thinking the system is stable.
However, when checking the system stability in the sense of Lyapunov, the system might be unstable due to the feedback connection of both circuits:
[ tex ] Z= \frac{Z1}{1+\frac{Z1}{Z2}} [ /tex ]
is non-passive in one region, Re{Z(jw)}<0, and oscillatories instabilities will happen when the system becomes non-passive, Re{Z(jw)}=0 or the Hamiltionian presents pure imaginary eigenvalues.

I want to know why modal analysis fails in the stability determination, and instead of looking at the equations of motion we need to look at the functions of energy to which Hamiltonian systems and passivity criteria are related.

Thank you very much in advance for your replies!

 

[gtoguy97]

Modal analysis fails in the stability determination because it does not take into account the feedback present in the system. In other words, modal analysis only considers the dynamics within each of the two circuits individually, without taking into account the interactions between them. The Lyapunov stability analysis, on the other hand, takes into account the feedback present in the system by considering the full system state, which includes the interactions between the two circuits. This is why passivity criteria and Hamiltonian systems are related to the Lyapunov stability analysis. Passivity criteria are used to ensure the stability of a system, while Hamiltonian systems are used to analyze the behavior of a system in terms of energy.