Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Difference between Lyapunov and linear stability criteria

  1. Feb 23, 2016 #1
    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!
     
  2. jcsd
  3. Feb 28, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Difference between Lyapunov and linear stability criteria
Loading...