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Is a passive LTI system will always be a stable(bibo and lyapunov) system? Or in other words Passivity of a LTI system implies stability?

If above statements are true then there must be some mathematical relation (description) should be there between passivity and stability, so anyone can help me about its mathematical description (if it exists) And if passivity doesn't implies stability then there must be some counter example to it to prove my assumption wrong.

So whether my assumption is wrong or there exists a mathematical relation between passivity and stability?

I asked this question here

https://electronics.stackexchange.com/questions/513052/stability-of-passive-networks

but I didn't find any useful answer so I posted it here

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2 Answers 2

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A counter-example would be an ideal LC-circuit, which is passive but not BIBO stable: if excited at its resonance frequency, the output grows without bounds. Such systems are sometimes called marginally stable, because they have poles on the stability boundary (the imaginary axis).

Note that if a system is not lossless, i.e., if there is a positive resistance such as in an RLC-circuit, the system is strictly passive, and as a consequence it is also BIBO-stable. I.e., strict passivity implies BIBO-stability (all transfer function poles are in the left half-plane).

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  • $\begingroup$ Thanks for response! So by this counter-example I understand that passive LTI systems May be bibo unstable but what about lyapunov stability ? Is there any other counter example by which we can say that passivity doesn't implies lyapunov stability also? $\endgroup$
    – user215805
    Jul 30, 2020 at 19:27
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    $\begingroup$ @user215805: I'm not an expert on Lyapunov stability (which is usually used for non-linear systems), but I think that all passive systems are Lyapunov stable, which also seems to be confirmed here. $\endgroup$
    – Matt L.
    Jul 30, 2020 at 19:32
  • $\begingroup$ what is the output of the passive LC circuit? say they're in series and the input is the applied voltage. is the output the shared current? yes, at the resonant frequency, the current would be infinite, but so also would a wire (or short circuit) result in infinite current. $\endgroup$
    – robert bristow-johnson
    Jul 30, 2020 at 19:53
  • $\begingroup$ @robertbristow-johnson: Yes, that would be one possible scenario. And a wire is also a lossless passive element, which is not BIBO stable if the transfer function is defined as relating "input" voltage to "output" current. $\endgroup$
    – Matt L.
    Jul 30, 2020 at 19:56
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If you add "dissipative" to your passivity requirement, then yes to both. A perfectly lossless LC system (or any other perfectly lossless resonant oscillator) is neither BIBO stable, nor is it Lyapunov stable.

(I had to look it up, but Lyapunov stability basically says that if the system is stable around an equilibrium point, then over time it'll settle out toward that equilibrium. A lossless resonator wouldn't do that).

If you use the definitions from nonlinear control, the system needs to be neither linear nor time-invariant. "Passive" in the Lyapunov sense means that the system always dissipates energy, no matter what -- eventually that energy runs out, at which point the system is at its equilibrium point.

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