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1

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 ...


2

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 ...


3

However when I look at the closed loop transfer function, I would say that this system is unstable for 𝐺𝐻=βˆ’1. In this case the transfer function becomes infinity so a bounded input will result in a unbounded (=infinity) output. This depends on your definition of stability. $GH = -1$ is called marginally stable because depending on how you look at it, it ...


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There are a few things I can note about your question. As far as I have always learned, the nyquist stability criterion is taken over the openloop transfer function. if you take the closed loop transfer function, you should count the encirclements of 0 instead (if i recall correctly). The formal definition of stability, as expressed by the Lyapunov's ...


3

I think your understanding is correct. The steady state response of an LTI system is the part of the response that is caused by a steady excitation at the input, like a sinusoidal signal, or any other periodic signal. The transient response is caused by changes at the input, like switching on a signal, or changing the parameters of a periodic input signal, ...


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