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This answer of mine shows that not every \$RLC\$ circuit reaches sinusoidal steady state at input frequency.

A well-known theorem states that:

If all natural frequencies of a uniquely solvable LTI circuit have negative real parts, then the circuit reaches a unique sinusoidal steady state at input frequency.

Is there any way to strengthen the condition of this theorem to be able to immediately, without computing the natural frequencies for each given circuit, determine a general class of \$RLC\$ circuits that reach sinusoidal steady state and thus can be analyzed using phasors?

I'm want a proof or a reference to a proof if one provides an answer.

I suspect the condition could have something to do with having at least one resistor and not having some resonance-prone \$LC\$ connections. But that's just a guess. I don't know how to show this or even how to properly formulate the correct condition.

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Is there any way to strengthen the condition of this theorem to be able to immediately, without computing the natural frequencies for each given circuit, determine a general class of RLC circuits that reach sinusoidal steady state

I would compute the Q-factor of the circuit. If the Q-factor is not \$\infty\$ (infinity) then it has a finite value and therefore will reach a steady state. Q-factor depends on the values of R, L and C and we don't need to compute anything else. When the Q-factor is \$\infty\$ its poles have no negative real value.

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  • \$\begingroup\$ I want to determine if the circuit reaches sinusoidal steady state at input frequency. Finite Q-factor could only indicate BIBO stability \$\endgroup\$
    – Sgg8
    Jan 15 at 15:11
  • \$\begingroup\$ As far as I know that condition does indicate AC sinusoidal stability. \$\endgroup\$
    – Andy aka
    Jan 15 at 15:15
  • \$\begingroup\$ Wow, interesting. Can you provide a proof/reference to a proof? \$\endgroup\$
    – Sgg8
    Jan 15 at 15:16
  • \$\begingroup\$ I can't provide a proof (and, I note that your question didn't ask for one). \$\endgroup\$
    – Andy aka
    Jan 15 at 15:18
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    \$\begingroup\$ @Sgg8 I parsed the question and the only text that was a question was what I embedded in my answer. Having said that I noticed what you said about what you'd like to have. In the process of thinking about this I focussed on Q-factor and realized that an infinite Q factor is the only implicit condition that doesn't allow the sinewave ever to settle down. Hence Q-factor requires no further proof other than you reading up on the strict definition realting energy input and output. \$\endgroup\$
    – Andy aka
    Jan 15 at 19:49

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