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Suppose a linear passive circuit is hooked up to an AC generator which outputs a sinusoidal voltage of angular frequency $\omega$.

How can we demonstrate rigorously that after a sufficient amount of time (after the transients have died out) the voltage across any of the individual circuit components will be a sinusoidal waveform of angular frequency $\omega$?


I think the idea is to show that the voltage across a circuit component satisfies a linear differential equation (*) with the constant term equal to the generator voltage $V=V_0\cos(\omega t)$. This is the "hard" part.

Then we know that there is a particular solution to (*) of the form $A\cos(\omega t + \phi)$. Next we should show that the homogenous solution either tends to zero or is a sinusoidal itself, and we'll be done. Now this is quite clear in the case of a second order LDE, but in general I don't know (can we even have circuits governed by LDE's of order higher than 2?).

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It's almost true; but not necessarily.
One thing I don't mention below concerning the networks your learning; the frequencies don't change! The amplitudes may change but the frequencies remain constant under the operations of integration/differentiation/propogation/ etc.. Thus you can study them in isolation, separate from one another, in typical cases.
If you do the Laplace transform of the passive dissipative linear network and all of the poles are in the left half plane then the only surviving pole should be the generator.
In other words: you break the network into parts surrounding a component signal of interest. And evaluate the transfer function and load function. If it's passive dissipative then all poles will be in the left half plane or on the imaginary axis.
But: if you stick a perfect resonant L/C element in there it might acquire some "tank" energy that would have to bleed off and I think I can imagine arrangements where that doesn't happen. Alternately if you don't have dissipative elements (including the generator) then the startup energy spectrum will be captured and endure; the poles will be on the imaginary axis.
From the Wikipedia Phasor page: https://en.wikipedia.org/wiki/Phasor
"An important additional feature of the phasor transform is that differentiation and integration of sinusoidal signals (having constant amplitude, period and phase) corresponds to simple algebraic operations on the phasors; the phasor transform thus allows the analysis (calculation) of the AC steady state of RLC circuits by solving simple algebraic equations (albeit with complex coefficients) in the phasor domain instead of solving differential equations (with real coefficients) in the time domain
"
This is why your teacher is bringing up phasor's. To avoid long avoidable differential calculus work. This actually leads into a more fundamental domain of Laplace and Fourier transforms. Before Heavyside introduced these operations (of which phasors are a derivation) pages were written containing innumerable differential equations cross-linked; almost impossible to extract meaning from.
I saw a page of from some of Maxwell's original work one time and it was an awesome collection of interconnected differential equations.
Unfortunately, I don't remember how to access it.

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  • $\begingroup$ I don't understand what you said here because I'm not familiar with the analysis of circuits using Laplace transform. Can you give me a concrete example of a passive linear circuit where the generator is a sinusoid and the steady state is not sinusoidal of the same frequency? $\endgroup$ Jan 13 '17 at 18:43
  • $\begingroup$ A series LC circuit with no resistance connected to a zero impedance signal source. When the signal generator is turned on transients exist. The transient frequency spectrum that matches the tank circuit will be stored and has nowhere to dissipate. $\endgroup$
    – rrogers
    Jan 13 '17 at 19:17
  • $\begingroup$ Alternately an LC tank circuit hooked to a current (infinite impedance) signal source. The Laplace transform captures the response of circuits to startup conditions/transients and allows detailed analysis. $\endgroup$
    – rrogers
    Jan 13 '17 at 19:19
  • $\begingroup$ I see. So essentially my teacher stated the result in the question, so that we could justify using phasor analysis to simplify calculations, etc. If indeed all the voltages and currents are sinusoids, then you can associate a phasor representation and all the calculations are justified. So if I understand correctly you should justify first that the steady state will be sinusoidal, no? $\endgroup$ Jan 13 '17 at 19:33
  • $\begingroup$ Not really, in what we call linear systems all typical excitations can be represented by "phasor analysis". That is to say, phasor's or isolated sinusoids are a good representation. There is a one to one correspondence between time domain analysis and phasor calculations. For some things, phasor analysis is easier. There are restrictions but you probably won't meet them. I have edited the answer little. $\endgroup$
    – rrogers
    Jan 15 '17 at 22:13

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