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Given a sinusoid input of frequency f, an RLC network, no matter how complex, should produce a sinusoidal output of frequency f, with possible attenuation and phase shift. At least that's my understanding of AC circuit theory, and confirmed here. It follows from the fact that, at a given frequency, any R, L, or C can be viewed as a complex impedance following the complex version of Ohm's law, E=IZ.

Yet, even these simple LC circuits change frequency, or produce beat frequencies:

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Even if we assume that the components have parasitic resistance, capacitance, or inductance, they still should form a passive linear RLC network.

How is it possible to explain the behavior shown? And, more importantly, if I can't use the classic AC circuit theory of complex impedances to analyze them, how do I analyze them? And how do I know that the classic AC circuit analysis will not apply here?

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  • \$\begingroup\$ An important consideration: Do you see this effect in reality, or only in simulation? \$\endgroup\$
    – Hearth
    Commented Jun 28 at 4:03
  • \$\begingroup\$ @Hearth I've only simulated them. In any event, I'd expect theory to be a good predictor of simulation. \$\endgroup\$
    – SRobertJames
    Commented Jun 28 at 4:08
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    \$\begingroup\$ I'm not saying it's definitely what's going on here, but simulation bugs are very possible. There may be a first-time-step transient that's exciting the resonant frequency. \$\endgroup\$
    – Hearth
    Commented Jun 28 at 4:16
  • \$\begingroup\$ Theoretical circuits are "sometimes" "buggy". Logically, in these cases, the simulator says "matrix singular" ... but it can "solve" when it adds a "big" resistor somewhere needed in the circuit. \$\endgroup\$
    – Antonio51
    Commented Jun 28 at 6:58
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    \$\begingroup\$ Note also that when you apply a sinusoid (in transient analysis) ... it is a "step sinusoid" that is applied, so the resonant frequency of the system is excited ... \$\endgroup\$
    – Antonio51
    Commented Jun 28 at 11:46

2 Answers 2

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There is no loss in the system, therefore the startup transient continues oscillating forever. (Due to numerical errors in the transient simulator, it won't actually go forever.) The measurement is the simple superposition of the homogeneous solution (the LC self-resonance) plus the source voltage (inhomogeneous solution). Try adding some ESR to the components and that'll dampen out.

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  • \$\begingroup\$ Interestingly enough, although adding 100 Ohm ESR did dampen things, they're still very different from the simple sinusoidal homogenous solution, even after running for quite some time... \$\endgroup\$
    – SRobertJames
    Commented Jun 28 at 19:53
  • \$\begingroup\$ To the first case, 16mH + 1.6nF, sqrt(L/C) = Zo = 3.16k and 100R gives a Q of 31; it'll take "30"s of cycles for the homo. to decay, say 3ms. At 3kR it'll be almost instant. \$\endgroup\$
    – Tim Williams
    Commented Jun 28 at 23:43
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How can these passive RLC circuits change a sinusoid's frequency?

There is no change. Only the "true" response to a "step sinusoid" of an infinite Q of a circuit.

And how do I know that the classic AC circuit analysis will not apply here?

The AC analysis does apply. Confirmed by an AC Analysis.

Here is a Maple sheet that shows your results.
Note that the voltage input is a "step sinusoid".
If you change the frequency input, the wave can also change as shown.

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