Exponential stability of a general \$RLC\$ circuit with sinusoidal input

Question

I conjecture that any \$RLC\$ circuit with \$n \ge 0\$ independent sinusoidal voltage sources of the same frequency is exponentially stable.

How to prove this or if this is not true, how to modify the condition and prove the corresponding result?

I find the general characteristic polynomial's roots not so easy to analyze to show that all of them have negative real parts.
Having an answer to this question would surely help, but I'd like to see any other approach, desirably as many other rigorous approaches as possible.

I tried to form a general system of ODEs using voltage node method as follows:

Say we have a reference node \$0\$ and nodes \$v_1, ... v_n\$. Consider an arbitrary node \$v_k\$. The equation for it will go into \$k\$-th row of the system matrix.

Define the following functions used to represent the currents:

$$f_R(R, v) = \frac{v}{R}$$ $$f_c(C, v) = C\frac{dv}{dt}$$ $$f_L(L, v) = \frac{1}{L}\int v dt$$ $$f_0(0, v) = 0$$

Then, using KCL, the \$k\$-th row will be:

$$-f_{\text{component_type}_1}(\text{component_type}_1, v1), \ ... \sum_{\text{component_type_i}} f(\text{component_type_i}, v_k), \ ...-f_{\text{component_type}_n}(\text{component_type}_n, vn), \ \sum_{i = 0}^m x_i$$,

where component types \$\in \{R, L, C, 0\}\$ depending on which component is directly connected between nodes \$v_i\$ and \$v_k\$ and \$x_i\$ are the sinusoidal sources connected directly to \$v_k\$.

The main difficulty in analyzing such a system is its generality. In order to eliminate the integral in \$f_L\$ functions I would have to differentiate the equation, but since I don't know whether the inductor is present or not, I have to differentiate all the equations, which further complicates the analysis.

Answer 1

Do you even need arcane math for that? Well perhaps you do if that's required in the homework assignment... Let me give a response based on elementary school physics and math :-)

A pure LC circuit, that has no losses, given an initial dose of energy, will resonate forever, at a constant amplitude. The energy just keeps circulating between L and C. Just an ideal sinewave of voltage and current (90 degrees phase shifted), never ending.

An RLC circuit, with all the components finite and non-zero, if primed with some energy to start with, will resonate at its nominal frequency (based on L and C) and will keep turning a part of the energy into heat, therefore the amplitude of its oscillation will decline along an exponential curve - converging to zero at time=infinity. It may be academically interesting to note the curve of energy loss within a single cycle of resonance, also depending on if the R is connected parallel or in series, but that's already wordplay.

Now if you add an energy (power) source, the conclusions should likely be: the RLC will not converge to zero, instead it will resonate forever at a constant amplitude, likely based on P = R * I^2 = U^2 / R. The resistive energy loss is equal to the energy supplied. The interesting question to me is, how the energy is coupled into the circuit = apart from source voltage or current, you also have to consider source impedance. Whose real=resistive component combines with your bare RLC circuit's "intrinsic" R. And, the style of the coupling circuit = using a coupling R, L, C, having a split L or C in the resonant cell, or a coupling winding on the L, or how. And if you're asking about multiple sources... the thing quickly gets "interesting" (ugly), you get a network of elementary R/L/C lumped components that you have to solve (Kirchhoff on phasors, or Laplace in the time domain). S-parameter analysis / matrix. Overall, the sum of energy that you supply needs to be dissipated by resistors, which is where an equilibrium of amplitude is achieved. Or rather, amplitudes, if you're speaking about sources of different frequencies, i.e. a spectrum.

Judging by the comments from @periblepsis, I guess I have completely misunderstood your question :-)

Answer 2

My statement is incorrect. Consider the following \$LC\$ circuit under sinusoidal input:

KVL states:

$$V(t) = V_c(t) + V_l(t)$$ $$V(t) = \frac{q}{C} + L \frac{di}{dt} \implies V'(t) = A\sin(\omega t) = Li''+\frac{i}{c}$$

The general solution to ODE is:

$$i(t) = i_h(t)+i_p(t)$$

\$i_p(t)\$ is of the form \$C_1 \sin(\omega t) + C_2 \cos(\omega t)\$

and \$i_h(t)\$ is of the form \$C_3 \sin(\omega_x t) + C_4 \cos(\omega_x t)\$,

where \$\omega_x ^2 = \frac{1}{LC}\$.

Thus,

$$i(t) = C_1 \sin(\omega t) + C_2 \cos(\omega t) + C_3 \sin(\omega_x t) + C_4 \cos(\omega_x t)$$

which is not the sinusoid of the same frequency as input.

The requirement about having a resistor is still not sufficient since we can have a resistor in parallel like that:

which wouldn't change anything.

In fact, this circuit is not even BIBO stable! That is because we can choose the circuit paramenters to cause resonance.