Electronic – Why, in a passive circuit with a sinusoidal input, do all voltages and currents have the same sinusoidal behavior as the input

circuit analysispassive-componentspassive-filter

I am familiar that in any circuit composed of linear passive elements and a sinusoidal input, all voltages and currents through and across any element will exhibit the same sinusoidal behavior and frequency as the input; that's how passive filters work in fact. But I can't figure out or find a concrete/straightforward proof for why this happens, if not plain observation.

Best Answer

I've been pouring my brains out and eventually I've found a nice mathematical approach to prove this and decided to answer my own question. In such a circuit, solving for any voltage/current across/through any component (I'll call that \$f\$) would always lead you to construct a differential equation that is always linear, with constant coefficients (due to linear properties of passive components) and non-homogeneous (due to the sinusoidal input). Such a differential equation will always take this form: $$a\frac{d^nf}{dt^n}+b\frac{d^{n-1}f}{dt^{n-1}}+...+j\frac{df}{dt}+kf=C\sin{(\omega t+\theta)}$$ where \$a...k\$ are constants (combinations of inductance, resistance, etc.), \$n\$ is the order of the differential equation (which reflects the number of energy storage elements in the circuit), and \$C\sin{(\omega t+\theta)}\$ is a generalized sinusoidal function that describes the input. A general solution to this differential equation will always take this form: $$f=\text{(general homogeneous solution)}+\text{(particular solution)}$$ where the particular solution \$=A\sin{(\omega t+\theta)}+B\cos{(\omega t+\theta)}\$ which is a sinusoidal function of the same frequency! Now, in AC circuit analysis, we are always looking at the circuit in steady state, when the homogeneous solution approaches zero (which inevitably happens because of resistances in the circuit).