How this calculator works
The output is taken across a capacitor in parallel with a resistive load, driven through one ideal series inductor by a zero-impedance source. The load damps the LC resonance. Changing it changes Q and the response even though the natural frequency stays fixed. The reported cutoff is the frequency where voltage gain is 1/sqrt(2) of DC gain. Only when Q = 1/sqrt(2), the Butterworth case, does that cutoff equal f0. Component solve modes target f0; they do not silently treat an arbitrary loaded cutoff as resonance. The table samples the resulting response. Coil resistance, capacitor ESR, source impedance and speaker impedance variation are omitted.
Formula
H(s) = 1/[LCs² + (L/R)s + 1]; f0 = 1/(2 pi sqrt(LC)); Q = R sqrt(C/L). For x = f/f0, |H| = 1/sqrt[(1-x²)² + (x/Q)²]. The -3 dB point solves x^4 + (1/Q²-2)x² - 1 = 0.
Worked example
With L = 1 H, C = 1 F and load R = 1 ohm, f0 = 0.159154943 Hz and Q = 1. Gain at f0 is 1 V/V, while the -3 dB cutoff is about 0.20245 Hz. Halving load resistance reduces Q to 0.5 without changing f0.
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Frequently asked questions
Why is 1/(2 pi sqrt(LC)) not always the cutoff?
That expression is the undamped natural frequency. A resistive load sets damping, so the half-power voltage point can move below or above it.
Does this model a bridge-tied class-D output directly?
It models a single series inductor with a shunt capacitor and load. Convert a differential network to its appropriate equivalent before using these component values.
Why can the gain exceed one?
A lightly damped LC network can magnify voltage near resonance through stored energy. This is voltage gain, not a claim of power generation.