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RC and RL use first-order −3 dB models. LC reports ideal resonance; its −3 dB point depends on loading and damping.
fRC = 1/(2πRC); fRL = R/(2πL); fLC = 1/(2π√(LC))
Calculate RC/RL low-pass and high-pass cutoff, reverse-solve components and plot response. Distinguish LC resonance from loaded −3 dB cutoff.
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Use fRC = 1/(2πRC); fRL = R/(2πL); fLC = 1/(2π√LC). Calculate RC/RL low-pass and high-pass cutoff, reverse-solve components and plot response. Distinguish LC resonance from loaded −3 dB cutoff.
Sources: wiki.analog.com [1]
fRC ≈ 159.154943 Hz. This is a calculated example under the model stated above; it is not a measured component specification.
Hz, Ω, F, H. Convert prefixes before substituting values; keep the input and output units attached to every number.
Sources: wiki.analog.com [1]
For the ideal first-order response, amplitude is 1/√2 ≈ 0.707 of the passband value. With equal impedances, this corresponds to half power.
Sources: wiki.analog.com [1]
No. Loading, topology and damping determine the transfer function. 1/(2π√LC) alone gives ideal resonance, not a universal −3 dB result.
Sources: wiki.analog.com [1]
Yes. They alter the effective network and gain. The calculator’s simple RC/RL model is not a substitute for a loaded transfer-function analysis.
Sources: wiki.analog.com [1]
Answers use the stated model and units. Examples are calculated values, not measurements or component ratings.
