Cutoff Frequency Calculator

The cutoff frequency (fc), also called corner frequency or half-power frequency, is the frequency at which a filter's output power drops to half (−3 dB) of its passband value. Below fc (low-pass) the signal passes; above fc it is increasingly attenuated.

fc = 1 / (2πRC), where R is resistance in ohms and C is capacitance in farads. The result is in hertz. Example: R = 1 kΩ, C = 100 nF → fc ≈ 1.59 kHz.

fc = R / (2πL), where R is resistance in ohms and L is inductance in henrys. Example: R = 100 Ω, L = 10 mH → fc ≈ 1.59 kHz.

Yes. For a first-order RC (or RL) filter the formula is identical — fc = 1/(2πRC). The difference is only which component the output is taken across: across the capacitor gives low-pass, across the resistor gives high-pass.

At fc the voltage ratio is 1/√2 ≈ 0.707, and 20·log₁₀(1/√2) ≈ −3.01 dB. Power is proportional to voltage squared, so the output power is exactly half — the "half-power point".

Exactly ∓45°: −45° for a first-order low-pass (output lags) and +45° for a first-order high-pass (output leads). Far from fc the phase approaches 0° in the passband and ∓90° in the stopband.

For first-order filters, yes — "cutoff frequency", "corner frequency", "−3 dB frequency", and "half-power frequency" all refer to fc. (For higher-order filters, definitions can diverge slightly.)

Pick one component at a convenient standard value, then solve for the other: C = 1/(2πR·fc) or R = 1/(2πC·fc). Use the reverse calculator above, then round to the nearest E12/E24 standard value and re-check fc.

No. For an ideal linear passive filter, fc depends only on the component values R, C (or L). Amplitude affects neither fc nor the shape of the response.

RC filters are cheaper, smaller, and more common for audio and signal filtering. RL filters suit higher-power or RF applications but inductors are bulkier, costlier, and have parasitic resistance/capacitance. Electrically, both are first-order with 20 dB/decade rolloff.

The signal is attenuated at 20 dB per decade (6 dB per octave): at 10×fc the output is roughly −20 dB (1/10 of passband voltage). It never reaches zero — a first-order filter rolls off gradually, it does not brick-wall.

Sweep a sine wave through the filter and find where the output voltage drops to 0.707× the low-frequency (passband) value. That frequency is fc. An oscilloscope or a sound card with sweep software works fine for audio-range filters.

τ = RC (or L/R for RL), and fc = 1/(2πτ). The time constant is the time for a step response to reach ~63.2% of its final value — the time-domain dual of the frequency-domain cutoff.

Yes, but two identical cascaded RC stages interact (loading), shifting fc and giving only ~12 dB/octave with a soft knee, not a true second-order response. For a proper second-order filter use an active stage (e.g. Sallen–Key) or buffer the stages.

A first-order filter changes slope by 20 dB/decade at fc: a low-pass goes from 0 dB/decade to −20 dB/decade; a high-pass goes from +20 dB/decade (rising out of the stopband) to flat 0 dB/decade in the passband.

Hz (fc) is standard for specifications and audio work. Angular frequency ωc = 2πfc (rad/s) is handier in math: the transfer functions use s = jω, so ωc = 1/(RC) or R/L directly. Convert with ωc = 2πfc.