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RC Filter Cutoff Calculator

Find the cutoff frequency of an RC filter.

How it is calculated

The cutoff frequency of a first-order RC filter is fc = 1 / (2π × R × C). R is the resistance in ohms, C the capacitance in farads, and the answer is in hertz. The tool applies no unit prefixes, so a 1 µF capacitor must be typed as 0.000001 (or 1e-6) and a 10 kΩ resistor as 10000.

At fc the output is about 70.7 percent of the input voltage, the −3 dB point. The same formula serves both filter types: in a low-pass filter signals below fc pass and higher ones are reduced, in a high-pass filter the reverse. Both values must be greater than zero. The tool assumes an unloaded, ideal filter, so a source resistance or load connected to it will move the real cutoff.

Worked example

Take a resistor of 1000 ohms and a capacitor of 0.000001 F (1 µF). Their product is R × C = 1000 × 0.000001 = 0.001 s. Then 2π × 0.001 = 0.006283, and the cutoff is 1 / 0.006283 = 159.15 Hz. In a low-pass filter this passes audio below roughly 159 Hz and attenuates what lies above it. Doubling either the resistance or the capacitance would halve the cutoff to about 79.58 Hz.

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Common mistakes

  • Typing the capacitance in microfarads (1 or 100) instead of farads. The tool expects 0.000001 for 1 µF, otherwise the result is off by a factor of a million.
  • Entering the resistance in kΩ as 10 rather than 10000. The tool does not read unit prefixes.
  • Expecting a sharp cliff at the cutoff. A first-order RC filter rolls off gradually, and at fc the signal is only reduced to about 70.7 percent of its voltage.
  • Forgetting the load and source resistance. Real circuits that drive or feed the filter shift the cutoff away from the ideal value.
  • Using the answer in rad/s. The formula gives hertz; the angular value would be 2π times larger.

About this calculator

An RC filter's cutoff frequency — the −3 dB point where the output power drops to half its passband value — is given by f_c = 1 / (2π·R·C), determined entirely by the resistor's resistance R and capacitor's capacitance C in the circuit. In a low-pass configuration, frequencies below f_c pass through largely unattenuated while higher frequencies are increasingly suppressed; in a high-pass configuration, the roles reverse.

Electronics engineers and hobbyists use this formula constantly when designing audio crossovers to split frequencies between speaker drivers, building signal-conditioning circuits to filter noise from sensor readings, and designing timing and coupling circuits. Electronics and electrical engineering students learn it as one of the first practical filter-design formulas, since it directly connects a circuit's component values to its frequency-domain behavior.

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Comments (1)

  • Editorial team

    Only the product of R and C sets the cutoff, so the same frequency can be reached with very different parts. A 10 kΩ resistor with 1 µF and a 1 kΩ resistor with 10 µF both give 15.92 Hz. The smaller resistor draws more current from whatever drives the filter, while the larger one is more easily disturbed by the next stage. What is your filter for, audio, a sensor input or a debounce, and which pair did you settle on?

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