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RC filter calculator

Cutoff frequency and time constant of a first-order RC filter, with the attenuation and phase at a frequency you choose and the full magnitude response.

Presets
kΩ
nF
kHz
Attenuation and phase are given at this frequency.
Cutoff frequency (−3 dB)
1.592kHz
VinR10 kΩC10 nFVoutCutoff fc1.592 kHz
Low-pass: R in series, C to ground.
Time constant τ = RC
100 µs
A step response reaches 63.2 % of its final value at τ.
Settling time, 5τ
500 µs
Within 0.7 % of final value (1 − e^−5 = 99.3 %).
Gain at the frequency of interest
-16.07 dB
Output / input at that frequency
0.1572
Phase at that frequency
-81 °
Low-pass magnitude response
-60-40-20010100100010,000100,0001,000,000Frequency (Hz)Gain (dB)
  • Low-pass
  • Chosen frequency
Method

How this is calculated

fc = 1 / ( 2π · R · C )
τ = R · C
Low-pass:  |H| = 1 / sqrt( 1 + (f / fc)² )      phase = −atan( f / fc )
High-pass: |H| = (f / fc) / sqrt( 1 + (f / fc)² )   phase = 90° − atan( f / fc )
Gain (dB) = 20 · lg |H|
Step response
The output reaches 1 − e⁻¹ = 63.2 % of a step at t = τ, and 1 − e⁻⁵ = 99.3 % at 5τ.
Roll-off
−3.01 dB at fc, then 20 dB per decade (6 dB per octave) beyond it.
Assumes
A zero-impedance source and an unloaded output. A load resistance in parallel with the output lowers the gain and moves fc; buffer it or include it in R.
FAQ

Questions

1/(2π × 10 kΩ × 10 nF) = 1.592 kHz. The time constant is 100 µs, and at 10 kHz a low-pass passes 0.157 of the input, which is −16.1 dB.

τ = RC is the time for the capacitor to cover 63.2 % of a step change. After 5τ it is within 0.7 % of the final value, the usual figure for "settled".

No. At fc the output is 0.707 of the input (−3 dB) and the phase is −45° for a low-pass, +45° for a high-pass. A first-order filter only falls 20 dB per decade, so a signal 10 times above fc is still at one tenth amplitude.

Use a second-order filter: an LC filter or an active Sallen-Key stage falls 40 dB per decade. Cascading two RC stages without a buffer loads the first one and softens the corner.

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