About the RC Filter Calculator
This RC filter calculator finds the cutoff (corner) frequency of a first-order resistor–capacitor filter from R and C, and shows how the filter responds at any test frequency you choose: the gain as a ratio and in decibels, the output voltage for a given input, the phase shift and the capacitor’s reactance. A Bode-style chart plots the gain over four decades around the cutoff.
It covers both arrangements. In a low-pass filter the resistor is in series and the capacitor goes to ground, passing low frequencies and attenuating high ones — used for anti-aliasing before an ADC, smoothing PWM into an analog voltage and removing hiss. In a high-pass filter the capacitor is in series and the resistor goes to ground, blocking DC and low-frequency rumble — used for AC coupling between amplifier stages.
Results assume an ideal source (zero output impedance) and an unloaded output (infinite load impedance). A real load in parallel with the output reduces the gain and shifts the cutoff upward, so buffer the filter or keep the load at least 10× larger than R.
With the default inputs, the cutoff frequency (−3 db) is 1,591.549 Hz. Change any value above to recalculate instantly.
How to use the rc filter calculator
- 1Choose low-pass or high-pass.
- 2Enter the resistor value in kilohms and the capacitor in nanofarads.
- 3Read the −3 dB cutoff frequency and the time constant.
- 4Enter a test frequency and input amplitude to see gain, output voltage and phase.
- 5Use the chart and table to check attenuation at other frequencies.
Formula and method
An RC filter is a voltage divider between a resistor and a capacitor whose reactance XC = 1/(2πfC) changes with frequency. At the cutoff frequency fc = 1/(2πRC) the reactance equals the resistance, the output is 1/√2 ≈ 70.7% of the input (−3.01 dB) and the phase shift is 45°.
For a low-pass filter the output is taken across the capacitor, so the gain falls as frequency rises; for a high-pass filter it is taken across the resistor and the gain rises with frequency. Far from cutoff a first-order filter rolls off at 20 dB per decade. Gain in decibels is 20·log10(Vout/Vin). Source and load impedances are assumed ideal.
- fc
- Cutoff (−3 dB) frequency in Hz
- R
- Resistance in ohms (entered in kΩ)
- C
- Capacitance in farads (entered in nF)
- f
- Signal (test) frequency
- |H|
- Voltage gain Vout/Vin
- φ
- Phase of the output relative to the input
Worked examples
Low-pass 10 kΩ / 10 nF, checked at 10 kHz
fc = 1 ÷ (2π × 10,000 × 10 nF) ≈ 1,591.5 Hz. At 10 kHz, f/fc ≈ 6.28, so the gain is 1/√(1 + 6.28²) ≈ 0.157 (−16.07 dB) with an −81° phase shift.
High-pass 1.6 kΩ / 1 µF to block 50 Hz hum
fc = 1 ÷ (2π × 1,600 × 1 µF) ≈ 99.5 Hz. At 50 Hz, f/fc ≈ 0.503, so only about 45% of the hum gets through (−6.95 dB), with the output leading by 63°.
At the cutoff frequency itself
1 kΩ and 100 nF give fc ≈ 1,591.5 Hz. Testing exactly at fc shows the textbook result: 70.7% output, −3.01 dB and a 45° phase lag.
Frequently asked questions
How do I calculate the cutoff frequency of an RC filter?+
Use fc = 1 / (2πRC) with R in ohms and C in farads. For example 10 kΩ and 10 nF gives 1 / (2π × 10,000 × 0.00000001) ≈ 1,592 Hz. The same formula applies to low-pass and high-pass versions.
Why is the cutoff called the −3 dB point?+
At fc the output voltage is 1/√2 ≈ 0.707 of the input, which is 20·log10(0.707) ≈ −3 dB. That is also the point where output power has halved.
How steep is an RC filter?+
A single RC stage is a first-order filter with a roll-off of 20 dB per decade (6 dB per octave). For steeper filtering, cascade buffered stages or use an active second-order filter such as a Sallen-Key.
How do I choose R and C for a given cutoff?+
Pick a convenient capacitor first, then calculate R = 1 / (2π × fc × C). Keep R between about 1 kΩ and 100 kΩ so it is large compared with the source impedance and small compared with the load.
Does the load affect an RC filter?+
Yes. For a low-pass filter a load resistor RL across the capacitor lowers the passband gain to RL/(R + RL) and raises the cutoff, because the effective resistance becomes R ∥ RL. Keep the load at least 10× R or add an op-amp buffer.