f_c

RC and RL Filter Cutoff Frequency Calculator

Find the −3 dB cutoff frequency and time constant of a first-order low-pass or high-pass filter.

Resistor RΩ Capacitor CF

How the circuit responds

Magnitude (dB)

Phase (°)

Step response (output after switch-on)

Formulas

RC circuit: f_c = (1)/(2πRC), τ = RC. RL circuit: f_c = (R)/(2πL), τ = (L)/(R). At the cutoff frequency the output is about 70.7% of the input (−3 dB), and a capacitor is more than 99% charged after about 5τ.

Have measured data? Compare it with theory in the lab data analyzer →

Try it: RC charge and discharge

Drag the sliders and the curve follows. The source switches on at t = 0 and drops to 0 V halfway across the chart.

Resistor R10 kΩ Capacitor C10 µF Source voltage V5 V

Hold the time axis, to compare R and C on one scale

Capacitor voltage after the switch closes

Time constant τ = R·C: 100 ms
5τ (99.3% charged): 500 ms
V_C at t = τ: 3.16 V
Current at switch-on: 500 µA
Energy at full charge: 125 µJ
V_C when the source turns off: 4.97 V

When the switch closes, the capacitor voltage rises fast at first and then slower and slower: the charging current is set by the voltage still left across R, and that shrinks as C fills up. After one time constant it has covered 63.2% of the way, after 5τ it is within 1%. When the source drops to 0 V, C discharges through the same R and falls to 36.8% of its starting value V₀ after another τ.

V_C(t) = V·(1 − exp(−(t)/(τ)))

V_C(t) = V₀·exp(−(t)/(τ))

Doubling R or C doubles τ. Tick "Hold the time axis" and drag a slider to watch the curve stretch; on auto the chart rescales once τ no longer fits the window.

Try it: a sine wave through the RC filter

Set the input frequency and compare the output (CH2) with the input (CH1): it gets smaller and slides along in time.

Resistor R10 kΩ Capacitor C100 nF Input frequency f150 Hz

Cutoff frequency f_c: 159 Hz
f ÷ f_c: 0.942
Gain (output ÷ input): ×0.728
Gain in dB: −2.76 dB
Phase of the output: −43.3°
Time shift on screen: 802 µs behind

Well below the cutoff, a low-pass filter passes the sine almost unchanged. Above it, the output falls by 20 dB for every tenfold rise in frequency and lags further behind, up to 90°. Right at f = f_c the output is 70.7% of the input and 45° behind. A high-pass filter does the opposite: it blocks low frequencies and its output leads the input.

|H| = (f_c)/(√f_c² + f²)

φ = −arctan(f)/(f_c)

Δt = (φ)/(360°)·T

These are the low-pass formulas. The shape of the sine never changes: a linear RC filter only scales it and shifts it in time. A square wave would come out rounded, because its harmonics are filtered by different amounts.

Worked example: 10 kΩ and 100 nF

Time constant: τ = R·C = 10 kΩ × 100 nF = 1 ms
Cutoff: f_c = (1)/(2π·R·C) = (1)/(2π × 10 kΩ × 100 nF) = 159 Hz
As a low-pass filter a 1 kHz signal comes out at 0.157 of the input (−16.1 dB).

Time constant and cutoff of common R and C pairs

R	C	τ = RC	f_c
1 kΩ	10 nF	10 µs	15.9 kHz
1 kΩ	100 nF	100 µs	1.59 kHz
1 kΩ	1 µF	1 ms	159 Hz
10 kΩ	10 nF	100 µs	1.59 kHz
10 kΩ	100 nF	1 ms	159 Hz
10 kΩ	1 µF	10 ms	15.9 Hz
100 kΩ	10 nF	1 ms	159 Hz
100 kΩ	100 nF	10 ms	15.9 Hz
100 kΩ	1 µF	100 ms	1.59 Hz

RC low-pass response versus frequency

(f)/(f_c)	Output ratio (×)	Gain	Phase
0.1	0.995	−0.04 dB	−5.7°
0.5	0.894	−1.0 dB	−26.6°
1	0.707	−3.0 dB	−45.0°
2	0.447	−7.0 dB	−63.4°
5	0.196	−14.1 dB	−78.7°
10	0.100	−20.0 dB	−84.3°
100	0.010	−40.0 dB	−89.4°

For a first-order RC low-pass. Paste a measured frequency response into the lab data analyzer to get the −3 dB point and an error table.

Using it in practice

Step by step

Choose RC or RL, then enter the two component values.
Read the cutoff frequency and the time constant. The cutoff is where the output has fallen to 70.7% (−3 dB) of the input.
Use the response graph to see how much a particular frequency is attenuated, not just the cutoff point.

Common mistakes

Expecting a sharp edge. A first-order filter rolls off only 20 dB per decade: a signal at ten times the cutoff is still at 10% amplitude. For steeper filtering, cascade stages or use an active filter.
Ignoring source and load. The source resistance adds to R, and a load in parallel with the capacitor lowers the output and moves the cutoff. Buffer the output if the load is not much larger than R.
Reading the time constant as the settling time. After one time constant a capacitor has reached only 63% of the step; it takes about five time constants to get within 1%.

Measuring it in the lab

Drive the filter with a sine wave from a function generator and sweep the frequency while watching input and output on an oscilloscope; the cutoff is where the output amplitude is 0.707 of the input and the phase shift is 45°. A 10× probe adds roughly 10 to 20 pF, which matters for filters built with small capacitors. Generators often have a 50 Ω output that adds to R, so include it when you compare with theory.

Frequently asked questions

Is the math different for low-pass and high-pass filters?

No, the cutoff formula is the same. In an RC circuit, taking the output across the capacitor gives a low-pass filter, and taking it across the resistor gives a high-pass filter.

What is the phase shift at the cutoff frequency?

A first-order RC low-pass filter's output lags the input by 45° at the cutoff frequency. A high-pass filter's output leads by 45°.