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RC low-pass filter

Calculate the cutoff, gain, phase, Bode response, and step response of a passive first-order RC low-pass filter.

Published 28. October 2026 · version 1

How it is calculated

For the passive first-order low-pass, $H(j\omega)=1/(1+j\omega RC)$ , $\tau=RC$ , and $f_c=1/(2\pi RC)$ . Magnitude and phase are $|H|=1/\sqrt{1+(f/f_c)^2}$ and $\phi=-\tan^{-1}(f/f_c)$ . The unit-step response is $v(t)/V_0=1-e^{-t/\tau}$ .

Assumptions

Ideal, linear resistor and capacitor; zero source impedance; infinite load impedance; no component tolerance, parasitics, noise, or dielectric effects.

Where it applies

Use for a single unloaded RC pole. Include source resistance in R and loading explicitly when the driving or following stage is not ideal. Frequency samples are logarithmically spaced, and the Bode charts have a logarithmic frequency axis.

References

Any standard linear-circuits text: first-order passive RC low-pass transfer function H(s) = 1/(1 + sRC).

Inputs

Input	Symbol	Unit	Default
Resistance	`R_in`	kΩ	1
Capacitance	`C_in`	nF	100
Frequency for gain and phase	`f_eval`	kHz	1
Bode range minimum	`f_min`	Hz	10
Bode range maximum	`f_max`	Hz	100000

Results

With the default inputs above, the calc gives:

Result	Value
Cutoff frequency	1,591.549 Hz
Time constant	0.1 ms
Gain magnitude at selected frequency	0.8467
Gain at selected frequency	-1.445 dB
Phase at selected frequency	-32.142 deg

Checked against sources

Each case below was calculated by this calc and agrees with the value from its source (within the tolerance where one is shown).

1 kΩ and 100 nF at 1 kHz

Source: Hand calculation: τ = (1000 Ω)(100×10⁻9 F) = 100 µs; fc = 1/(2π·100 µs) = 1591.549 Hz; at 1 kHz, f/fc = 0.628319, |H| = 1/√(1 + 0.628319²) = 0.846733 and phase = −atan(0.628319) = −32.1419°.

Inputs: the default inputs above.

Result	Source	This calc
Time constant	0.1 ms	0.1 ms
Cutoff frequency	1,592 Hz	1,592 Hz
Gain magnitude at selected frequency	0.8467	0.8467
Phase at selected frequency	-32.142 deg	-32.142 deg

Chosen frequency equals cutoff

Source: Hand calculation: with R = 1000 Ω and C = 159.154943 nF, τ = 159.154943 µs and fc = 1/(2πRC) = 1000.000001 Hz; at f ≈ fc, |H| = 1/√2 = 0.707106781, gain = 20 log10(1/√2) = −3.010300 dB, and phase = −atan(1) = −45°.

Inputs: Capacitance = 159.2 nF; the others as above.

Result	Source	This calc
Cutoff frequency	1,000 Hz	1,000 Hz
Gain magnitude at selected frequency	0.7071	0.7071
Gain at selected frequency	-3.01 dB	-3.01 dB
Phase at selected frequency	-45 deg	-45 deg