Smoothing a Signal with an RC Filter

A resistor and a capacitor make the simplest filter. At low frequencies the capacitor barely conducts, so the output follows the input; at high frequencies it shorts the signal to ground, so the output shrinks and lags behind. The crossover is the cutoff frequency, 1/(2πRC).

How it works

Impedance and frequency response

In one sentence

For a series resistor feeding a capacitor, with the output across $C$ , the steady-state response to $V_{in}\cos\omega t$ is $|H|V_{in}\cos(\omega t + \phi)$ with $H(\omega) = \frac{1}{1 + j\omega RC}$ , $|H| = (1 + (\omega RC)^2)^{-1/2}$ and $\phi = -\arctan(\omega RC)$ . The cutoff is $\omega_c = 1/(RC)$ .

General formula

{\htmlData{sym=0}{H(\omega)}} = \frac{1}{1 + {\htmlData{sym=1}{j}}{\htmlData{sym=2}{\omega}} {\htmlData{sym=3}{R}}{\htmlData{sym=4}{C}}}, \qquad |H| = \frac{1}{\sqrt{1 + ({\htmlData{sym=2}{\omega}} {\htmlData{sym=3}{R}}{\htmlData{sym=4}{C}})^2}}, \qquad {\htmlData{sym=5}{\phi}} = -\arctan({\htmlData{sym=2}{\omega}} {\htmlData{sym=3}{R}}{\htmlData{sym=4}{C}}), \qquad {\htmlData{sym=6}{f_c}} = \frac{1}{2\pi {\htmlData{sym=3}{R}}{\htmlData{sym=4}{C}}}

The capacitor needs time to charge. Slow wiggles give it time to follow; fast ones reverse before it can charge much, so they barely show up across it.

What the symbols mean

$H(\omega)$
Transfer function
Output phasor divided by input phasor at angular frequency ω.
$j$
Imaginary unit
√−1, written j in circuits (the i of math class).
$\omega$
Angular frequency
2π times the frequency in hertz.
$R$
Resistance
The resistor's value.
$C$
Capacitance
The capacitor's value.
$\phi$
Phase
How far the output lags behind the input (negative means behind).
$f_c$
Cutoff frequency
Where the gain has dropped to 0.707 (−3 dB).

With phasors, the capacitor has impedance $Z_C = \frac{1}{j\omega C}$ , so the resistor and capacitor form a voltage divider: $H = \frac{Z_C}{R + Z_C} = \frac{1}{1 + j\omega RC}$ .
The output's size is $|H|$ times the input's and it lags by $\arctan(\omega RC)$ .
At the cutoff $\omega RC = 1$ the gain is $1/\sqrt2 \approx 0.707$ (−3 dB) and the lag is 45°; well above it the gain falls by a factor of 10 for every factor of 10 in frequency (−20 dB per decade).
The lab plots three periods of the input (gray) and the output (teal) for the chosen frequency, $R$ in kΩ and $C$ in nF.

Learn the math

Phasors and ImpedanceE&M · Lesson 23, AC Circuits Resonance and FiltersE&M · Lesson 25, AC Circuits