RC lowpass filter

An RC lowpass filter is the simplest filter in electronics: a resistor $R$ in series with the signal path and a capacitor $C$ from the output node to ground, with the output taken across the capacitor. It passes low-frequency signals almost unchanged and increasingly attenuates high-frequency ones — the canonical Lowpass filter.

The two fundamental RC filters: a series resistor with a shunt capacitor forms a lowpass; swapping the two gives a highpass.

Intuition: the capacitor as a frequency-dependent resistor

The whole behaviour follows from how a capacitor’s impedance changes with frequency (see Capacitive reactance). The impedance of the capacitor is

$Z_C=\frac{1}{j\omega C}$

where $\omega$ is the angular frequency, $C$ the capacitance, and $j=\sqrt{-1}$. Its magnitude is $X_C=1/(2\pi fC)$.

• At low frequency ($\omega \to 0$): $Z_C\to \infty$. The capacitor looks like an open circuit. No current flows through $R$, so no voltage is dropped across $R$, and essentially all of $V_{in}$ appears at the output. Gain $\approx 1$.
• At high frequency ($\omega \to \infty$): $Z_C\to 0$. The capacitor looks like a short to ground. It clamps the output node near zero regardless of the input. Gain $\to 0$.

In between there is a frequency where $|Z_C|$ equals $R$; that is the crossover between “passes” and “blocks”. That is the Cutoff frequency.

(Note: a handwritten note on the slide writes $X_C=1/\sqrt{2\pi fC}$ with a stray square-root sign. This is a slide typo acknowledged in the notes — the correct reactance magnitude is $X_C=1/(2\pi fC)$, no square root.)

Transfer function — derive it

The circuit is just a voltage divider between $R$ and $Z_C$, with the output across $Z_C$. The Transfer function is the ratio of output phasor to input phasor:

$H(j\omega )=\frac{V_{out}}{V_{in}}=\frac{Z_C}{R+Z_C}=\frac{\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}$

Multiply top and bottom by $j\omega C$ to clear the fraction:

$H(j\omega )=\frac{1}{j\omega RC+1}=\frac{1}{1+j\omega RC}$

Check the limits against the intuition above: at $\omega =0$, $H=1$ (passes). As $\omega \to \infty$, $H\to 0$ (blocks). The magnitude is

$|H(j\omega )|=\frac{1}{\sqrt{1+(\omega RC{)}^2}}$

Cutoff frequency

The corner of the response is where the magnitude has fallen to $1/\sqrt{2}\approx 0.707$ of its low-frequency value — the −3 dB point (half power). Set $(\omega RC{)}^2=1$:

${\omega }_c=\frac{1}{RC},f_c=\frac{1}{2\pi RC}$

with ${\omega }_c$ in rad/s and $f_c$ in Hz. At $\omega ={\omega }_c$ the capacitor’s reactance exactly equals $R$, the output is $0.707V_{in}$, and the phase is $-45°$. Worked example: with $R=1\text{k}\Omega$ and $C=0.1\mu \text{F}$,

$f_c=\frac{1}{2\pi (10^3)(10^{-7})}=\frac{1}{2\pi \times 10^{-4}}\approx 1.59\text{kHz}$

Signals well below $1.59\text{kHz}$ pass; signals well above it are attenuated.

Rolloff: 20 dB/decade

Far above $f_c$, $(\omega RC{)}^2\gg 1$, so $|H|\approx 1/(\omega RC)$ — the magnitude is inversely proportional to frequency. Every factor-of-10 increase in frequency divides the output by 10, i.e. drops it by $20\text{dB}$. So above the cutoff the response rolls off at a constant $-20\text{dB/decade}$, the signature slope of a first-order filter on a Bode plot.

Where you meet it

This network is everywhere, often without being labelled a filter. The parasitic capacitance at the output of an amplifier stage forms an RC lowpass with the stage’s output resistance, and that is what limits the amplifier’s high-frequency bandwidth. The complementary network — swap $R$ and $C$ — is the RC highpass filter, which is the same circuit you use as a Coupling capacitor between amplifier stages.

The lowpass RC filter: series $R$ then shunt $C$, $V_{out}$ across $C$; $-3$ dB at $f_c=1/(2\pi RC)$, then $-20$ dB/decade.