NapkinCalc

Electronics 1 — Filters & Op-Amps

The RC low-pass filter

continues from lesson 1 — values defined earlier in the course stay live here

ELI5: one resistor + one capacitor = the simplest useful circuit there is. Slow (low-frequency) signals pass through; fast (high-frequency) ones get shorted to ground by the cap. The dividing line is the cutoff f_c = 1/(2π·R·C), where the output is down exactly 3 dB.

Rf:=1000ohmR_{f} := 1000 ohm = 1000 ohm
Cf:=1e−6FC_{f} := 1e-6 F = 0.000001 F 1 µF
fc=12⋅π⋅Rf⋅Cff_{c} = \frac{1}{2 \cdot \pi \cdot R_{f} \cdot C_{f}} = 159.15 Hz cutoff ≈ 159 Hz
1 / sqrt(1 + (x / 159.2)^2)
0.20.40.60.850010001500

filter response |H| vs frequency (Hz) — 0.707 at the cutoff, sliding down after

✓ pass abs(11+1−0.7071)<0.001\mathrm{abs}\left(\frac{1}{\sqrt{1 + 1}} - 0.7071\right) < 0.001 AT the cutoff: exactly 1/√2 — the −3 dB point

Real-world hook: swap R and C and you get a high-pass instead — together they're the tone knob on a guitar amp, the bass/treble controls on a stereo, and the anti-aliasing filter in front of every audio converter.

Try it yourself: the cutoff of a low-pass filter with R = 2000 Ω and C = 1 µF (1e-6 F)? (f_c = 1/(2π·R·C).)

fcyou:=1/(2∗π∗2000ohm∗1e−6F)fc_{you} := 1 / (2 * \pi * 2000 ohm * 1e-6 F) = 79.577 Hz ✏️ Your turn: compute 1/(2π·R·C) with R = 2000 Ω, C = 1e-6 F. The result is in hertz.
✓ pass abs(fcyou−1/(2∗π∗2000ohm∗1e−6F))<0.1Hzabs(fc_{you} - 1 / (2 * \pi * 2000 ohm * 1e-6 F)) < 0.1 Hz green when your cutoff frequency is correct