04 // Interactive Formula Compendium // ENGINEERINGf_c = 1 / (2π · R · C), τ = R · C
RC Filter Cutoff Frequency & Time Constant Formula
The first-order RC filter cutoff frequency f_c = 1/(2πRC) marks the −3.01 dB corner where capacitive reactance X_C = 1/(2πfC) equals resistance R, attenuating output voltage to 1/√2 (70.71%) with a −45° phase shift.
Formula Reference
Main Formula
f_c = 1 / (2π · R · C), τ = R · C
Rearranged Forms
|H(jf)| = 1 / √(1 + (f / f_c)²)
φ(f) = −arctan(f / f_c)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| fc | −3 dB Cutoff Frequency | Hz |
| R | Series Resistance | Ω |
| C | Capacitance | F |
Inputs & Parameters
[Ω]
[F]
Calculated Result (−3 dB Cutoff Frequency)
fc =1,591.549Hz
Formula Notes & Background
- Voltage divider with Z_C = 1/(jωC) gives H(jω) = 1/(1 + jωRC). Setting |H(jω_c)|² = ½ yields ω_c·RC = 1 ⇒ f_c = 1/(2πRC).
- Above f_c, magnitude rolls off asymptotically at −20 dB/decade (−6.02 dB/octave).
Step-by-Step Examples
#1: Audio anti-aliasing RC low-pass network with R = 10 kΩ and C = 10 nF (1.0 × 10⁻⁸ F).
R = 10,000 ΩC = 1.0×10⁻⁸ F
τ = 10⁴ × 10⁻⁸ = 100 μs; f_c = 1 / (2π × 10⁻⁴)
→ f_c = 1,591.55 Hz (1.59 kHz)
#2: Time required for a 5 V step input to charge the capacitor to 99.3% of final voltage.
τ = R·C = 100 μs
1 − e^(−5) = 0.99326 ⇒ t = 5τ = 5 × 100 μs
→ t = 500 μs (0.50 ms)
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Frequently Asked Questions
Why is the cutoff called the −3 dB point?+
At f = f_c, output amplitude is V_in/√2, so power (∝ V²) drops to ½: 10·log₁₀(0.5) = −3.0103 dB.
How many time constants τ does it take to fully charge an RC circuit?+
1τ reaches 63.21%, 3τ reaches 95.02%, and 5τ reaches 99.33% (considered settled in engineering design).
How do I switch an RC low-pass filter into a high-pass filter?+
Swap the positions of R and C (take output across R); the −3 dB corner frequency f_c = 1/(2πRC) remains identical.