Hong Life CalculatorsUnderstanding the RC / RL filter calculator — concept and principle
A first-order RC filter separates slow signals from fast ones with one resistor and one capacitor. Taking the output across the capacitor gives a low-pass filter (it removes noise and high-frequency hiss); taking it across the resistor gives a high-pass filter (it blocks DC and slow drift).
How it is calculated
- Cut-off frequency: f_c = 1 / (2πRC), with R in ohms and C in farads (the C input is in nanofarads, so the calculator multiplies by 10⁻⁹).
- Low-pass gain: G_LP = −10·log₁₀(1 + (f/f_c)²) dB.
- High-pass gain: G_HP = −10·log₁₀(1 + (f_c/f)²) dB.
- Low-pass phase: φ = −arctan(f/f_c), in degrees. The high-pass phase is 90° plus this value.
For an RL filter the same formulas apply with f_c = R / (2πL), since the time constant becomes τ = L/R instead of τ = RC.
The principle
A capacitor's impedance is 1/(j2πfC): large at low frequency, small at high frequency. The resistor and capacitor form a frequency-dependent voltage divider. At low frequency the capacitor is almost an open circuit, so nearly all the voltage appears across it (low-pass passes). At high frequency it is almost a short, so the voltage appears across the resistor (high-pass passes). At f = f_c the two impedances have equal magnitude, the output is 1/√2 of the input (−3.01 dB, half power) and the phase is −45°. Far above f_c the low-pass gain falls by about 20 dB per decade (6 dB per octave), because a single reactive element gives a first-order response.
Worked example
- Inputs: R = 1000 Ω, C = 100 nF = 1.0 × 10⁻⁷ F, test frequency f = 5000 Hz.
- RC = 1000 × 10⁻⁷ = 1.0 × 10⁻⁴ s, so f_c = 1 / (2π × 10⁻⁴) = 1,591.549431 Hz.
- Ratio f/f_c = 5000 / 1591.55 = 3.1416.
- Low-pass gain: −10·log₁₀(1 + 9.8696) = −10.36213738 dB.
- High-pass gain: −10·log₁₀(1 + 1/9.8696) = −0.4191399285 dB.
- Low-pass phase: −arctan(3.1416) = −72.34321285°.
| R (Ω) | Cut-off frequency (Hz) |
|---|---|
| 250 | 6,366.197724 |
| 500 | 3,183.098862 |
| 750 | 2,122.065908 |
| 1,000 | 1,591.549431 |
| 1,250 | 1,273.239545 |
| 1,500 | 1,061.032954 |
| 2,000 | 795.7747155 |
How to read the result
A gain of −10.4 dB means the 5 kHz signal leaves a low-pass filter at about 30% of its input amplitude, while the same signal passes a high-pass filter almost untouched (−0.42 dB is about 95%). A useful rule: one decade above f_c the low-pass gain is about −20 dB (10%), two decades about −40 dB (1%).
Common mistakes
- Entering C in microfarads when the field expects nanofarads (1 µF = 1000 nF), which shifts f_c by a factor of 1000.
- Forgetting that source impedance and load resistance add to or sit in parallel with R and change the real cut-off.
Frequently asked questions
Why is the cut-off called the −3 dB point?
At f_c the magnitude is 1/√2 ≈ 0.707, and 20·log₁₀(0.707) ≈ −3.01 dB.
What is the time constant and how does it relate to f_c?
τ = RC. A step input reaches about 63% of its final value after one τ, and f_c = 1/(2πτ). Here τ = 0.1 ms.
Sources: P. Horowitz and W. Hill, The Art of Electronics, 3rd ed., Cambridge University Press, Chapter 1 · A. S. Sedra and K. C. Smith, Microelectronic Circuits, Oxford University Press