What is the RC filter cutoff?

How it works

An RC filter is a resistor R and a capacitor C in series, driven by a sine wave. The capacitor's reactance is X_C = 1 ÷ (2πfC), so the series impedance is Z = √(R² + X_C²) and the current is V₀ ÷ Z. The output is the voltage across one part:

• Low-pass (output across C): Vout ÷ Vin = X_C ÷ Z = 1 ÷ √(1 + x²).
• High-pass (output across R): Vout ÷ Vin = R ÷ Z = x ÷ √(1 + x²).

Here x = f ÷ f_c = 2πfRC. Both outputs are 1 ÷ √2 of the input where X_C = R, which gives the cutoff frequency:

• f_c = 1 ÷ (2πRC), and the cutoff angular frequency ω_c = 1 ÷ (RC).
• R = 1 ÷ (2πf_cC) and C = 1 ÷ (2πf_cR) when you find R or C.
• The time constant is τ = RC (1 ÷ (2πf_c) when R or C is found).

The page always shows f_c in Hz, ω_c in rad/s, τ in s, R in Ω and C in F. With a signal frequency f, it also shows:

• Gain Vout ÷ Vin from the two formulas above.
• Gain in decibels: 20 log₁₀ of the gain, which is −10 log₁₀(1 + x²) for a low-pass filter and −10 log₁₀(1 + 1 ÷ x²) for a high-pass filter (−3.0103 dB at the cutoff).
• Phase of the output against the input, in degrees: −arctan x for a low-pass filter (the output lags) and +arctan(1 ÷ x) for a high-pass filter (the output leads). At the cutoff these are −45° and +45°.
• Capacitive reactance at f: X_C = 1 ÷ (2πfC), in Ω.

Units (exact): 1 kΩ = 1,000 Ω; 1 MΩ = 10⁶ Ω; 1 pF = 10⁻¹² F; 1 nF = 10⁻⁹ F; 1 µF = 10⁻⁶ F; 1 mF = 10⁻³ F; 1 kHz = 1,000 Hz; 1 MHz = 10⁶ Hz; 1 GHz = 10⁹ Hz.

Rules

• Resistance from 0.001 Ω to 10¹² Ω; cutoff and signal frequencies from 10⁻⁶ Hz to 10¹² Hz. Limits are in ohms and hertz and include their ends; a value typed in another unit is converted first.
• The capacitance is a number from 10⁻⁶ to 10⁶ in the unit chosen beside it.
• The signal frequency is optional; without it the page shows no gain, phase or reactance.
• A result too large or too small to show as a number gives no answer. Within these limits it does not happen.

Output format

The answer line reads f_c = a u, R = a u or C = a u for the value found. The number a has 6 significant figures, rounded half up, trailing zeros dropped, with thousands commas, and below 10⁻⁶ or from 10¹⁵ up it is written as a × 10ⁿ (R = 1.59155 × 10⁻⁷ Ω). The unit u is the largest of GHz, MHz, kHz and Hz (for f_c), GΩ, MΩ, kΩ and Ω (for R), or F, mF, µF, nF and pF (for C) in which the value, rounded to 6 significant figures, is at least 1; a smaller value uses Hz, Ω or pF. Every other number shows to 6 significant figures, below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ.

Typed values are read back as the exact decimals you typed. R × C, ω_c from R and C, and x = f ÷ f_c when the cutoff is typed are exact fractions; the gain is then the square root of an exact fraction, rounded once, and at x = 1 the phase is exactly ±45°. Steps with π are 64-bit floats.

Worked examples by hand

A cutoff from R and C. R = 1 kΩ and C = 100 nF: RC = 1,000 × 0.0000001 = 0.0001 s, ω_c = 1 ÷ 0.0001 = 10,000 rad/s, and f_c = 10,000 ÷ 2π = 1,591.55 Hz, shown as f_c = 1.59155 kHz.

R for a cutoff. A 10 µF capacitor and a 60 Hz cutoff: R = 1 ÷ (2π × 60 × 0.00001) = 265.258 Ω. OpenStax Example 15.1 finds X_C = 265 Ω for the same capacitor at 60 Hz, as it must: at the cutoff, X_C = R. At a 60 Hz signal, x = 1, so a high-pass filter passes √(1 ÷ 2) = 0.707107 of the input, 20 log₁₀ 0.707107 = −3.0103 dB, leading by 45°.

C for a cutoff. R = 10 kΩ and f_c = 1 kHz: C = 1 ÷ (2π × 1,000 × 10,000) = 1.59155 × 10⁻⁸ F, shown as C = 15.9155 nF.

Gain and phase. The 1 kΩ, 100 nF low-pass filter at f = 10 kHz: x = 2π × 10,000 × 0.0001 = 2π = 6.28319, gain = 1 ÷ √(1 + 39.4784) = 0.157177, −10 log₁₀ 40.4784 = −16.0722 dB, and the phase is −arctan 6.28319 = −80.9569°.