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RC Low-Pass Filter Gain Calculator

Voltage gain and dB of an RC low-pass filter.

How it is calculated

Enter resistance R in ohms, capacitance C in farads (1 µF = 0.000001, 100 nF = 0.0000001) and frequency f in hertz. The tool computes x = 2π × f × R × C, then the voltage gain gain = 1 ÷ √(1 + x²) (output voltage divided by input voltage, unloaded) and dB = 20 × log₁₀(gain). R and C must be above zero and f cannot be negative; at f = 0 the gain is exactly 1.00 (0.00 dB). Gain is rounded to two decimals, so very small values show as 0.00 even though the dB figure is still meaningful. It gives magnitude only, with no phase shift, and assumes the filter drives no load.

Worked example

A filter with R = 10,000 Ω and C = 0.0000001 F (100 nF) is fed a 1,000 Hz signal. Its cutoff would be 1 ÷ (2π × 10,000 × 0.0000001) ≈ 159 Hz. Now x = 2π × 1,000 × 10,000 × 0.0000001 = 6.283. Gain: 1 ÷ √(1 + 6.283²) = 1 ÷ 6.362 = 0.16. In decibels: 20 × log₁₀(0.1572) = -16.07 dB. At 1 kHz this filter lets through about 16% of the input voltage.

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Common mistakes

  • Typing capacitance in microfarads or nanofarads. The field is farads, so 100 nF must be entered as 0.0000001; typing 100 gives a huge capacitor and almost no gain.
  • Entering the frequency in kHz. The field is in hertz, so 1 kHz is 1000.
  • Confusing gain with power ratio. The dB value here is for voltage, 20·log₁₀, so halving the voltage is about −6 dB, not −3 dB.
  • Expecting −3 dB at any frequency. The −3 dB point (gain 0.71) occurs only at the cutoff frequency 1 ÷ (2π × R × C).
  • Ignoring the load. The formula assumes an unloaded output, so a low-impedance load after the filter will change the real response.

About this calculator

For a first-order RC low-pass filter, the voltage gain magnitude at frequency f is |H(f)| = 1 / √(1 + (f/fc)²), where the cutoff frequency fc = 1/(2πRC). Expressed logarithmically, that's gain(dB) = 20 log₁₀|H(f)| — at the cutoff itself the gain is always 0.707 (−3 dB), and it falls off at roughly 20 dB per decade beyond that.

Audio engineers use this exact math to design tone controls and anti-aliasing filters, electronics hobbyists use it to pick R and C values that tame high-frequency noise on a signal line, and it's a staple circuit-analysis exercise because it shows so clearly how a filter's response depends on both the component values and the signal frequency.

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Comments (1)

  • Editorial team

    One filter stage can be much gentler than people expect. With 10 kΩ and 1 µF, a 50 Hz hum still passes at a gain of 0.30, only about 10 dB down, and 60 Hz gives 0.26. To clean up a slow sensor signal you usually need a cutoff well below the interference, or more than one stage. Which frequency are you trying to remove, and what values did you try?

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