Library / Electrical / ELE-40
Electrical
Gain of a first-order RC low-pass filter
Ratio of output amplitude to input amplitude for an unloaded RC low-pass filter, taken across the capacitor.
- Ref
- ELE-40
- Rev
- v1
- Field
- Electrical
- Result
- |HLP|
The amplitude gain of a first-order RC low-pass filter at frequency f is |H| = 1/√(1 + (f/fc)²). It is 1 at low frequency, 0.707 (−3 dB) at the cutoff frequency fc and falls by a factor of ten per decade well above it. The phase lag grows from 0° to −90°, reaching −45° at fc.
Ratio of output amplitude to input amplitude for an unloaded RC low-pass filter, taken across the capacitor. It is 1 at low frequency, 0.707 at the cutoff and falls steadily above it.
Calculator
- Gain in decibels, 20 log₁₀|H|−6.99 dB
- Phase of the output relative to the input−63.43 °
- Gain at the cutoff frequency0.7071
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- At 2,000 Hz, 2 times the cutoff frequency, a low-pass filter passes 44.7 % of the amplitude (−6.99 dB) and lags the input by 63.4°.
- The roll-off is 20 dB per decade, so a signal ten times above the cutoff is reduced to about 10 %: a single RC stage is a gentle filter and noise only a decade away is not removed completely.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| fc Cutoff frequency | 1 to 100,000 Hz | 5e-4 | 0.9998 | 0.9993 | +10% | +7.76 % |
| f Signal frequency | 1 to 100,000 Hz | 0.01 | 1 | 0.99 | +10% | −7.47 % |
Sweep of Signal frequency
| Signal frequency (Hz) | Low-pass gain |
|---|---|
| 1 | 1 |
| 4.217 | 1 |
| 17.78 | 0.9998 |
| 74.99 | 0.9972 |
| 316.2 | 0.9535 |
| 1,334 | 0.5999 |
| 5,623 | 0.1751 |
| 23,710 | 0.04213 |
| 100,000 | 0.01 |
Sweep one variable
Variables and how each one contributes
Low-pass gain
What the equation gives you. Everything else feeds it.
Signal frequency
The signal frequency, in Hz. Lowers the gain: higher frequencies are attenuated more. Raising it by 200 Hz (to 2,200 Hz) moves the result by −7.47 %, from 0.4472 to 0.4138.
Cutoff frequency
The filter’s cutoff frequency, 1/(2πRC), in Hz. Raises the gain: a higher cutoff lets more of the signal through. Raising it by 100 Hz (to 1,100 Hz) moves the result by +7.76 %, from 0.4472 to 0.4819.
Worked example with the values above
- Frequency ratio: f/fc = 2,000 / 1,000 = 2.
- Under the root: 1 + 2² = 5.
- |H| = 1 / √5 = 0.4472.
Assumptions and limits
- A first-order RC low-pass filter, driven from a low-impedance source into a high-impedance load.
- Steady-state sinusoidal signals.
Common mistakes
- Forgetting to square the frequency ratio.
- Using the gain in decibels as an amplitude ratio.
Statistics and uncertainty
Near the corner the gain changes by about 0.7 % for each 1 % error in f/fc, so the filter’s attenuation at a given frequency is known only as well as the corner frequency, which is set mainly by the capacitor tolerance.
Where it is used
- Predicting how much noise a sensor filter removes at a given frequency.
- Checking the attenuation at an alias or interference frequency.
Maths you need
- Cutoff frequency: where the gain is 0.707 of the passband value
- Decibel: 20 log₁₀ of an amplitude ratio