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RC low pass filter operation
Short answer
A filter passes desired frequencies and blocks others. In an RC low-pass filter, series R and shunt C pass low frequencies (Xc large) and attenuate high frequencies (Xc small). Transfer function H(jω)=1/(1+jωRC), cut-off fc=1/(2πRC), gain falls at -20 dB/decade above fc.
What do you mean by a filter circuit? Explain the operation of RC low pass filter with its transfer function and frequency response.
A filter is a circuit that passes signals of some frequencies and blocks (attenuates) signals of other frequencies. It is built from resistors, capacitors, inductors or active devices.
RC low pass filter
A low pass filter (LPF) passes low frequencies and attenuates high frequencies. A resistor R is in series with the signal and a capacitor C is connected across the output.
Vin o----[ R ]----+----o Vout
|
=== C
|
GND o-------------+----o GND
Operation
- Capacitive reactance is Xc = 1/(2πfC). At low frequency Xc is very large, so almost no current flows and the capacitor takes nearly the whole input voltage: Vout ≈ Vin.
- At high frequency Xc is very small, so the capacitor acts almost like a short circuit and Vout ≈ 0.
- So low frequencies pass and high frequencies are blocked.
Transfer function
Using the voltage divider with phasors (s = jω):
H(jω) = Vout/Vin = (1/jωC) / (R + 1/jωC) = 1 / (1 + jωRC)
Let the cut-off frequency be ωc = 1/RC, that is fc = 1/(2πRC). Then
H(jω) = 1 / (1 + j(ω/ωc))
- Magnitude: |H| = 1 / √(1 + (ω/ωc)²)
- Phase: φ = −tan⁻¹(ω/ωc)
Frequency response
| Frequency | Magnitude | Phase |
|---|---|---|
| f = 0 (dc) | 1 (0 dB) | 0° |
| f = fc | 1/√2 = 0.707 (−3 dB) | −45° |
| f ≫ fc | ≈ fc/f, falls at −20 dB/decade | → −90° |
|H|
1 |------------.
| `.
0.707|- - - - - - - -`.
| `--.
+-----------|---------------- f
fc
- The gain is flat (1) in the pass band, 0 to fc.
- At fc the output power is half of the maximum, so fc is called the half-power or −3 dB frequency.
- Beyond fc the gain falls at 20 dB per decade (6 dB per octave).
- The output lags the input; the lag increases from 0° to 90°.
Example: R = 1.59 kΩ, C = 0.1 µF gives fc = 1/(2π × 1590 × 0.1×10⁻⁶) ≈ 1 kHz.
Uses: removing noise from signals, power supply ripple filtering, anti-aliasing before an ADC.
✓ Answer checked by Claude Opus 5.5 (high effort)
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