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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

FrequencyMagnitudePhase
f = 0 (dc)1 (0 dB)0°
f = fc1/√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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