Passive RC Low Pass Filter

Audio crossover networks, anti-aliasing filters before ADCs, and noise filters on power rails all use the passive RC low pass filter. It passes DC and low frequencies while attenuating high frequencies with a -20 dB/decade slope.

Core Concept

The RC low pass filter works because the capacitor's impedance ZC = 1/(jωC) decreases as frequency increases. At low frequencies, ZC is large and most of the input voltage appears across it (the output). At high frequencies, ZC becomes a near short-circuit and the output voltage drops to nearly zero.

The cutoff frequency fc (also called the -3dB frequency or corner frequency) is the frequency at which the output power falls to half the input power, which means the output voltage falls to 1/sqrt(2) = 0.707 of the input voltage. At fc, the resistive and capacitive impedances are equal: R = 1/(2π*fc*C).

Above fc, the gain rolls off at -20 dB per decade (or -6 dB per octave). This first-order rolloff is the signature of a single RC section. The phase of Vout lags Vin by 45 degrees exactly at fc, and approaches -90 degrees at very high frequencies. Cascading multiple RC sections increases the rolloff rate but also shifts the -3 dB point.

Key Equations

Cutoff frequency: f_c = 1 / (2π * R * C) in Hz

Transfer function magnitude: |H(jω)| = 1 / sqrt(1 + (f/f_c)^2)

Output voltage magnitude: |Vout| = |Vin| / sqrt(1 + (f/f_c)^2)

At f = fc: |Vout| = |Vin| / sqrt(2) = 0.707 * |Vin| (which is -3 dB)

Phase shift: φ = -arctan(f / f_c) (output lags input)

Time constant: τ = R * C in seconds; fc = 1/(2π*τ)

Given:
  R    = 10 kΩ = 10,000 Ω
  C    = 100 nF = 100e-9 F
  Vin  = 2 V (peak)
  f    = 500 Hz  (test frequency)

Why this formula:
  Find output voltage at 500 Hz using transfer function magnitude.

Step 1 - Find cutoff frequency:
  f_c = 1 / (2π * R * C)
      = 1 / (2π * 10000 * 100e-9)
      = 1 / (2π * 1e-3)
      = 1 / 6.2832e-3
      = 159.2 Hz

Step 2 - Find f/f_c:
  f / f_c = 500 / 159.2 = 3.14

Step 3 - Transfer function magnitude:
  |H| = 1 / sqrt(1 + (3.14)^2)
      = 1 / sqrt(1 + 9.86)
      = 1 / sqrt(10.86)
      = 1 / 3.296
      = 0.3034

Step 4 - Output voltage:
  |Vout| = 2 * 0.3034 = 0.607 V  (peak)

Step 5 - Gain in dB:
  Gain = 20 * log10(0.3034) = 20 * (-0.518) = -10.35 dB

Final Answer:
  |Vout| = 0.607 V peak at 500 Hz, gain = -10.35 dB

Exam Tip: GATE frequently asks for attenuation at a specific frequency, or to find R or C given fc. Remember fc = 1/(2πRC). At f = fc the output is -3 dB (0.707 of input). At f = 10*fc the output is approximately -20 dB (0.1 of input). A common mistake is confusing the -3 dB frequency with the frequency where gain is exactly 0 dB; the gain is 0 dB in the passband, not at fc.

Key Properties

• First-order filter: single RC section gives -20 dB/decade rolloff above fc.
• Cutoff frequency fc = 1/(2πRC); for R = 10 kΩ and C = 100 nF, fc = 159.2 Hz.
• At f = fc: output is 0.707 times input (-3 dB), and phase lag is exactly 45 degrees.
• Time constant τ = RC determines how quickly the capacitor charges; fc = 1/(2πτ).
• Output impedance is frequency-dependent: at low frequencies it looks like R, at high frequencies it falls (C dominates).
• Passive filter: no amplification. Passband gain is 0 dB (unity); output never exceeds input.
• Used as anti-aliasing filter before ADC, noise filter on sensor signals, and audio treble cut filter.

Quick Revision

• fc = 1/(2πRC). Memorize this.
• At f < fc: gain ≈ 0 dB (passband).
• At f = fc: gain = -3 dB, phase = -45 degrees.
• At f > fc: gain rolls off at -20 dB/decade.
• |H(jω)| = 1/sqrt(1+(f/fc)^2). Output voltage = Vin * |H|.
• Phase: φ = -arctan(f/fc). Lags up to -90 degrees.
• Time constant τ = RC = 1/(2πfc).
• Exam trap: Students confuse the -3 dB point with the start of the stopband. The filter does not become ideal at f = fc; it is already attenuating by 3 dB there. The transition from passband to stopband is gradual, not a sharp cutoff.