Passive RC High Pass Filter

Every audio preamplifier blocks DC offset while passing the audio signal. The passive RC high-pass filter does exactly that using one resistor and one capacitor.

Core Concept

At DC and very low frequencies, the capacitor's reactance Xc = 1/(2πfC) is very large. Most of the input voltage drops across the capacitor, leaving almost nothing at the output. The circuit blocks low-frequency signals.

As frequency increases, Xc falls. At the cut-off frequency fc, Xc equals R. The output voltage is 0.707 times the input, which is a -3 dB drop. Above fc the filter passes signals with little attenuation.

A practical use is the input coupling stage of an op-amp amplifier like the TL071, where a 1 µF capacitor and 10 kΩ resistor set fc = 15.9 Hz. This blocks the DC bias while passing audio above 16 Hz.

Key Equations

Cut-off frequency: fc = 1 / (2πRC) where R is in ohms and C is in farads. At fc, output power is half the input power.

Voltage transfer function: H(jω) = jωRC / (1 + jωRC) where ω = 2πf.

Magnitude response: |H| = ωRC / sqrt(1 + (ωRC)^2). At f >> fc, |H| → 1. At f << fc, |H| → 0.

Phase response: φ = 90° - arctan(ωRC). At fc, phase shift is +45 degrees. Below fc, phase approaches +90 degrees.

Roll-off in stopband: -20 dB/decade or equivalently -6 dB/octave for a first-order RC filter.

Given:
  R = 10 kΩ = 10,000 Ω
  C = 1 µF = 1 × 10^-6 F
  Vin = 2 V (peak), f = 10 Hz

Why this formula:
  At any frequency, voltage divider between Xc and R gives Vout.

Formula:
  Vout/Vin = R / sqrt(R^2 + Xc^2)
  Xc = 1 / (2πfC)

Substitution:
  Xc = 1 / (2π × 10 × 1×10^-6)
     = 1 / (62.83 × 10^-6)
     = 15,915 Ω = 15.9 kΩ

Calculation:
  Vout/Vin = 10,000 / sqrt(10,000^2 + 15,915^2)
           = 10,000 / sqrt(1×10^8 + 2.53×10^8)
           = 10,000 / sqrt(3.53×10^8)
           = 10,000 / 18,788
           = 0.532

Final Answer:
  Vout = 0.532 × 2 V = 1.064 V peak
  Note: f = 10 Hz is below fc = 15.9 Hz, so attenuation is expected.

Exam Tip: GATE frequently tests phase shift at fc. For a high-pass RC filter, the output leads the input by exactly +45 degrees at fc. Students often confuse this with the low-pass case where the output lags by -45 degrees at fc. Also remember: the -3 dB frequency and the phase-shift frequency are the same point, fc = 1/(2πRC). Roll-off is always -20 dB/decade for first order, not -40 dB/decade.

Key Properties

• Passes frequencies above fc and attenuates frequencies below fc with a -20 dB/decade roll-off.
• At fc = 1/(2πRC), output is -3 dB (0.707 × Vin) and phase lead is exactly +45 degrees.
• Capacitive reactance Xc = 1/(2πfC) decreases as frequency increases, reducing the voltage drop across C.
• A 100 nF capacitor with 15.9 kΩ sets fc = 100 Hz, suitable for speech-band filtering above 100 Hz.
• The filter introduces a phase lead at all frequencies below and near fc. Above fc, phase lead approaches 0 degrees.
• Loading effect: connecting a low-impedance load across Vout shifts the effective R and changes fc.
• Cascading two identical RC stages gives a second-order response but the -3 dB frequency shifts down by a factor of sqrt(2^(1/0.5) - 1) ≈ 0.644.

Quick Revision

• RC high-pass filter passes high frequencies and blocks DC and low frequencies.
• Cut-off frequency fc = 1/(2πRC). Unit of fc is Hz.
• At fc: attenuation = -3 dB, |H| = 0.707, phase = +45 degrees.
• Transfer function H(jω) = jωRC/(1 + jωRC). This is a first-order high-pass response.
• Stopband roll-off is -20 dB/decade or -6 dB/octave.
• Phase shift varies from +90 degrees at DC to 0 degrees at very high frequency.
• Doubling R or doubling C each halves fc.
• Exam trap: Students write Vout = Xc/(R+Xc) × Vin (the low-pass formula) instead of Vout = R/(R+Xc) × Vin for the high-pass case. In the HPF, the output is taken across R, not across C.