Filter Frequency Response

Every real filter changes its gain as frequency changes, and that variation is its frequency response. A Butterworth low-pass filter built around a TL071 op-amp will pass a 100 Hz audio signal cleanly while attenuating a 10 kHz noise component by over 40 dB.

Core Concept

A filter is a frequency-selective network. Components like resistors and capacitors create an impedance that changes with frequency, so the ratio of output voltage to input voltage, called transfer function H(jω), is frequency-dependent.

For a first-order RC low-pass filter, the capacitor's impedance drops at high frequencies, pulling the output down. The cutoff frequency fc = 1/(2πRC) marks where the output power falls to half its passband value, which corresponds to a gain drop of 0.707 or -3 dB.

Higher-order filters built around op-amps like the TL071 produce steeper roll-off. A second-order Butterworth filter falls at -40 dB per decade beyond fc, while a first-order passive RC falls at only -20 dB per decade.

Key Equations

Cutoff frequency: fc = 1 / (2π R C) where R is in ohms, C in farads, fc in hertz.

Transfer function magnitude for first-order low-pass: |H(jω)| = 1 / sqrt(1 + (f/fc)^2)

Gain in dB: AdB = 20 * log10(|H|). At f = fc, this gives AdB = 20 * log10(0.707) = -3.01 dB

Roll-off rate for an nth-order filter: -20n dB/decade beyond fc.

Given:
  R = 10 kΩ = 10,000 Ω
  C = 15.9 nF = 15.9 × 10^-9 F
  Input frequency = 2 kHz
  Find: cutoff frequency and gain at 2 kHz

Why this formula:
  First-order RC filter cutoff is fc = 1/(2πRC)

Formula:
  fc = 1 / (2π × R × C)
  |H| = 1 / sqrt(1 + (f/fc)^2)

Substitution:
  fc = 1 / (2π × 10000 × 15.9×10^-9)
  fc = 1 / (2π × 1.59×10^-4)

Calculation:
  2π × 1.59×10^-4 = 9.99×10^-4
  fc = 1 / 9.99×10^-4 = 1001 Hz ≈ 1 kHz

  f/fc = 2000/1000 = 2
  |H| = 1 / sqrt(1 + 4) = 1 / sqrt(5) = 1 / 2.236 = 0.447
  AdB = 20 × log10(0.447) = 20 × (-0.350) = -7.0 dB

Final Answer:
  fc ≈ 1 kHz
  Gain at 2 kHz = 0.447 (-7.0 dB)

Exam Tip: GATE frequently tests whether students confuse the -3 dB frequency with the frequency at which gain becomes zero. At fc, gain is 0.707, not zero. Also, roll-off rate is -20n dB/decade for an nth-order filter, so a fourth-order filter gives -80 dB/decade. Many students incorrectly write -40 dB/decade for all multi-order filters.

Key Properties

• A first-order RC low-pass filter with R = 1 kΩ and C = 159 nF gives fc = 1 kHz exactly.
• At the cutoff frequency, gain magnitude is 1/√2 ≈ 0.707 and phase shift is -45° for a low-pass filter.
• Butterworth filters have a maximally flat passband, meaning no ripple before fc, while Chebyshev filters trade passband ripple for a steeper roll-off.
• Active filters using TL071 or LM741 can provide gain in the passband and avoid loading effects that degrade passive RC filters.
• A high-pass filter's transfer function is the complement of a low-pass filter, with the same fc formula but attenuating below fc instead of above.
• Band-pass filters combine a high-pass and low-pass stage; their bandwidth BW = fH - fL and quality factor Q = fc / BW.

Quick Revision

• fc = 1/(2πRC) for a first-order RC filter.
• At f = fc, |H| = 0.707 and gain = -3 dB.
• Phase at fc is -45° for low-pass, +45° for high-pass.
• Roll-off: -20 dB/decade per filter order beyond fc.
• Butterworth: flat passband. Chebyshev: equiripple passband, steeper roll-off.
• Active filters use op-amps to prevent loading and add gain.
• Doubling R or C halves fc; halving either one doubles fc.
• Exam trap: students write the roll-off of a second-order filter as -20 dB/decade instead of -40 dB/decade, forgetting to multiply by the filter order.