Active filters: low-pass, high-pass, band-pass and notch

Why it matters

Every data-acquisition channel needs filtering: an anti-aliasing low-pass before the ADC, a high-pass to strip a sensor's DC offset, a band-pass to pick out a vibration frequency, a notch to remove 50 Hz mains pickup from an ECG or strain signal. Active filters built from op-amps, resistors and capacitors do this without inductors, with gain, and with outputs that can drive the next stage without upsetting the filter.

Key ideas

Why active. Passive RC filters lose signal and their response changes when loaded; RLC filters need inductors, which are bulky and non-ideal at low frequencies. An op-amp buffers the RC network (high input, low output impedance), can add gain, and lets second-order responses be built from R and C alone. Limits: the op-amp needs a supply, adds noise and offset, and works only well below its gain-bandwidth product and slew-rate limit.

Filter types.

• Low-pass (LPF): passes 0 to f_H. Anti-aliasing, smoothing.
• High-pass (HPF): passes above f_L. Removes DC and drift.
• Band-pass (BPF): passes f_L to f_H around a centre f₀. A wide-band BPF is an HPF cascaded with an LPF; a narrow-band BPF (Q above about 10) uses a single multiple-feedback stage.
• Band-stop / notch: rejects a narrow band; the twin-T notch is the classic 50 Hz rejector.

Order and roll-off. An n-th-order filter rolls off at 20n dB/decade (6n dB/octave) beyond the cut-off. First-order sections use one RC; second-order sections (Sallen-Key, multiple-feedback) use two. Higher orders are built by cascading first- and second-order sections.

Response shapes. The Butterworth response is maximally flat in the passband and is −3 dB at f_c for any order; it is the default in instrumentation. Chebyshev gives a steeper transition at the cost of passband ripple; Bessel gives the most constant group delay (cleanest pulse shape) but the gentlest roll-off.

First-order sections. A non-inverting LPF puts R in series and C to ground at the op-amp's non-inverting input; the op-amp sets the passband gain 1 + R_f/R₁. Swap R and C for an HPF.

Second-order Sallen-Key (equal-component). Two equal R and two equal C give f₀ = 1/(2πRC); the amplifier gain K sets Q = 1/(3 − K). For a Butterworth response Q = 0.707, so K = 1.586. If K reaches 3 the circuit oscillates.

Quality factor and bandwidth. For band-pass and notch filters Q = f₀/BW, and the centre of a band-pass is the geometric mean of its edges: f₀ = √(f_L·f_H).

Formulas

f_c = 1 / (2π·R·C) — cut-off of a first-order section or centre of an equal-component second-order section (Hz).

A_F = 1 + R_f/R₁ — passband gain of a non-inverting section.

|H(f)| = A_F / √(1 + (f/f_H)^(2n)) — n-th-order Butterworth low-pass; for a high-pass replace f/f_H by f_L/f.

Roll-off = 20·n dB/decade — beyond the cut-off.

Q = 1 / (3 − K) — equal-component Sallen-Key; K = 1.586 gives Butterworth (Q = 0.707).

f₀ = √(f_L·f_H), BW = f_H − f_L, Q = f₀/BW — band-pass and notch.

n ≥ log₁₀(10^(A/10) − 1) / (2·log₁₀(f/f_c)) — Butterworth order needed for attenuation A (dB) at frequency f.

f_N = 1 / (2π·R·C) — twin-T notch frequency (R, R, R/2 and C, C, 2C).

Worked examples

Example 1 (standard: first-order LPF). Design a first-order non-inverting low-pass filter with f_H = 1 kHz and passband gain 2, using C = 0.01 µF. Find R, R_f and R₁, and the gain at 5 kHz.

1. R = 1/(2π·f_H·C) = 1/(2π × 1000 × 10⁻⁸) = 15.9 kΩ (use 15.8 kΩ, 1 %).
2. A_F = 1 + R_f/R₁ = 2, so R_f = R₁; choose 10 kΩ each.
3. Gain at 5 kHz: |H| = A_F/√(1 + (5)²) = 2/√26 = 0.392 (14.1 dB below the passband gain).

Answer: R ≈ 15.9 kΩ, R_f = R₁ = 10 kΩ, |H(5 kHz)| ≈ 0.39.

Example 2 (GATE level: Sallen-Key and order). (a) An equal-component Sallen-Key low-pass uses R = 10 kΩ and C = 10 nF. Find f₀ and the amplifier gain for a Butterworth response; give R_f if R₁ = 10 kΩ. How far below the passband is the response at 10f₀? (b) What Butterworth order gives at least 30 dB attenuation at three times the cut-off?

1. (a) f₀ = 1/(2πRC) = 1/(2π × 10⁴ × 10⁻⁸) = 1.59 kHz.
2. Butterworth needs Q = 0.707: K = 3 − 1/Q = 3 − 1.414 = 1.586.
3. K = 1 + R_f/R₁ → R_f = 0.586 × 10 kΩ = 5.86 kΩ.
4. Second-order Butterworth at 10f₀: 1/√(1 + 10⁴) ≈ 0.01, i.e. 40 dB below the passband.
5. (b) n ≥ log₁₀(10³ − 1)/(2·log₁₀3) = 3.0/0.954 = 3.14, so n = 4.

Answer: f₀ ≈ 1.59 kHz, K = 1.586, R_f ≈ 5.86 kΩ, −40 dB at 10f₀; a fourth-order filter is needed for 30 dB at 3f_c.

Example 3 (band-pass). A wide-band band-pass filter has f_L = 200 Hz and f_H = 1 kHz. Then f₀ = √(200 × 1000) = 447 Hz, BW = 800 Hz and Q = 447/800 = 0.56 (wide-band, so a cascade of HPF and LPF is appropriate).

Common mistakes

• Taking the band-pass centre as the arithmetic mean (f_L + f_H)/2 instead of the geometric mean.
• Assuming a cascade of two identical first-order sections is still −3 dB at the original f_c (it is −6 dB there).
• Forgetting that a Sallen-Key gain of 3 or more makes the circuit oscillate.
• Using a filter too close to the op-amp's GBW, where its response deviates from the formulas.
• Assuming every filter is −3 dB at its quoted cut-off; that holds for Butterworth, while a Chebyshev cut-off is usually quoted at the ripple-band edge.
• Confusing roll-off per decade (20 dB) with per octave (6 dB) per order.

For GATE IN

Expect cut-off frequency and component calculations, identification of filter type from a circuit or transfer function, gain at a given frequency, Q and bandwidth, Sallen-Key gain for Butterworth, and minimum order for a given attenuation. Practise reading the type directly from H(s): numerator s² means high-pass, s means band-pass, s² + ω₀² means notch.

Quick check

1. A first-order HPF has R = 500 Ω and C = 10 µF. Cut-off?
2. A notch filter has f₀ = 50 Hz and Q = 5. Bandwidth?
3. Roll-off of a fourth-order Butterworth LPF in dB/decade?
4. Gain needed in an equal-component Sallen-Key for Q = 1?

Answers: 1. 31.8 Hz. 2. 10 Hz. 3. 80 dB/decade. 4. 2.