Butterworth Filter Design

An anti-aliasing filter before an ADC must have the flattest possible pass-band. The Butterworth filter is the standard answer: it sacrifices roll-off sharpness to achieve a pass-band with no ripple at all.

Core Concept

The Butterworth filter is defined by its maximally flat magnitude response in the pass-band. There is no ripple above or below the -3 dB point in the pass-band. This flatness comes from a specific pole placement on a circle in the s-plane, equally spaced in angle.

The poles of an nth-order Butterworth low-pass filter lie on a circle of radius ωc in the left half of the s-plane. For odd n, one pole is on the negative real axis. For even n, poles come in complex conjugate pairs. Each pair forms a second-order section that can be realized as a Sallen-Key stage with specific Q values.

Realizing a 4th-order Butterworth low-pass with fc = 1 kHz requires two cascaded Sallen-Key stages using TL071 op-amps. Stage 1 has Q = 0.5412 and stage 2 has Q = 1.3066. These specific Q values are tabulated in filter design tables and produce the overall maximally flat response.

Key Equations

Magnitude response: |H(jω)|^2 = 1 / (1 + (ω/ωc)^(2n)) where n is the filter order and ωc = 2πfc.

At ω = ωc: |H| = 1/sqrt(2) = 0.707 for all orders. This is the -3 dB point regardless of n.

Roll-off in stopband: -20n dB/decade or -6n dB/octave. A 4th-order Butterworth rolls off at -80 dB/decade.

Order required for a given attenuation: n = log(10^(As/10) - 1) / (2 × log(fs/fc)) where As is the required stopband attenuation in dB at frequency fs.

Pole angles for nth-order Butterworth: θk = π/2 + (2k-1)π/(2n) for k = 1, 2, ..., n. All poles lie on a circle of radius ωc.

Given:
  Filter type: Butterworth low-pass
  Pass-band cut-off: fc = 1 kHz (-3 dB)
  Required attenuation: 40 dB at fs = 10 kHz

Why this formula:
  Determine minimum order n to meet stopband spec.
  n = log10(10^(As/10) - 1) / (2 × log10(fs/fc))

Formula:
  n = log10(10^(40/10) - 1) / (2 × log10(10000/1000))

Substitution:
  10^(40/10) = 10^4 = 10,000
  10,000 - 1 = 9,999
  log10(9,999) ≈ 4.0
  log10(10000/1000) = log10(10) = 1

Calculation:
  n = 4.0 / (2 × 1)
    = 4.0 / 2
    = 2.0

Final Answer:
  n = 2 (round up to nearest integer)
  A 2nd-order Butterworth provides exactly 40 dB at 10× fc (one decade into stopband).
  Actual attenuation check: 20×2×log10(10) = 40 dB. Confirmed.

Exam Tip: GATE frequently asks you to find the minimum filter order for a given stopband attenuation. Use n = log(sqrt(10^(As/10) - 1)) / log(Ωs) where Ωs = fs/fc is the normalized stopband frequency. Always round n UP to the next integer. A common mistake is rounding down, which gives less attenuation than required. Also remember: Butterworth is maximally flat (no ripple), Chebyshev trades pass-band ripple for sharper roll-off, and Bessel is maximally flat in group delay.

Key Properties

• Maximally flat magnitude response in the pass-band: |H|^2 = 1/(1 + (ω/ωc)^(2n)). Zero ripple in pass-band.
• All Butterworth filters of any order pass through exactly -3 dB at ω = ωc.
• Roll-off is -20n dB/decade. 1st order: -20 dB/decade. 4th order: -80 dB/decade.
• Poles lie on a circle in the s-plane, equally spaced at angles of π/n radians. This ensures maximally flat response.
• Higher-order Butterworth has steeper roll-off but more phase distortion and group delay variation.
• Sallen-Key topology is standard for even-order Butterworth: two 2nd-order sections for 4th order, using TL071 op-amps.
• Compared to Chebyshev, Butterworth needs a higher order to achieve the same stopband attenuation at a given frequency.

Quick Revision

• Butterworth: maximally flat pass-band, no ripple. |H|^2 = 1/(1 + (ω/ωc)^(2n)).
• All orders cross -3 dB at ωc. This is a defining property.
• Roll-off = -20n dB/decade.
• n = ceil(log(sqrt(10^(As/10)-1)) / log(Ωs)) to find minimum order.
• Poles on circle of radius ωc, angles at θk = 90 + (2k-1)×90/n degrees.
• Better than Bessel for magnitude response. Worse than Chebyshev for roll-off sharpness.
• Used in anti-aliasing, audio crossover, and data acquisition front ends where pass-band flatness is critical.
• Exam trap: Students round filter order n DOWN instead of UP. If the calculation gives n = 2.3, the correct answer is n = 3, not n = 2. Using n = 2 fails to meet the stopband attenuation specification.