How it works
The Butterworth response is defined as |H(jω)|² = 1 / (1 + (ω/ωc)^(2N)), where N is the filter order. At ω = ωc, |H| = 1/√2 = −3 dB always, regardless of N. As N increases, the transition from passband to stopband steepens: a 1st-order filter falls at −20 dB/decade; each additional order adds −20 dB/decade. Poles of the Nth-order Butterworth filter lie equally spaced on a unit circle in the left half of the s-plane at angles 90° + (2k−1)·180°/N. The Sallen-Key topology implements a 2nd-order section using one op-amp in non-inverting unity-gain configuration with two R and two C elements, requiring Q = 1/√2 = 0.707 for Butterworth response.
Key points to remember
Butterworth filters have maximally flat passband — no ripple — but relatively gradual roll-off compared to Chebyshev or elliptic designs of the same order. A 3rd-order Butterworth has poles at −1, −0.5 ± j0.866 (normalised), realised as a first-order RC section cascaded with a 2nd-order Sallen-Key section. Required filter order: N ≥ log[(10^(A_s/10)−1)/(10^(A_p/10)−1)] / (2·log(ωs/ωp)), where A_s is stopband attenuation and A_p is passband ripple in dB. Cascading op-amp stages requires buffer amplifiers between sections to prevent loading effects from altering pole frequencies.
Exam tip
Anna University analog electronics exams almost always give you passband ripple, stopband attenuation, and the two edge frequencies and ask you to calculate Butterworth filter order — memorise the order formula and practice the log calculations until they are mechanical.