Butterworth Filters

The Butterworth filter is the most fundamental IIR filter approximation, prized for its monotonically decreasing magnitude response with no ripple in either the passband or stopband. It is often called the maximally flat filter because it achieves the flattest possible passband response for a given filter order.

For GATE and university exams, Butterworth filter design involves determining the order N and cutoff frequency from passband and stopband specifications, then computing pole locations and building the transfer function. Understanding the geometric structure of Butterworth poles is essential for exam problems.

Core Concept Explanation

The Butterworth filter solves a specific optimization problem: among all rational transfer functions of order N, which one has the flattest magnitude response near zero frequency? The answer is the Butterworth approximation, where the magnitude squared function takes the form |H(jΩ)|^2 = 1 / (1 + (Ω/Ω_c)^(2N)).

This function equals 1 at Ω = 0, equals 0.5 (which is -3 dB in power) at Ω = Ω_c regardless of N, and decreases monotonically for Ω greater than Ω_c. The higher the order N, the steeper the transition from passband to stopband, but the cutoff point remains fixed at Ω_c.

The poles of a Butterworth filter of order N are evenly spaced in angle on a circle of radius Ω_c in the s-plane. They lie at angles theta_k = pi/2 + (2k-1)*pi/(2N) for k = 1, 2, ..., N. Only the poles in the left half plane are selected to form a stable filter.

Mathematical Expression

The magnitude squared response and the minimum order formula are the two most exam-relevant expressions for Butterworth filters.

|H(jΩ)|^2 = 1 / (1 + (Ω/Ω_c)^(2N))

The minimum order N required to satisfy passband attenuation A_p at Ω_p and stopband attenuation A_s at Ω_s is:

N >= log(sqrt((10^(A_s/10) - 1)/(10^(A_p/10) - 1))) / log(Ω_s / Ω_p)

Always round N up to the nearest integer. Then compute the actual cutoff Ω_c by back-substituting to satisfy the passband specification exactly.

Practical Understanding

Butterworth filters are preferred when phase linearity and absence of ripple are more important than achieving a very sharp transition with a low order filter. Audio applications, measurement instruments, and analog-to-digital front ends commonly use Butterworth designs because the smooth roll-off causes less group delay distortion near cutoff.

The main limitation is that for a given order, Butterworth provides a less sharp transition than Chebyshev or elliptic filters. If a steep transition with a small number of poles is needed, Chebyshev or elliptic approximations are more efficient.

Given:
Passband: A_p = 3 dB at Omega_p = 1000 rad/s
Stopband: A_s = 20 dB at Omega_s = 5000 rad/s

Why this formula applies:
Need minimum order Butterworth satisfying both edge specifications.

Formula:
N >= log(sqrt((10^(A_s/10) - 1)/(10^(A_p/10) - 1))) / log(Omega_s / Omega_p)

Substitution:
Numerator: 10^(20/10) - 1 = 100 - 1 = 99
Denominator: 10^(3/10) - 1 = 2 - 1 = 1
sqrt(99/1) = 9.95
log(9.95) = 0.998
log(5000/1000) = log(5) = 0.699

Calculation:
N >= 0.998 / 0.699 = 1.43

Final Answer with units:
N = 2 (round up to nearest integer)
A 2nd order Butterworth filter satisfies both specifications.

Exam Tip: In GATE, after computing N, always verify by substituting back into the magnitude expression. If A_p = 3 dB is specified, then Omega_c = Omega_p because the 3 dB point is always at Omega_c for Butterworth. If A_p is not 3 dB, compute Omega_c separately using the passband constraint.

Mechanism Summary

• Magnitude squared: |H|^2 = 1/(1+(Omega/Omega_c)^(2N)). Always -3dB at Omega_c.
• Poles are on a circle of radius Omega_c, evenly spaced by 360/N degrees. Only LHP poles kept.
• Order N determines rolloff steepness. Rolloff slope in stopband is -20N dB/decade.
• No ripple anywhere: response is monotonically decreasing from passband to stopband.
• Order formula: N >= log(sqrt((10^(As/10)-1)/(10^(Ap/10)-1)))/log(Ωs/Ωp). Round up.
• If Ap = 3 dB, then Omega_c = Omega_p. Otherwise, compute Omega_c from the passband edge.

Quick Revision

• Butterworth = maximally flat. No ripple in passband or stopband.
• |H(jΩ)|^2 = 1/(1+(Ω/Ω_c)^2N). Response is always -3 dB at Ω_c.
• Stopband rolloff rate = -20N dB/decade.
• Poles on circle of radius Ω_c, equally spaced. Only left half plane poles selected.
• Order selection: use log formula, always round N up.
• Trap: if A_p is not 3 dB, do not assume Ω_c = Ω_p. Compute Ω_c separately.
• Compared to Chebyshev: less sharp transition for same N, but no ripple and better phase behavior.