An engineer needs a lowpass filter where the passband response must be as flat as possible — zero ripple across all passband frequencies — and she can tolerate a gradual transition to the stopband. A second engineer needs the sharpest possible transition at the same filter order, even if it introduces passband ripple. Which filter type should each engineer choose?
ABoth should use Butterworth filters — Butterworth is optimal for all design goals
BFirst engineer: Butterworth (maximally flat passband, monotonic rolloff); Second engineer: Chebyshev or elliptic (sharper rolloff at same order, at the cost of passband or stopband ripple)
CFirst engineer: Chebyshev (flatter passband); Second engineer: Butterworth (steeper rolloff)
DFirst engineer: elliptic (lowest ripple); Second engineer: Butterworth (sharpest rolloff)
The Butterworth filter's defining property is maximally flat magnitude response in the passband — no ripple, monotonically decreasing from DC to the cutoff. This flatness comes at a cost: for a given filter order N, Butterworth has a less steep transition band than Chebyshev (which allows equiripple in the passband) or elliptic (which allows ripple in both bands). Chebyshev and elliptic filters achieve sharper rolloff at the same order by trading passband or stopband flatness. The choice among filter families is always a tradeoff between flatness and rolloff steepness.
Question 2 Multiple Choice
A 3rd-order Butterworth lowpass filter has a cutoff frequency of 1 kHz. At 10 kHz (one decade above cutoff), what is the approximate attenuation?
A20 dB, since one decade always produces 20 dB of attenuation regardless of filter order
B60 dB, since each order contributes 20 dB/decade and three orders give 60 dB/decade
C3 dB, since the −3 dB point is at the cutoff frequency
D120 dB, since each pole contributes 40 dB/decade in the rolloff region
The Butterworth rolloff rate is 20N dB/decade, where N is the filter order. For N=3, that is 60 dB/decade. One decade above the cutoff (10 kHz vs. 1 kHz) gives approximately 60 dB of attenuation. This follows directly from |H(jω)| ≈ (ω_c/ω)^N for ω ≫ ω_c — at 10× the cutoff, the magnitude is (1/10)^3 = 10^−3, which is −60 dB. Option A confuses the rolloff rate of a first-order filter with an order-independent rule; option D (40 dB/pole) is a misconception that arises from confusing double poles with single poles.
Question 3 True / False
All Butterworth lowpass filters, regardless of order, have exactly −3 dB attenuation at their cutoff frequency ω_c.
TTrue
FFalse
Answer: True
This follows directly from the Butterworth magnitude formula: |H(jω_c)|² = 1/(1 + (ω_c/ω_c)^(2N)) = 1/(1+1) = 1/2, so |H(jω_c)| = 1/√2 ≈ 0.707 ≈ −3.01 dB for any N. This is a defining property of the Butterworth design: the −3 dB point is always exactly at the cutoff frequency, regardless of order. The order N only controls how quickly the response rolls off beyond ω_c, not where the −3 dB point is. This makes the cutoff frequency a well-defined, order-independent design parameter.
Question 4 True / False
A higher-order Butterworth filter generally achieves steeper stopband attenuation than a lower-order Chebyshev filter of any type.
TTrue
FFalse
Answer: False
Chebyshev filters achieve steeper rolloff than Butterworth filters of the same order, precisely because they allow ripple in the passband (Type I) or stopband (Type II). The passband ripple is the 'payment' for the extra rolloff sharpness. So a 3rd-order Chebyshev Type I can have steeper attenuation at a given stopband frequency than a 4th- or even 5th-order Butterworth, depending on the allowed ripple level. The order comparison only holds within the same filter family. Across families, the tradeoff is always flatness vs. transition-band sharpness, and higher order alone does not overcome a family's fundamental design tradeoff.
Question 5 Short Answer
Why do Butterworth filters place all their poles on a circle in the left-half s-plane, and how does this geometric arrangement produce the maximally flat magnitude response?
Think about your answer, then reveal below.
Model answer: Placing poles evenly spaced on a circle of radius ω_c in the left-half s-plane produces a magnitude response |H(jω)|² = 1/(1 + (ω/ω_c)^(2N)) — a function that equals exactly 1 at ω=0, equals 1/2 at ω=ω_c, and rolls off monotonically with no ripple. The 'maximally flat' property means the first 2N−1 derivatives of |H(jω)|² with respect to ω are zero at ω=0, making the passband deviate from unity as slowly as possible. Geometrically, the even angular spacing of poles on the circle distributes the rolloff contribution uniformly — no single pole dominates, and there are no resonant peaks or troughs. The poles must be in the left-half plane to ensure stability.
Understanding why the circle placement works requires connecting the s-plane pole locations to the frequency response. Along the imaginary axis (jω), the magnitude of the transfer function is the product of distances from jω to each zero divided by distances to each pole. When poles are evenly spaced on a circle, the denominator polynomial factors to produce exactly the Butterworth magnitude formula. The result is elegant: a purely geometric constraint (poles on a circle) produces an optimal mathematical property (maximally flat passband). This is why the Butterworth is often the first filter design introduced — the geometry and the math reinforce each other.