Active filter order vs roll-off: A two-pole (second-order) active filter exhibits a stopband roll-off of −20 dB/decade. Is this statement accurate?

Introduction / Context:
Filter order determines asymptotic attenuation slope beyond the cutoff region. Each reactive pole contributes an additional −20 dB/decade (or −6 dB/octave) in the stopband. Understanding this principle is core to predicting selectivity and choosing between first- and second-order topologies in active RC designs such as Sallen–Key or multiple-feedback filters.

Given Data / Assumptions:

• “Two-pole” means a second-order low-pass, high-pass, band-pass, or band-stop realized actively.
• Asymptotic stopband behavior is considered (well beyond cutoff).
• Standard linear, time-invariant response without special equalization.

Concept / Approach:
For a first-order (one-pole) filter, the far-stopband slope is −20 dB/decade. For a second-order (two-pole) filter, slopes add to −40 dB/decade. This is independent of whether the active form is Sallen–Key, multiple-feedback, or biquad. While passband ripple/peaking depends on Q or damping ratio, the asymptotic slope is set by order alone.

Step-by-Step Solution:

Verification / Alternative check:
Bode magnitude plots of standard 2nd-order Butterworth, Bessel, or Chebyshev sections all show a −40 dB/decade asymptote. Cascading two first-order stages demonstrates the same cumulative slope.

Why Other Options Are Wrong:
“True for unity-gain Sallen–Key” is false; gain does not change the asymptotic order. “Only at cutoff” confuses the local slope near the corner with the far-stopband asymptote. High-pass and low-pass of the same order share the same magnitude slope magnitude in their respective stopbands.

Common Pitfalls:
Equating peaking or damping with slope; mistaking the −3 dB point behavior for the asymptotic roll-off far from cutoff.

Final Answer:
Incorrect