For a two-pole (second-order) active high-pass filter stage, what is the magnitude of its asymptotic roll-off rate in the stopband?

Introduction:Roll-off rate describes how quickly a filter attenuates frequencies in the stopband. Recognizing roll-off versus filter order is fundamental when selecting between first- and second-order stages to meet attenuation targets at a given frequency offset from the cutoff.

Given Data / Assumptions:

• Two-pole (second-order) active high-pass filter.
• Asymptotic stopband slope is requested (far from cutoff).
• Standard definitions: first order → 20 dB/decade; each additional pole adds another 20 dB/decade.

Concept / Approach:A filter’s order N gives the stopband slope magnitude ≈ 20 * N dB/decade. Thus, a second-order stage yields 40 dB/decade. For a high-pass, the stopband lies below the cutoff; the magnitude slope is the same as for a low-pass above cutoff, differing only in frequency region and sign convention.

Step-by-Step Solution:Identify order: N = 2 (two poles).Compute slope magnitude: 20 * N = 40 dB/decade.Interpretation: every 10× decrease in frequency within the stopband yields ≈ 40 dB more attenuation (asymptotically).Near cutoff, the slope is gentler; the 40 dB/dec holds well in the far stopband.

Verification / Alternative check:Compare with a first-order RC high-pass which gives ~20 dB/decade in the stopband; cascading two such sections (or using a biquad) doubles the slope to ~40 dB/decade, matching second-order theory.

Why Other Options Are Wrong:

• −40 dB/decade: Same magnitude but sign; many texts quote magnitude only. The question asks for the magnitude.
• 20/−20 dB/decade: First-order behavior, not second order.
• 60 dB/decade: Would require a third-order (three-pole) filter.

Common Pitfalls:

• Confusing sign convention (increase vs decrease) with magnitude; design typically uses magnitude.
• Assuming the exact slope at cutoff equals the asymptotic slope; it does not.

Final Answer:40 dB/decade