Filter selectivity — “The less selective a filter is, the steeper the slope (sharper the roll-off) at its cutoff frequencies.” Is this statement accurate?

Difficulty: Easy

Correct Answer: False

Explanation:


Introduction / Context:
Selectivity describes how well a filter separates passband from stopband. It is tightly linked to roll-off steepness, transition bandwidth, and quality factor (Q). This question tests whether you can correctly relate selectivity and slope at the cutoff region.


Given Data / Assumptions:

  • Standard low-pass/high-pass/band-pass/band-stop filters.
  • Slope discussed in dB/decade or dB/octave per order.


Concept / Approach:
Greater selectivity corresponds to sharper transition regions and steeper skirts. Higher-order filters (or higher Q for resonant responses) achieve steeper roll-off, which improves selectivity. Therefore, a filter that is “less selective” has gentler slopes and broader transition regions, not steeper ones.


Step-by-Step Solution:

1) Define selectivity: ability to attenuate unwanted frequencies near the cutoff.2) Recognize slope per order: first order ~20 dB/decade, second order ~40 dB/decade, etc.3) Higher order → steeper slope → more selective, not less.4) Conclude the statement as written is false.


Verification / Alternative check:
Bode plots of Butterworth/Chebyshev/Elliptic filters show that tighter selectivity comes with steeper roll-off (or ripples/zeros) near cutoff.


Why Other Options Are Wrong:

“True/only for active/only below 100 Hz/only for high-Q”: selectivity principles are frequency-scale invariant and do not depend on “active vs passive” per se (though implementation choices affect practical limits).


Common Pitfalls:
Confusing passband flatness with selectivity; assuming “less selective” equals “more slope,” which reverses the real relationship.


Final Answer:
False

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