Selectivity versus bandwidth: In tuned circuits and filters, does greater selectivity imply a wider bandwidth, or the opposite?
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ACorrect
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BIncorrect
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CTrue only when Q < 1
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DTrue only for active filters with op-amps
Answer
Correct Answer: Incorrect
Explanation
Introduction / Context:Selectivity describes how well a circuit discriminates between a wanted frequency and nearby unwanted frequencies. Bandwidth is the range of frequencies passed (for band-pass) or rejected (for band-stop) within a defined amplitude criterion (often −3 dB points). Understanding their inverse relationship is vital when specifying filters or RF stages.
Given Data / Assumptions:
- Standard definitions: bandwidth BW = f_H − f_L at the chosen reference (e.g., −3 dB), center frequency f0 = sqrt(f_H * f_L).
- Quality factor Q = f0 / BW for band-pass networks.
- Linear, time-invariant response; symmetrical response for simplicity.
Concept / Approach:Higher selectivity means the response is more sharply peaked (band-pass) or deeper/narrower (notch), which mathematically corresponds to a higher Q and therefore a smaller bandwidth. Stated differently, for a fixed center frequency f0, increasing Q compresses BW = f0 / Q. Hence the claim that greater selectivity implies wider bandwidth is backward and must be rejected.
Step-by-Step Solution:
Define Q = f0 / BW for a band-pass filter. Greater selectivity ⇒ higher Q. Higher Q ⇒ BW = f0 / Q becomes smaller. Therefore, greater selectivity implies narrower (not wider) bandwidth.Verification / Alternative check:Plot responses for two band-pass filters with the same f0 but different Q. The higher-Q curve is visibly narrower, confirming the inverse relation.
Why Other Options Are Wrong:Correct: reverses the true relationship. Q < 1 and active-only qualifiers: the selectivity/BW relationship stems from definitions, not from active vs passive implementation or Q below/above 1.
Common Pitfalls:Confusing peak gain with bandwidth; assuming added stages always broaden the passband (they typically narrow it when tuned identically).
Final Answer:Incorrect