Difficulty: Easy
Correct Answer: Somewhat worse when the receiving frequency is raised
Explanation:
Introduction / Context:
Selectivity characterizes how well a receiver rejects adjacent channels. With tuned circuits of roughly constant quality factor Q, the absolute bandwidth increases with frequency, so selectivity can degrade at higher tuned frequencies unless additional measures are taken.
Given Data / Assumptions:
Concept / Approach:
For a resonant circuit, fractional bandwidth ≈ 1/Q. Thus absolute bandwidth BW ≈ f0 / Q. As the center frequency f0 increases and Q remains the same, BW increases proportionally, making adjacent-channel rejection worse (poorer selectivity) at the high-frequency end of the band.
Step-by-Step Solution:
Assume Q is roughly constant with tuning.Compute BW: BW ≈ f0 / Q; as f0 rises, BW rises.Wider BW means more adjacent energy passes → worse selectivity.Therefore, selectivity is somewhat worse at higher receiving frequencies.
Verification / Alternative check:
Many broadcast receivers employ additional tracking circuits or stagger-tuned IF stages to keep effective selectivity more uniform across the band; without these, selectivity degrades at the high end.
Why Other Options Are Wrong:
“Better when raised”: Contradicts BW ∝ f0.
“Better at IF only”: IF selectivity may be fixed, but the question concerns RF tuning.
“Unchanged”: Only true if designs compensate Q or employ filters whose fractional BW changes appropriately.
Common Pitfalls:
Confusing Q (dimensionless) with absolute bandwidth; higher frequency with same Q means wider absolute BW and thus poorer selectivity.
Final Answer:
Somewhat worse when the receiving frequency is raised
Discussion & Comments