Passband requirement of tuned circuits in an AM broadcast receiver For standard AM broadcast reception, the composite RF/IF passband of the tuned circuits should be approximately equal to which value to pass the transmitted audio without excessive distortion?

Introduction / Context:
AM broadcast signals carry two sidebands around the carrier. The required receiver passband is tied to the highest transmitted audio frequency to avoid cutting off significant information while still rejecting adjacent-channel interference.

Given Data / Assumptions:

• Standard AM program audio bandwidth is up to about 10 kHz in many textbook examples (varies by region and practice; often less in crowded bands).
• Both upper and lower sidebands must be passed around the carrier.
• IF (e.g., 455 kHz) is the center of selectivity shaping, not the passband width itself.

Concept / Approach:
If the highest audio frequency is f_a (≈ 10 kHz in many illustrative cases), the required RF/IF passband to pass both sidebands is approximately 2 * f_a (about 20 kHz). This ensures minimal amplitude distortion of the modulating audio while maintaining reasonable adjacent-channel rejection.

Step-by-Step Solution:
Let f_a ≈ 10 kHz → sidebands at carrier ± 0…10 kHz.To pass both sidebands, the IF passband ≈ 2 * f_a ≈ 20 kHz.Selectivity can be tailored, but the fundamental width is set by the double-sideband content.

Verification / Alternative check:
Service data often cites AM IF bandwidths on the order of 6–12 kHz for narrow/communications and up to ~20 kHz for high-fidelity AM, aligning with the 2 * f_a guideline.

Why Other Options Are Wrong:
455 kHz / >455 kHz / 1455 kHz: These are center frequencies or unrelated large values, not passband widths.

2 kHz: Much too narrow; would severely cut audio frequencies.

Common Pitfalls:
Confusing the IF center frequency (e.g., 455 kHz) with the passband width required to pass both sidebands.

Final Answer:
20 kHz