Deep Dive: G8B06
The correct answer is D: 16 kHz. The total bandwidth of an FM phone transmission having 5 kHz deviation and 3 kHz modulating frequency is 16 kHz. FM bandwidth ≈ 2(deviation + modulating frequency) = 2(5 + 3) = 16 kHz (Carson's rule). For amateur radio operators, this is a fundamental FM calculation. Understanding this helps when operating FM.
Why Other Answers Are Wrong
Option A (3 kHz): Incorrect. 3 kHz is just the modulating frequency, not the total bandwidth - FM bandwidth is wider due to deviation. 3 kHz is too narrow. Option B (5 kHz): Incorrect. 5 kHz is just the deviation, not the total bandwidth - FM bandwidth includes both deviation and modulating frequency. 5 kHz is too narrow. Option C (8 kHz): Incorrect. 8 kHz is deviation + modulating frequency, not the total bandwidth - FM bandwidth ≈ 2(deviation + modulating frequency). 8 kHz is too narrow.
Exam Tip
FM bandwidth = 2(deviation + modulating frequency) = 2(5 + 3) = 16 kHz. Think 'F'M 'B'andwidth = 'F'ull 'B'andwidth = '2'×(deviation + modulating). Carson's rule: bandwidth ≈ 2(Δf + fm). Not 3 kHz, not 5 kHz, not 8 kHz - just 16 kHz.
Memory Aid
FM bandwidth = 2(deviation + modulating frequency) = 16 kHz. Think 'F'M 'B'andwidth = '2'×(deviation + modulating). Carson's rule: bandwidth ≈ 2(Δf + fm). Standard FM bandwidth calculation.
Real-World Example
An FM phone transmission: Deviation = 5 kHz, modulating frequency = 3 kHz. Total bandwidth ≈ 2(5 + 3) = 16 kHz (Carson's rule). FM bandwidth is approximately twice the sum of deviation and modulating frequency. This is why FM uses more bandwidth than SSB - FM bandwidth is wider.
Source & Coverage
Question Pool: 2023-2027 Question Pool
Subelement: G8B
Reference: 2023-2027 Question Pool · G8 - Signals and Emissions
Key Concepts
Verified Content
Question from the official FCC General Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the G8B topic.