Deep Dive: E4C06
The correct answer is D: 13 dB. How much does increasing a receiver's bandwidth from 50 Hz to 1,000 Hz increase the receiver's noise floor is 13 dB. Noise increases 10×log10(bandwidth ratio) = 10×log10(1000/50) = 10×log10(20) ≈ 13 dB. For amateur radio operators, this is important for receiver performance. Understanding this helps when selecting bandwidths.
Why Other Answers Are Wrong
Option A (3 dB): Incorrect. 3 dB is too low - bandwidth increased 20×, so noise increases about 13 dB. 3 dB is too low. Option B (5 dB): Incorrect. 5 dB is too low - bandwidth increased 20×, so noise increases about 13 dB. 5 dB is too low. Option C (10 dB): Incorrect. 10 dB is too low - bandwidth increased 20×, so noise increases about 13 dB. 10 dB is too low.
Exam Tip
Bandwidth increase noise floor = 10×log10(bandwidth ratio). Think 'B'andwidth 'I'ncrease = '1'0×log10('R'atio). 50 Hz to 1000 Hz = 20× increase = 10×log10(20) ≈ 13 dB. Not 3 dB, not 5 dB, not 10 dB - just 13 dB.
Memory Aid
Bandwidth increase noise floor = 10×log10(bandwidth ratio). Think 'B'andwidth 'I'ncrease = '1'3 dB for 20×. 50 Hz to 1000 Hz = 20× increase = 13 dB. Important for receiver performance.
Real-World Example
Increasing receiver bandwidth from 50 Hz to 1,000 Hz: The noise floor increases by 10×log10(1000/50) = 10×log10(20) ≈ 13 dB. Wider bandwidth means more noise. This is the increase - 13 dB noise floor increase.
Source & Coverage
Question Pool: 2024-2028 Question Pool
Subelement: E4C
Reference: 2024-2028 Question Pool · E4 - Amateur Practices
Key Concepts
Verified Content
Question from the official FCC Extra Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the E4C topic.