How much does increasing a receiver’s bandwidth from 50 Hz to 1,000 Hz increase the receiver’s noise floor?
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.
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 Application
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.
Key Concepts
Why Other Options 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.
题目解析
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.
考试技巧
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.
记忆口诀
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.
实际应用示例
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.
错误选项分析
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.
知识点
Receiver bandwidth, 50 Hz to 1,000 Hz, Noise floor increase, 13 dB
Verified Content
Question from official FCC Extra Class question pool. Explanation reviewed by licensed amateur radio operators.