What is the approximate bandwidth of a 13-WPM International Morse Code transmission?
The correct answer is C: 52 Hz. The approximate bandwidth of a 13-WPM International Morse Code transmission is 52 Hz. A general rule of thumb is that CW bandwidth is approximately 4 times the sending speed in WPM.
Calculation: 13 WPM × 4 = 52 Hz. This rule accounts for the keying speed and typical rise/fall times. Faster sending speeds require more bandwidth because the dots and dashes are shorter, creating faster transitions. The bandwidth also depends on the keying waveform shape (rise and fall times). For 13 WPM, approximately 52 Hz bandwidth is typical. This is why CW signals can be very narrow - they're much narrower than voice signals.
Exam Tip
CW bandwidth ≈ 4 × WPM. Remember: Approximate CW bandwidth is about 4 times the sending speed in WPM. For 13 WPM, bandwidth ≈ 13 × 4 = 52 Hz.
Memory Aid
"**C**W **B**andwidth = **4** × **W**PM (think 'CB = 4×WPM', 13 WPM = 52 Hz)"
Real-World Application
You're sending CW at 13 words per minute. The bandwidth is approximately 52 Hz (13 × 4). This narrow bandwidth allows many CW signals to fit in a small frequency range. Your receiver needs a filter wide enough to pass this bandwidth, typically 500 Hz or wider for comfortable copying.
FCC Part 97.3Key Concepts
Why Other Options Are Wrong
Option A: Incorrect. 13 Hz would be just the sending speed, not accounting for the bandwidth needed for the keying transitions.
Option B: Incorrect. 26 Hz would be 2 × WPM, but the typical rule is approximately 4 × WPM for CW bandwidth.
Option D: Incorrect. 104 Hz would be 8 × WPM, which is too wide. The typical bandwidth is about 4 × WPM.
题目解析
The correct answer is C: 52 Hz. The approximate bandwidth of a 13-WPM International Morse Code transmission is 52 Hz. A general rule of thumb is that CW bandwidth is approximately 4 times the sending speed in WPM. Calculation: 13 WPM × 4 = 52 Hz. This rule accounts for the keying speed and typical rise/fall times. Faster sending speeds require more bandwidth because the dots and dashes are shorter, creating faster transitions. The bandwidth also depends on the keying waveform shape (rise and fall times). For 13 WPM, approximately 52 Hz bandwidth is typical. This is why CW signals can be very narrow - they're much narrower than voice signals.
考试技巧
CW bandwidth ≈ 4 × WPM. Remember: Approximate CW bandwidth is about 4 times the sending speed in WPM. For 13 WPM, bandwidth ≈ 13 × 4 = 52 Hz.
记忆口诀
**C**W **B**andwidth = **4** × **W**PM (think 'CB = 4×WPM', 13 WPM = 52 Hz)
实际应用示例
You're sending CW at 13 words per minute. The bandwidth is approximately 52 Hz (13 × 4). This narrow bandwidth allows many CW signals to fit in a small frequency range. Your receiver needs a filter wide enough to pass this bandwidth, typically 500 Hz or wider for comfortable copying.
错误选项分析
Option A: Incorrect. 13 Hz would be just the sending speed, not accounting for the bandwidth needed for the keying transitions. Option B: Incorrect. 26 Hz would be 2 × WPM, but the typical rule is approximately 4 × WPM for CW bandwidth. Option D: Incorrect. 104 Hz would be 8 × WPM, which is too wide. The typical bandwidth is about 4 × WPM.
知识点
CW bandwidth, Morse code, WPM, Bandwidth calculation
Verified Content
Question from official FCC Extra Class question pool. Explanation reviewed by licensed amateur radio operators.