Deep Dive: E8C05
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.
Why Other Answers 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.
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 Example
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.
Source & Coverage
Question Pool: 2024-2028 Question Pool
Subelement: E8C
Reference: FCC Part 97.3
Key Concepts
Verified Content
Question from the official FCC Extra Class pool. Explanation reviewed by licensed amateur radio operators and mapped to the E8C topic.