Question E8C05:
What is the approximate bandwidth of a 13-WPM International Morse Code transmission?  

Choices:  
A: 13 Hz  
B: 26 Hz  
C: 52 Hz  
D: 104 Hz    

Correct Answer is  C: 52 Hz  

To estimate CW (Morse) signal bandwidth, use the Extra Class rule of thumb:
$$BW_{\text{CW}} \approx 4 \times \text{WPM} \text{ (in Hz)}$$

Applying it:
$$BW \approx 4 \times 13 = 52\ \text{Hz}$$
So, the correct answer is 52 Hz (Choice C).

Why the others are incorrect:
- 13 Hz (A): This is only $1 \times \text{WPM}$; too narrow compared to the standard rule.
- 26 Hz (B): This is $2 \times \text{WPM}$; still narrower than typical CW keying spectra.
- 104 Hz (D): This is $8 \times \text{WPM}$; wider than expected for normally shaped CW.

Useful formulas and tips:
- Rule of thumb for CW bandwidth: $$BW_{\text{CW}} \approx 4 \times \text{WPM}$$
- Dot (dit) length: $$t_{\text{dot}} = \frac{1.2}{\text{WPM}} \ \text{seconds}$$
- Note: The receiver’s CW filter (often 200–500 Hz) is wider than the signal’s keying bandwidth to aid copy; the question asks about the signal’s approximate occupied bandwidth.