Question E8D04:
What is the primary effect of extremely short rise or fall time on a CW signal?  

Choices:  
A: More difficult to copy  
B: The generation of RF harmonics  
C: The generation of key clicks  
D: More difficult to tune    

Correct Answer is  C: The generation of key clicks  

Correct answer: C — The generation of key clicks

1) Why C is correct
- Extremely short rise/fall times mean the RF carrier is turned on/off almost instantaneously. That abrupt amplitude change contains lots of high-frequency components (large dA/dt), which spread the CW signal’s spectrum into “far-out” sidebands. These wide, impulsive sidebands are heard on nearby frequencies as “key clicks.”
- In frequency terms, a faster edge makes a wider spectrum. A useful rule of thumb is:
  $$B \approx \frac{0.35}{t_r}$$
  where $t_r$ is the 10–90% rise (or fall) time in seconds. Smaller $t_r$ ⇒ larger $B$ ⇒ stronger, farther-out clicks.

2) Why the other choices are incorrect
- A: More difficult to copy  
  On-frequency readability is usually not the primary issue; the main problem is adjacent-channel interference. Your signal may sound harsh, but copying you on your frequency is still generally OK.
- B: The generation of RF harmonics  
  Harmonics are at $2f_0, 3f_0$, etc. Key clicks are sidebands around $f_0$ caused by abrupt amplitude modulation, not new carriers at multiples of $f_0$.
- D: More difficult to tune  
  Tuning a CW signal is not primarily affected by keying edge speed. The issue is spectral splatter into adjacent channels, not the ability to find the tone.

3) Formulas and rules of thumb
- Bandwidth vs. rise time:
  $$B \approx \frac{0.35}{t_r}$$
- For a simple RC keying envelope shaper (10–90%):
  $$t_r \approx 2.2\,RC$$
- Practical guideline: Use shaped keying with $t_r$ and $t_f$ of about 4–8 ms to keep key clicks low. Longer edges narrow the spectrum and reduce adjacent-channel interference.