Question E8B04:
What is the modulation index of an FM phone signal having a maximum carrier deviation of plus or minus 6 kHz if the highest modulating frequency is 2 kHz?  

Choices:  
A: 0.3  
B: 3  
C: 0.6  
D: 6    

Correct Answer is  B: 3  

To find the FM modulation index:
$$\beta = \frac{\Delta f}{f_m}$$
where:
- $\Delta f$ = peak frequency deviation
- $f_m$ = highest modulating frequency

Compute:
$$\beta = \frac{6\ \text{kHz}}{2\ \text{kHz}} = 3$$
So the correct answer is B: 3.

Why the others are incorrect:
- A (0.3): Wrong by a factor of 10. Modulation index is a ratio, so unit conversions (kHz to MHz) don’t change it; it remains 3, not 0.3.
- C (0.6): Also too small; likely from an incorrect extra scaling or averaging. Proper calculation gives 3.
- D (6): Uses peak-to-peak deviation (12 kHz) in the numerator. The formula uses peak deviation (6 kHz), not peak-to-peak.

Useful rule of thumb:
- FM modulation index: $$\beta = \frac{\Delta f}{f_m}$$
- Carson’s Rule for FM bandwidth: $$BW \approx 2(\Delta f + f_m)$$