Question E8B03:
What is the modulation index of an FM phone signal having a maximum frequency deviation of 3000 Hz either side of the carrier frequency if the highest modulating frequency is 1000 Hz?  

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

Correct Answer is  A: 3  

To find the FM modulation index for a single-tone signal, use:
$$\beta = \frac{\Delta f}{f_m}$$
where:
- $\Delta f$ = peak frequency deviation (Hz)
- $f_m$ = highest modulating frequency (Hz)

Compute it:
$$\beta = \frac{3000}{1000} = 3$$
So the correct answer is 3 (Choice A).

Why the other choices are incorrect:
- 0.3 (Choice B): Would result from $\Delta f = 300$ Hz with $f_m = 1000$ Hz, or $3000/10000$. Not this case.
- 6 (Choice C): Would require $\Delta f = 6000$ Hz with $f_m = 1000$ Hz. Our deviation is only 3000 Hz.
- 0.6 (Choice D): Would require $\Delta f = 600$ Hz with $f_m = 1000$ Hz. Not this case.

Useful formulas and rules of thumb:
- FM modulation index (single tone): $$\beta = \frac{\Delta f}{f_m}$$
- Carson’s Rule for approximate FM bandwidth: $$B \approx 2(\Delta f + f_m)$$
  This is not needed to answer the question but is useful for FM bandwidth estimation.