*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?*  

*Correct Answer is  A: 3*  

### AI Response: (may contain errors)  

To determine the modulation index of an FM signal, we use the formula:

$$
\text{Modulation Index} = \frac{\Delta f}{f_m}
$$

where:
- $\Delta f$ is the maximum frequency deviation (in Hz).
- $f_m$ is the highest modulating frequency (in Hz).

For this question:
- $\Delta f = 3000 \, \text{Hz}$
- $f_m = 1000 \, \text{Hz}$

Plug these values into the formula:

$$
\text{Modulation Index} = \frac{3000}{1000} = 3
$$

### Explanation of Choices:

- **A: 3**  
  This is the correct answer because the calculation shows that the modulation index is 3.

- **B: 0.3**  
  This is incorrect. A modulation index of 0.3 would imply a much smaller frequency deviation relative to the modulating frequency.

- **C: 6**  
  This is incorrect. A modulation index of 6 would require a frequency deviation of 6000 Hz, which is not the case here.

- **D: 0.6**  
  This is incorrect. A modulation index of 0.6 would imply a frequency deviation of only 600 Hz, which is less than the given 3000 Hz.

### Key Points:
- The modulation index in FM is a measure of how much the carrier frequency varies with the modulating signal.
- A higher modulation index indicates a greater frequency deviation relative to the modulating frequency.