*Question G8B06:
What is the total bandwidth of an FM phone transmission having 5 kHz deviation and 3 kHz modulating frequency?*  

*Correct Answer is  D: 16 kHz*  

### AI Response: (may contain errors)  

To determine the total bandwidth of an FM phone transmission, we can use Carson's Rule, which states:

$$
BW = 2 \times (f_{\Delta} + f_m)
$$

where:
- $BW$ = total bandwidth
- $f_{\Delta}$ = frequency deviation (in kHz)
- $f_m$ = modulating frequency (in kHz)

In this case:
- The frequency deviation $f_{\Delta} = 5 \text{ kHz}$
- The modulating frequency $f_m = 3 \text{ kHz}$

Now, substituting these values into the formula:

$$
BW = 2 \times (5 \text{ kHz} + 3 \text{ kHz}) = 2 \times 8 \text{ kHz} = 16 \text{ kHz}
$$

### Explanation of the Correct Answer:
- The correct answer is **16 kHz** because we applied Carson's Rule correctly, which accounts for both the frequency deviation and the modulating frequency. 

### Why the Other Choices are Incorrect:
- **A: 3 kHz**: This only considers the modulating frequency and ignores the frequency deviation, which is not how bandwidth is calculated for FM signals.
- **B: 5 kHz**: This only considers the frequency deviation and does not include the modulating frequency, which is also incorrect.
- **C: 8 kHz**: This is the sum of the frequency deviation and the modulating frequency but does not account for the factor of 2 in Carson's Rule, which is necessary for FM bandwidth calculation.

### Summary:
- Use Carson's Rule for FM bandwidth: 
$$
BW = 2 \times (f_{\Delta} + f_m)
$$
- For this question, the total bandwidth is **16 kHz**.