Question E8B05:
What is the deviation ratio of an FM phone signal having a maximum frequency swing of plus or minus 5 kHz if the highest modulation frequency is 3 kHz?  

Choices:  
A: 6  
B: 0.167  
C: 0.6  
D: 1.67    

Correct Answer is  D: 1.67  

To find the deviation ratio, use the definition:

$$\text{Deviation Ratio} = \frac{\Delta f_{\max}}{f_{m,\max}}$$

- $\Delta f_{\max}$ = maximum frequency deviation (use the peak value, not peak-to-peak)
- $f_{m,\max}$ = highest modulating (audio) frequency

Compute it:
$$\text{DR} = \frac{5\ \text{kHz}}{3\ \text{kHz}} = \frac{5}{3} \approx 1.67$$

Therefore, the correct answer is D: 1.67.

Why the others are incorrect:
- A: 6 — Too large; this results from misusing a factor of 2 (e.g., using total “swing” or mixing with Carson’s Rule). Deviation ratio is just $\Delta f / f_m$, not $2\Delta f / f_m$.
- B: 0.167 — Off by a factor of 10 and inverted; could come from dividing $f_m$ by an incorrect “swing” value or a unit slip. Not the correct definition.
- C: 0.6 — This is the inverse, $f_m / \Delta f = 3/5$. The deviation ratio is $\Delta f / f_m$.

Useful formulas and notes:
- Deviation ratio: $$\text{DR} = \frac{\Delta f}{f_m}$$
- For a single-tone FM signal, the modulation index $\beta$ equals the deviation ratio: $$\beta = \frac{\Delta f}{f_m}$$
- Carson’s Rule for occupied bandwidth (helpful context): $$B \approx 2(\Delta f + f_m)$$
  For this case, $B \approx 2(5 + 3) = 16\ \text{kHz}$.
- Always use peak deviation (±5 kHz ⇒ $\Delta f = 5$ kHz), not peak-to-peak swing.