Discussion
A phase modulator of \( 0.5 \ \mathrm{rad/V} \) is driven by \( 1.2 \ \mathrm{V} \) peak. Its index is
Answer: C
Multiplying out:
\[ \beta_p = 0.5 \times 1.2 = 0.6 \ \mathrm{rad} \]
so 0.6 radians, which sits just beyond the narrowband regime and would give a bandwidth somewhat above twice the message frequency. 2.4 divides the amplitude by the sensitivity. 1.7 inverts it the other way. And 0.42 corresponds to no correct combination. Because the answer is a pure number of radians, no message frequency was needed, which is the signature of a PM calculation.
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