Discussion
For \( \beta_p = 6 \) and \( f_m = 2 \ \mathrm{kHz} \), the occupied bandwidth is
Answer: B
Applying the rule:
\[ B = 2 \times 2(1+6) = 28 \ \mathrm{kHz} \]
giving 28 kHz. 24 kHz omits the unity term, which matters at any index. 14 kHz drops the factor of two. And 12 kHz is the equivalent frequency deviation, \( \beta_p f_m \), which is a component of the answer rather than the answer. Working the equivalent deviation out first and then applying Carson's rule keeps the two angle schemes on the same footing. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.
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