Discussion
A PM signal with index 4 carries a \( 2.5 \ \mathrm{kHz} \) tone. Its equivalent deviation is
Answer: A
Multiplying index by message frequency:
\[ \Delta f = 4 \times 2.5 = 10 \ \mathrm{kHz} \]
giving 10 kHz, from which Carson's rule yields a bandwidth of 25 kHz. 1.6 kHz divides frequency by index. 0.625 kHz divides the other way and slips a factor. And 4 kHz reports the index as a frequency. The two-step route - equivalent deviation, then bandwidth - keeps PM arithmetic on the same footing as FM's.
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