Discussion
An FM broadcast station deviates \( \pm 75 \ \mathrm{kHz} \) with audio to \( 15 \ \mathrm{kHz} \), in a \( 200 \ \mathrm{kHz} \) channel. The guard band is about
Answer: C
Carson's rule gives 180 kHz, so
\[ 200 - 180 = 20 \ \mathrm{kHz} \]
leaving about 20 kHz of guard band, which accommodates the Bessel tail beyond Carson's ninety-eight per cent and any frequency error. 125 kHz would follow from comparing the channel with twice the deviation alone. 50 kHz and 5 kHz do not follow from the figures. The allocation is a good example of an engineering rule and a regulatory channel plan being designed together.
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