Discussion
A receiver's intermediate frequency bandwidth is \( 8 \ \mathrm{kHz} \). The highest audio frequency it can reproduce is about
Answer: D
The bandwidth must accommodate both sidebands, so
\[ f_{max} = \frac{8}{2} = 4 \ \mathrm{kHz} \]
giving 4 kHz, which is why AM broadcasting sounds bandwidth-limited. 8 kHz treats the whole passband as if it were one sideband. 16 kHz doubles rather than halves. And 2 kHz would follow from dividing twice. The factor of two between channel bandwidth and audio bandwidth runs through every AM calculation and is worth checking in both directions.
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