Discussion
A receiver tuned to \( 1000 \ \mathrm{kHz} \) uses a \( 455 \ \mathrm{kHz} \) intermediate frequency with high-side injection. Its image frequency is
Answer: B
The oscillator sits above the signal, and the image sits above the oscillator by the same offset:
\[ \begin{aligned} f_{o} &= 1000 + 455 = 1455 \\ f_{i} &= f_{o} + 455 = 1910 \ \mathrm{kHz} \end{aligned} \]
so 1910 kHz, which is also the signal plus twice the intermediate frequency. 1455 kHz is the oscillator itself. 545 kHz is the signal minus twice the intermediate frequency, the image for low-side injection. And 2000 kHz is simply the second harmonic. Working from the oscillator outwards, rather than memorising a formula, keeps the two injection cases straight.
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