Q130 / 150 · MHT CET 2020 (13 Oct Shift 1) · Physics

A signal of frequency $3 \mathrm{kHz}$ is amplitude modulated on a carrier wave of frequency $2 \cdot 5 \mathrm{MHz}$. The upper and lower sideband frequencies in the resultant signal are

  • A
    $2500 \mathrm{kHz}, 2503 \mathrm{kHz}$
  • Correct answer
    B
    $2503 \mathrm{kHz}, 2497 \mathrm{kHz}$
  • C
    $2 \cdot 5 \mathrm{MHz}, 3 \mathrm{kHz}$
  • D
    $5 \cdot 5 \mathrm{MHz}, 0 \cdot 5 \mathrm{MHz}$

Solution

$\begin{aligned} f_{R}=f_{C}+f_{m} &=2500 \mathrm{KHz}+3 \mathrm{kz} \\ &=2503 \mathrm{KHz} \\ f_{R}=f_{C}-f_{m} &=2500 \mathrm{kHz}-3 \mathrm{kHz} \\ &=2497 \mathrm{kHz} \end{aligned}$

Hence, the correct answer is option: 'B'.

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