JEE PYQ: Communication System Question 30
Question 30 - 2019 (11 Jan 2019 Shift 1)
An amplitude modulated signal is given by $V(t) = 10[1 + 0.3 \cos(2.2 \times 10^4 t)] \sin(5.5 \times 10^5 t)$. Here $t$ is in seconds. The sideband frequencies (in kHz) are, [Given $\pi = 22/7$]
(1) 1785 and 1715
(2) 178.5 and 171.5
(3) 89.25 and 85.75
(4) 892.5 and 857.5
Show Answer
Answer: (3)
Solution
$f_c = \frac{5.5 \times 10^5}{2\pi} \approx 87.5$ kHz, $f_m = \frac{2.2 \times 10^4}{2\pi} \approx 3.5$ kHz. Upper $= 87.5 + 1.75 = 89.25$ kHz, Lower $= 87.5 - 1.75 = 85.75$ kHz.
BETA
