JEE PYQ: Communication System Question 20
Question 20 - 2020 (02 Sep 2020 Shift 1)
An amplitude modulated wave is represented by the expression $v_m = 5(1 + 0.6 \cos 6280t) \sin(211 \times 10^4 t)$ volts. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively:
(1) $\frac{3}{2}$ V, 5 V
(2) $\frac{5}{2}$ V, 8 V
(3) 5 V, 8 V
(4) 3 V, 5 V
Show Answer
Answer: (2)
Solution
$\mu = 0.6$, $A_c = 5$. $A_{max} = A_c(1+\mu) = 5(1.6) = 8$. $A_{min} = A_c(1-\mu) = 5(0.4) = 2$. So $\frac{5}{2}$ V and 8 V.
BETA
