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JEE Main 202226 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 26 June
Shift: Evening Shift
Subject: Physics
Chapter: Communication System

JEE Main 2022 · 26 June · Evening Shift · Physics · Communication System

A sinusoidal wave y (t) = 40sin (10 × 10 ^ 6 ) is amplitude...

A sinusoidal wave

y

\left(t\right) = 40sin

\left(10 \times {10}^{6}\pit\right)

is amplitude modulated by another sinusoidal wave

x\left(t\right) = 20\sin \left(1000\pit\right)

. The amplitude of minimum frequency component of modulated signal is

Options

A $0 . 5$
B $0 . 25$
C $20$
D $10$

Correct answer

D. $10$

Step-by-step solution

Here, modulation index, $\mu = \dfrac{Am}{Ac} = \dfrac{20}{40} = \dfrac{1}{2}$

Modulated amplitude $= \dfrac{\muAc}{2}$

$= \dfrac{1}{2} \times \dfrac{40}{2} = 10$

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