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JEE Main 202126 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Morning Shift
Subject: Physics
Chapter: Communication System

JEE Main 2021 · 26 February · Morning Shift · Physics · Communication System

The maximum and minimum amplitude of an amplitude modulated wave is...

The maximum and minimum amplitude of an amplitude modulated wave is

16V

and

8V

respectively. The modulation index for this amplitude modulated wave is

x \times {10}^{- 2}

. The value of

x

is _______. (Round off your answer to the nearest integer)

Correct answer

0

Step-by-step solution

Modulation index $= \dfrac{{A}_\max - {A}_\min}{{A}_\max + {A}_\min}$

$= \dfrac{16 - 8}{16 + 8} = \dfrac{8}{24} = \dfrac{1}{3} = 0 . 33$

$x \times {10}^{- 2} = 0 . 33$

So,

x=33

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