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QJEEM21Phd4f3: JEE Main 2021 Physics - Class 12 JEEM21Phd4f3 Solution: The maximum amplitude for an amplitude modulated wave is found…

The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ____________.

Step-by-Step Verified Solution

Amax = Ac + Am = 12

Amin = Ac −- Am = 3

⇒\Rightarrow Ac = 152{{15} \over 2} & Am = 92{9 \over 2}

modulation index = AmAc=9/215/2=0.6{{{A_m}} \over {{A_c}}} = {{9/2} \over {15/2}} = 0.6

⇒\Rightarrow x = 1

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