Verified Solution • 2025–26 Session moderate

QJEEM16Phedd2: JEE Main 2016 Physics - Class 12 JEEM16Phedd2 Solution: A modulated signal C m (t) has the form C m (t) = 30 sin 300 t…

A modulated signal Cm(t) has the form Cm(t) = 30 sin 300 π\pi t + 10 (cos 200 π\pi t −cos 400 π\pi t). The carrier frequency c, the modulating frequency (message frequency) f ω\omega , and the modulation index μ\mu are respectively given by :

Multiple Choice Options

  • A fc = 200 Hz; f ω\omega = 50 Hz; μ\mu = 12{1 \over 2}
  • B fc = 150 Hz; f ω\omega = 50 Hz; μ\mu = 23{2 \over 3}
  • C fc = 150 Hz; f ω\omega = 30 Hz; μ\mu = 13{1 \over 3}
  • D fc = 200 Hz; f ω\omega = 30 Hz; μ\mu = 12{1 \over 2}

Step-by-Step Verified Solution

Given,

Cm(t) = 30 sin 300 π\pi t + 10 (cos 200 π\pi t −- cos 400 π\pi t)

Standard equation of amplitude modulated wave,

Cm(t) = Ac sin ( ω\omega ct) −- μAc2{{\mu {A_c}} \over 2} cos (ωc+ωm)\left( {{\omega _c} + {\omega _m}} \right) t + μAc2{{\mu {A_c}} \over 2} cos (ωc−ωm)\left( {{\omega _c} - {\omega _m}} \right) t

By comparing we get,

Ac = 30 V

ω\omega c = 300 π\pi

⇒\Rightarrow    2 π\pi fc = 300 π\pi

⇒\Rightarrow    fc = 150 Hz

ω\omega c −- ω\omega s = 200 π\pi

⇒\Rightarrow    2 π\pi (fc −- f ω\omega ) = 200 π\pi

⇒\Rightarrow    fc −- f ω\omega = 100 Hz

∴\therefore    f ω\omega = 150 −- 100 = 50 Hz

and μAc2{{\mu {A_c}} \over 2} = 10

⇒\Rightarrow     μ\mu ×\times 302{{30} \over 2} = 10

⇒\Rightarrow     μ\mu = 1015{{10} \over {15}} = 23{2 \over 3}

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