QJEEM19Pha783: JEE Main 2019 Physics - Class 12 JEEM19Pha783 Solution: An amplitude modulated signal is given by V(t) = 10[1 +…

Question

Verified Solution • 2025–26 Session moderate

An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2

\times

10⁴ t)] sin(5.5

\times

10⁵ t). Here t is in seconds. The sideband frequencies (in kHz) are, [Given

\pi

= 22/7]

Accepted Answer

V(t) = 10 +

{3 \over 2}

[2cos A sinB]

= 10 +

{3 \over 2}

[sin(A+B)

-

sin(A

-

B)]

= 10+

{3 \over 2}

[sin (57.2

\times

10⁴ t)

-

sin(52.8

\times

10⁴ t)]

\omega

₁ = 57.2

\times

10⁴ = 2

\pi

f₁

f₁ =

{{57.2 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}} = 9.1 \times {10^4}

\simeq

91KHz

f₂ =

{{52.8 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}}

\simeq

84 KHz

Side band frequency are

f₁ = f_c

-

f_w =

{{52.8 \times {{10}^4}} \over {2\pi }}

\simeq

85.00 kHz

f₂ = f_c + f_w =

{{57.2 \times {{10}^4}} \over {2\pi }}

\simeq

90.00 kHz

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