JEE Main 17 March 2021 Shift 2 Solved Paper

Given, carrier signal,
C(t)=25‌sin(2.512×1010t)
Message signal,
m(t)=5‌sin(1.57×108t)
Angular frequency of the message signal,
ωm=1.57×108rads−1
Frequency of the message signal,
of the message fm=‌

ωm

2π

Substituting the values in the above equation, we get fm=‌

1.57×108

2π

Hz
Bandwidth of the modulated signal is
β=2fm
Substituting the values in the above equation, we get
β=2(‌

1.57×10B

2π

)
β=50MHz
Hence, the bandwidth of the modulated signal is 50MHz.