JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Semiconductors

In Amplitude Modulation, the message signal

V_{m} (t) = 10 sin (2 π \times 1 0^{5} t)

volts and carrier signal

V_{c} (t) = 20 sin (2 π \times 1 0^{7} t)

volts. The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is

α kHz

. The value of

α

is:

(A)200 kHz✓ Correct
(B)100 kHz
(C)0
(D)50 kHz

Answer: (A)

Solution

In amplitude modulation, the bandwidth is twice the message frequency: Bandwidth =

2 f_{m}

. The message signal has angular frequency

ω_{m} = 2 π \times 1 0^{5}

rad/s, so

f_{m} = 1 0^{5}

Hz = 100 kHz. The modulated signal contains the carrier frequency and two sidebands at

(f_{c} - f_{m})

and

(f_{c} + f_{m})

. The bandwidth is the difference between the upper and lower sideband frequencies: Bandwidth =

(f_{c} + f_{m}) - (f_{c} - f_{m}) = 2 f_{m} = 2 \times 100 = 200

kHz. Therefore,

α = 200

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Semiconductors

Solution

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