Semiconductors — JEE Main · 2022 · 29 Jul 2022 · Shift 2 Question with Solution

A modulating signal

2 sin (6.28 \times 1 0^{6}) t

is added to the carrier signal

4 sin (12.56 \times 1 0^{9}) t

for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passed through a band pass filter. The bandwidth of the output signal of band pass filter will be ______ MHz.

Answer: 2

Solution

In amplitude modulation, the modulating frequency is

f_{m} = \frac{6.28 \times 1 0 ^{6}}{2 π} = 1

MHz and carrier frequency is

f_{c} = \frac{12.56 \times 1 0 ^{9}}{2 π} = 2000

MHz.

When passed through a square law device, the output contains frequencies:

2 f_{c}

f_{c} + f_{m}

f_{c}

f_{c} - f_{m}

2 f_{m}

f_{m}

.

The band pass filter centered around

f_{c}

allows only:

f_{c} + f_{m}

f_{c}

, and

f_{c} - f_{m}

.

The bandwidth is the difference between the highest and lowest frequencies:

Bandwidth = (f_{c} + f_{m}) - (f_{c} - f_{m}) = 2 f_{m} = 2 \times 1 = 2 MHz

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Semiconductors

Solution

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