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For the modulated signal x(t) = m(t) cos (2πfct), the message signal m(t) = 4 cos (1000πt) and the carrier frequency fc is 1 MHz. The signal x(t) is passed through a demodulator, as shown in the figure below. The output y(t) of the demodulator is
  • a)
    cos(460πt)
  • b)
    cos(920πt)
  • c)
    cos(1000πt)
  • d)
    cos(540πt)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
For the modulated signal x(t) = m(t) cos (2πfct), the message signa...
Concept: Low pass filter allow the low frequency components ( up to Bandwidth of filter) to pass through it.
Calculation:
x(t) = m(t) cos (2πfct)
m(t) = 4 cos (1000 πt)
The output of the multiplier(s(t)) will be:
s(t) = x(t) cos (2π(fc + 40)t)
s(t) = m(t) cos (2πfct) cos (2π(fc + 40)t)
s(t) = 2 cos (100 πt)⋅cos (2π 40t) + 2 cos (1000 πt) cos (2π(2fc + 40)t
Again using the same trigonometric identify, the above equation can be written as:
s(t) = cos (1000πt – 80πt) + cos (1000πt + 80πt) + cos (1000πt – 2π(2fc + 40)t) + cos (1000πt + 2π (2fc + 40)t)
s(t) = cos (920πt) + cos (1080πt) + cos (1000πt – 2π (2fc - 40)t + cos(1000πt + 2π (2fc + 40)t)
The frequency specimen of s(t) can be drawn as:
The output of the low pass filter will contain frequencies of 460 Hz only as shown:
∴ y(t) = cos (920 πt)
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Most Upvoted Answer
For the modulated signal x(t) = m(t) cos (2πfct), the message signa...
Concept: Low pass filter allow the low frequency components ( up to Bandwidth of filter) to pass through it.
Calculation:
x(t) = m(t) cos (2πfct)
m(t) = 4 cos (1000 πt)
The output of the multiplier(s(t)) will be:
s(t) = x(t) cos (2π(fc + 40)t)
s(t) = m(t) cos (2πfct) cos (2π(fc + 40)t)
s(t) = 2 cos (100 πt)⋅cos (2π 40t) + 2 cos (1000 πt) cos (2π(2fc + 40)t
Again using the same trigonometric identify, the above equation can be written as:
s(t) = cos (1000πt – 80πt) + cos (1000πt + 80πt) + cos (1000πt – 2π(2fc + 40)t) + cos (1000πt + 2π (2fc + 40)t)
s(t) = cos (920πt) + cos (1080πt) + cos (1000πt – 2π (2fc - 40)t + cos(1000πt + 2π (2fc + 40)t)
The frequency specimen of s(t) can be drawn as:
The output of the low pass filter will contain frequencies of 460 Hz only as shown:
∴ y(t) = cos (920 πt)
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For the modulated signal x(t) = m(t) cos (2πfct), the message signal m(t) = 4 cos (1000πt) and the carrier frequency fcis 1 MHz. The signal x(t) is passed through a demodulator, as shown in the figure below. The output y(t) of the demodulator isa)cos(460πt)b)cos(920πt)c)cos(1000πt)d)cos(540πt)Correct answer is option 'B'. Can you explain this answer?
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