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An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]
Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.
- a)1001 kHz, 1 MHz
- b)1000 Hz, 100 kHz
- c)1001 Hz, 1 MHz
- d)1000 kHz, 1000 Hz
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
An angle modulated wave is given as follows x(t) = 50 cos[2π ×...
Angle of the signal is given by, θ(t) = 2π × 106t + 0.001 cos 2π(500)t
Instantaneous frequency is given by ωi(t)=dθ(t)/dt = 2π × 106 − π sin 1000πt
⇒ fi(t) = 106 − 0.5 sin 1000πt
At t = 2ms fi(0.002) = 106 − 0.5 sin 2π = 106 Hz = 1M Hz
Frequency deviation: |Δω| = π
Message signal is of the form, m(t) = Am cos 1000πt
Thus, ωm = 1000π
Bandwidth of message signal is given by B = 500Hz
Deviation ratio is given as, β = Δω/ωm = 0.001
By Carson's rule, the bandwidth of the angle modulated signal is given by, BW = 2(β + 1)B = 1001 Hz
Most Upvoted Answer
An angle modulated wave is given as follows x(t) = 50 cos[2π ×...
Bandwidth Calculation:
Bandwidth of an angle modulated signal is given by the formula:
Bandwidth = 2 * (Δf + fm)
Where, Δf is the frequency deviation and fm is the maximum frequency of the modulating signal.
Given:
Δf = 500 Hz
fm = 500 Hz
Calculations:
Bandwidth = 2 * (500 + 500)
Bandwidth = 2 * 1000
Bandwidth = 2000 Hz or 2 kHz
Instantaneous Frequency:
The instantaneous frequency of an angle modulated signal can be calculated as:
Instantaneous Frequency = fc + kf * m(t)
Where, fc is the carrier frequency, kf is the frequency sensitivity constant, and m(t) is the modulating signal.
Given:
Carrier frequency, fc = 106 MHz = 106,000 kHz
Frequency sensitivity constant, kf = 0.001
Modulating signal, m(t) = 2π(500)t
Calculations at t = 2ms:
m(0.002) = 2π(500)(0.002) = 2π Hz = 6.2832 Hz
Instantaneous Frequency = 106,000 + 0.001 * 6.2832
Instantaneous Frequency = 106,000 + 6.2832
Instantaneous Frequency = 106,006.2832 kHz or approximately 1 MHz
Therefore, the correct answer is option C) 1001 Hz, 1 MHz.
Bandwidth of an angle modulated signal is given by the formula:
Bandwidth = 2 * (Δf + fm)
Where, Δf is the frequency deviation and fm is the maximum frequency of the modulating signal.
Given:
Δf = 500 Hz
fm = 500 Hz
Calculations:
Bandwidth = 2 * (500 + 500)
Bandwidth = 2 * 1000
Bandwidth = 2000 Hz or 2 kHz
Instantaneous Frequency:
The instantaneous frequency of an angle modulated signal can be calculated as:
Instantaneous Frequency = fc + kf * m(t)
Where, fc is the carrier frequency, kf is the frequency sensitivity constant, and m(t) is the modulating signal.
Given:
Carrier frequency, fc = 106 MHz = 106,000 kHz
Frequency sensitivity constant, kf = 0.001
Modulating signal, m(t) = 2π(500)t
Calculations at t = 2ms:
m(0.002) = 2π(500)(0.002) = 2π Hz = 6.2832 Hz
Instantaneous Frequency = 106,000 + 0.001 * 6.2832
Instantaneous Frequency = 106,000 + 6.2832
Instantaneous Frequency = 106,006.2832 kHz or approximately 1 MHz
Therefore, the correct answer is option C) 1001 Hz, 1 MHz.
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An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.a)1001 kHz, 1 MHzb)1000 Hz, 100 kHzc)1001 Hz, 1 MHzd)1000 kHz, 1000 HzCorrect answer is option 'C'. Can you explain this answer?
Question Description
An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.a)1001 kHz, 1 MHzb)1000 Hz, 100 kHzc)1001 Hz, 1 MHzd)1000 kHz, 1000 HzCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.a)1001 kHz, 1 MHzb)1000 Hz, 100 kHzc)1001 Hz, 1 MHzd)1000 kHz, 1000 HzCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.a)1001 kHz, 1 MHzb)1000 Hz, 100 kHzc)1001 Hz, 1 MHzd)1000 kHz, 1000 HzCorrect answer is option 'C'. Can you explain this answer?.
An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.a)1001 kHz, 1 MHzb)1000 Hz, 100 kHzc)1001 Hz, 1 MHzd)1000 kHz, 1000 HzCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.a)1001 kHz, 1 MHzb)1000 Hz, 100 kHzc)1001 Hz, 1 MHzd)1000 kHz, 1000 HzCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An angle modulated wave is given as follows x(t) = 50 cos[2π × 106t + 0.001 cos 2π(500)t]Find the bandwidth of the signal and the instantaneous frequency of the signal at t = 2ms respectively.a)1001 kHz, 1 MHzb)1000 Hz, 100 kHzc)1001 Hz, 1 MHzd)1000 kHz, 1000 HzCorrect answer is option 'C'. Can you explain this answer?.
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