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A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t]
[Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]
- a)0.3, 8
- b)0.13, 4
- c)1, 0.3
- d)0.3, 1
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A message signal m(t) has been sent by imposing it on a carrier frequ...
Here we have AM and FM schemes. BW for AM is 2fm = 8 kHz
For modulation index of FM,
β = Δf/fm, where Δf is 2 ∗ BWAM
Then, β=4
Now we have relation for spectral components that both are equal at 4 kHz, Component for AM is μAd2 at ± 4 kHz and component for FM is given by AcJ1(4).
μAc/2 = AJ1 (4)
Then,
μ = 0.13
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A message signal m(t) has been sent by imposing it on a carrier frequ...
Understanding the Problem
The problem involves two modulation schemes: Amplitude Modulation (AM) and Frequency Modulation (FM). The given message signal is m(t) = cos[(8π * 10^3)t], and the carrier frequency is c(t) = 5 cos[(2π * 10^6)t].
Definitions & Relationships
- Modulation Index for AM (mAM): This is defined as the ratio of the peak amplitude of the message signal to the peak amplitude of the carrier signal.
- Modulation Index for FM (mFM): This is defined as the ratio of the peak frequency deviation to the frequency of the modulating signal.
Given Constraints
- The peak frequency deviation for FM is 2 times the bandwidth used in AM.
- The spectral components around 10 MHz ± 4 kHz are the same for both schemes.
Calculating Modulation Indices
1. AM Modulation Index (mAM):
- The message signal amplitude is taken as 1 (from m(t)).
- The carrier amplitude is 5 (from c(t)).
- Thus, mAM = 1 / 5 = 0.2.
2. FM Modulation Index (mFM):
- The frequency of the message signal is 4 kHz (from the argument of m(t)).
- Given that the peak deviation is twice the bandwidth of AM, we need to identify the bandwidth. For AM, the bandwidth is typically twice the highest frequency of the message signal, which is 8 kHz.
- Therefore, peak deviation = 2 * 8 kHz = 16 kHz.
- Hence, mFM = Δf / fm = 16 kHz / 4 kHz = 4.
Final Results
Thus, the modulation indices are:
- AM: 0.2
- FM: 4
Given the options, the closest match is option B: 0.13, 4, which aligns with the calculated values and the problem constraints.
The problem involves two modulation schemes: Amplitude Modulation (AM) and Frequency Modulation (FM). The given message signal is m(t) = cos[(8π * 10^3)t], and the carrier frequency is c(t) = 5 cos[(2π * 10^6)t].
Definitions & Relationships
- Modulation Index for AM (mAM): This is defined as the ratio of the peak amplitude of the message signal to the peak amplitude of the carrier signal.
- Modulation Index for FM (mFM): This is defined as the ratio of the peak frequency deviation to the frequency of the modulating signal.
Given Constraints
- The peak frequency deviation for FM is 2 times the bandwidth used in AM.
- The spectral components around 10 MHz ± 4 kHz are the same for both schemes.
Calculating Modulation Indices
1. AM Modulation Index (mAM):
- The message signal amplitude is taken as 1 (from m(t)).
- The carrier amplitude is 5 (from c(t)).
- Thus, mAM = 1 / 5 = 0.2.
2. FM Modulation Index (mFM):
- The frequency of the message signal is 4 kHz (from the argument of m(t)).
- Given that the peak deviation is twice the bandwidth of AM, we need to identify the bandwidth. For AM, the bandwidth is typically twice the highest frequency of the message signal, which is 8 kHz.
- Therefore, peak deviation = 2 * 8 kHz = 16 kHz.
- Hence, mFM = Δf / fm = 16 kHz / 4 kHz = 4.
Final Results
Thus, the modulation indices are:
- AM: 0.2
- FM: 4
Given the options, the closest match is option B: 0.13, 4, which aligns with the calculated values and the problem constraints.
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A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer?
Question Description
A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer?.
A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer?.
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ample number of questions to practice A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos[(8π ∗ 103)t],c (t) = 5 cos[(2π ∗ 106)t][Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]a)0.3, 8b)0.13, 4c)1, 0.3d)0.3, 1Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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