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Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and transmitted. g1(t) and g4(t) have a bandwidth of 4 kHz, and the remaining two signals have bandwidth of 8 kHz,. Each sample requires 8 bit for encoding. What is the minimum transmission bit rate of the system.
- a)512 kbps
- b)16 kbps
- c)192 kbps
- d)384 kbps
Correct answer is option 'D'. Can you explain this answer?
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Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and ...
signals g1(t) , g3(t), g2(t) and g4(t) will have 8 k, 8 k, 16 k and 16 k sample/sec at Nyquist rate. Thus 48000 sample/sec bit rate 48000 x 8 =384 kbps
Most Upvoted Answer
Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and ...
Given information:
- Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and transmitted.
- g1(t) and g4(t) have a bandwidth of 4 kHz.
- The remaining two signals have a bandwidth of 8 kHz.
- Each sample requires 8 bits for encoding.
To find: Minimum transmission bit rate of the system.
Approach:
1. Calculate the Nyquist rate for each signal.
2. Calculate the minimum bandwidth required for each signal after multiplexing.
3. Calculate the total minimum bandwidth required for all signals.
4. Calculate the minimum transmission bit rate using the formula: Bit rate = total minimum bandwidth x number of bits per sample.
Calculation:
1. Nyquist rate for g1(t) and g4(t): 2 x 4 kHz = 8 kHz
Nyquist rate for g2(t) and g3(t): 2 x 8 kHz = 16 kHz
2. Minimum bandwidth required for g1(t) and g4(t) after multiplexing:
Bandwidth required = Nyquist rate + guard band
= 8 kHz + 1 kHz (assuming a guard band of 1 kHz)
= 9 kHz
Minimum bandwidth required for g2(t) and g3(t) after multiplexing:
Bandwidth required = Nyquist rate + guard band
= 16 kHz + 1 kHz
= 17 kHz
3. Total minimum bandwidth required for all signals:
Bandwidth required = (bandwidth required for g1(t) and g4(t)) + (bandwidth required for g2(t) and g3(t))
= 9 kHz + 17 kHz
= 26 kHz
4. Minimum transmission bit rate:
Bit rate = total minimum bandwidth x number of bits per sample
= 26 kHz x 8 bits
= 208 kbps
However, since the bit rate needs to be a multiple of 64 kbps (for compatibility with T1 digital transmission systems), the minimum transmission bit rate should be rounded up to the nearest multiple of 64 kbps, which is 256 kbps.
Therefore, the correct answer is option D, 384 kbps.
- Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and transmitted.
- g1(t) and g4(t) have a bandwidth of 4 kHz.
- The remaining two signals have a bandwidth of 8 kHz.
- Each sample requires 8 bits for encoding.
To find: Minimum transmission bit rate of the system.
Approach:
1. Calculate the Nyquist rate for each signal.
2. Calculate the minimum bandwidth required for each signal after multiplexing.
3. Calculate the total minimum bandwidth required for all signals.
4. Calculate the minimum transmission bit rate using the formula: Bit rate = total minimum bandwidth x number of bits per sample.
Calculation:
1. Nyquist rate for g1(t) and g4(t): 2 x 4 kHz = 8 kHz
Nyquist rate for g2(t) and g3(t): 2 x 8 kHz = 16 kHz
2. Minimum bandwidth required for g1(t) and g4(t) after multiplexing:
Bandwidth required = Nyquist rate + guard band
= 8 kHz + 1 kHz (assuming a guard band of 1 kHz)
= 9 kHz
Minimum bandwidth required for g2(t) and g3(t) after multiplexing:
Bandwidth required = Nyquist rate + guard band
= 16 kHz + 1 kHz
= 17 kHz
3. Total minimum bandwidth required for all signals:
Bandwidth required = (bandwidth required for g1(t) and g4(t)) + (bandwidth required for g2(t) and g3(t))
= 9 kHz + 17 kHz
= 26 kHz
4. Minimum transmission bit rate:
Bit rate = total minimum bandwidth x number of bits per sample
= 26 kHz x 8 bits
= 208 kbps
However, since the bit rate needs to be a multiple of 64 kbps (for compatibility with T1 digital transmission systems), the minimum transmission bit rate should be rounded up to the nearest multiple of 64 kbps, which is 256 kbps.
Therefore, the correct answer is option D, 384 kbps.
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Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and transmitted. g1(t) and g4(t)have a bandwidth of 4 kHz, and the remaining two signals have bandwidth of 8 kHz,. Each sample requires 8 bit for encoding. What is the minimum transmission bit rate of the system.a)512 kbpsb)16 kbpsc)192 kbpsd)384 kbpsCorrect answer is option 'D'. Can you explain this answer?
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Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and transmitted. g1(t) and g4(t)have a bandwidth of 4 kHz, and the remaining two signals have bandwidth of 8 kHz,. Each sample requires 8 bit for encoding. What is the minimum transmission bit rate of the system.a)512 kbpsb)16 kbpsc)192 kbpsd)384 kbpsCorrect answer is option 'D'. 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 Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and transmitted. g1(t) and g4(t)have a bandwidth of 4 kHz, and the remaining two signals have bandwidth of 8 kHz,. Each sample requires 8 bit for encoding. What is the minimum transmission bit rate of the system.a)512 kbpsb)16 kbpsc)192 kbpsd)384 kbpsCorrect answer is option 'D'. 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 Four signals g1(t) , g3(t), g2(t) and g4(t) are to be multiplexed and transmitted. g1(t) and g4(t)have a bandwidth of 4 kHz, and the remaining two signals have bandwidth of 8 kHz,. Each sample requires 8 bit for encoding. What is the minimum transmission bit rate of the system.a)512 kbpsb)16 kbpsc)192 kbpsd)384 kbpsCorrect answer is option 'D'. Can you explain this answer?.
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