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A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).
- a)2.99
- b)3.01
Correct answer is between '2.99,3.01'. Can you explain this answer?
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Most Upvoted Answer
A symbol stream contains alternate QPSK and 16-QAM symbols. If symbol...
To calculate the raw (uncoded) data rate, we need to consider the modulation schemes used for the QPSK and 16-QAM symbols and the transmission rate of the symbol stream.
1. QPSK Modulation:
QPSK (Quadrature Phase Shift Keying) is a modulation scheme that uses four different phase shifts to represent different symbols. Each symbol carries two bits of information. So, the data rate for QPSK modulation is equal to the symbol rate divided by 2.
2. 16-QAM Modulation:
16-QAM (Quadrature Amplitude Modulation) is a modulation scheme that uses both amplitude and phase to represent symbols. Each symbol carries four bits of information. So, the data rate for 16-QAM modulation is equal to the symbol rate multiplied by 4.
3. Calculation:
Given that the symbol stream contains alternate QPSK and 16-QAM symbols and is transmitted at a rate of 1 mega-symbol per second, we can calculate the raw (uncoded) data rate as follows:
- The symbol rate for QPSK modulation is 0.5 mega-symbols per second (1 mega-symbol per second divided by 2).
- The symbol rate for 16-QAM modulation is 0.5 mega-symbols per second (1 mega-symbol per second divided by 2).
- The data rate for QPSK modulation is 1 mega-bit per second (0.5 mega-symbols per second multiplied by 2 bits per symbol).
- The data rate for 16-QAM modulation is 2 mega-bits per second (0.5 mega-symbols per second multiplied by 4 bits per symbol).
- Since the symbol stream contains alternate QPSK and 16-QAM symbols, the average data rate is the average of the data rates for QPSK and 16-QAM modulation, which is (1 + 2) / 2 = 1.5 mega-bits per second.
- Rounding off to one decimal place, the raw (uncoded) data rate is 1.5 mega-bits per second, which falls between the given options of 2.99 and 3.01.
Therefore, the correct answer is 1.5 mega-bits per second, which is between 2.99 and 3.01.
1. QPSK Modulation:
QPSK (Quadrature Phase Shift Keying) is a modulation scheme that uses four different phase shifts to represent different symbols. Each symbol carries two bits of information. So, the data rate for QPSK modulation is equal to the symbol rate divided by 2.
2. 16-QAM Modulation:
16-QAM (Quadrature Amplitude Modulation) is a modulation scheme that uses both amplitude and phase to represent symbols. Each symbol carries four bits of information. So, the data rate for 16-QAM modulation is equal to the symbol rate multiplied by 4.
3. Calculation:
Given that the symbol stream contains alternate QPSK and 16-QAM symbols and is transmitted at a rate of 1 mega-symbol per second, we can calculate the raw (uncoded) data rate as follows:
- The symbol rate for QPSK modulation is 0.5 mega-symbols per second (1 mega-symbol per second divided by 2).
- The symbol rate for 16-QAM modulation is 0.5 mega-symbols per second (1 mega-symbol per second divided by 2).
- The data rate for QPSK modulation is 1 mega-bit per second (0.5 mega-symbols per second multiplied by 2 bits per symbol).
- The data rate for 16-QAM modulation is 2 mega-bits per second (0.5 mega-symbols per second multiplied by 4 bits per symbol).
- Since the symbol stream contains alternate QPSK and 16-QAM symbols, the average data rate is the average of the data rates for QPSK and 16-QAM modulation, which is (1 + 2) / 2 = 1.5 mega-bits per second.
- Rounding off to one decimal place, the raw (uncoded) data rate is 1.5 mega-bits per second, which falls between the given options of 2.99 and 3.01.
Therefore, the correct answer is 1.5 mega-bits per second, which is between 2.99 and 3.01.
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Community Answer
A symbol stream contains alternate QPSK and 16-QAM symbols. If symbol...
Given : Symbol rate = 1 mega symbol per second = 106 symbols per second
For QPSK, M = 4
data rate = 106 x log24
data rate = 2 x106 bps .
For 16 QAM, M = 16
data rate = 106 x log216
data rate = 4 x 106 bps
As the symbol stream contains alternate QPSK and 16 QAM symbol, the overall data rate will be
Hence, the correct answer is 3.
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A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. Can you explain this answer?
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A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. 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 symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. 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 symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. Can you explain this answer?.
A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. 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 symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. 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 symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. Can you explain this answer?.
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A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. Can you explain this answer?, a detailed solution for A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. Can you explain this answer? has been provided alongside types of A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).a)2.99b)3.01Correct answer is between '2.99,3.01'. Can you explain this answer? theory, EduRev gives you an
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