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A discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.
- a)1.181 bits and 2.362 bits respectively
- b)1.097 bits and 2.194 bits respectively
- c)1.17 bits and 1.17 bits respectively
- d)None of these.
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A discrete memoryless source with source alphabet ∅={S0,S1,S2} and so...
To calculate the entropy of a discrete memoryless source, we can use the formula:
H(X) = - Σ P(xi) log2 P(xi)
where H(X) is the entropy of the source and P(xi) is the probability of symbol xi in the source alphabet.
Calculating the entropy of the source:
Given source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}, we can calculate the entropy as follows:
H(X) = - (0.7 * log2(0.7) + 0.15 * log2(0.15) + 0.15 * log2(0.15))
≈ - (0.7 * (-0.5146) + 0.15 * (-2.737) + 0.15 * (-2.737))
≈ - (-0.3602 + 0.41055 + 0.41055)
≈ 1.181 bits
Therefore, the entropy of the source is approximately 1.181 bits.
Calculating the entropy of the second-order extension of the source:
To calculate the entropy of the second-order extension of the source, we need to consider the joint probabilities of pairs of symbols in the source alphabet.
In the second-order extension, we have a new source alphabet ∅^2 = {S0S0, S0S1, S0S2, S1S0, S1S1, S1S2, S2S0, S2S1, S2S2}. The joint probabilities can be calculated using the source statistics.
For example, P(S0S1) = P(S0) * P(S1) = 0.7 * 0.15 = 0.105.
Using the joint probabilities, we can calculate the entropy of the second-order extension as follows:
H(X^2) = - Σ P(xi,xj) log2 P(xi,xj)
where H(X^2) is the entropy of the second-order extension and P(xi,xj) is the joint probability of symbols xi and xj.
Calculating the entropy of the second-order extension:
H(X^2) = - (P(S0S0) * log2(P(S0S0)) + P(S0S1) * log2(P(S0S1)) + ... + P(S2S2) * log2(P(S2S2)))
Performing the calculations using the joint probabilities, we find:
H(X^2) ≈ 2.362 bits
Therefore, the entropy of the second-order extension of the source is approximately 2.362 bits.
Hence, the correct answer is option 'A': 1.181 bits for the entropy of the source and 2.362 bits for the entropy of the second-order extension of the source.
H(X) = - Σ P(xi) log2 P(xi)
where H(X) is the entropy of the source and P(xi) is the probability of symbol xi in the source alphabet.
Calculating the entropy of the source:
Given source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}, we can calculate the entropy as follows:
H(X) = - (0.7 * log2(0.7) + 0.15 * log2(0.15) + 0.15 * log2(0.15))
≈ - (0.7 * (-0.5146) + 0.15 * (-2.737) + 0.15 * (-2.737))
≈ - (-0.3602 + 0.41055 + 0.41055)
≈ 1.181 bits
Therefore, the entropy of the source is approximately 1.181 bits.
Calculating the entropy of the second-order extension of the source:
To calculate the entropy of the second-order extension of the source, we need to consider the joint probabilities of pairs of symbols in the source alphabet.
In the second-order extension, we have a new source alphabet ∅^2 = {S0S0, S0S1, S0S2, S1S0, S1S1, S1S2, S2S0, S2S1, S2S2}. The joint probabilities can be calculated using the source statistics.
For example, P(S0S1) = P(S0) * P(S1) = 0.7 * 0.15 = 0.105.
Using the joint probabilities, we can calculate the entropy of the second-order extension as follows:
H(X^2) = - Σ P(xi,xj) log2 P(xi,xj)
where H(X^2) is the entropy of the second-order extension and P(xi,xj) is the joint probability of symbols xi and xj.
Calculating the entropy of the second-order extension:
H(X^2) = - (P(S0S0) * log2(P(S0S0)) + P(S0S1) * log2(P(S0S1)) + ... + P(S2S2) * log2(P(S2S2)))
Performing the calculations using the joint probabilities, we find:
H(X^2) ≈ 2.362 bits
Therefore, the entropy of the second-order extension of the source is approximately 2.362 bits.
Hence, the correct answer is option 'A': 1.181 bits for the entropy of the source and 2.362 bits for the entropy of the second-order extension of the source.
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Community Answer
A discrete memoryless source with source alphabet ∅={S0,S1,S2} and so...
The entropy of the source is
= 0.3602+0.4105+0.4105
=1.181 bits
the entropy of second-order extension of the source is
H(S2)=2×1.181
=2.362 bits
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A discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.a)1.181 bits and 2.362 bits respectivelyb)1.097 bits and 2.194 bits respectivelyc)1.17 bits and 1.17 bits respectivelyd)None of these.Correct answer is option 'A'. Can you explain this answer?
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A discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.a)1.181 bits and 2.362 bits respectivelyb)1.097 bits and 2.194 bits respectivelyc)1.17 bits and 1.17 bits respectivelyd)None of these.Correct answer is option 'A'. 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 discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.a)1.181 bits and 2.362 bits respectivelyb)1.097 bits and 2.194 bits respectivelyc)1.17 bits and 1.17 bits respectivelyd)None of these.Correct answer is option 'A'. 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 discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.a)1.181 bits and 2.362 bits respectivelyb)1.097 bits and 2.194 bits respectivelyc)1.17 bits and 1.17 bits respectivelyd)None of these.Correct answer is option 'A'. Can you explain this answer?.
A discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.a)1.181 bits and 2.362 bits respectivelyb)1.097 bits and 2.194 bits respectivelyc)1.17 bits and 1.17 bits respectivelyd)None of these.Correct answer is option 'A'. 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 discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.a)1.181 bits and 2.362 bits respectivelyb)1.097 bits and 2.194 bits respectivelyc)1.17 bits and 1.17 bits respectivelyd)None of these.Correct answer is option 'A'. 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 discrete memoryless source with source alphabet ∅={S0,S1,S2} and source statistics {0.7, 0.15, 0.15}Calculate the entropy of source and the entropy of second-order extension of the source.a)1.181 bits and 2.362 bits respectivelyb)1.097 bits and 2.194 bits respectivelyc)1.17 bits and 1.17 bits respectivelyd)None of these.Correct answer is option 'A'. Can you explain this answer?.
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