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Consider a discrete memoryless source having a set of 5 letters each occurring with equal probability. If three letters are encoded at a time into a binary sequence, then what will be the efficiency of the binary code? (Answer up to three decimal place)
    Correct answer is '0.994'. Can you explain this answer?
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    Consider a discrete memoryless source having a set of 5 letters each ...
    Calculation of Efficiency of Binary Code for Discrete Memoryless Source with 5 Letters

    Step 1: Calculate the number of possible combinations for 3 letters to be encoded into a binary sequence.

    - Since each letter occurs with equal probability, the probability of choosing any one letter is 1/5.

    - Therefore, the number of possible combinations for 3 letters is 5 x 5 x 5 = 125.

    Step 2: Calculate the number of bits required to represent each combination.

    - Since there are 2 possible values for each bit (0 or 1), the number of bits required to represent each combination is log2(125) ≈ 6.98.

    - Since we can't use fractional bits, we need to round up to 7 bits.

    Step 3: Calculate the total number of bits required to encode all possible combinations.

    - There are 125 possible combinations, and each combination requires 7 bits.

    - Therefore, the total number of bits required is 125 x 7 = 875 bits.

    Step 4: Calculate the average number of bits per letter.

    - Since we are encoding 3 letters at a time, the average number of bits per letter is 875 / (3 x 5) = 58.33 bits/letter.

    Step 5: Calculate the efficiency of the binary code.

    - Efficiency = (number of bits in original message) / (number of bits in encoded message)

    - Since each letter is equally likely, the entropy of the source is log2(5) ≈ 2.32 bits/letter.

    - Therefore, the number of bits in the original message is 3 x 2.32 = 6.96 bits.

    - The efficiency of the binary code is 6.96 / 58.33 ≈ 0.994.

    Conclusion: The efficiency of the binary code for a discrete memoryless source with 5 letters each occurring with equal probability, when three letters are encoded at a time into a binary sequence, is 0.994.
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    Community Answer
    Consider a discrete memoryless source having a set of 5 letters each ...
    The source entropy is
    In case of encoding three letters at a time, we have 53 = 125 possible sequences.
    We need [log2 125] + 1 = [6.96] + 1 = 6 + 1 = 7 bits/3 letter
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    Consider a discrete memoryless source having a set of 5 letters each occurring with equal probability. If three letters are encoded at a time into a binary sequence, then what will be the efficiency of the binary code? (Answer up to three decimal place)Correct answer is '0.994'. Can you explain this answer?
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    Consider a discrete memoryless source having a set of 5 letters each occurring with equal probability. If three letters are encoded at a time into a binary sequence, then what will be the efficiency of the binary code? (Answer up to three decimal place)Correct answer is '0.994'. 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 Consider a discrete memoryless source having a set of 5 letters each occurring with equal probability. If three letters are encoded at a time into a binary sequence, then what will be the efficiency of the binary code? (Answer up to three decimal place)Correct answer is '0.994'. 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 Consider a discrete memoryless source having a set of 5 letters each occurring with equal probability. If three letters are encoded at a time into a binary sequence, then what will be the efficiency of the binary code? (Answer up to three decimal place)Correct answer is '0.994'. Can you explain this answer?.
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