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A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is as
- a)6000 bits/sec
- b)4500 bits/sec
- c)3000 bits/sec
- d)1500 bits/sec
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A source generates three symbols with probabilities 0.25, 0.25, 0.50 a...
Use Shannon Fano coding.
(0.25) is encoded as 01
Number of bits/sec= 0.25 x 3000 x 2 + 0.25 x 3000 x 2 + 0.5 x 3000
45000 bits/sec.Most Upvoted Answer
A source generates three symbols with probabilities 0.25, 0.25, 0.50 a...
Explanation:
Given, the probabilities of three symbols are 0.25, 0.25, and 0.50.
The most efficient source encoder would be the Huffman coding technique.
Huffman Coding:
Huffman coding is a lossless data compression technique that assigns codes to characters based on their frequency of occurrence. The characters that occur more frequently are assigned shorter codes, while those that occur less frequently are assigned longer codes.
Calculation:
The average bit rate of the Huffman coding technique can be calculated as follows:
Average bit rate = (Probability of symbol 1 x Length of code for symbol 1) + (Probability of symbol 2 x Length of code for symbol 2) + (Probability of symbol 3 x Length of code for symbol 3)
Using the Huffman coding technique, the length of code for symbol 1 and symbol 2 would be 2 bits, while the length of code for symbol 3 would be 1 bit.
Therefore, the average bit rate would be:
Average bit rate = (0.25 x 2) + (0.25 x 2) + (0.50 x 1) = 0.5 + 0.5 + 0.5 = 1.5 bits per symbol
Since the rate of generation of symbols is 3000 symbols per second, the total bit rate would be:
Total bit rate = 3000 x 1.5 = 4500 bits per second
Thus, the most efficient source encoder would have an average bit rate of 4500 bits per second.
Hence, the correct option is (b) 4500 bits/sec.
Given, the probabilities of three symbols are 0.25, 0.25, and 0.50.
The most efficient source encoder would be the Huffman coding technique.
Huffman Coding:
Huffman coding is a lossless data compression technique that assigns codes to characters based on their frequency of occurrence. The characters that occur more frequently are assigned shorter codes, while those that occur less frequently are assigned longer codes.
Calculation:
The average bit rate of the Huffman coding technique can be calculated as follows:
Average bit rate = (Probability of symbol 1 x Length of code for symbol 1) + (Probability of symbol 2 x Length of code for symbol 2) + (Probability of symbol 3 x Length of code for symbol 3)
Using the Huffman coding technique, the length of code for symbol 1 and symbol 2 would be 2 bits, while the length of code for symbol 3 would be 1 bit.
Therefore, the average bit rate would be:
Average bit rate = (0.25 x 2) + (0.25 x 2) + (0.50 x 1) = 0.5 + 0.5 + 0.5 = 1.5 bits per symbol
Since the rate of generation of symbols is 3000 symbols per second, the total bit rate would be:
Total bit rate = 3000 x 1.5 = 4500 bits per second
Thus, the most efficient source encoder would have an average bit rate of 4500 bits per second.
Hence, the correct option is (b) 4500 bits/sec.
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A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is asa)6000 bits/secb)4500 bits/secc)3000 bits/secd)1500 bits/secCorrect answer is option 'B'. Can you explain this answer?
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A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is asa)6000 bits/secb)4500 bits/secc)3000 bits/secd)1500 bits/secCorrect 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 source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is asa)6000 bits/secb)4500 bits/secc)3000 bits/secd)1500 bits/secCorrect 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 source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is asa)6000 bits/secb)4500 bits/secc)3000 bits/secd)1500 bits/secCorrect answer is option 'B'. Can you explain this answer?.
A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is asa)6000 bits/secb)4500 bits/secc)3000 bits/secd)1500 bits/secCorrect 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 source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is asa)6000 bits/secb)4500 bits/secc)3000 bits/secd)1500 bits/secCorrect 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 source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is asa)6000 bits/secb)4500 bits/secc)3000 bits/secd)1500 bits/secCorrect answer is option 'B'. Can you explain this answer?.
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