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Unformatted text preview: 1 ECE 534: Elements of Information Theory Midterm Exam I (Spring 2007) Problem 1 [20 pts] What are the relations ( ≥ , = , ≤ ) between the following pairs of expressions? Explain why. 1) H (5 X ) and H ( X ) ; 2) H ( X  X 1 ) and H ( X  X 1 ,X 1 ) ; 3) H ( X 1 ,X 2 ,...,X n ) and H ( c ( X 1 ,X 2 ,...,X n )) , where c ( x 1 ,x 2 ,...,x n ) is the Huffman codeword assigned to ( x 1 ,x 2 ,...,x n ) ; 4) H ( X,Y ) and H ( X ) + H ( Y ) . 5) H ( X n  X 1 ,...,X n 1 ) and H ( X n +1  X 1 ,...,X n ) , where ( X 1 ,X 2 ,...,X n ,... ) is a stationary random process. Solution: (Provided by Bhanu Kamandla and Ishita Basu) 1) H (5 X ) = H ( X ) , since entropy depends on the probability of the variable not on the value of the variable. Another explanation is that f ( x ) = 5 x is an invertible function of x . 2) H ( X  X 1 ) ≥ H ( X  X 1 ,X 1 ) , since conditioning reduces entropy. 3) H ( X 1 ,X 2 ,...,Xn ) = H ( c ( X 1 ,X 2 ,...,X n )) , since Huffman code is uniquely decodable, that is c ( X 1 ,...,X n ) is an invertible function. 4) H ( X,Y ) = H ( X ) + H ( Y  X ) ≤ H ( X ) + H ( Y ) (chain rule and conditioning reduces entropy) 5) H ( X n  X 1 ,...,X n 1 ) = H ( X n +1  X 2 ,...,X n ) Since X 1 ,X 2 ,...,X n is a stationary random pro cess. H ( X n +1  X 1 ,X 2 ,...,X n ) ≤ H ( X n +1  X 2 ,...,X n ) , since conditioning reduces entropy. Therefore H ( X n  X 1 ,...,X n 1 ) ≥ H ( X n +1  X 1 ,X 2 ,...,X n ) . Problem 2 [20 pts] (a) Find a binary Huffman code for the following random variable: X = parenleftBigg x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 . 35 0 . 26 0 . 19 0 . 07 0 . 04 0 . 04 0 . 03 0 . 02 parenrightBigg (b) Find a ternary Huffman code for the above source.(b) Find a ternary Huffman code for the above source....
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This note was uploaded on 10/24/2011 for the course ELECTRICAL ECE 571 taught by Professor Kelly during the Spring '11 term at University of Illinois, Urbana Champaign.
 Spring '11
 Kelly

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