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Unformatted text preview: University of Illinois Fall 2005 ECE 556: Solutions to the First MidSemester Exam 1 . ( a ) Applying the result wt( x y ) = wt( x ) + wt( y ) 2 wt( x y ) to codewords x and y , we use the fact that wt( x ), wt( y ), and wt( x y ) are multiples of 4 to deduce that wt( x y ) is even. Hence x y for any two codewords, C C , and the rate of C is 1 2 . ( b ) Two binary vectors of odd Hamming weight sum to a vector of even Hamming weight. Now, the minimum Hamming weight of a code is the smallest number of columns of the paritycheck matrix H that sum to 0. Since the columns are distinct vectors of odd Hamming weight, two columns cannot sum to zero. The sum of three columns has odd Hamming weight, and hence cannot be zero. Thus, the minimum Hamming weight must be at least 4. 2 . ( a ) If f ( x ) = x 3 + x 2 +2 were reducible over GF(3), it would have at least one linear factor, which must be x, ( x 1) , or ( x 2) and correspondingly we would have f (0) = 0 or f (1) = 0 , or f (2) = 0. But, f (0) = f (1) = f (2) = 2, and hence f ( x ) is irreducible over GF(3)....
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This note was uploaded on 02/08/2012 for the course ECE 556 taught by Professor Milenkovic,o during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Milenkovic,O

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