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Unformatted text preview: c 2 . We denote the distance between these two codewords by ( ) 2 1 , c c d . By definition, . ( ) min 2 1 , d c c d ≥ Let the given bit string be x . Since we can always flip ( ) x c d , 1 bits to change c 1 into x , and then flip bits to change x into c ( 2 , c x d ) 2 , we must have ( ) ( ) ( ) 2 1 2 1 , , , c c d c x d x c d ≥ + . Combining the above two results, we have ( ) ( ) min 2 1 , , d c x d x c d ≥ + . If , then we must have ( ) 2 / , min 1 d x c d < ( ) min 2 , d c x d > . b) If the number of errors is fewer than d min / 2, then according to the result of part (a), the received bit string must be closer to the original codeword than to any other codewords....
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This note was uploaded on 05/01/2011 for the course ELECTRICAL EE3015 taught by Professor Albertsun during the Spring '11 term at City University of Hong Kong.
- Spring '11