Exercise-1.sol - THE UNIVERSITY OF WESTERN ONTARIO...

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Unformatted text preview: THE UNIVERSITY OF WESTERN ONTARIO DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING ECE-433b-COMMUNICATION THEORY EXERCISE # 1 Solutions 1. For k =3 The parity bit 1 b is chosen based on 1 3 2 1 = b m m m , where is used to denote modulo-2 addition. Message bits 3 2 1 m m m Parity bit 1 b Codeword ( 1 3 2 1 b m m m ) 000 0000 001 1 0011 010 1 0101 011 0110 100 1 1001 101 1010 110 1100 111 1 1111 The code, C , therefore consists of 8 code words given by: C={ , 1010 1001 , 0110 , 0101 , 0011 , 0000 6 5 4 3 2 1 = = = = = = C C C C C C 1111 , 1100 8 7 = = C C } (a) Distance of the code C = min d = min , j i j i ( 29 } , { j i C C d =2 (b) Code Rate = n k = 4 3 (c) Percentage Bandwidth Expansion = 100 1 - k n = 100 1 3 4 - =33% The distance between two code words ) , ( j i C C d between the pair of code vectors i C and j C is defined as the number of locations in which their respective elements differ. For example, distance between elements differ....
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This note was uploaded on 06/15/2008 for the course COMPUTER S 853 taught by Professor Choi during the Spring '08 term at Seoul National.

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Exercise-1.sol - THE UNIVERSITY OF WESTERN ONTARIO...

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