Exercise-2.sol

# Exercise-2.sol - iii Compute the syndrome for the received...

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THE UNIVERSITY OF WESTERN ONTARIO FACULTY OF ENGINEERING ECE-433b-COMMUNICATION SYSTEMS EXERCISE # 2 Solution 1. Let ij x be the bit in row i and column j . Then the th i horizontal parity check is = j ij i x h where the summation is modulo-2. Summing both sides of the above equation in modulo-2 over the rows i , we obtain = j i ij i i x h , This shows that the modulo-2 sum of all the horizontal parity checks is the same as the modulo-2 sum of all data bits. The corresponding argument on columns verifies the vertical parity checks is the same. 2. Any pattern of the form ----110---- ----011---- ----101---- will fail to be detected by horizontal and vertical parity checks. That is, for any three rows 2 1 , i i and 3 i and any three columns 2 1 , j j , and 3 j , a pattern of six errors in positions ) , ( ), , ( ), , ( ), , ( ), , ( 1 3 3 2 2 2 2 1 1 1 j i j i j i j i j i and ) , ( 3 3 j i will fail to be detected. 3. Consider a (7,4) code whose generator matrix is = 1 0 0 0 0 1 1 0 1 0 0 1 1 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1 G i. Find all the code vectors of the code ii. Find the parity check matrix of the code.

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Unformatted text preview: iii. Compute the syndrome for the received vector 1101101. Is this a valid code vector? iv. What is the error correcting capability of the code? v. What is the error-detecting capability of the code? i) c=m G Messages Code Vectors 0000 0000000 0001 1100001 0010 0110010 0011 1010011 0100 1010100 0101 0110101 0110 1100110 0111 0000111 1000 1111000 1001 0011001 1010 1001010 1011 0101011 1100 0101100 1101 1001101 1110 0011110 1111 1111111 ii) [ ] T k n P I H -= = 1 1 1 1 1 1 1 1 1 1 1 1 iii) s = r T H = [ ] 1101101 1 1 1 1 1 1 1 1 1 1 1 1 = [ ] 010 Since the syndrome is not [ ] 000 , 1101101 is not a valid code word. iv) 3 min = d . Therefore the error correcting capability is 1 v) Error detecting capability is 2....
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Exercise-2.sol - iii Compute the syndrome for the received...

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