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In effect, at the end of every calculation, the answer is taken modulo 2.For instance, in standard arithmetic, x 4 G = 11112 . The correct code word c 4 is found by reducing each calculation modulo 2. In M atlab , this is done with mod(x4*g,2) where x4=[1,1] and g is defined as in [link] . In modulo 2 arithmetic, 1 represents any odd number and0 represents any even number. This is also true for negative numbers so that, for instance, - 1 = + 1 and - 4 = 0 .

After transmission, the received signal y is multiplied by H T . If there were no errors in transmission, then y is equal to one of the four code words c i . With H defined as in [link] , c 1 H T = c 2 H T = c 3 H T = c 4 H T = 0 , where the arithmetic is binary, and where 0 means thezero vector of size 1 by 3 (in general, 1 by ( n - k ) ). Thus y H T = 0 and the received signal is one of the code words.

However, when there are errors, y H T 0 , and the value can be used to determine the most likely errorto have occurred. To see how this works, rewrite

y = c + ( y - c ) c + e ,

where e represents the error(s) that have occurred in the transmission. Note that

y H T = ( c + e ) H T = c H T + e H T = e H T ,

since c H T = 0 . The value of e H T is used by looking in the syndrome [link] . For example, suppose that the symbol x 2 = 01 is transmitted using the code c 2 = 01011 . But an error occurs in transmission so that y = 11011 is received. Multiplication by the parity check matrix gives y H T = e H T = 101 . Looking this up in the syndrome table shows that the most likely error was 10000.Accordingly, the most likely code word to have been transmitted was y - e = 11011 - 10000 = 01011 , which is indeed the correct code word c 2 .

Syndrome Table for the binary ( 5 , 2 ) code with generator matrix [link] and parity check matrix [link]
Syndrome e H T Most lik Most likely error e
000 00000
001 00001
010 00010
011 01000
100 00100
101 10000
110 11000
111 10010

On the other hand, if more than one error occurred in a single symbol, then the (5,2) code cannot necessarilyfind the correct code word. For example, suppose that the symbol x 2 = 01 is transmitted using the code c 2 = 01011 but that two errors occur in transmission so that y = 00111 is received. Multiplication by the parity check matrix gives y H T = e H T = 111 . Looking this up in the syndrome table shows that the most likely error was 10010.Accordingly, the most likely symbol to have been transmitted was y - e = 00111 + 10010 = 10101 , which is the code word c 3 corresponding to the symbol x 3 , and not c 2 .

The syndrome table can be built as follows. First, take each possible single error pattern, that is, each ofthe n = 5 e 's with exactly one 1, and calculate e H T for each. As long as the columns of H are nonzero and distinct, each error pattern corresponds to a different syndrome.To fill out the remainder of the table, take each of the possible double errors (each of the e 's with exactly two 1's) and calculate e H T . Pick two that correspond to the remaining unused syndromes.Since there are many more possible double errors n ( n - 1 ) = 20 than there are syndromes ( 2 n - k = 8 ), these are beyond the ability of the code to correct.

The M atlab program blockcode52.m shows details of how this encoding and decoding proceeds. The first part defines the relevantparameters of the ( 5 , 2 ) binary linear block code: the generator g , the parity check matrix h , and the syndrome table syn . The rows of syn are ordered so that the binary digits of e H T can be used to directly index into the table.

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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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