# tut6_sol - If r = 1 1 1 0 1 1 0 (second MSB in error), feed...

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1 EE 4212 Information and Coding Solution to Tutorial 6 A (7,3) linear block code has generator polynomial g ( X ) = 1 + X + X 2 + X 4 (a) The code is a valid cyclic code because the following two criteria are satisfied. (i) g ( X )is a factor of X n +1 since n =7 and X 7 +1 = (1+ X + X 3 ) (1+ X + X 2 + X 4 ) (ii) Degree of g ( X ) = n k = 7 – 3 = 4. (b) m = 1 0 0 => m ( X ) = 1 U ( X ) = m ( X ) g ( X ) = g ( X ) = 1 + X + X 2 + X 4 U = 1 1 1 0 1 0 0 (c) m = 1 0 0 => m ( X ) = 1 and X n-k m ( X ) = X 4 r ( X ) = X n-k m ( X ) mod g ( X ) = X 4 mod 1+ X + X 2 + X 4 = 1 + X + X 2 U ( X ) = r ( X ) + X n-k m ( X ) = 1 + X + X 2 + X 4 U = 1 1 1 0 1 0 0

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2 (d) 0000100 0 0 0 0 000010 0 0 0 0 00001 0 0 0 0 0000 1 0 0 0 000 0 1 0 0 00 0 0 1 0 0 0 0 0 1 1 1 1 0 => Remainder r ( X ) = 1 + X + X 2 0000100 X 2 X 3 X 1 g 1 g 2 g 4 g 0
3 (e) If e = 0000001 (i.e., MSB in error) => S ( X ) = X 6 mod g ( X ) = 1+ X + X 3 => check the syndrome parttern 1101
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Unformatted text preview: If r = 1 1 1 0 1 1 0 (second MSB in error), feed into the Meggit decoder. Switch at UP position. Shift Input ( r ) 1 X X 2 X 3 Quotient 1 1110110 0 0 0 0 2 111011 0 0 0 0 3 11101 1 0 0 0 4 1110 1 1 0 0 5 111 0 1 1 0 6 11 1 0 1 1 0 ( X 2 ) 7 1 0 0 1 1 1 ( X 1 ) 8 0 1 1 1 1 ( X ) Remainder = 0 1 1 1, not equal to 1101 =&gt; MSB of r is not in error. Switch move to DOWN position, input becomes e n-1 and continue to shift. &lt;= Bingo ! &lt;= S =0000, No more error =&gt; r ˆ = 111010 0, single bit error corrected. Shift Input ( e n-1 ) 1 X X 2 X 3 9 1 1 1 0 1 10 0 0 0 0 0 Syndrome Calculator Pattern Recognizer (AND gate) 7-bit Buffer r Switch r 0 r 1 … r n-1 e n-1 e n-1...
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## This note was uploaded on 02/26/2011 for the course EE 4011 taught by Professor Wadewe during the Spring '11 term at City University of Hong Kong.

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tut6_sol - If r = 1 1 1 0 1 1 0 (second MSB in error), feed...

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