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# HW06 - University of Illinois Spring 2011 ECE 361 Problem...

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University of Illinois Spring 2011 ECE 361: Problem Set 6: Solutions Erasure Channels; Gaussian Channel Capacity 1. [Single erasure correcting code] A simple “overall parity-check code” with generator matrix G = 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 suffices. Note that the transmitted codeword is [ u 0 , u 1 , u 2 , u 3 , u 0 u 1 u 2 u 3 ] = [ c 0 , c 1 , c 2 , c 3 , c 4 ] where c 4 is the overall parity-check bit. If c 4 is erased, no problems! [ c 0 , c 1 , c 2 , c 3 , ?] = [ u 0 , u 1 , u 2 , u 3 , ?] and the four data bits are received without error. But if one of the other bits is erased, then it can be obtained as the XOR sum of the four unerased bits. For example, if the receiver gets [ c 0 , c 1 , ? , c 3 , c 4 ], then c 0 c 1 c 3 c 4 = u 0 u 1 u 3 ( u 0 u 1 u 2 u 3 ) = u 2 , the erased data bit! Of course, the unerased data bits are just c 0 , c 1 , and c 3 . 2. [Decoding a Reed-Solomon on an erasure channel] (a) We have to interpolate ˆ d ( z ) of degree 3 through the 4 points (0 , 1) , (1 , 3) , (2 , 0) , (4 , 0). But since x i 3 = x i 4 = 0, the corresponding terms disappear from the sum defining ˆ d ( z ), and we have ˆ d ( z ) = 1 × ( z - 1)( z - 2)( z - 4) (0 - 1)(0 - 2)(0 - 4) + 3 × ( z - 0)( z - 2)( z - 4) (1 - 0)(1 - 2)(1 - 4) = ( z - 2)( z - 4) z - 1 6 + 3 z 3 = ( z - 2)( z - 4) since z/ 6 ≡ - z and - 1 / 6 = 1 modulo 7 = z 2 - 6 z + 8 z 2 + z + 1 modulo 7 . We can cross-check the arithmetic calculations by computing ˆ d ( i ) for i = 0 , 1 , 2 , 4 to verify that we do get 1 , 3 , 0 , 0 respectively Since ˆ d ( z ) = 1 + z + z 2 + 0 z 3 must equal d ( z ), we have that the

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HW06 - University of Illinois Spring 2011 ECE 361 Problem...

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