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Unformatted text preview: University of California San Diego Fall 2008 ECE 259A: Solutions to the Midterm Exam
Problem 1. a. Assume to the contrary that some of the vectors have even weight and some have odd weight. Specifically, suppose we have x, y, z in our set, with wt x , wt y even and wt z odd. Then d x, z wt x wt z 2wt x z 1 mod 2 where x y denotes a vector that is nonzero only in those positions where both x and y are nonzero. This is a contradiction, since d x, y d x, z d. b. It is obvious from (1) that the distance between any two vectors of the same parity (that is, either both even weight or both odd weight) is even. Hence d must be even by (a). wt y c. Without loss of generality, let 0, x, y be the three vectors. Then wt x junction with d x, y wt x wt y 2wt x y d, implies that wt x y x y. Since wt u d 2, it is clear that u is at distance d 2 from 0, x, y. It remains Let u to show that such a vector u is unique. Assume to the contrary that there exists another vector v at distance d 2 from 0, x, y. Then d v, 0 d implies wt v d 2. Furthermore, wt x d, wt v d 2, and d v, x d 2 together imply that v i 1 only if xi 1. In a similar manner, we can show that vi 1 only if yi 1. Hence v x y u. Problem 2.
1 b. Since there is a 11 correspondence between syndromes and cosets of , it would suffice to construct a paritycheck matrix such that each of the errorpatterns in 2m 1 has a different synt 1i 0n i and require that Hei is the binary representation drome. If, for instance, we let ei of i, then the corresponding paritycheck matrix H may be constructed recursively as follows: H 0001000 0101010 1111111 Note that the minimum distance of the resulting code is 2, and hence on a BSC it cannot correct even single errors. c. The standardarray decoding scheme may be applied in this case as well, with the only difference that the 2m vectors in 2m 1 should be chosen as the cosetleaders for the 2 m cosets of . There will be a unique leader in each coset, by the construction of . h 1 , h2 , h n , such that i 1 hi s j where d. In general, we need to construct the matrix H s1 , s2 , . . . s2m 1 are all the nonzero binary mtuples in some fixed order. One way to satisfy the above property is to take h 1 s1 and hi si si 1 for i 2, 3, . . . n. j 1 ) 2 98272 1 3)5(32 6 242 )' 1 0(" a. We must have e error patterns in 2m 2n m or dim e for all distinct e, e are not always correctable. Hence n m. 2m 1 , 2m 1 otherwise some of the n 2 , which implies I d x, y wt x wt y 2wt x y 0 mod 2 H 4GF4 4 BC 1 D E @A 1 2 6 1 ) & % $#" ! 1 ) (1) (2) 1 62 1 32 d, in cond 2. Problem 3. It is easy to see that W is just the total weight of the 2 k n matrix M, having the codewords of as its rows. The fact that the distance of the dual code is 2 implies that there are no columns in the generator matrix of that are entirely zero. This, in turn, means that the weight of each column of M is exactly 2k 1 , and hence W n2k 1 . Problem 4. First, observe that all the columns in the paritycheck matrix H are nonzero and distinct. This implies that that the minimum distance of is at least 3. Now let and observe that h3 h 4 h 5 0. Hence, the minimum distance of is exactly 3. We claim that the covering radius of is 2. Recall that the covering radius is equal to the maximum number of columns of H required to represent every nonzero syndrome. There are 15 nonzero syndromes altogether. Thirteen of them are columns of H, and the other two can be represented as follows: 1 1 0 0 1 0 0 0 0 1 0 0 h1 h2 1 0 1 0 1 0 0 0 13 This proves that 2 and implies that every vector in the ambient space 2 is at distance at most 2 from . It follows that adjoining any such vector to will reduce its minimum distance. BU C @TA 8 BU C @TA BU C @TA S FG4 44 BU C @TA R H h1 , h2 , h13 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 1 0 h1 h3 VUBC P Q @TA 1 BU C @TA BU C @TA 1 ...
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 Fall '08
 Prof.AlexanderVardy

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