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Unformatted text preview: University of California San Diego ECE 259A: Solutions to Problem Set #1
1. Suppose that a vector y corrupted by at most errors and erasures was observed at the channel output. Delete (puncture) from both y and the positions where erasures have occurred, to is at least d 2 1. Hence, we can obtain y and . The minimum distance of correct the errors in y to recover x , which stands for the transmitted codeword x with the positions erased. Since d x 1 , x2 d for any x1 , x2 , the unique codeword in which agrees with x in all the unerased positions is x. 2. There are precisely three binary linear MDS codes: the n, 1, n repetition code, the n, n 1, 2 evenweight code, and the n, n, 1 code equal to the entire space 2n . Let be a n, k, n k 1 binary linear code with 2 k n 2, and let G Ik A be a systematic generator matrix for . Since d n k 1 each row in G has weight n k 1, and therefore A must be n k all1 matrix. Hence, adding any two rows of G produces a codeword of weight 2. ak Yet, for k n 2 we must have d n k 1 2. Note: The trivial code consisting of a single codeword is sometimes also considered MDS. However, the minimum distance is not welldefined for this code. Question: Are there any nonlinear binary MDS codes with 2 k n 2? (Hint: no, use arguments similar to those employed in the proof of the Singleton bound.) 3. (a) By elementary row operations: (b) A paritycheck matrix in systematic form is given by: H A
t (c) The minimum distance is at most 2, since G contains a row of weight 2. As all the rows of A are distinct, any linear combination of the rows of G has weight at least 2. Hence d 2. d d  since we are appending an extra coordinate to a code with distance d. d d 1  since we are appending only one coordinate. d is even  since the weight of all the codewords in is even. is given by: The paritycheck matrix for 0 ( ## H H && & 33 443 1 1 1 1 0 0 . . . 0 '&&% ) ) "## #$ 4. Clearly is a n , k , d follows from: code, where n n 1, k k, and d d 1. The last equality ( "#$ ! I4 100 001 101 010 1000 0100 0010 0001 '&% G I3 A 100 1010 010 0001 001 0110 1 1 2 1 n 5. Note that for any x, y y wt x wt y 2wt x y , where x y is 2 we have wt x the vector having 1's at those positions where both x and y have 1's. Hence the weight of x y is even if and only if either both x and y have even weight, or both have odd weight. Let e and o denote the sets of all codewords in a binary linear code with even and odd weights, respectively. If o is empty there is nothing to prove, hence assume o , and let x o. Then y y z e : x o . This shows that o e . Similarly z o : x e , which implies e . Hence o . Now let 0 and 1 denote the sets of all codewords o e in with 0 and 1 at the given position, respectively. Proceed as before. Assuming x , 1 we have y y z 0 : x 1 and z 1 : x 0. 6. Let d be odd, and let be an n, M, d code with M A 2 n, d . By appending an overall even parity check bit to each codeword of , we obtain an n 1, M, d 1 code. This shows that A2 n, d . Now let be a n 1, M, d 1 code with M A 2 n 1, d 1 . A2 n 1, d 1 Consider the set of positions where two codewords of , at distance d 1 from each other, differ. By deleting (puncturing) any position in this set, we obtain an n, M, d code. This implies A2 n, d A2 n 1, d 1 . The optional part (b) of the problem is actually a very old conjecture in coding theory. See, for example, the article by Bernard Elspas, "A conjecture on binary nongroup codes," in the IEEE T RANSACTIONS ON I NFORMATION T HEORY, pp. 599600, October 1965. The extra credit promised is A for the course (at least!). 7. (a) By definition: where the summation in (1) is modulo 2. Let b 1 , b 2 , . . . b k be the rows of a generator matrix G for . Since any x is a linear combination of b 1 , b2 , . . . bk , we have: Yet (2) is just another way of writing: v This shows that G is a paritycheck matrix for (c) Using part (a), we have dim n . (d) By (a) and (b), a paritycheck matrix for paritycheck matrix for the original code . is a generator matrix for 8. All the possible errorpatterns have syndrome Hy t s and it is easy to verify that the syndrome of e 0k s is H 0k s t s, provided H is systematic. Since wt e t and all the vectors of weight t are unique coset leaders, there couldn't be another "possible" errorpattern of weight t in the same coset with e and y. A conclusion here is that if the code is systematic (i.e. the first k bits of a codeword are equal to the information bits), then wt s t indicates that all the channel errors have occurred in the last n k bits, which may be simply discarded in this case. This fact is used in decoding techniques known as error trapping. A A Hence the vectors a 1 , a 2 , . . . a n k belong to linearly independent, and since dim n . Since H is fullrank these vectors are k they form a basis for .  which is a I (b) Let a 1 , a 2 , . . . a n k be the rows of H. Then by the definition of x : ai x 0 for i 1, 2, . . . n k rank G n H n 2 : Gvt 0 k. we have: H A A A I A 3 9 A A C B A 3 v n 2 : bi v 0 for i 1, 2, . . . k H 333 G44F 3 E C AB D 9 v n 2 : x x1 , x2 , . . . xn , x v x1 v1 xn vn 0 7 @6 786 9 5 5 9 C B A 9 9 P (1) (2) 9. The standard array for the code at hand is given by: code coset coset coset coset coset coset coset 00 00000 00001 00010 00100 01000 10000 00101 10100 01 10011 10010 10001 10111 11011 00011 10110 00111 10 01110 01111 01100 01010 00110 11110 01011 11010 11 11101 11100 11111 11001 10101 01101 11000 01001 The syndrome of 00111 with respect to H is 111 t . Note that there are two vectors of the minimum weight 2 in the coset corresponding to 111 t . Hence, using the MLD strategy, the vector 00111 may be decoded either as 10011 or as 01110 . R Q dca ihgfeXb`YS
000 001 010 100 110 011 101 111 R XUVUST)Q W SR ...
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 Fall '08
 Prof.AlexanderVardy

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