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SolvedMidterm08

# SolvedMidterm08 - EE 231E Channel Coding Instructor Rick...

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EE 231E Spring 08 Midterm Channel Coding Tuesday, May 6, 2008 Instructor: Rick Wesel 82+10 pts, 110 minutes SOLUTIONS: 1. (10 pts) Proof of Singleton. State and prove the Singleton Bound: Solution: The Singleton bound states that d min n - k + 1. The proof is as follows: For an alphabet of size q , there are q n possible sensewords. Consider a sub-word using k - 1 of the n symbol positions. There are q k - 1 possible sub-word values, but q k possible codewords. Hence at least two codewords share the same sub-word value. These two codewords differ in at most n - ( k - 1) symbol positions, proving the bound. 1

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2. (10 pts) Galois Field 2 3 For GF (2 3 ), a primitive polynomial is x 3 + x + 1. Generate representations for the powers of α as polynomials in α that are degree-two or less. Then, give the degree- two-or-less polynomial representation of α 3 + α 5 . α j Polynomial in α of degree less than 3 α α α 2 α 2 α 3 α + 1 α 4 α 2 + α α 5 α 2 + α + 1 α 6 α 2 + 1 α 7 1 α 3 + α 5 = α + 1 + α 2 + α + 1 (1) = α 2 (2) 3. (10 pts) Factors and Blocklengths. (a) Which polynomials of the form x n - 1 are factors (over GF (2)) of x 63 - 1? Solution: The values of n that are factors of 63 are the only ones that work since this implies the existence of an order- n element. x - 1 (3) x 3 - 1 (4) x 7 - 1 (5) x 9 - 1 (6) x 21 - 1 (7) x 63 - 1 (8) (b) What are the possible blocklengths of cyclic codes with symbols from GF (64)? Solution: For G ( x ) to generate a cyclic code of blocklength n it must be a factor of x n - 1. This is only possible when x n - 1 has all of its roots in the extension field. Hence the possible blocklengths are all factors of 63: { 1 , 3 , 7 , 9 , 21 , 63 } 2
4. (10 pts) BCH Code Design. The following table lists the conjugates and their corresponding minimal polynomials for GF(64). Table 1: Cyclotomic cosets (conjugates) and their minimal polynomials for GF(64). Conjugacy Class Minimal Polynomial 1 x + 1 α, α 2 , α 4 , α 8 , α 16 , α 32 x 6 + x + 1 α 3 , α 6 , α 12 , α 24 , α 33 , α 48 x 6 + x 4 + x 2 + x + 1 α 5 , α 10 , α 17 , α 20 , α 34 , α 40 x 6 + x 5 + x 2 + x + 1 α 7 , α 14 , α 28 , α 35 , α 49 , α 56 x 6 + x 3 + 1 α 9 , α 18 , α 36 x 3 + x 2 + 1 α 11 , α 22 , α 25 , α 37 , α 44 , α 50 x 6 + x 5 + x 3 + x 2 + 1 α 13 , α 19 , α 26 , α 38 , α 41 , α 52 x 6 + x 4 + x 3 + x + 1 α 15 , α 30 , α 39 , α 51 , α 57 , α 60 x 6 + x 5 + x 4 + x 2 + 1 α 21 , α 42 x 2 + x + 1 α 23 , α 29 , α 43 , α 46 , α 53 , α 58 x 6 + x 5 + x 4 + x + 1 α 27 , α 45 , α 54 x 3 + x + 1 α 31 , α 47 , α 55 , α 59 , α 61 , α 62 x 6 + x 5 + 1 Design a binary BCH code with length n = 63 for design distance δ = 8. You may express the generator polynomial as a product of minimal polynomials. To get full credit you should find the highest rate code possible.

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SolvedMidterm08 - EE 231E Channel Coding Instructor Rick...

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