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BCH Codes: Problem Set 7 Solutions from University of Illinois, Fall 2004, Assignments of Electrical and Electronics Engineering

University of Illinois - Urbana-ChampaignElectrical and Electronics Engineering

The solutions to problem set 7 of the bch codes course offered by the university of illinois during the fall 2004 semester. It includes the calculation of generator polynomials, rates, and minimum distances for various bch codes.

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Pre 2010

Uploaded on 02/24/2010

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University of Illinois Fall 2004
ECE 556/CS 577/MATH 579: Solutions to Problem Set 7
1. Let αdenote a primitive element of GF(25), and let Mi(x) denote the minimal poly-
nomial of αiover GF(2). Then, the the conjugacy constraints imply that
•M1(x) = M2(x) = M4(x) = M8(x) = M16(x)
•M3(x) = M6(x) = M12(x) = M24 (x) = M17(x)
•M5(x) = M10(x) = M20 (x) = M9(x) = M18(x)
•M7(x) = M14(x) = M28 (x) = M25(x) = M19 (x)
•M11(x) = M22 (x) = M13(x) = M26 (x) = M21(x)
•M15(x) = M30 (x) = M29(x) = M27 (x) = M23(x)
(a) Thus, we have the following generator polynomials and rates:
t g(x) rate
1M1(x) 26/31
2M1(x)M3(x) 21/31
3M1(x)M3(x)M5(x) 16/31
4M1(x)M3(x)M5(x)M7(x) 11/31
5M1(x)M3(x)M5(x)M7(x) 11/31
6M1(x)M3(x)M5(x)M7(x)M11(x) 6/31
7M1(x)M3(x)M5(x)M7(x)M11(x) 6/31
8M1(x)M3(x)M5(x)M7(x)M11(x)M15(x) 1/31
Note that the 4-error-correcting BCH code is also the 5-error-correcting BCH
code, and similarly for t= 6 and t= 7. The 8-error-correcting code is actually
the repetition code that can correct 15 errors. Such serendipitious increases in
error-correcting capability are often seen in BCH codes of long block lengths.
(b) Since g(x) = M5(x)M7(x)M11(x)M15(x)M0(x), it has 7 consecutive powers of
αas roots, viz. α25 ,α26, . . . , α30,α31 = 1. Hence, the BCH bound on the
minimum distance is 8. The actual minimum distance of the code happens to
be 12. Exercise: are the Hartmann-Tzeng or Roos bounds tighter than the BCH
bound?
(c) Since g(x) = M3(x)M7(x)M11(x)M15(x)M0(x), it has 11 consecutive powers of α
as roots, viz. α21,α22 , . . . , α30,α31 = 1. Hence, the BCH bound on the minimum
distance is 12. The actual minimum distance is also 12.
(d) If we write the exponents of conjugate elements as m-bit binary (base-2) numbers,
we discover the pleasing property that the m-bit representations are cyclic shifts
of one another. For example, the exponents of the roots of M5are
001012= 5,010102= 10,101002= 20,010012= 9,100102= 18.
More generally, the exponent of α1+2(m+1)/2is of the form 000. . . 0100 . . . 01 where
there are (m−3)/2 zeroes preceding the leftmost 1 and (m−1)/2 zeroes between
pf2

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Download BCH Codes: Problem Set 7 Solutions from University of Illinois, Fall 2004 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!

University of Illinois Fall 2004

ECE 556/CS 577/MATH 579: Solutions to Problem Set 7

  1. Let α denote a primitive element of GF(2^5 ), and let Mi(x) denote the minimal poly- nomial of αi^ over GF(2). Then, the the conjugacy constraints imply that - M 1 (x) = M 2 (x) = M 4 (x) = M 8 (x) = M 16 (x) - M 3 (x) = M 6 (x) = M 12 (x) = M 24 (x) = M 17 (x) - M 5 (x) = M 10 (x) = M 20 (x) = M 9 (x) = M 18 (x) - M 7 (x) = M 14 (x) = M 28 (x) = M 25 (x) = M 19 (x) - M 11 (x) = M 22 (x) = M 13 (x) = M 26 (x) = M 21 (x) - M 15 (x) = M 30 (x) = M 29 (x) = M 27 (x) = M 23 (x)

(a) Thus, we have the following generator polynomials and rates: t g(x) rate 1 M 1 (x) 26 / 31 2 M 1 (x)M 3 (x) 21 / 31 3 M 1 (x)M 3 (x)M 5 (x) 16 / 31 4 M 1 (x)M 3 (x)M 5 (x)M 7 (x) 11 / 31 5 M 1 (x)M 3 (x)M 5 (x)M 7 (x) 11 / 31 6 M 1 (x)M 3 (x)M 5 (x)M 7 (x)M 11 (x) 6 / 31 7 M 1 (x)M 3 (x)M 5 (x)M 7 (x)M 11 (x) 6 / 31 8 M 1 (x)M 3 (x)M 5 (x)M 7 (x)M 11 (x)M 15 (x) 1 / 31

Note that the 4-error-correcting BCH code is also the 5-error-correcting BCH code, and similarly for t = 6 and t = 7. The 8-error-correcting code is actually the repetition code that can correct 15 errors. Such serendipitious increases in error-correcting capability are often seen in BCH codes of long block lengths. (b) Since g(x) = M 5 (x)M 7 (x)M 11 (x)M 15 (x)M 0 (x), it has 7 consecutive powers of α as roots, viz. α^25 , α^26 ,... , α^30 , α^31 = 1. Hence, the BCH bound on the minimum distance is 8. The actual minimum distance of the code happens to be 12. Exercise: are the Hartmann-Tzeng or Roos bounds tighter than the BCH bound? (c) Since g(x) = M 3 (x)M 7 (x)M 11 (x)M 15 (x)M 0 (x), it has 11 consecutive powers of α as roots, viz. α^21 , α^22 ,... , α^30 , α^31 = 1. Hence, the BCH bound on the minimum distance is 12. The actual minimum distance is also 12. (d) If we write the exponents of conjugate elements as m-bit binary (base-2) numbers, we discover the pleasing property that the m-bit representations are cyclic shifts of one another. For example, the exponents of the roots of M 5 are

001012 = 5, 010102 = 10, 101002 = 20, 010012 = 9, 100102 = 18.

More generally, the exponent of α1+

(m+1)/ 2 is of the form 000... 0100... 01 where there are (m − 3)/2 zeroes preceding the leftmost 1 and (m − 1)/2 zeroes between

the two 1s. Thus, α1+

(m−1)/ 2 is a conjugate of α1+

(m+1)/ 2

. For the faint of heart, here is another argument:

( α1+

(m+1)/ 2 )^2 (m−1)/^2 = α^2

m+2(m−1)/ 2 = α1+

(m−1)/ 2

since α^2

m = α. To find the BCH bound, note that the highest power of α that is a root of h(x) is 2m−^1 + 2(m−1)/^2 , and hence g(x) has as roots the 2m−^1 − 2 (m−1)/^2 − 1 consecutive powers

2 m−^1 + 2(m−1)/^2 + 1, 2 m−^1 + 2(m−1)/^2 + 2,... , 2 m^ − 1

giving a BCH bound on the minimum distance of 2m−^1 − 2 (m−1)/^2. The maximum weight (and hence maximum distance) in this code is 2m−^1 + 2(m−1)/^2. The inner product of {+1, − 1 } sequences is n− 2 d where d is the Hamming distance between the sequences, and hence the correlations are bounded between 2m^ − 1 − 2(2m−^1 − 2 (m−1)/^2 ) = 2(m+1)/^2 − 1 and 2m^ − 1 − 2(2m−^1 + 2(m−1)/^2 ) = − 2 (m+1)/^2 − 1, i.e., the magnitude is at most 2(m+1)/^2 + 1.

  1. (a) Since α^2 i^ is a conjugate of αi, g(x) is a product of at most t different minimal polynomials. Since each such minimal polynomial has degree m (or a divisor thereof,) the maximum possible degree of g(x) is mt, and the rate of the code is at least (n − mt)/n = 1 − (t/n) · log 2 (n + 1). (b) If t/n is constant, the lower bound on the rate decreases as n increases (and soon becomes negative, and therefore useless). But, we can improve the bound to R ≥ 0 since the rate cannot be negative.
  2. The matrix M 3 with extension can be triangularized via row operations to give

M 3 =

α^14 α^13 1 α^11 α^13 1 α^11 α^5 1 α^11 α^5

α^14 α^13 1 α^11 0 α^11 α^10 0 α^10 α^2 α^11

α^14 α^13 1 α^11 0 α^11 α^10 0 0 α^11 α^10

giving that λ 1 = α^10 /α^11 = α^14 , α^11 λ 2 + α^10 λ 1 = α^11 λ 2 + α^10 α^14 = 1 giving λ 2 = α^11 and similarly, λ 3 = α^14 which is the same result as in the book.

  1. The syndrome sequence is S 1 , S 2 ,... , S 6 = α^5 , α^10 , α^2 , α^5 , α^0 , α^4. Hence we triangulate

M 3 =

α^5 α^10 α^2 α^10 α^2 α^5 α^2 α^5

α^5 α^10 α^2 0 α^8 α^13 0 α^13 α^3

α^5 α^10 α^2 0 α^8 α^13 0 0 0

M 2 =

[

α^5 α^10 α^2 α^10 α^2 α^5

]

[

α^5 α^10 α^2 0 α^8 α^13

]

giving λ 1 = α^5 and λ 2 = α^3 , and therefore X 1 = α, X 2 = α^2. The PGZ algorithm now solves Y 1 α + Y 2 α^2 = S 1 = α^5 and Y 1 α^2 + Y 2 α^4 = S 2 = α^10 to get Y 1 = Y 2 = 1. The Forney formula needs the computation of ω(z) = λ(z)S(z) mod z^6 = α^5. Since λ′(z) = α^5 , we have that ω(z)/λ′(z) = 1, as it ought to be: the error values are both 1.