On the minimum distance of LDPC codes based on repetition codes and permutation matrices

Transcription

1 On the minimum distance of LDPC codes based on repetition codes and permutation matrices Fedor Ivanov Institute for Information Transmission Problems, Russian Academy of Science XV International Workshop on Algebraic and Combinatorial Coding Theory (ACCT) June, 2016 Albena, Bulgaria

2 Outline Definitions and notation Circulant matrices Auxiliary statements Code structure Lower bound on the minimum distance of proposed codes Simulation and numerical results Conclusion

3 Definitions and notation - I Notation Under R(n 0 ) we shall assume [n 0, 1, n 0 ] (n 0 > 1) repetition code of length n 0 and minimum distance d min = n 0.

4 Definitions and notation - I Notation Under R(n 0 ) we shall assume [n 0, 1, n 0 ] (n 0 > 1) repetition code of length n 0 and minimum distance d min = n 0. Notation Under GF m (2) (m > 1, m N) we shall assume a vector space of length m vectors over GF (2).

5 Definitions and notation - I Notation Under R(n 0 ) we shall assume [n 0, 1, n 0 ] (n 0 > 1) repetition code of length n 0 and minimum distance d min = n 0. Notation Under GF m (2) (m > 1, m N) we shall assume a vector space of length m vectors over GF (2). Notation Let y GF m (2), then under y we shall assume hamming weight of y.

6 Definitions and notation - II Notation Let y GF m (2), then under supp(y) we shall assume a support of y, i. e. supp(y) = {j : y j = 1}.

7 Definitions and notation - II Notation Let y GF m (2), then under supp(y) we shall assume a support of y, i. e. supp(y) = {j : y j = 1}. Notation Let y GF m (2), p Z,then under the set p + supp(y) we shall assume: p + supp(y) = {j + p mod m : y j = 1}.

8 Circulant matrices Definition Let m > 1, m N and I is a m m unity matrix. Let us choose an arbitrary p Z, then under I p we shall assume a matrix of p-times right cyclic shift of columns (or rows) of I.

9 Circulant matrices Definition Let m > 1, m N and I is a m m unity matrix. Let us choose an arbitrary p Z, then under I p we shall assume a matrix of p-times right cyclic shift of columns (or rows) of I. Matrix I p is an circulant with column and row weights 1. Also it is evident that I mk = I for all k Z.

10 Circulant matrices Definition Let m > 1, m N and I is a m m unity matrix. Let us choose an arbitrary p Z, then under I p we shall assume a matrix of p-times right cyclic shift of columns (or rows) of I. Matrix I p is an circulant with column and row weights 1. Also it is evident that I mk = I for all k Z. Moreover: I p1 I p2 = I p1 +p 2 mod m, I t p = I tp1 mod m, in particular if p 1 N, 0 p 1 m, then I 1 p 1 = I m p1. It is easy to note that the set I m = {I p : p Z} of m m matrices I p is a cyclic group with generator I 1.

11 Auxiliary statements If c = yi p, and supp(y) is the support of y, then supp(c) = p + supp(y).

12 Auxiliary statements If c = yi p, and supp(y) is the support of y, then supp(c) = p + supp(y). Lemma If I p I m, y GF m (2), y = w, and supp(y) = p + supp(y) then pw 0 mod m.

13 Auxiliary statements If c = yi p, and supp(y) is the support of y, then supp(c) = p + supp(y). Lemma If I p I m, y GF m (2), y = w, and supp(y) = p + supp(y) then pw 0 mod m. Corollary If y GF m (2), y = w, p Z and m Z is prime, then supp(y) = p + supp(y) only when w = m or w = 0.

14 Code structure - I Let us consider a parity-check matrix of H b of R(n 0 ): H b =

15 Code structure - I Let us consider a parity-check matrix of H b of R(n 0 ): Choose: H b = m > 1, m N

16 Code structure - I Let us consider a parity-check matrix of H b of R(n 0 ): Choose: m > 1, m N H b = k 0 > 0, k 0 N

17 Code structure - I Let us consider a parity-check matrix of H b of R(n 0 ): Choose: m > 1, m N H b = k 0 > 0, k 0 N (n 0 1)k 2 arbitraty matrices I pj, p j N, j = 1..2(n 0 1)k 2 0 from I m

18 Code structure - II P 11 P 1k = Q i P k0 1 P k0 k 0

19 Code structure - II P 11 P 1k = Q i P k0 1 P k0 k 0 Q i = I pi1 I pi2 I pi3... I pik0 I pi(k0 I +1) pi(k0 I +2) pi(k0... I +3) pi(2k0 ) I pi(k 2 0 k 0 +1) I pi(k 2 0 k 0 +2) I pi(k 2 0 k 0 +3)... I pik 2 0

20 Code Structure - III H = Q 1 Q n Q 2 0 Q n Q n Q 2(n0 1)

21 Code Structure - III H = Q 1 Q n Q 2 0 Q n Q n Q 2(n0 1) Size of H is mk 0 (n 0 1) mkn 0

22 Code Structure - III H = Q 1 Q n Q 2 0 Q n Q n Q 2(n0 1) Size of H is mk 0 (n 0 1) mkn 0 All rows have weight 2k 0

23 Code Structure - III H = Q 1 Q n Q 2 0 Q n Q n Q 2(n0 1) Size of H is mk 0 (n 0 1) mkn 0 All rows have weight 2k 0 Weights of first mk 0 columns are k 0 (n 0 1), other columns have weight k 0

24 Code Structure - IV We will consider matrix H as a parity-check matrix of LDPC code. Thus, choosing an arbitrary numbers m > 1, k 0 > 0 and 2(n 0 1)k 2 0 random elements from the group I m one can determine an ensemble of LDPC codes with the length n = mk 0 n 0. Let us denote this ensemble as E RC (m, k 0, n 0 ).

25 Code Structure - IV We will consider matrix H as a parity-check matrix of LDPC code. Thus, choosing an arbitrary numbers m > 1, k 0 > 0 and 2(n 0 1)k 2 0 random elements from the group I m one can determine an ensemble of LDPC codes with the length n = mk 0 n 0. Let us denote this ensemble as E RC (m, k 0, n 0 ). Definition An arbitrary code C E RC (m, k 0, n 0 ), will be called a LDPC code based on R(n 0 ) and permutation matrices.

26 Lower bound on the minimum distance of code from E RC (m, k 0, n 0 ) - auxiliary results Lemma Let C E RC (m, k 0, n 0 ) then for all k 0, n 0, (expect the case when simultaneously k 0 is even, and n 0 is odd) and for any c C: c is even.

27 Lower bound on the minimum distance of code from E RC (m, k 0, n 0 ) - auxiliary results Lemma Let C E RC (m, k 0, n 0 ) then for all k 0, n 0, (expect the case when simultaneously k 0 is even, and n 0 is odd) and for any c C: c is even. Lemma Let H is a parity-check matrix of code C from the ensemble E RC (m, k 0, n 0 ). If H has girth greater than 4, then d min (C) 4.

28 Lower bound on the minimum distance of code from E RC (m, k 0, n 0 ) - auxiliary results Lemma Let C E RC (m, k 0, n 0 ) then for all k 0, n 0, (expect the case when simultaneously k 0 is even, and n 0 is odd) and for any c C: c is even. Lemma Let H is a parity-check matrix of code C from the ensemble E RC (m, k 0, n 0 ). If H has girth greater than 4, then d min (C) 4. Lemma Let H is the parity-check matrix of code C from E RC (m, 2, n 0 ). If this matrix is free of cycles of length 4 and m > 5 is prime number, then d min (C) 8.

29 Lower bound on the minimum distance of code from E RC (m, k 0, n 0 ) - main result Theorem Let H is the parity-check matrix of code C from E RC (m, 2, k 0 ), and, moreover, let at least one sub-matrix (Q i Q n0 +i 1) of H (i = 1..n 0 1) is free of cycles of length 8, then d min (C) 10.

30 Lower bound on the minimum distance of code from E RC (m, k 0, n 0 ) - main result Theorem Let H is the parity-check matrix of code C from E RC (m, 2, k 0 ), and, moreover, let at least one sub-matrix (Q i Q n0 +i 1) of H (i = 1..n 0 1) is free of cycles of length 8, then d min (C) 10. Corollary Let H is the parity-check matrix of code C from E RC (m, 2, n 0 ), where n 0 > 4 and m > 5 is prime. If H is free of cycles of length 4 then d min (C) 10.

31 Simulation Results - Setup AWGN channel

32 Simulation Results - Setup AWGN channel BPSK modulation

33 Simulation Results - Setup AWGN channel BPSK modulation Sum-Product decoding algorithm

34 Simulation Results - Setup AWGN channel BPSK modulation Sum-Product decoding algorithm 50 iterations

35 Simulation Results - Setup AWGN channel BPSK modulation Sum-Product decoding algorithm 50 iterations

36 Simulation Results - Setup AWGN channel BPSK modulation Sum-Product decoding algorithm 50 iterations Table: Code constructions m n 0 k 0 n R d min

37 Numerical results for n = 56 EbNo P b, error rate N err, proposed D(N err ), proposed N err, PEG N err, ACE

38 Numerical results for n = 88 EbNo P b, error rate N err, proposed D(N err ), proposed N err, PEG N err, ACE

39 Simulation Results, FER versus E b /N o n = FER 10 2 proposed ACE, regular, l=3 PEG, irregular 10 3 gallager, l=3 Gallager, l=6 PEG, regular, l= E b /N o

40 Simulation Results, FER versus E b /N o n = FER 10 2 proposed ACE, regular, l=3 PEG, irregular 10 3 Gallager, l=3 Gallager, l=6 PEG, regular, l= E b /N o

41 Simulation Results, FER versus E b /N o n = FER 10 2 proposed 10 3 PEG, irregular ACE, regular, l=3 PEG, regular, l= E b /N o

42 Conclusion 1 New ensemble of low-rate LDPC codes was suggested

43 Conclusion 1 New ensemble of low-rate LDPC codes was suggested 2 A lower bound on minimal distance of proposed codes was obtained

44 Conclusion 1 New ensemble of low-rate LDPC codes was suggested 2 A lower bound on minimal distance of proposed codes was obtained 3 Simulation and numerical results allow us to conclude that proposed codes have an excellent performance even for very small code lengths

45 Thank you for your attention!