Hybrid LDPC codes and iterative decoding methods - i3s

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Hybrid LDPC codes and iterative decoding methods - i3s

2.7 Proofs of theorems in Chapter 2 89

To shorten the notations we can omit the index of iteration t. Moreover, in the remainder

of this proof, we choose to use simpler notations although not fully rigorous: R (j,l)

denotes a message going into a check node of degree j in G(q l ) while R (i,k) denotes a

message going out of a variable of degree i in G(q k ). However, there is not ambiguity in

the following thanks to the unique use of indexes i, j, k, l and we always precise of which

nature is a message.

The n th component of a message coming from a variable of degree i in G(q k ) is denoted

by R n

(i,k) . The n th component of the initial message going into a variable in G(q k ) is

denoted by R n

(0)(k)

. The n th component of a message going into a degree i variable in

G(q k ) is denoted by L (i,k)

n . The data pass, through a variable node of degree n in G(q k ),

is translated by

R (i,k)

n

= R (0)(k)

n

∏i−1

p=1

L (i,k)

n

Let R (k)

t denote the average message going out of a variable node in G(q k ). By noting

that the messages L (i,k) are i.i.d. when (i, k) is set, we have:

D a (R (k)

t ) = ∑ i

= ∑ i

= ∑ i

1

Π(i|k)

q k − 1

1

Π(i|k)

q k − 1

1

Π(i|k)

q k − 1

q k


n=1

q k


n=1

q k


n=1



E⎝

∏ i−1

√ R(0)(k) n p=1 L(i,k) n


R (0)(k) i−1

0 p=1 L(i,k) 0



E⎝

√ R(0)(k) n

R (0)(k)

0



E⎝

√ R(0)(k) n

R (0)(k)

0

⎞ ⎛


⎠ E⎝




√ L(i,k) n

L (i,k)

0

⎠ D a (L (i,k) )

The last step is obtained thanks to the LM-invariance of L (i,k) . Finally we get:



i−1

D a (R (k)

t ) = D a (R (0)(k) ) ∑ i

Π(i|k)D a (L (i,k) ) (2.33)

Moreover, if we consider two LM-invariant vectors L (k) and L (l) , where L (k) is the random

truncation of L (l) , it is clear that D a (L (k) ) = D a (L (l) ). Hence:

D a (L (i,k) ) = ∑ j,l

Π(j, l|i, k)D a (L (j,l) ) (2.34)

where L (j,l) is the message going out of a check node of degree j in G(q l ).

Let us recall the result of equation (68) in [48]:

1 − D(L t ) ≥ ∑ d

ρ d (1 − D(R t )) d−1 + O ( D(R t ) 2)

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