Hybrid LDPC codes and iterative decoding methods - i3s

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Hybrid LDPC codes and iterative decoding methods - i3s

84 Chapitre 2 : Hybrid LDPC Codes

Thus

P(W ×A = w) = e w Aa

P(W = (w +Aa ) ×A−1 )Π i/∈Im(A) δ w i ,∞ (2.28)

But we note that:

P(W ×A = w +Aa ) = P(W = (w +Aa ) ×A−1 )Π j /∈Im(A) δ w +Aa

j ,∞

(2.29)

• We first examine the case w Aa = ∞:

P(W ×A = w +Aa ) ≠ 0 ⇒ ∀i /∈ Im(A), = ∞

(

But, if y = LDR −1 (w), w +Aa y

i = log Aa

y Aa+i

), and since y Aa = 0 because w Aa =

∞, we cannot have w i

+Aa = ∞, ∀i /∈ Im(A). Hence we have w Aa = ∞ ⇒

P(W ×A = w +Aa ) = 0. This proof by contradiction ensures that equation (2.26) is

fulfilled when w Aa = ∞.

w +Aa

i

• Then we examine the case w Aa = −∞:

But w Aa = log

(

y0

w i = log

( )

y 0

y i

P(W ×A = w) ≠ 0 ⇒ ∀i /∈ Im(A),

y Aa

)

w i = ∞

= −∞ implies that y 0 = 0. Hence we cannot have

for all i ∈ [0, q 2 − 1]. Hence we have w Aa = −∞ ⇒ P(W ×A =

w) = 0. This proof by contradiction ensures that equation (2.26) is fulfilled when

w Aa = −∞.

• Finally we examine the case w Aa /∈ {−∞, ∞}:

In this case, for all j ∈ [0, q 2 − 1], δ w

+Aa

j ,∞ = δ w Aa+j −w Aa ,∞ = δ wAa+j ,∞. For

all i ∈ [0, q 2 − 1], if i /∈ Im(A), then ∃j /∈ Im(A): i = Aa + j. Therefore

{i ∈ [0, q 2 − 1]s.t.i /∈ Im(A)} = {j ∈ [0, q 2 − 1]s.t.Aa + j /∈ Im(A)}. We finally

obtain:

Π j /∈Im(A) δ w +Aa

j ,∞ = Π i/∈Im(A) δ w i ,∞

The above equality allows to insert equation (2.29) into equation (2.28). We can

now conclude that, when w Aa /∈ {−∞, ∞}, equation (2.26) is satisfied.

This completes the proof of the first part of lemma 6.

We now prove that any truncation of a symmetric random vector remains symmetric.

We have to prove that

∀a ∈ [0, q 1 − 1], P(W ×A−1 = w) = e wa P(W ×A−1 = w +a ) (2.30)