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Decoding Collided Acknowledgement Transmission in IEEE 802.11 ...

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which shows that c is a column of MN. S<strong>in</strong>ce Ω+(G1, c) =<br />

Ω−(G1, c), by (3) we thus have F (G1, c) = 0.<br />

Proposition 1: Consider a certa<strong>in</strong> system with N stations<br />

where N ≥ 1. For any G1, G2 ⊂ G and G1, G2 = ∅, −→ G1 = −→ G2<br />

iif G1 = G2.<br />

Proof: By def<strong>in</strong>ition given <strong>in</strong> (4), we can easily see that<br />

if G1 = G2, then −→ G1 = −→ G2. In the follow<strong>in</strong>g, we shall prove<br />

that if G1 = G2, then −→ G1 = −→ G2.<br />

We will exam<strong>in</strong>e two cases: (i) |G1| is an odd <strong>in</strong>teger; (ii)<br />

|G1| is an even <strong>in</strong>teger. Without loss of generality, we assume<br />

that |G1| ≥ |G2|.<br />

Case (i): |G1| is an odd <strong>in</strong>teger. Let g = |G1|. Based on<br />

lemma 1, there exists a column c <strong>in</strong> MN such that<br />

and hence accord<strong>in</strong>g to (4), we also have<br />

−→<br />

G1(c) = 1.<br />

F (G1, c) = 1, (6)<br />

This column will have the follow<strong>in</strong>g property<br />

Ω+(G1, c) = g+1<br />

2<br />

Ω−(G1, c) = Ω+(G1, c) − 1 = g−1<br />

2<br />

Ω+(G ′ 1, c) = Ω−(G ′ 1, c) = N−g<br />

2<br />

with any permutation.<br />

If G1 = G2, there exists K1 = G1 \ G2 = ∅ and K2 =<br />

G2 \ G1 where |K1| = k1 ≥ 1 and |K2| = k2 ≥ 0. It is clear<br />

that K1 ⊆ G1, K2 ⊆ G ′ 1 imply<br />

1 ≤ k1 ≤ g, k2 ≤ N − g.<br />

where g is an odd <strong>in</strong>teger and N − g is an even <strong>in</strong>teger.<br />

With the above condition, there exists a permutation with<strong>in</strong><br />

c such that<br />

Ω+(K1, c) = ⌊ k1<br />

g+1<br />

2 ⌋ + 1 ≤ 2<br />

Ω−(K1, c) = ⌊ k1−1 g−1<br />

2 ⌋ ≤ 2<br />

Ω+(K2, c) = ⌊ k2 N−g<br />

2 ⌋ ≤ 2<br />

Ω−(K2, c) = ⌊ k2+1 N−g<br />

2 ⌋ ≤ 2<br />

that confirms K1 ⊆ G1, K2 ⊆ G ′ 1 given (7). The immediate<br />

result gives<br />

and<br />

which yield<br />

F (K1, c) =<br />

F (K2, c) =<br />

1, k1 = 1, 3, ...<br />

2, k1 = 2, 4, ...<br />

−1, k2 = 1, 3, ...<br />

0, k2 = 0, 2, 4, ...<br />

(7)<br />

(8)<br />

(9)<br />

F (K1, c) − F (K2, c) ≥ 1. (10)<br />

S<strong>in</strong>ce G2 = (G1 \ K1) ∪ K2, with (6) and (10), we obta<strong>in</strong><br />

F (G2, c) = F (G1, c) − F (K1, c) + F (K2, c) ≤ 0<br />

and hence −→ G2(c) = 0. S<strong>in</strong>ce −→ G1(c) = 1, there exists a column<br />

<strong>in</strong> MN such that −→ G2(c) = −→ G2(c), which is sufficient to show<br />

that −→ G1 = −→ G2.<br />

Case (ii): |G1| is an even <strong>in</strong>teger. Let g = |G1|. Based on<br />

lemma 2, there exists a column c <strong>in</strong> MN such that<br />

and hence accord<strong>in</strong>g to (4), we also have<br />

−→<br />

G1(c) = 0.<br />

F (G1, c) = 0 (11)<br />

This column will have the follow<strong>in</strong>g property<br />

Ω+(G1, c) = Ω−(G1, c) = g<br />

2<br />

Ω+(G ′ 1, c) = N−g+1<br />

2<br />

Ω−(G ′ 1, c) = N−g−1<br />

2<br />

(12)<br />

with any permutation.<br />

If G1 = G2, there exist K1 = G1 \ G2 = ∅ and K2 =<br />

G2 \ G1 where |K1| = k1 ≥ 1 and |K2| = k2 ≥ 0. It is clear<br />

that K1 ⊆ G1, K2 ⊆ G ′ 1 imply<br />

1 ≤ k1 ≤ g, k2 ≤ N − g.<br />

where g is an nonzero even <strong>in</strong>teger and N −g is an odd <strong>in</strong>teger.<br />

We first exclude the condition where k2 = 0. With the above<br />

condition, there exists a permutation with<strong>in</strong> c such that<br />

Ω+(K1, c) = ⌊ k1 g<br />

2 ⌋ ≤ 2<br />

Ω−(K1, c) = ⌊ k1+1 g<br />

2 ⌋ ≤ 2<br />

Ω+(K2, c) = ⌊ k2<br />

N−g+1<br />

2 ⌋ + 1 ≤ 2<br />

Ω−(K2, c) = ⌊ k2−1 N−g−1<br />

2 ⌋ ≤ 2<br />

that confirms K1 ⊆ G1, K2 ⊆ G ′ 1 given (12). The immediate<br />

result gives<br />

and<br />

which yield<br />

F (K1, c) =<br />

F (K2, c) =<br />

−1, k1 = 1, 3, ...<br />

0, k1 = 2, 4, ...<br />

1, k2 = 1, 3, ...<br />

2, k2 = 2, 4, ...<br />

(13)<br />

(14)<br />

F (K2, c) − F (K1, c) ≥ 1. (15)<br />

For k2 = 0 which implies G2 ⊂ G1, s<strong>in</strong>ce G2 cannot be<br />

empty, k1 = g − |G2| must be ≤ g, there exists another permutation<br />

<strong>in</strong> c such that Ω+(K1, c) < Ω−(K1, c) ≤ Ω−(G1, c)<br />

giv<strong>in</strong>g<br />

F (K1, c) =<br />

−1, k1 = 1, 3, ...<br />

−2, k1 = 2, 4, ...<br />

(16)<br />

and F (K2, c) − F (K1, c) ≥ 1 as <strong>in</strong> (15) s<strong>in</strong>ce F (K2, c) = 0.<br />

Given that G2 = (G1 \ K1) ∪ K2, with (11) and (15), we<br />

obta<strong>in</strong><br />

F (G2, c) = F (G1, c) − F (K1, c) + F (K2, c) ≥ 1<br />

and hence −→ G2(c) = 1. S<strong>in</strong>ce −→ G1(c) = 0, there exists a column<br />

<strong>in</strong> MN such that −→ G2(c) = −→ G2(c) which is sufficient to show<br />

that −→ G1 = −→ G2.<br />

With the above two cases, we have proven that if G1 = G2,<br />

then −→ G1 = −→ G2. Together with the earlier establishment that if<br />

G1 = G2, then −→ G1 = −→ G2, we conclude that for any G1, G2 ⊂<br />

G and G1, G2 = ∅, −→ G1 = −→ G2 iif G1 = G2.

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