Interference-Free Transmission for X channels - Interference Cancellation Using Space-Time Processing and Precoding Design - page 82

B 2 D 23 ⊥ D 22 . Design precoder to make D 14 ⊥ D 11 , D 14 ⊥ D 12 , D 24 ⊥ D 21 ,

D 23 .
2. At time slot 2, design precoder B 1

D 24 ⊥ D 22 , D 24 ⊥

. Design precoder A 1

to make D 12 U G 1 ( 1 )

D 12 . Design precoder B 2 to make D 14 ⊥

to make D 11 ⊥ D 11 , D 14 ⊥ D 12 , D 24 ⊥ D 21 ,

A 2 D 24 ⊥ D 22 . Design precoder to make D 13 ⊥ D 12 , D 13 ⊥ D 11 , D 23 ⊥ D 21 ,

D 24 .
3. At time slot 3, design precoder A 2

D 23 ⊥ D 22 , D 23 ⊥

. Design precoder B 2 to make D 23 U H 2 ( 1 )

D 23 . Design precoder A 1 to make D 11 ⊥

to make D 24 ⊥ D 13 , D 11 ⊥ D 14 , D 21 ⊥ D 23 ,

B 1 D 21 ⊥ D 24 . Design precoder to make D 12 ⊥ D 13 , D 12 ⊥ D 14 , D 12 ⊥ D 11 ,

D 24 .
4. At time slot 4, design precoder B 2

D 22 ⊥ D 23 , D 22 ⊥

. Design precoder A 2 to make D 24 U G 2 ( 1 )

D 24 . Design precoder B 1 to make D 12 ⊥ to make D 23 ⊥ D 13 , D 12 ⊥ D 14 , D 22 ⊥ D 23 ,

A 1

D 22 ⊥ D 24 . Design precoder to make D 11 ⊥ D 12 , D 11 ⊥ D 13 , D 11 ⊥ D 14 ,

D 24 .
Note that the design method at each time slot is similar. The key is that we change
the design order for C 1 , S 1 , C 2 , S 2 at different time slots. At time slot 1, we should
design precoder for C 1 , then for S 1 , then for C 2 , finally for S 2 . At time slot 2, we
should design precoder for S 1 , then for C 1 , then for S 2 , finally for C 2 . At time slot
3, we should design precoder for C 2 , then for S 2 , then for C 1 , finally for S 1 .Attime
slot 4, we should design precoder for S 2 , then for C 2 , then for S 1 , finally for C 1 .
In what follows, we prove that our proposed scheme can provide full diversity
for each codeword. We only provide the proof for codewords c 11 , c 12 , c 13 , c 14 .The
proof for other codewords is similar. The diversity is defined as

D 21 ⊥ D 23 , D 21 ⊥

log P e
log

d =− lim

ρ →∞

(5.50)

ρ

where ρ denotes the SNR and P e represents the probability of error. If we let

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

e 1
e 2
e 3
e 4

c 11
c 12
c 13
c 14

c 11

⎝ ⎠ ⎝ ⎠ − ⎝ ⎠

c 12

e = = denote the error vector, based on Eq. ( 5.42 ),

c 13

c 14

the pairwise error probability (PEP) for c 11 , c 12 , c 13 , c 14 can be written as [ 2 ]

⎛ ⎞

2
F

ρ || H 1 e ||

⎝ ⎠

P ( c → c | H 1 ) = Q

4

⎛ ⎞

exp

e †

† H 1 e

e †

† H 1 e

ρ ( H 1 )

− ρ

( H 1 )

⎝ ⎠ ≤

= Q

4 4

exp

− ρ 4 = (5.51)