Interference Cancellation and Detection for More than Two Users - Interference Cancellation Using Space-Time Processing and Precoding Design - page 39

⎛ ⎞

ρ || HR

2
F

ε ||

| H

⎝ ⎠

P ( d → d ) = Q

4

⎛ ⎞

|| H

† R †

ρ ε ||

2 R

ε

⎝ ⎠

= Q

4

exp

|| H

† R † 2 R

− ρ ε

|| ε

≤ (3.32)

4

⎛ ⎞

γ 1
γ 2
γ 3
γ 4

⎝ ⎠

and H in Equation ( 3.27 )into( 3.32 ), Nowwe assume R ε = . Substituting R ε

we have

exp − ρ 4

| H

P ( d → d ) ≤ (3.33)

where

2 2

ζ = | γ 1 +

k 1 γ 2 + k 2 γ 3 + k 3 γ 4 |

+ |

k 3 γ 1 − γ 2 + k 1 γ 3 − k 2 γ 4 |

+

a b

2 2

c | k 2 γ 1 + k 3 γ 2 + γ 3 + k 1 γ 4 | + d | k 1 γ 1 − k 2 γ 2 + k 3 γ 3 − γ 4 | (3.34)

Further, we have

exp

2

− ρ ·

a | γ 1 + k 1 γ 2 + k 2 γ 3 + k 3 γ 4 |

| H P ( d → d ) ≤

4

exp − ρ · i = 1 |

h 1 (

2 2

i , 1 ) | | γ 1 + k 1 γ 2 + k 2 γ 3 + k 3 γ 4 |

= (3.35)

4

Note that

4 4 4

1
4 ||

1
4 ||

1
4

h 1 (

2

H 1 v ||

2 2 2 2

1 |

i , 1 ) | =|| ≥ H 1 V || = H 1 || =

1 |

h 1 ( i , j ) |

i = i = 1 j =

(3.36)

So we have

E exp

− ρ · i = 1 j = 1 |

2

· ζ h 1 ( i , j ) |

| H P ( d → d ) = E [ P ( d → d ) ]≤

16

1

= (3.37)

16
j

+ (ρζ /

1 [

1 16 ) ]

=