Digital Signal Processing Reference
In-Depth Information
For the sake of simplicity, we consider 2 users each with 2 transmit antennas and
one receiver with 3 receive antennas. The approach for a general case of
N
transmit
and
M
receive antennas is similar. Our approach is to select 2 of the 3 receive anten-
nas and use the scheme in Sect.
2.1
for the selected antennas. Nowwe will present our
selection criterion. Note that by using the scheme proposed in Sect.
2.1
,asshown
in (
2.72
), the term that determines diversity is
i
=
1
j
=
1
|
Φ(
|
λ
i
,
j
|
2
2
.We
j
,
1
)
|
Φ
=
V
1
R
T
where
V
1
is constant and
know
is the error
matrix.For a given constellation, the unitary rotation matrix
R
is chosen opti-
mally and is fixed. So we can always find
ε
=[
Φ(
1
,
1
), Φ(
2
,
1
)
]
ε
φ
1
=
min
|
Φ(
1
,
1
)
|
,
i
=
j
and
d
i
d
j
∀
,
ϕ
=
i
=
1
j
=
1
|
φ
j
|
|
λ
i
,
j
|
2
.Dif-
ferent choice of receive antennas will lead to different
λ
i
,
j
and thus different
2
φ
2
=
min
|
Φ(
2
,
1
)
|
,
i
=
j
. Now we define
d
i
d
j
∀
,
.To
pick 2 out of 3 antennas, we have 3 choices. We call the scenario that receive anten-
nas 1 and 2 are chosen Case 1, the scenario that receive antennas 1 and 3 are chosen
Case 2, and, the scenario that receive antennas 2 and 3 are chosen Case 3. The corre-
sponding
ϕ
ϕ
k
=
i
=
1
j
=
1
|
φ
j
|
|
λ
k
i
2
2
,
k
3. Our
selection criterion is to pick the two receive antennas of Case
i
whose corresponding
ϕ
i
is the largest among all the three cases. In other words, if
ϕ
for each case is given by
j
|
=
1
,
2
,
,
,
then we choose the two antennas corresponding to Case
i
. Obviously, by this method,
we can achieve interference cancellation for each user. In what follows, we prove
that we can also achieve full diversity for each user.
We first present the proof for User 1. Let us assume the channel for User 1 is
ϕ
i
=
max
{
ϕ
1
,ϕ
2
,ϕ
3
}
⎛
⎞
h
11
h
12
h
21
h
22
,
h
11
h
12
h
21
h
22
h
31
h
32
⎝
⎠
. The channels for User 1 in Cases 1, 2, 3 are
H
1
=
H
=
h
11
h
12
h
31
h
32
, and
H
3
=
h
21
h
22
h
31
h
32
, respectively. Without loss of generality,
H
2
=
let us assume
i
and the two receive antennas in
case
i
is selected. By our selection criterion, we know that
=
argmax
{
ϕ
1
,ϕ
2
,ϕ
3
}∈{
1
,
2
,
3
}
ϕ
1
+
ϕ
2
+
ϕ
3
3
≤
ϕ
i
≤
ϕ
1
+
ϕ
2
+
ϕ
3
(2.96)
where
2
(
|
λ
i
2
+|
λ
i
2
2
(
|
λ
i
2
+|
λ
i
2
ϕ
i
=|
φ
1
|
11
|
21
|
)
+|
φ
2
|
12
|
22
|
)
(2.97)
Now, let us define
2
(
|
λ
11
2
+|
λ
21
2
+|
λ
21
2
δ
1
=|
φ
1
|
|
|
|
)
2
(
|
λ
1
2
+|
λ
1
2
+|
λ
2
2
+|
φ
2
|
12
|
22
|
22
|
)
(2.98)
2
(
|
λ
3
2
+|
λ
3
2
+|
λ
2
2
δ
2
=|
φ
1
|
11
|
21
|
11
|
)
2
(
|
λ
12
2
+|
λ
22
2
+|
λ
12
2
+|
φ
2
|
|
|
|
)
(2.99)
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