Digital Signal Processing Reference
In-Depth Information
J
,
M
and
N
are all even. By some simple antenna or user removals, our proposed
scheme can also be used when not all of
J
,
M
and
N
are even.
3.4.1 More Transmit Antennas, i.e., N
>
J
=
M
First, we consider the case
N
>
J
=
M
. For simplification, we assume
N
=
6,
4. Also we take User 1 for example. Each user transmits QOSTBCs and
the precoder will be a 6
J
=
M
=
4 matrix. Similar to Equation (
3.7
), to make User 1 and
User 2 orthogonal to each other, we design precoders such that
×
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
∗
a
2
(
a
1
(
1
,
1
)
1
,
1
)
a
2
(
a
1
(
,
)
,
)
2
1
2
1
a
2
(
a
1
(
3
,
1
)
3
,
1
)
=
η
H
1
H
2
(3.42)
a
2
(
a
1
(
4
,
1
)
4
,
1
)
a
2
(
a
1
(
5
,
1
)
5
,
1
)
a
2
(
a
1
(
6
,
1
)
6
,
1
)
where
H
1
and
H
2
are all 4
6 matrices. Since
H
1
and
H
2
are not square matrices,
we cannot take the inverse as in Equation (
3.10
). Instead, we multiply both sides of
Equation (
3.42
)by
H
2
and
×
H
2
H
2
)
−
1
resulting in
(
H
2
H
2
)
−
1
H
2
H
1
Q
=
(
(3.43)
Then we can calculate the singular value decomposition of
Q
and use the same
method used in Sect.
3.1
to design the precoders. For the sake of brevity, we do not
include the decoding and the proof of full diversity. They are similar in nature to
what we presented earlier for users with 4 transmit antennas.
3.4.2 More Receive Antennas, i.e., M
>
J
=
N
For the case of
M
N
receive antennas with the best
channel conditions among all
M
receive antennas for User
k
at time slot
k
. In what
follows, we illustrate our selection criterion and prove that it provides full diversity.
For simplification, we assume
M
>
J
=
N
, we can pick the
J
=
=
6,
J
=
N
=
4.
We assume the channel matrix for User
k
,
k
=
1
,...,
4, is
H
k
=[
h
k
(
i
,
j
)
]
6
×
4
(3.44)
Then we pick the 4 rows in
H
k
with the largest norms and put them in a matrix
H
k
.
So we have
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