Digital Signal Processing Reference
In-Depth Information
2 K
presented in Sect. 4.3 , the diversity of our scheme is M
·
min
(
N
,
L
)
or M
·
min
(
N
,
)
.
In order to achieve a diversity of MN , we need
1. L
l og 2 N .
2. The rank of matrix
N , i.e., K
Υ 1 to be N .
M . In this case, we assume the channel
matrices and precoders for Users 1 and 2 are given by ( 4.62 ) and ( 4.63 ). We can
use the same method as discussed before to maximize
Now we consider the case that N
<
v h |
|
. However, if we want to
1
2 , like ( 4.54 ), we need to design precoders to make
maximize
|
cos
θ
hg |
H M × 1 = η ·
B 1
G M × N ·
(
1
) N × 1
(4.65)
which means the equivalent signal vectors of the two users are orthogonal to each
other. In the above equation, we need to determine N unknown parameters by M
equations. Since N
M , the number of equations is greater than the number of
unknown parameters. Therefore, even with prefect feedback, we cannot find these
unknown parameters to satisfy the equations. In other words, since we do not have
enough dimensions for precoders, we cannot make v i g orthogonal to v i h .
In order to make our proposed scheme extendable to the case of M
<
>
N , we can
choose N receive antennas among all M receive antennas as follows:
In time slot 1, we can choose the N receive antennas such that
H new || F is
maximized, where H new is the new channel matrix with N transmit antennas and
the selected N receive antennas. Once the number of receive antennas is equal to
the number of transmit antennas, the same method used in Sect. 4.3 can be used to
determine the codebook and precoders for Users 1 and 2. At time slot 2, we choose
the N receive antennas such that
||
G new || F is maximized, where G new is the new
channel matrix with N transmit antennas and the selected N receive antennas. Then
we design the codebook and precoders for Users 1 and 2 using the same method in
the case that M
||
N .
In order to show that we can achieve full diversity for each user using the above
proposed method, we consider ( 4.44 ). By ( 4.44 ), we know
=
HU Υ 1 Υ 1 V
2
2
= |
Υ 1 |
|
H
Υ 1 |
v h |
2
|
L
L
j = 1 ( | λ Υ 1
2 i = 1 |
h ij |
2
|
)
j
=
L
2 j = 1 i = 1 |
| λ Υ min |
h ij |
2
L
2 j = 1 i = 1 |
| λ Υ min |
2
h ij |
=
L
| λ Υ min |
2
2
F
||
H new ||
=
.
(4.66)
L
 
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