Digital Signal Processing Reference
In-Depth Information
Since Pr
{
d
1
→
d
1
|
σ
=
0
}≤
1 and 1
−
Pr
{
σ
=
0
}≤
1, we have
Pr
(
d
1
→
d
1
)
≤
Pr
{
d
1
→
d
1
|
σ
=
0
}
(
1
−
Pr
{
σ
=
0
}
)
+
Pr
{
σ
=
0
}
≤
Pr
{
d
1
→
d
1
|
σ
=
0
}+
Pr
{
σ
=
0
}
(3.55)
Note that when
0, we can follow the steps in Sect.
3.3
to detect the signals of
User 1 and by the same technique used in Sect.
3.3
, we can easily derive
σ
=
ρζ
24
−
24
=
τ
1
ρ
−
24
Pr
{
d
1
→
d
1
|
σ
=
0
}≤
(3.56)
where
τ
1
is a constant. Further, from (
3.51
), we know that
}≤
τ
2
ρ
−
24
+
τ
3
ρ
−
24
+
τ
4
ρ
−
24
(3.57)
Pr
{
σ
=
0
}≤
Pr
{
σ
1
=
0
}+
Pr
{
σ
2
=
0
}+
Pr
{
σ
3
=
0
where
τ
2
,
τ
3
,
τ
4
are all constants. Substituting (
3.56
) and (
3.57
)in(
3.55
), we get
d
1
)
≤
(τ
1
+
τ
2
+
τ
3
+
τ
4
)ρ
−
24
Pr
(
d
1
→
(3.58)
Using (
3.58
), it is easy to show that the diversity
d
≥
24. Also, it is easy to show that
the diversity
d
24. So the diversity for User 1 is still full diversity. Similarly, we
can show that all the other users can also achieve full diversity. In addition, since all
the receive antennas are used in the decoding efficiently, it is obvious that the coding
gain will be increased.
≤
3.4.3 More Users, i.e., J
>
M
=
N
In this section, we consider the case that
J
>
M
=
N
. For simplification, we assume
J
=
6,
M
=
N
=
4. First, we assume User
k
transmits codewords
⎛
⎝
⎞
⎠
c
k
1
c
k
2
c
k
3
c
k
4
c
k
5
c
k
6
c
k
2
c
k
3
c
k
4
c
k
5
c
k
6
c
k
1
c
k
3
c
k
4
c
k
5
c
k
6
c
k
1
c
k
2
c
k
4
c
k
5
c
k
6
c
k
1
c
k
2
c
k
3
c
k
5
c
k
6
c
k
1
c
k
2
c
k
3
c
k
4
c
k
6
c
k
1
c
k
2
c
k
3
c
k
4
c
k
5
c
k
=
(3.59)
)
]
4
×
4
and
A
l
k
=
Channel matrix and precoder for User
k
are given by
H
k
=[
h
k
(
i
,
j
a
k
(
[
)
]
4
×
6
, respectively, where
l
denotes the time slot. Note that we can only have
four orthogonal directions at most since there are four receive antennas. In order to
get the orthogonal structure, we let
i
,
j
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