Digital Signal Processing Reference
In-Depth Information
is an isotropically distributed unitary vector.
The intuitive meaning of an isotropically distributed complex unit vector is that it is
equally likely to point in any direction in complex space. Therefore, the problem to
design a codebook to maximize
It has been shown in [
6
] that
V
H
(
1
)
v
h
|
becomes how to pack one-dimensional subspaces
of a complex space known as Grassmannian line packing [
7
]. In other words, it is
the problem of finding a set of
L
1
one-dimensional subspaces in the complex space
that maximize the minimum distance between any pair of subspaces in the set.
The problem of finding optimal line packings using analytical or numerical meth-
ods is not new [
7
-
10
]. We utilize the existing methodologies in the literature to design
a codebook for User 1 in time slot 1.
Now we summarize the procedures to construct our codebook for User 1 in time
slot 1:
1. For
K
bits of feedback, find
L
1
=
|
2
K
two-by-one unit norm complex vectors
which can maximize the minimum distance between any pair of vectors in the
two-dimensional complex space. We denote all these vectors as
ψ
i
,
i
=
1
,...,
L
1
.
1
2
K
matrices satisfying
2. Create a codebook
Υ
1
that contains
L
1
=
Υ
1
[
i
]=
√
2
[
ψ
i
,ψ
i
]
.
It is easy to check that the created codebook satisfies all the conditions we need.
Therefore,
v
h
|
can be maximized if User 1 adopts the above codebook.
In what follows, we will show that if User 2 adopts the above codebook,
|
1
|
cos
θ
hg
|
v
h
|
1
will also be maximized. By (
4.36
), we know that once
are max-
imized at the same time, the coding gain will be maximized. Therefore, the above
codebook is the optimal codebook that both User 1 and User 2 should adopt in time
slot 1.
First, note that in order to maximize
|
and
|
cos
θ
hg
|
1
,by(
4.36
), we need
v
g
=
η
v
h
, i.e.,
|
θ
hg
|
cos
−
(
h
11
h
21
or
(
h
21
)
∗
−
(
h
11
)
∗
g
21
)
∗
g
11
=
η
η
=
(4.54)
g
11
)
∗
g
21
(
where
η
is a constant. Further, we have
(
h
21
)
∗
−
(
h
11
)
∗
g
11
g
12
g
21
g
22
b
11
b
21
η
=
b
11
b
21
g
11
g
12
g
21
g
22
−
1
(
h
21
)
∗
−
(
h
11
)
∗
or
=
η
.
(4.55)
Since the norm of
b
11
is 1, we have
b
21
g
11
g
12
g
21
g
22
−
1
(
h
21
)
∗
−
(
h
11
)
∗
b
11
b
21
F
=
.
(4.56)
g
11
g
12
g
21
g
22
−
1
(
h
21
)
∗
−
(
h
11
)
∗
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