Digital Signal Processing Reference
In-Depth Information
⎛
⎝
⎞
⎠
+
c
1
(
E
s
(
h
12
)
∗
−
(
h
11
)
∗
(
(
(4.13)
y
1
)
∗
g
12
)
∗
−
(
g
11
)
∗
n
1
)
∗
c
2
=
y
2
)
∗
(
h
22
)
∗
−
(
h
21
)
∗
(
g
22
)
∗
−
(
g
21
)
∗
n
2
)
∗
(
s
1
(
s
2
Equations (
4.12
) and (
4.13
) are the input-output relationship of our system at the
first two time slots.
4.2 Interference Cancellation Precoding and Decoding
In this section, we will show the property that our codebooks should possess in order
to achieve our first goal, i.e., interference cancellation.
4.2.1 Precoding
First, in time slot 1, by Eq. (
4.12
),
c
1
,
c
2
,
s
1
,
s
2
are transmitted along four equivalent
channel vectors
H
1
,
H
1
,
G
1
,
G
1
, respectively. Suppose that we want
to remove the signals of User 2, we can find a 2-by-1 complex vector
g
satisfying
g
†
(
1
)
(
2
)
(
1
)
(
2
)
G
1
G
1
0. Then by simply multiplying both sides of Eq. (
4.12
)by
g
†
,
we can remove the signals of User 2. This is our basic idea to achieve the interference
cancellation.
However, since
G
1
g
†
(
1
)
=
(
2
)
=
,
G
1
(
1
)
(
2
)
are 2-by-1 complex vectors, a non-zero com-
G
1
G
1
plex vector
g
2
×
1
that satisfies
g
†
g
†
(
1
)
=
(
2
)
=
0 does not exist unless
G
1
)
=
α
G
1
is a constant. Therefore, in order to cancel the inter-
ference from User 2, we need
G
1
(
1
(
2
)
, where
α
)
=
α
G
1
.Tomake
G
1
)
=
α
G
1
(
1
(
2
)
(
1
(
2
)
, our
precoders
A
1
and
B
1
should have the following properties:
A
1
A
1
B
1
B
1
(
1
)
=
(
2
),
(
1
)
=
(
2
),
(4.14)
i.e.,
a
11
a
21
a
12
a
22
b
11
b
21
b
12
b
22
=
,
=
.
(4.15)
Since we choose a matrix in the codebook
Υ
1
as the precoder for User 1 and a matrix
in the codebook
Υ
2
as the precoder for User 2, Eq. (
4.14
) results in:
Υ
1
[
i
]
(
1
)
=
Υ
1
[
i
]
(
2
),
Υ
2
[
j
]
(
1
)
=
Υ
2
[
j
]
(
2
),
(4.16)
i.e., the two columns of any matrix in codebooks
Υ
2
should be the same.
From Eqs. (
4.10
), (
4.11
), and (
4.15
), it is easy to see that the resulted
G
1
Υ
1
and
,
G
1
(
1
)
(
2
)
satisfy
G
1
)
=
G
1
(
1
(
2
)
, i.e.,
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