Digital Signal Processing Reference
In-Depth Information
3.2 Decoding
Using our precoders, Equation (
3.5
) becomes
⎛
⎝
⎞
⎠
+
⎛
⎝
⎞
⎠
+
⎛
⎝
⎞
⎠
+
⎛
⎝
⎞
⎠
)
+
c
1
c
2
c
3
c
4
s
1
s
2
s
3
s
4
t
1
t
2
t
3
t
4
z
1
z
2
z
3
z
4
E
s
(
y
=
H
1
H
2
H
3
H
4
n
(3.20)
where
⎛
⎞
h
i
(
k
1
h
i
(
k
2
h
i
(
k
3
h
i
(
1
,
1
)
1
,
1
)
1
,
1
)
1
,
1
)
h
i
(
k
1
h
i
(
k
2
h
i
(
k
3
h
i
(
⎝
,
)
,
)
,
)
,
)
⎠
2
1
2
1
2
1
2
1
h
i
(
k
1
h
i
(
k
2
h
i
(
k
3
h
i
(
3
,
1
)
3
,
1
)
3
,
1
)
3
,
1
)
h
i
(
k
1
h
i
(
k
2
h
i
(
k
3
h
i
(
4
,
1
)
4
,
1
)
4
,
1
)
4
,
1
)
h
i
(
))
∗
−
(
h
i
(
))
∗
h
i
(
))
∗
−
h
i
(
))
∗
k
1
(
1
,
1
1
,
1
k
3
(
1
,
1
k
2
(
1
,
1
h
i
(
))
∗
−
(
h
i
(
))
∗
h
i
(
))
∗
−
h
i
(
))
∗
k
1
(
2
,
1
2
,
1
k
3
(
2
,
1
k
2
(
2
,
1
h
i
(
))
∗
−
(
h
i
(
))
∗
h
i
(
))
∗
−
h
i
(
))
∗
k
1
(
3
,
1
3
,
1
k
3
(
3
,
1
k
2
(
3
,
1
h
i
(
))
∗
−
(
h
i
(
))
∗
h
i
(
))
∗
−
h
i
(
))
∗
k
1
(
4
,
1
4
,
1
k
3
(
4
,
1
k
2
(
4
,
1
H
i
=
(3.21)
k
2
h
i
(
k
3
h
i
(
h
i
(
k
1
h
i
(
1
,
1
)
1
,
1
)
1
,
1
)
1
,
1
)
k
2
h
i
(
k
3
h
i
(
h
i
(
k
1
h
i
(
2
,
1
)
2
,
1
)
2
,
1
)
2
,
1
)
k
2
h
i
(
k
3
h
i
(
h
i
(
k
1
h
i
(
3
,
1
)
3
,
1
)
3
,
1
)
3
,
1
)
k
2
h
i
(
k
3
h
i
(
h
i
(
k
1
h
i
(
4
,
1
)
4
,
1
)
4
,
1
)
4
,
1
)
h
i
(
))
∗
−
h
i
(
))
∗
h
i
(
))
∗
−
(
h
i
(
))
∗
k
3
(
1
,
1
k
2
(
1
,
1
k
1
(
1
,
1
1
,
1
h
i
(
))
∗
−
h
i
(
))
∗
h
i
(
))
∗
−
(
h
i
(
))
∗
k
3
(
2
,
1
k
2
(
2
,
1
k
1
(
2
,
1
2
,
1
h
i
(
))
∗
−
h
i
(
))
∗
h
i
(
))
∗
−
(
h
i
(
))
∗
k
3
(
3
,
1
k
2
(
3
,
1
k
1
(
3
,
1
3
,
1
h
i
(
))
∗
−
h
i
(
))
∗
h
i
(
))
∗
−
(
h
i
(
))
∗
k
3
(
4
,
1
k
2
(
4
,
1
k
1
(
4
,
1
4
,
1
Here
y
and
n
are the same with
y
and
n
in Equation (
3.5
). Note that using
ou
r
pr
ec
od
ers, each column of array
H
1
is orthogonal to each column of matrices
H
2
,
H
3
,
H
4
.
In ord
er
to decode symbols fromUser 1, we multiply both sides of Equation (
3.20
)
by array
H
1
to achieve
⎛
⎝
⎞
⎠
+
c
1
c
2
c
3
c
4
E
s
H
1
H
1
H
1
y
H
1
n
=
(3.22)
Note that the noise elements of
H
†
1
n
are correlated with covariance matrix
H
†
1
H
1
.
We can
w
h
ite
n this noise vector by multiplying both sides of Equation (
3.22
)bythe
matrix
H
†
1
2
as follows
1
H
1
)
−
(
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