Digital Signal Processing Reference
In-Depth Information
Multi−User, 2 bits/sec/Hz
10
0
Beamforming using 256QAM
Interference alignment using QPSK
Our scheme using QPSK
10
−1
10
−2
10
−3
10
−4
10
−5
10
−6
10
15
20
25
30
35
40
Signal to Noise Ratio (dB)
Fig. 5.7
Simulation results for 2 users each with 6 transmit antennas and 2 receivers each with 4
receive antennas
Each orthogonal relationship corresponds to one equation. Since at each of the
J
r
−
1
receivers, there are
J
t
useful
J
t
J
r
×
1 signal vectors, we need to satisfy
J
t
·
(
J
r
−
1
)
equations.
At Receiver
J
r
, the signal vector of
C
J
t
J
r
is a useful signal and its direction should
be orthogonal to all other signal vectors. Since there are
J
t
·
(
J
r
−
1
)
interference signal
vectors and
J
t
−
1 useful signal vectors in the space, we will have
J
t
·
(
J
r
−
)
+
J
t
−
1
1
=
J
t
·
J
r
−
1 equations to satisfy.
Therefore, in order to solve all these equations, it is easy to see that we only need
[
J
t
·
(
J
r
−
1
)
+
J
t
·
J
r
−
1
]+
1
=
J
t
·
(
2
·
J
r
−
1
)
transmit antennas which lead to
. We need one more
unknown parameter to make these orthogonal equations have a solution.
Following the logic of the last section, if Scheme II is used, we need
N
J
t
·
(
2
·
J
r
−
1
)
unknown parameters, i.e.,
N
≥
J
t
·
(
2
·
J
r
−
1
)
≥
M
·
(
J
t
. In what follows, we show that the minimum number of needed transmit
antennas for Scheme II is equal to or higher than that of Scheme I, i.e.
N
J
r
−
1
)
+
≥
J
t
·
(
2
·
J
r
−
1
)
. In Scheme II,
M
≥
J
t
·
J
r
, which results in
N
≥
M
·
(
J
r
−
1
)
+
J
t
≥
J
r
J
t
·
J
r
·
(
J
r
−
1
)
+
J
t
=
J
t
·
(
−
J
r
+
1
)
≥
J
t
·
(
2
·
J
r
−
1
)
. Only when
J
r
=
2
and
M
J
r
, both of these two methods need the same minimum number of
transmit antennas. In all other cases, Scheme I will need less minimum number
of transmit antennas. From another perspective, this means that when the number of
transmit and receive antennas is fixed and both Scheme I and Scheme II can work,
=
J
t
·
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