13
10-9
10-7
10-6
10-5
10-4
10-3
10-2
10-1
100
0 5 10 15 20
Outage probability
Average SNR (dB)
Outage capacity is 2 bps/Hz
(d) The following plots are of the “success probability” 1 −Pout vs. Cfor
10-7
10-6
10-5
10-4
10-3
10-2
10-1
100
2 3 4 5 6 7 8 9
Outage capacity (bps/Hz)
1 antenna
2 antennas
4 antennas
8 antennas
10-7
10-6
10-5
10-3
10-1
5 6 7 8 9 10 11 12
Outage capacity (bps/Hz)
1 antenna
(e) Setting Pout = 0.1, we plot Cvs. γfor NT= 1,2,4,8.
0
1
2
3
4
5
6
7
8
Outage capacity (bps/Hz)
Outage probability is 0.1
1 antenna
2 antennas
4 antennas
8 antennas
Problem 15.13
An (NT, NR) = (2, NR) MIMO system employs the Alamouti code with
15
01 10 10 01 11 00
s1s2s3s4s5s6
Problem 15.14
We wish to show that the decision metrics in equations (15.4-25) and (15.4-
22) are equivalent. The received signals are
and
ˆs1=y1h∗
11 +y∗
2h12
16
Signalling interval Antenna 1 Antenna 2
1s1=v2s2=v4
2−s∗
2=v3s∗
1=v3
The decision metric µ(s1) may be expressed as
µ(s1) = 2Re{y1h∗
11s∗
1+y∗
2h12s∗
1} − |s1|2(|h11|2+|h12|2)
We substitute for ˆs1= (|h11|2+|h12|2)s1+h∗
11n1+h12n∗
2into µ(s1) and
obtain
µ(s1) = 2Re{s∗
1(|h11|2+|h12|2)s1/2 + (h∗
11n1+h12n∗
2)}
Problem 15.15
Since
GHG=|s1|2+|s2|2+|s3|2I3
17
and the received signals are
y1=h11s1+h12s2+h13s3+n1
the maximum likelihood detector bases its decisions on the estimates ˆs1,ˆs2,ˆs3,
where
ˆs1=h∗
11y1+h12y∗
2−h13y∗
3
For example, if we substitute for y1, y2and y3in the estimate ˆs1, we obtain
ˆs1= (|h11|2+|h12|2+|h13|2)s1+h∗
11n1+h∗
12n∗
2−h13n∗
3
Problem 15.16
Gis given by equation (15.4-42). The received signal vector is
y=Gh+n=G[h11 h12 h13 h14]T+ [n1n2. . . n8]T
The received signal components are:
y1=h11s1+h12s2+h13s3+h14s4+n1
3+h13s∗
2+h14s∗
1+n8
18
Since GHG=P4
i=1 |si|2I4, the maximum-likelihood detector bases its deci-
sions on the estimates:
ˆs1=h∗
11y1+h∗
12y2+h∗
13y3+h∗
14y4+h11y∗
5+h12y∗
6+h13y∗
7+h14y∗
8
ˆs2=h∗
12y1−h∗
11y2−h∗
14y3+h∗
13y4+h12y∗
5−h11y∗
6−h14y∗
7+h13y∗
8
Problem 15.17
From equation (15.3-2), the output of the MRC is
µj=hH
jyj=rEb
NT
sjkhjk2
F+hH
jnj, j = 1,2,…,NT
For BPSK modulation, assume sj=±1 and consider the case of the signal
transmitted from antenna 1. Then, with s1= 1,
µ1=rEb
NT
s1kh1k2
F+hH
1n1
¯γcNR(NR−1)!γNR−1
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19
Problem 15.18
For the SIMO (1,2) system, with transmitted energy √Eband AWGN, we
simply have the performance of a conventional dual diversity system with
MRC. Hence, from equation (13.4-15), with L= 2, we have
PSIMO =1
4(1 −µ)2(2 + µ)
Problem 15.19
20
+
+
s1
s∗
1
2
h1
h2
c1
c1
c2
c2
(b)
y= (c1s1−c2s∗
Hence,
(c) One advantage is that the two symbols are transmitted and detected in
the same time interval, so the channel is not required to be constant over
Problem 15.20
(a)
y=Hs+n
21
The ICD computes
H−1y=H−1Hs+H−1n=s+H−1n
where γkis the SNR for the kth received symbol. The SNR is variable
because the noise variance may differ among the NTsymbols.
(d) If ˆ
s=WHy, then the noise component is WHn. Hence
Rw=E[WHnnHW] = NoWHW
Problem 15.21
H=
0.4 0.5
0.7 0.3
HHH=
0.41 0.43
0.43 0.58
(a) The eigenvalues of HHHare found as follows:
22
where Λ= diag(λ1, λ2). Therefore, Umay be determined by computing the
eigenvectors of HHH. Thus
0.41 −λ0.43
0.43 0.58 −λ
q1
q2
=
0
0
With λ=λ1, we find that the normalized eigenvectors are
0.63485
0.77263
Hence,
HHH=
0.65 0.41
0.41 0.34
23
(b) When the channel is known at the transmitter and the receiver, the
MIMO channel is equivalent to two parallel SISO channels whose outputs
ks
k=1
with the constraint that σ2
1s+σ2
2s=NT= 2. The optimum power allocation
can be determined by maximizing
J(σ2
1s, σ2
2s) =
2
X
k=1
log21 + γ
2λkσ2
ks−1
µ 2
X
k=1
σ2
ks −2!
ks =µ−2
γλk
(c) When His known at the receiver only, the capacity is given as
C=
2
X
log21 + Es
2No
λi
Problem 15.22
The capacity of the MISO (2,1) channel which is known at the receiver only,
with AWGN, is given as
CMISO = log2 1 + Es
2No
2
X
24
When the MISO (2,1) channel is known at the transmitter, the capacity
of the channel is (see problem 15.10b),
Problem 15.23
(a)
s∗
1−s2−s3s∗
4
s1s2s3s4
25
Since GHGis not a diagonal matrix, the code is not orthogonal.
(b) The received signals in the four signal intervals are:
y1=h11s1+h12s2+h13s3+h14s4+n1
y2=−h11s∗
2+h12s∗
1−h13s∗
4+h14s∗
3+n2
p(y1, y2, y3, y4|H,s)∼ |y1−(h11s1+h12s2+h13s3+h14s4)|2
+|y2−(−h11s∗
2+h12s∗
1−h13s∗
4+h14s∗
3)|2