Spring 2017 MIMO Communication Systems
Department of Electrical and Computer Engineering
National Chiao Tung University
Exam I
Mar. 23, 2017
1:20 pm ∼ 3:20 pm
Do not open the exam until you are told to do so.
SOLUTION
Problem
1
2
3
4
Total
Points
20
25
25
30
100
Your Score
Read all of the following information before starting the exam:
• This exam is closed book, but you are allowed to bring a calculator and one A4-sized
“hand-written” sheet that contains the information you need during the exam.
• Show all your work clearly and in order if you want to get full credit. I reserve the
right to take off points if I cannot see how you arrived at your answer (even if your
final answer is correct).
• This exam has 4 problems (4 pages) and is worth 100 points. It is your responsibility
to make sure that you have all of the pages!
• Good luck!
1
Problem 1 (20%)
(a) Explain under what condition(s) a wireless channel undergoes “frequency-selective fading”. What is the main drawback of frequency-selective fading? How to avoid this
drawback?
(b) Explain the channel coherence time and bandwidth of a wireless channel. What are the
advantages of knowing them in designing a wireless system?
(c) Explain what the channel-inversion power is and its advantages and drawbacks.
(d) Find the coverage probability (area) for a microcellular system where path loss follows
the simplified model with γ = 3, d0 = 1, and K = 0 dB and there is also log normal
shadowing with σ = 4 dB. Assume a cell radius of 100m, a transmit power of 80 mW,
and a minimum received power requirement of Pmin = −100 dBm. (Using Q(x) =
R ∞ − y2
√1
e 2 dy function to express your answer.)
2π x
Sol: The received mean power is
γ
d0
P γ = Pt K
= 0.08 × 10−6 . = 8 × 10−8 = −71 dB.
d
Pmin − P γ
= 14.7575.
σ
10γ log10 e
= 3.2572.
b=
σψ (dB)
2−2ab
2 − 2ab
2
c = Q(a) + e b Q
≈ 1.
b
a=
Problem 2 (25%)
Let the joint probability p(x, y) of two discrete random variables X and Y is given in the
following table:
X\Y
0
1
2
0
1
6
1
12
1
12
1
1
6
0
1
6
2
1
6
1
12
1
12
For example, the probability of X = Y = 0 is p(0, 0) = 16 . Find the following:
(a) H(X) and H(Y )
Sol: Since pX (0) = pX (1) = pX (2) = 13 , H(X) = log2 3 = 1.585. Since pY (0) = 12 ,
pY (1) = 61 and pY (2) = 13 , H(Y ) = 1.459 bits
2
(b) H(X|Y ) and H(Y |X)
Sol:
H(X|Y ) = pY (0)H(X|Y = 0) + pY (1)H(X|Y = 1) + pY (2)H(X|Y = 2)
1
1 1
1
1
1
(log2 3) + (log2 2) +
log2 4 + log2 4 + log2 2
=
2
6
3 4
4
2
= 1.459
H(Y |X) = pX (0)H(Y |X = 0) + pX (1)H(Y |X = 1) + pX (2)H(Y |X = 2)
1 1
1
1
1
=
log2 4 + log2 4 + log2 2 + log2 2
3 4
4
2
3
1
1
1 1
log2 4 + log2 4 + log2 2
+
3 4
4
2
4
=
3
(c) H(Y ) − H(Y |X)
Sol: H(Y ) − H(Y |X) = 1.459 − 1.333 = 0.126.
(d) H(X, Y ) and I(X; Y )
Sol:
1
1
log2 12 × 4 = 2.918.
H(X, Y ) = log2 6 × 4 +
6
12
I(X; Y ) = H(X) − H(X|Y ) = 1.585 − 1.459 = 0.126.
(e) D(p(x|y)||p(x)) and D(p(y|x)||p(y)). Is D(p(x|y)||p(x)) = D(p(y|x)||p(y))?
Sol:
D(p(x|y)||p(x)) =
X
p(x|y) log2
p(x|y)
= H(X) − H(X|Y ) = I(X; Y ) = 0.126.
p(x)
p(y|x) log2
p(y|x)
= H(Y ) − H(Y |X) = I(X; Y ) = 0.126.
p(y)
x,y
D(p(y|x)||p(y)) =
X
x,y
So D(p(x|y)||p(x)) = D(p(y|x)||p(y)) in this case since they both represent the mutual
information between X and Y .
Problem 3 (25%)
In order to improve the performance of cellular systems, multiple base stations can receive
the signal transmitted from a given mobile unit and combine these multiple signals either by
selecting the strongest one or summing the signals together, perhaps with some optimized
weights. This typically increases SNR and reduces the effects of shadowing. Combining of
signals received from multiple base stations is called macrodiversity, and in this problem we
explore the benefits of this technique.
3
Consider a mobile at the midpoint between two base stations in a cellular network. The
received signals (in dBW) from the base stations are given by
Pr,1 = W + Z1 ,
Pr,2 = W + Z2 ,
where Z1,2 are N (0, σ2 ) random variables. We define outage with macrodiversity to be the
event that both Pr,1 and Pr,2 fall below a threshold T .
(a) Interpret the terms W , Z1 , Z2 in Pr,1 and Pr,2 .
Sol:
W=average received power
Zi = shadowing over link i
Pr,i = Received power in dBW, which is Gaussian with mean W, variance σ 2
(b) If Z1 and Z2 are independent, show that the outage probability is given by
Pput = [Q(∆/σ)]2 ,
where ∆ = W − T is the fade margin at the mobile’s location.
Sol: Since Z1 and Z2 are independent,
2 2
∆
W −T
= Q
.
Pout = P[Pr,1 < T ∩Pr,2 < T ] = P[Pr,1 < T ]P[Pr,2 < T ] = Q
σ
σ
(c) Now suppose Z1 and Z2 are correlated in the following way:
Z1 = aY1 + bY
Z2 = aY2 + bY,
where Y , Y1 , Y2 are independent N (0, σ2 ) random variables, and a, b are such that
a2 + b2 = 1. Show that
2
Z −∞
∆ + byσ
1
2
√
e−y /2 dy
Q
Pout =
|a|σ
2π
+∞
Sol:
Z
∞
P[Pr,1 ≤ T, Pr,2 < T |Y = y]fY (y) dy.
Pout =
−∞
Note that for a given Y = y, Pr,1 ∼ N (W + by, a2 σ 2 ) and Pr,1 is independent of Pr,2 .
2
y2
W + by − T
1
=
Q
·√
e− 2σ2 dy
|a|σ
2πσ
−∞
2
Z ∞
y2
1
∆ + byσ
√
=
Q
e− 2 dy
|a|σ
2π
−∞
Z
Pout
∞
4
Problem 4 (30%)
Consider a cellular system where the transmit power falloff with distance follows the following
formula
α
d0
Pr = Pt H
,
d
where Pt is the transmit power, Pr is the received power, α is the path loss exponent, d0
denotes the near-field reference distance, and H is the channel power gain that is a discrete
random variable with the following distribution:
p(H = 1) = 0.4, p(H = 0.75) = 0.3, p(H = 0.5) = 0.2, p(H = 0.25) = 0.1.
Assume a user at a distance d = 500m from the base station (BS) has an average transmit
power P t = 100 mW, d0 = 50m, α = 4 and a receiver noise power of 0.1 mW. Suppose the
BS wants to transmit information to the user and both the BS and the user have channel
side (state) information.
(a) Compute the distribution of the received SNR with the average transmit power P t .
Sol: Suppose the noise power is σn2 and thus the SNR for H = Hj is given by
γj =
P t dα0 Hj
Pr
=
,
σn2
dα σn2
The four values of γj ’s are listed as follows:

γ1 = 0.1,



γ = 0.075,
2
γ=

γ3 = 0.05,



γ4 = 0.025,
j = 1, 2, . . . , 4.
if
if
if
if
H
H
H
H
=1
= 0.75
.
= 0.5
= 0.25
Therefore, p(γ = γ1 ) = 0.4, p(γ = γ2 ) = 0.3, p(γ = γ3 ) = 0.2, p(γ = γ4 ) = 0.1.
(b) Derive the optimal power control policy for the user’s channel with the average transmit
power constraint and its corresponding Shannon capacity per unit bandwidth.
Sol: The Shannon capacity of this channel per unit bandwidth is given by
C=
4
X
log2 (1 + γj ) p(γ = γj ) =
j=1
4
X
log2
j=1
and it is subject to the average power constraint
Lagrangian function of Pt (γj ) can be written as
J(Pt ) =
4
X
log2
j=1
Pt (γj )γj
1+
Pt
Pt (γj )γj
1+
Pt
P4
p(γ = γj ) − λ
j=1
4
X
p(γ = γj ),
Pt (γ)p(γ = γj ) ≤ P t . The
!
Pt (γj )p(γ = γj ) − P t
j=1
and we further have
∂J
γj
= p(γ = γj )
− λ = 0,
∂Pt (γj )
ln 2(P t + γj Pt (γj ))
5
which yields
Pt (γj )
=
Pt
1
1
−
(ln 2)λP t γj
+
=
1
1
−
γ0 γj
+
,
where γ0 = (ln 2)λP t and (x)+ = max{0, x}. γ0 can be found by the average transmit
power constraint, i.e.,
+
4 X
1
1
−
p(γ = γj ) = 1.
γ0 γj
j=1
Substituting the above power control policy into C results in
+ !
4
X
γj
−1
p(γ = γj ).
C=
log2 1 +
γ
0
j=1
(c) Find the capacity per unit bandwidth of the user’s channel when the following channelinversion power control is used:
Pt (γj )
β
= ,
γj
Pt
j = 1, . . . , 4.
P P (γ )
Sol: Since j tP tj p(γ = γj ) = 1, that means β = 1/E[1/γ] and thus the zero-outage
capacity per unit bandwidth is C = log2 (1 + β). Also, we can calculate
β=
0.4
0.3
0.2
0.1
+
+
+
0.1 0.075 0.05 0.025
So C = log2 (1 + β) = 0.08746.
6
−1
=
1
.
16