Download Solutions to HW 1 - Venugopal V. Veeravalli

ECE 559
Fall 2007
Handout # 4
September 20, 2007
SOLUTIONS TO HOMEWORK ASSIGNMENT 1
1. Vanderkulk’s Lemma. The complex random variable Z = X + jY is zero mean and Gaussian but not
necessarily proper. Show that
E exp(jνZ) = exp(−ν 2 EZ 2 /2) ,
where ν can be assumed to be real (though this is not really necessary). This result is known as
Vanderkulk’s lemma and is similar to the characteristic function result for a real Gaussian random
variable. Note that this gives the interesting result that E exp(jνZ) = 1 when Z is proper complex
Gaussian.
Solution. By definition
∆
pZ (z) = pX,Y (x, y) = pX|Y (x|y)pY (y)
Since X and Y are zero mean and jointly Gaussian, X is conditionally Gaussian given Y , with
conditional mean and variance:
σX ρ y
2
m1 (y) =
, σ12 = σX
(1 − ρ2 )
σY
where ρ = E[XY ]/σX σY .
Now, we have the following result on the characteristic function of Gaussian random variables. If V
is a real-valued Gaussian random variable with mean m and variance σ 2 , then
E[ejθV ] = ejθm −θ
2 σ 2 /2
(1)
Note that (1) holds even when θ is complex.
Now,
E[ejνZ ] = E[ejνX−νY ] = EY EX|Y [ejνX ] e−νY
Applying (1) to the inner expectation, we get
EX|Y [ejνX ] = ejνm1 (Y )−ν
2 σ 2 /2
1
which implies that
jνZ
Now apply (1) to Y with θ = ν
jνZ
E[e
]=e
2
ν 2 σ1
2
E[e
σX ρ
σY
2
− ν2
e
“
2
ν 2 σ1
2
jν
EY e
]=e
+ j and we get
”2
σX ρ
2
+j σY
σY
= e−
“
σX ρ
+j
σY
ν2
2 +σ 2
(σX
Y
2
” Y
−2ρσX σY j)
= e−
ν 2 E[Z 2 ]
2
2. (Optional) Properness of a PCG Vector. Prove Result B.2 in the class notes: Let Y = Y I + jY Q be
proper complex and Gaussian, i.e. Y I , Y Q are jointly Gaussian. Then
pY (y) := pY I Y Q (y I , y Q )
o
n
1
exp −(y − mY )† Σ−1
= n
Y (y − mY )
π |ΣY |
Note: This problem is optional, and is meant for brave souls who wish to explore the dark world of
..
Matrix Algebra ⌣ The solution can be found in the paper by Neeser and Massey.
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3. Cellular Area Reliability. Read Section 2.5.1 of the notes on area reliability and derive equation
(2.29), i.e., show that
exp b22 − 2a
2
b
Farea = Q(a + b ln R) +
1 − Q a + b ln R −
.
R2
b
Also derive equation (2.31), i.e., show that
2
b2
2
−1
1 − Q Q (Fedge ) −
b
− 2b Q−1 (Fedge )
Farea = Fedge + e e
Solution:
Farea
2
= 2
R
Z
R
0
2
Q(a + b ln ρ)ρdρ =
bR2
Z
a+b ln R
Q(t)e
2(t−a)
b
−∞
2 − 2a
e b
dt =
bR2
Z
a+b ln R
2t
Q(t)e b dt
−∞
Defining I to be the last integral in the previous equation, we get
Z a+b ln R
2t
Q(t)e b dt
I=
−∞
b
b 2t
ln R
+ √
= Q(t) e b |a+b
−∞
2
2 2π
a
Z
a+b ln R
t2
2t
e− 2 e b dt
−∞
(t− 2 )2
Z
b
2(a+b ln R)
b
b 22 a+b ln R e− 2
b
b
√
= Q(a + b ln R)e
dt
+ e
2
2
2π
−∞ 2
b
b 22
2 2a
= Q(a + b ln R)R e b + e b 1 − Q a + b ln R −
2
2
b
where in a we used integration by parts. Substituting this result back into the previous equation
2
Farea
2a
e b2 − b
2 − 2a
b
e
I = Q(a + b ln R) +
=
bR2
R2
2
1 − Q a + b ln R −
b
.
From the definition of Fedge we have a + b ln R = Q−1 (Fedge ). Thus
2
e− b
e− b (Q
=
R2
2a
−1 (F
)
edge )−b ln R
R2
and
2
b2
− 2b Q−1 (Fedge )
Farea = Fedge + e e
2
= e− b (Q
−1 (F
1 − Q(Q
−1
2 ln R
)e
edge )
R2
2
(Fedge ) − )
b
Note that this is independent of R and a.
4. Signal strength prediction. Consider a mobile that is moving on a straight line path (not necessarily
radial) at a constant velocity v. In order to make handoff decisions, the mobile periodically takes
pilot power measurements from neighboring BS’s. Let us assume that these power measurements
are averaged to remove multipath fluctuations, so the resulting sampled measurements only have a
median component and shadow fading. The k-th sample value of the pilot power (in dBm) from a
particular BS is given by:
Pr,k [dBm] = P̄r (dk ) + Zk = At − B log dk + Zk ,
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where dk is the distance from the BS at the k-th sampling time, and At includes the transmitted pilot
power. Note that the dk values are not necessarily equally spaced.
Let us assume isotropic shadow fading with exponential ACF. Since the velocity vector is constant,
the random process {Zk , k = 1, 2, . . .} is a stationary first-order auto-regressive (AR) process with
E[Zk Zk+m ] = σZ2 a|m| ,
where a = exp(−vts /Dc ), ts is the sampling time, and Dc is the correlation distance.
Handoff decisions are often based on signal strength prediction. Our goal here is to find the MMSE
predictor of Pr,k+1 based on Pr,1 , Pr,2 , . . . , Pr,k .
(a) Under the assumption that the dk values are known, show that
MMSE
P̂r,k+1
= E[Pr,k+1 |Pr,1 , Pr,2 , . . . , Pr,k ] = aPr,k + (1 − a)At − B log
dk+1
dak
and that the corresponding mean-squared error
MSE = Var[Pr,k+1 |Pr,1 , Pr,2 , . . . , Pr,k ] = (1 − a2 )σZ2 .
Hint: You don’t need to solve the Yule-Walker equations to find the MMSE solution in this
special case. Use the AR-1 property of the {Zk } to find the solution directly.
Solution:
Pr,k+1 = At + B log dk+1 + Zk+1
= At + B log dk+1 + aZk + σZ
p
1 − a2 Wk
= a (At + B log dk + Zk ) + (1 − a)At + B log dk+1 − aB log dk + σZ
Thus,
Pr,k+1 = aPr,k + (1 − a)At + B log
p
1 − a2 Wk
p
dk+1
+ σZ 1 − a2 Wk .
a
dk
By the AR-1 model construction, Wk is independent of (Pr,k , . . . , Pr,1 ). This implies that Pr,k+1
given Pr,k is independent of (Pr,k−1 , . . . , Pr,1 ), and
E [Pr,k+1 |Pr,k , . . . , Pr,1 ] = E [Pr,k+1 |Pr,k ] = aPr,k + (1 − a)At + B log
dk+1
.
dak
Similarly,
p
2
2
2
Var [Pr,k+1 |Pr,k , . . . , Pr,1 ] = Var [Pr,k+1 |Pr,k ] = E σZ 1 − a
Wk = σZ2 (1 − a2 )
(b) Discuss how you might address the prediction problem if the dk values were unknown.
Solution: The prediction problem is considerably harder when dk is unknown. One way to approach the problem is to use a velocity estimate based on measurements of doppler frequency
to first obtain an estimate of the distance ∆ρ between consecutive samples. We may then use a
“curve fitting” technique to obtain an estimate of dk based on all of the signal strength measurements. Thus, the best estimate of Pr,k+1 based on (Pr,k , . . . , Pr,1 ) will be a function of all the
measurements, not just the most recent one.
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5. Moments of lognormals. Suppose X is a lognormal random variable with mean mX and second
moment δX , and suppose Y = 10 log X has mean mX and variance σY2 .
(a) Show that
mX = exp
(βσY )2
2
exp(βmY ) and
δX = exp 2(βσY )2 exp(2βmY ),
where β = ln(10)/10.
Solution: The characteristic function of Y is given by
φY (s) = E[esY ] = esmY +
2
s2 σY
2
Thus,
mX = E[eβY ] = φY (β) = eβmY +
2
β 2 σY
2
and
δX = E[X 2 ] = φY (2β) = e2βmY +2β
2 σ2
Y
.
(b) Also, show that
mY = 20 log mX − 5 log δX ,
σY2 =
and that
1
(10 log δX − 20 log mX ).
β
Solution: Taking the log of the previous equations, we get
βσY2
ln mX
=
+ mY
β
2
ln δX
= 2βσY2 + 2mY .
=
β
10 log mX =
10 log δX
Solving for mY and σY2 , we obtain
mY = 20 log mX − 5 log δX
1
σY2 = (10 log δX − 20 log mX ) .
β
6. Outage with macrodiversity. Consider a mobile at the midpoint between two base stations in a cellular
network. The received signals (in dB-W) from the base stations are given by
Pr,1 = At − B log(D/2) + Z1 ,
Pr,2 = At − B log(D/2) + Z2 ,
where Z1 and Z2 are N (0, σ 2 ) random variables.
We define outage with macrodiversity to be the event that both Pr,1 and Pr,2 fall below a pre-specified
threshold Pthresh .
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(a) If Z1 and Z2 are independent, show that the outage probability is given by
Pout
where
2
∆
,
= Q
σ
∆
∆ = At − B log(D/2) − Pthresh
is the fade margin at the edge of the cell.
Solution:
Pout = P({Pr,1 < Pthresh } ∩ {Pr,2 < Pthresh }) = P(Pr,1 < Pthresh )P(Pr,2 < Pthresh )
= P (Z1 < Pthresh − At − B log(D/2)) P (Z2 < Pthresh − At − B log(D/2))
2
∆
= P (Z1 < −∆) P (Z2 < −∆) = Q
σ
(b) Now suppose Z1 and Z2 are correlated in the following way.
Z1 = aY1 + bY
and
Z2 = aY2 + bY,
where Y , Y1 and Y2 are independent N (0, σ 2 ) random variables, and a and b are such that
a2 + b2 = 1.
Show that
Z ∞ 2
∆ + byσ 2 e−y /2
√
dy.
Pout =
Q
aσ
2π
−∞
Solution:
Pout = P({aY1 + bY < −∆} ∩ {aY2 + bY < −∆})
Z ∞
P({aY1 + by < −∆} ∩ {aY2 + by < −∆})pY (y)dy
=
−∞
Z ∞
P(aY1 < −∆ − by)P(aY2 < −∆ − by)pY (y)dy
=
−∞
=
Z
∞
−∞
1
√
2πσ
2
2 − w2
Z ∞ 2
∆ + bwσ
∆ + by
e 2
− y2
√ dw
Q
e 2σ dy =
Q
aσ
aσ
2π
−∞
which is the desired expression.
√
(c) Compare the outage probabilities of (i) and (ii) for the special case of a = b = 1/ 2, σ = 8
and ∆ = 5. (Use a numerical integration routine for Pout of (ii).)
Solution:
Pout,1 = 0.0707
Pout,2 = 0.1316
(2)
Thus, correlation reduces diversity gain.
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V.
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7. Squared-envelope correlation for isotropic fading:
(a) Prove that the squared-envelope covariance function for a Rayleigh fading process {E(t)} is
given by:
Cα2 (τ ) = |RE (τ )|2
Hint: It may be easier to first show that Rα2 (τ ) = |RE (τ )|2 + 1. You may find the following
result to be useful: if X1 , X2 , X3 , X4 are jointly Gaussian zero-mean random variables, then
E[X1 X2 X3 X4 ] = E[X1 X2 ]E[X3 X4 ] + E[X1 X3 ]E[X2 X4 ] + E[X1 X4 ]E[X2 X3 ].
Solution: In particular, when X1 = X2 and X3 = X4 we get
E[X1 X1 X3 X3 ] = E[X1 X1 ]E[X3 X3 ] + 2 (E[X1 X3 ])2 .
Thus
Rα2 (τ ) = E
2
2
(t + τ ) EI2 (t) + EQ
(t)
EI2 (t + τ ) + EQ
2
2
2
2
(t + τ )EI2 (t)]
(t)] + E[EQ
(t + τ )EQ
(t)] + E[EI2 (t + τ )EQ
= E[EI2 (t + τ )EI2 (t)] + E[EQ
2
(t)]
= 2E[EI2 (t + τ )EI2 (t)] + 2E[EI2 (t + τ )EQ
= 2 E[EI2 (t + τ )]E[EI2 (t)] + 2 (E[EI (t + τ )EI (t)])2
2
+ 2 E[EI2 (t + τ )]E[EQ
(t)] + 2 (E[EI (t + τ )EQ (t)])2
1
1
2
2
+ 2(REI (τ )) + 2
+ 2(REI EQ (τ ))
=2
4
4
= 1 + 4 (REI (τ ))2 + (REI EQ (τ ))2 = 1 + |RE (τ )|2
(b) Prove that the squared-envelop covariance function for a Ricean fading process {E(t)} is:
2 h
h
ii
1
Cα2 (τ ) =
|RĚ (τ )|2 + 2κRe RĚ (τ )e−j2πfmax τ cos θ0 ,
κ+1
where κ is the Rice factor and θ0 is the angle of arrival of the specular component.
Solution:
Rα2 (τ ) = E α2 (t)α2 (t + τ ) = E [E(t)E ∗ (t)E(t + τ )E ∗ (t + τ )]
q
q
jφ0 (t)
−jφ0 (t)
∗
2
2
= E[ β0 e
+ 1 − β0 Ě(t)
β0 e
+ 1 − β0 Ě (t)
q
q
β0 e−jφ0 (t+τ ) + 1 − β02 Ě ∗ (t + τ ) ]
β0 ejφ0 (t+τ ) + 1 − β02 Ě(t + τ )
a
= β04 + (1 − β02 )2 Rα̌2 α̌2 (τ ) + β02 (1 − β02 )E[α̌2 (t)] + β02 (1 − β02 )E[α̌2 (t + τ )]
β02 (1 − β02 )ej(φ0 (t)−φ0 (t+τ )) E[Ě ∗ (t)Ě(t + τ )] + β02 (1 − β02 )e−j(φ0 (t)−φ0 (t+τ )) E[Ě(t)Ě ∗ (t + τ )]
= β04 + (1 − β02 )2 (1 + |RĚ (τ )|2 ) + 2β02 (1 − β02 )
h
i
+ 2β02 (1 − β02 )Re E[Ě ∗ (t)Ě(t + τ )]e−j(φ0 (t+τ )−φ0 (t))
i
h
= 1 + (1 − β02 )2 |RĚ (τ )|2 + 2β02 (1 − β02 )Re E[Ě ∗ (t)Ě(t + τ )]e−j(φ0 (t+τ )−φ0 (t))
2
2
i
h
1
1
2
=1+
|RĚ (τ )| + 2
κRe RĚ (τ )e−j2πfmax τ cos θ0
1+κ
1+κ
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6
Note that there are 16 terms in the expectation. Of these 16 terms, 8 involve only one phase term
φ0 . Since the specular component is independent of the diffuse component, and the marginal
distribution of the phase is uniform, these eight terms go to zero. Of the six terms containing
two phase terms, two vanish because they involve phase of the form φ0 (t) + φ0 (t + τ ). Thus, 6
terms are left in a.
2 h
ii
h
2
1
2
|RĚ (τ )|2 + 2κRe RĚ (τ )e−j2πfmax τ cos θ0
Cα2 = Rα2 − E[α (t)] =
1+κ
8. Power Spectrum of Rayleing Fading Process. Consider a Rayleigh fading environment with angular
power density p(θ).
(a) Show that:
SE (f ) =
Z
π
0
[p(θ) + p(−θ)]δ(f − fmax cos θ)dθ
(b) Now show that SE (f ) has the closed-form expression:

−1
)) + p(− cos−1 (f /fmax ))
 p(cos (f /fmax
p
2 − f2
SE (f ) =
fmax

0
for |f | < fmax
otherwise
(c) Specialize this result for isotropic Rayleigh fading, i.e., p(θ) = 1/2π.
Solution:
SE (f ) =
Z
∞
Z
π
j2πfmax τ cos θ −j2πf τ
p(θ)e
e
−π
Z
Z−∞
π
p(θ)δ(f − fmax cos θ)dθ =
=
=
−π
Z fmax
−fmax
=
dθdτ =
p(θ)
−π
Z
∞
e−j2π(f −fmax τ cos θ) dτ dθ
−∞
0
[p(θ) + p(−θ)] δ(f − fmax cos θ)dθ
p(cos−1 (u/fmax )) + p(− cos−1 (u/fmax )) δ(u − f )
2 −f 2
fmax
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π
π

 [p(cos−1 (f /fmax√))+p(− cos−1 (f /fmax ))]
0
Z
if |f | < fmax ,
du
fmax
p
1 − (u/fmax )2
otherwise.
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