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《wireless communications》习题解答
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更新于2023-03-03
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本书是《wireless communications》的习题解答,内中包含了几乎所有的习题答案,部分还有matlab代码实现!
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Chapter 1
1. In case of an accident, there is a high chance of getting lost. The transportation cost is very high each
time. However, if the infrastructure is set once, it will be very easy to use it repeatedly. Time for
wireless transmission is negligible as signals travel at the speed of light.
2. Advantages of bursty data communication
(a) Pulses are made very narrow, so multipaths are resolvable
(b) The transmission device needs to be switched on for less time.
Disadvantages
(a) Bandwidth required is very high
(b) Peak transmit power can be very high.
3. P
b
= 10
−12
1
2γ
= 10
−12
γ =
10
12
2
= 5 × 10
11
(very high)
4. Geo: 35,786 Km above earth ⇒ RT T =
2×35786×10
3
c
= 0.2386s
Meo: 8,000- 20,000 Km above earth ⇒ RT T =
2×8000×10
3
c
= 0.0533s
Leo: 500- 2,000 Km above earth ⇒ RT T =
2×500×10
3
c
= 0.0033s
Only Leo satellites as delay = 3.3ms < 30ms
5.
6. optimum no. of data user = d
optimum no. of voice user = v
Three different cases:
Case 1: d=0, v=6
⇒ revenue = 60.80.2 = 0.96
Case 2: d=1, v=3
revenue = [prob. of having one data user]×(revenue of having one data user)
+ [prob. of having two data user]×(revenue of having two data user)
+ [prob. of having one voice user]×(revenue of having one voice user)
+ [prob. of having two voice user]×(revenue of having two voice user)
+ [prob. of having three or more voice user]×(revenue in this case)
⇒ 0.5
2
µ
2
1
¶
× $1 + 0.5
2
× $1 +
µ
6
1
¶
0.8 × 0.2
5
× $0.2 +
µ
6
2
¶
0.8
2
× 0.2
4
× $0.4+
·
1 −
µ
6
1
¶
0.8 × 0.2
5
× $0.2 −
µ
6
2
¶
0.8
2
× 0.2
4
× $0.4
¸
× $0.6
⇒ $1.35
Case 3: d=2, v=0
revenue =2 × 0.5 = $1
So the best case is case 2, which is to allocate 60kHz to data and 60kHz to voice.
7.
8. 1. Hand-off becomes a big problem.
2. Inter-cell interference is very high and should be mitigated to get reasonable SINR.
3. Infrastructure cost is another problem.
9. Smaller the reuse distance, larger the number of users who can use the same system resource and so
capacity (data rate per unit bandwidth) increases.
10. (a) 100 cells, 100 users/cell ⇒ 10,000 users
(b) 100 users/cell ⇒ 2500 cells required
100km
2
Area/cell
= 2500cells ⇒
Area
cell
= .04km
2
(c) From Rappaport or iteration of formula, we get that 100
channels
cell
⇒ 89
channels
cell
@P
b
= .02
Each subscriber generates
1
30
of an Erlang of traffic.
Thus, each cell can support 30 × 89 = 2670 subscribers
Macrocell: 2670 × 100 ⇒ 267, 000 subscribers
Microcell: 6,675,000 subscribers
(d) Macrocell: $50 M
Microcell: $1.25 B
(e) Macrocell: $13.35 M/month ⇒ 3.75 months approx 4 months to reco op
Microcell: $333.75 M/month ⇒ 3.75 months approx 4 months to recoop
11. One CDPD line : 19.2Kbps
average W
imax
∼ 40M bps
∴ number of CDPD lines ∼ 2 × 10
3
Chapter 2
1.
P
r
= P
t
·
√
G
l
λ
4πd
¸
2
λ = c/f
c
= 0.06
10
−3
= P
t
·
λ
4π10
¸
2
⇒ P
t
= 4.39KW
10
−3
= P
t
·
λ
4π100
¸
2
⇒ P
t
= 438.65KW
Attenuation is very high for high frequencies
2. d= 100m
h
t
= 10m
h
r
= 2m
delay spread = τ =
x+x
0
−l
c
= 1.33×
3. ∆φ =
2π(x
0
+x−l)
λ
x
0
+ x − l =
p
(h
t
+ h
r
)
2
+ d
2
−
p
(h
t
− h
r
)
2
+ d
2
= d
s
µ
h
t
+ h
r
d
¶
2
+ 1 −
s
µ
h
t
− h
r
d
¶
2
+ 1
d À h
t
, h
r
, we need to keep only first order terms
∼ d
1
2
s
µ
h
t
+ h
r
d
¶
2
+ 1
−
1
2
s
µ
h
t
− h
r
d
¶
2
+ 1
=
2(h
t
+ h
r
)
d
∆φ ∼
2π
λ
2(h
t
+ h
r
)
d
4. Signal nulls occur when ∆φ = (2n + 1)π
2π(x
0
+ x − l)
λ
= (2n + 1)π
2π
λ
h
p
(h
t
+ h
r
)
2
+ d
2
−
p
(h
t
− h
r
)
2
+ d
2
i
= π(2n + 1)
p
(h
t
+ h
r
)
2
+ d
2
−
p
(h
t
− h
r
)
2
+ d
2
=
λ
2
(2n + 1)
Let m = (2n + 1)
p
(h
t
+ h
r
)
2
+ d
2
= m
λ
2
+
p
(h
t
− h
r
)
2
+ d
2
square both sides
(h
t
+ h
r
)
2
+ d
2
= m
2
λ
2
4
+ (h
t
− h
r
)
2
+ d
2
+ mλ
p
(h
t
− h
r
)
2
+ d
2
x = (h
t
+ h
r
)
2
, y = (h
t
− h
r
)
2
, x −y = 4h
t
h
r
x = m
2
λ
2
4
+ y + mλ
p
y + d
2
⇒ d =
s
·
1
mλ
µ
x − m
2
λ
2
4
− y
¶¸
2
− y
d =
s
µ
4h
t
h
r
(2n + 1)λ
−
(2n + 1)λ
4
¶
2
− (h
t
− h
r
)
2
, n ∈ Z
5. h
t
= 20m
h
r
= 3m
f
c
= 2GHz λ =
c
f
c
= 0.15
d
c
=
4h
t
h
r
λ
= 1600m = 1.6Km
This is a good radius for suburban cell radius as user density is low so cells can be kept fairly large.
Also, shadowing is less due to fewer obstacles.
6. Think of the building as a plane in R
3
The length of the normal to the building from the top of Tx antenna = h
t
The length of the normal to the building from the top of Rx antenna = h
r
In this situation the 2 ray model is same as that analyzed in the book.
7. h(t) = α
1
δ(t − τ) + α
2
δ(t − (τ + 0.22µs))
G
r
= G
l
= 1
h
t
= h
r
= 8m
f
c
= 900MHz, λ = c/f
c
= 1/3
R = −1
delay spread =
x + x
0
− l
c
= 0.022 × 10
−6
s
⇒
2
q
8
2
+
¡
d
2
¢
2
− d
c
= 0.022 × 10
−6
s
⇒ d = 16.1m
∴ τ =
d
c
= 53.67ns
α
1
=
µ
λ
4π
√
G
l
l
¶
2
= 2.71 ×10
−6
α
2
=
µ
λ
4π
√
RG
r
x + x
0
¶
2
= 1.37 ×10
−6
8. A program to plot the figures is shown below. The power versus distance curves and a plot of the
phase difference between the two paths is shown on the following page. From the plots it can be seen
that as G
r
(gain of reflected path) is decreased, the asymptotic behavior of P
r
tends toward d
−2
from
d
−4
, which makes sense since the effect of reflected path is reduced and it is more like having only a
LOS path. Also the variation of power before and around dc is reduced because the strength of the
reflected path decreases as G
r
decreases. Also note that the the received power actually increases with
distance up to some point. This is because for very small distances (i.e. d = 1), the reflected path is
approximately two times the LOS path, making the phase difference very small. Since R = -1, this
causes the two paths to nearly cancel each other out. When the phase difference becomes 180 degrees,
the first local maxima is achieved. Additionally, the lengths of both paths are initially dominated by
the difference between the antenna heights (which is 35 meters). Thus, the powers of both paths are
roughly constant for small values of d, and the dominant factor is the phase difference between the
paths.
clear all;
close all;
ht=50;
hr=15;
f=900e6;
c=3e8;
lambda=c/f;
GR=[1,.316,.1,.01];
Gl=1;
R=-1;
counter=1;
figure(1);
d=[1:1:100000];
l=(d.^2+(ht-hr)^2).^.5;
r=(d.^2+(ht+hr)^2).^.5;
phd=2*pi/lambda*(r-1);
dc=4*ht*hr/lambda;
dnew=[dc:1:100000];
for counter = 1:1:4,
Gr=GR(counter);
Vec=Gl./l+R*Gr./r.*exp(phd*sqrt(-1));
Pr=(lambda/4/pi)^2*(abs(Vec)).^2;
subplot(2,2,counter);
plot(10*log10(d),10*log10(Pr)-10*log10(Pr(1)));
hold on;
plot(10*log10(dnew),-20*log10(dnew));
plot(10*log10(dnew),-40*log10(dnew));
end
hold off
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