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首页无线通信基础(Davaid Tse版)答案
无线通信基础(Davaid Tse版)答案
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国外大牛写的无线通信基础这本书的答案,很详细,值得无线通信方向的同学学习
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Chapter 1
Solutions to Exercises
1
![](https://csdnimg.cn/release/download_crawler_static/5611949/bg2.jpg)
Chapter 2
Solutions to Exercises
Exercise 2.1. 1. Let r(t) =
p
r
2
0
+ (vt)
2
+ 2r
0
vt cos(φ). Then,
E
r
(f, t, r(t), θ, ψ)) =
<[α(θ, ψ, f) exp{j2πf(1 − r(t)/c)}]
r(t)
.
Moreover, if we assume that r
0
À vt, then we get that r(t) ≈ r
0
+ vt cos(φ).
Thus, the doppler shift is fv cos(φ)/c.
2. Let (x, y, z) be the position of the mobile in Cartesian coordinates, and (r, ψ, θ)
the position in polar coordinates. Then
(x, y, z) = (r sin θ cos ψ, r sin θ sin ψ, r cos θ)
(r, ψ, θ) =
³
p
x
2
+ y
2
+ z
2
, arctan(y/x), arccos(z/
p
x
2
+ y
2
+ z
2
)
´
˙
ψ =
x ˙y − ˙xy
x
2
+ y
2
˙
θ = −
˙zr − z ˙r
r
2
p
1 − (z/r)
2
We see that
˙
ψ is small for large x
2
+ y
2
. Also
˙
θ is small for |z/r| < 1 and r large.
If |r/z| = 1 then θ = 0 or θ = π and v <= r|
˙
θ| so v/r large assures that
˙
θ is
small. If r is not very large then the variation of θ and ψ may not be negligible
within the time scale of interest even for moderate speeds v. Here large depends
on the time scale of interest.
Exercise 2.2.
E
r
(f, t) =
α cos [2πf (t − r(t)/c)]
2d − r(t)
+
2α [d − r(t)] cos [2πf (t − r(t)/c)]
r(t)[2d − r(t)]
−
α cos [2πf (t + (r(t) − 2d)/c)]
2d − r(t)
2
![](https://csdnimg.cn/release/download_crawler_static/5611949/bg3.jpg)
Tse and Viswanath: Fundamentals of Wireless Communication 3
=
2α sin [2πf (t − d/c)] sin [2πf (r(t) − d) /c]
2d − r(t)
+
2α [d − r(t)] cos [2πf (t − r(t)/c)]
r(t)[2d − r(t)]
(2.1)
where we applied the identity
cos x − cos y = 2 sin
µ
x + y
2
¶
sin
µ
y − x
2
¶
We observe that the first term of (2.1) is similar in form to equation (2.13) in the
notes. The second term of (2.1) goes to 0 as r(t) → d and is due to the difference in
propagation losses in the 2 paths.
Exercise 2.3. If the wall is on the other side, both components arrive at the mobile
from the left and experience the same Doppler shift.
E
r
(f, t) =
<[α exp{j2π[f(1 − v/c)t − fr
0
/c]}]
r
0
+ vt
−
<[α exp{j2π[f(1 − v/c)t − f(r
0
+ 2d)/c]}]
r
0
+ 2d + vt
We have the interaction of 2 sinusoidal waves of the same frequency and different
amplitude.
Over time, we observe the composition of these 2 waves into a single sinusoidal
signal of frequency f (1−v/c ) and constant amplitude that depends on the attenuations
(r
0
+ vt) and (r
0
+ 2d + vt) and also on the phase difference f2d/c.
Over frequency, we observe that when f2d/c is an integer both waves interfere
destructively resulting in a small received signal. When f2d/c = (2k + 1)/2, k ∈ Z
these waves interfere constructively resulting in a larger received signal. So when f
is varied by c/4d the amplitude of the received signal varies from a minimum to a
maximum.
The variation over frequency is similar in nature to that of section 2.1.3, but since
the delay spread is different the coherence bandwidth is also different.
However there is no variation over time because the Doppler spread is zero.
Exercise 2.4. 1. i) With the given information we can compute the Doppler spread:
D
s
= |f
1
− f
2
| =
fv
c
|cos θ
1
− cos θ
2
|
from which we can compute the coherence time
T
c
=
1
4D
s
=
c
4fv|cos θ
1
− cos θ
2
|
ii) There is not enough information to compute the coherence bandwidth, as it
depends on the delay spread which is not given. We would need to know the
difference in path length to compute the delay spread T
d
and use it to compute
W
c
.
![](https://csdnimg.cn/release/download_crawler_static/5611949/bg4.jpg)
Tse and Viswanath: Fundamentals of Wireless Communication 4
2. From part 1 we see that a larger angular range results in larger delay spread and
smaller coherence time. Then, in the richly scattered environment the channel
would show a smaller coherence time than in the environment where the reflectors
are clustered in a small angular range.
Exercise 2.5. 1.
r
1
=
p
r
2
+ (h
s
− h
r
)
2
= r
p
1 + (h
s
− h
r
)
2
/r
2
≈ r(1 +
(h
s
− h
r
)
2
2r
2
)
r
2
=
p
r
2
+ (h
s
+ h
r
)
2
= r
p
1 + (h
s
+ h
r
)
2
/r
2
≈ r(1 +
(h
s
+ h
r
)
2
2r
2
)
r
2
− r
1
≈
(h
s
+ h
r
)
2
− (h
s
− h
r
)
2
2r
=
h
2
s
+ h
2
r
+ 2h
s
h
r
− h
2
s
− h
2
r
+ 2h
s
h
r
2r
=
2h
s
h
r
r
Therefore b = 2h
s
h
r
.
2.
E
r
(f, t) ≈
Re[α[exp{j2π(ft − fr
1
/c)] − exp{j2π(ft − fr
2
/c)]]
r
1
=
Re[α[exp{j2π(ft − fr
1
/c)][1 − exp(j2πf(r
1
− r
2
)/c)]
r
1
≈
Re[α[exp{j2π(ft − fr
1
/c)][1 − exp(j2πf/c ∗ b/r)]
r
1
≈
Re[α[exp{j2π(ft − fr
1
/c)][1 − (1 − j2πf/c ∗b/r)]
r
1
=
2πf|α|b
cr
2
<[j exp(j∠α) exp[j2π(ft −f r
1
/c)]]
= −
2πf|α|b
cr
2
sin[2π(ft − fr
1
/c) + ∠α]]
Therefore β = 2πf|α|b/c.
3.
1
r
2
=
1
r
1
+ (r
2
− r
1
)
=
1
r
1
[1 + (r
2
− r
1
)/r
1
]
≈
1
r
1
µ
1 −
r
2
− r
1
r
1
¶
≈
1
r
1
µ
1 −
b
r
2
1
¶
Therefore if we don’t make the approximation of b) we get another term in
the expansion that decays as r
−3
. This term is negligible for large enough r as
compared to β/r
2
.
![](https://csdnimg.cn/release/download_crawler_static/5611949/bg5.jpg)
Tse and Viswanath: Fundamentals of Wireless Communication 5
Exercise 2.6. 1. Let f
2
be the probability density of the distance from the origin
at which the photon is absorbed by exactly the 2nd obstacle that it hits. Let x
be the location of the first obstacle, then
f
2
(r) = P {photon absorbed by 2nd obstacle at r}
=
Z
x
P {absorbed by 2nd obstacle at r | not absorbed by 1st obstacle at x}
× P {not absorbed by 1st obstacle at x} dx
Since the obstacle are distributed according to poisson process which has mem-
oryless distances between consecutive points, the first term inside the integral is
f
1
(r − x). The second term is the probability that the first obstacle is at x and
the photon is not absorbed by it. Thus, it is given by (1 − γ)q(x). Thus,
f
2
(r) =
Z
∞
x=−∞
(1 − γ)q(x)f
1
(r − x)dx
2. Similarly, we observe that f
k+1
(r) is given by
f
k+1
(r) =
Z
x
P {absorbed by (k + 1)th obst at r | not absorbed by 1st obst at x}
× P {not absorbed by 1st obstacle at x} dx
=
Z
∞
x=−∞
(1 − γ)q(x)f
k
(r − x)dx (2.2)
3. Summing up (2.2) for k = 1 to ∞, we get:
∞
X
k=2
f
k
(r) =
Z
∞
x=−∞
(1 − γ)q(x)
Ã
∞
X
k=1
f
k
(r − x)
!
dx
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