无线通信基础（Davaid Tse版）答案

Chapter 1

Solutions to Exercises

1

Chapter 2

Solutions to Exercises

Exercise 2.1. 1. Let r(t) =

p

r

2

0

+ (vt)

2

+ 2r

0

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

, arctan(y/x), arccos(z/

p

x

2

+ y

2

+ z

2

)

´

r

2

p

2

We see that

ψ is small for large x

2

+ y

2

. Also

˙

If |r/z| = 1 then θ = 0 or θ = π and v <= r|

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

−

α cos [2πf (t + (r(t) − 2d)/c)]

2d − r(t)

2

Tse and Viswanath: Fundamentals of Wireless Communication 3

=

2α sin [2πf (t − d/c)] sin [2πf (r(t) − d) /c]

2d − r(t)

where we applied the identity

sin

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

(f, t) =

<[α exp{j2π[f(1 − v/c)t − fr

0

/c]}]

+ vt

<[α exp{j2π[f(1 − v/c)t − f(r

r

We have the interaction of 2 sinusoidal waves of the same frequency and different

amplitude.

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:

s

= |f

1

− f

2

| =

fv

|cos θ

1

− cos θ

2

|

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

.

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

(h

s

− h

)

2

r

2

p

+ (h

+ h

r

)

2

/r

2

(h

)

(h

+ h

r

)

r

2

2r

h

2

s

+ h

2

r

+ 2h

2h

r

Therefore b = 2h

.

1

1

2

1

Tse and Viswanath: Fundamentals of Wireless Communication 5