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CS421/S03
Solutions to Homework Four
Chapter 14 (7/e)
1.
Problem 14.1
a.
The number of clusters
M
= 16.
The total available bandwidth in each direction is 20 MHz. Therefore, the total number of
channels in each way is
20 MHz/30 kHz
= 666, and each cell can have 666/N
channels. Because each system is duplicated 16 times, the total number of channels will
be 16
×
666 = 10656. So, the number of simultaneous communications that can be
supported by each system is 10656.
b.
Total number of channels available is K = 666. For a frequency reuse factor
N
, each cell can use
k
CE
=
K
/
N
= 666/
N
channels.
For
N
= 4,
k
CE
= 166 channels.
For
N
= 7,
k
CE
= 95 channels.
For
N
= 12,
k
CE
= 55 channels.
For
N
= 19,
k
CE
= 35 channels.
c.
The number of cells covered is 16
×
N
. For
N
= 4, area = 64 cells; For
N
= 7, area
=112
cells; For
N
= 12, area = 192 cells; For
N
= 19, area = 304 cells.
d.
From part b, we know the number of channels that can be carried per cell for each
system. The total number of channels available is just 100 times that number, for a result
of 16600, 9500, 5500, 3500, respectively.
2
.
Problem 14.4
(12.5
×
10
6
– 2(10
×
10
3
))/(30
×
10
3
) = 416
Chapter 15
.
3.
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 Spring '08
 Shen

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