CHAPTER 10
CELLULAR WIRELESS NETWORKS
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10.1 Hexagon
10.2 For frequency reuse in a cellular system, the same set of frequencies are used in
multiple cells, with these cells separated from one another by enough distance to
avoi
The total data transmission rate is 5 11.25 = 56.25 Mbps
Efficiency = 56.25/60 = 0.9375
9.8 The transponder must carry a total data bit rate of 15 + 10 + 5 = 30 Mbps. Thus,
each frame carries 30 Mbps 0.001 s = 30 kb The three preamble and guard times
take
= 20 log [(4 4 10 7 )/(2.727 x 102) = 205.3 dB
received_power = 10 + 17 + 52.3 205.3 = 126 dBW
9.5 The total bandwidth is 500 MHz. The channel bandwidth is 12 36 = 432 MHz. So
the overhead is (500 - 432)/500) 100% = 13.6%
9.6 a. The data rate R = 2 QPSK b
equator, the coverage near the north and south poles is inadequate. For these
regions, better communication can be achieved by LEO or HEO satellites. HEO
satellites have the additional advantage that they spend most of their time at the
high part of their
CHAPTER 9
SATELLITE COMMUNICATIONS
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9.1 Coverage area: global, regional, or national. The larger the area of coverage, the
more satellites must be involved in a single networked system. Service type: fixed
service satellite (
M = log2 (N )
8.25 a.
0
1
2
3
4
5
6
7
0
0
1
2
3
4
5
6
7
0
0
1
2
3
4
5
6
7
0
b.
c.
8.26 Let t1 = time to transmit a single frame
t1 =
1024 bits
= 1. 024 m sec
106 bps
The transmitting station can send 7 frames without an acknowledgment. From the
beginning
b. The encoder is in state a = 00 before transmission and after transmission of the
last information bit two zero bits are transmitted. The sequence of states
traversed is abdcbcbdcb. The output sequence is 10 11 10 01 01 01 11 10 01 00
8.22 a. Because on
8.9
a.
00000
10101
01010
00000
0
3
2
10101
2
0
5
01010
2
5
0
000000
010101
101010
110110
000000
0
3
3
4
b.
010101
3
0
6
6
101010
3
6
0
3
110110
4
6
3
0
8.10 a. p(v|w) = d(w,v)(1 ) (n d(w,v)
n d
1 d
p (v|w1 ) d (1 )
b. If we write di = d(w i ,v), then
=
n
8.7
a. The multiplication of M(X) by X16 corresponds to shifting M(X) 16 places and
thus providing the space for a 16-bit FCS. The addition of XkL(X) to X16M(X)
inverts the first 16 bits of G(X) (one's complements). The addition of L(X) to
R(X) inverts al
8.4 At the conclusion of the data transfer, just before the CRC pattern arrives, the shift
register should contain the identical CRC result. Now, the bits of the incoming
CRC are applied at point C 4 (Figure 8.3). Each 1 bit will merge with a 1 bit
(exclu
Pr [single bit not in error] = 1 103 = 0.999
Pr [8 bits not in error] = (1 103)8 = (0.999)8 = 0.992
Pr [at least one error in frame] = 1 (1 103)8 = 0.008
b. Pr [at least one error in frame] = 1 (1 103)10 = 1 (0.999)10 = 0.01
8.3 a.
b.
Shift
1
2
3
4
5
6
7
CHAPTER 8
CODING AND ERROR CONTROL
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8.1 A parity bit appended to an array of binary digits to make the sum of all the binary
digits, including the parity bit, always odd (odd parity) or always even (even
parity).
8.2 The CRC
It is easily seen that a bitwise multiplication of any two rows produces 0. For
example, row 3 multiplied by row 4 equals 1 + (1) + 1 + (1) + 1 + (1) + 1 + (1) =
0.
7.14 a. 8
b. 8
c.
A output (data = 1)
B output (data = 1)
Received
Receiver codeword
Multi
7.9 Let us start with an initial seed of 1. The first generator yields the sequence:
1, 6, 10, 8, 9, 2, 12, 7, 3, 5, 4, 11, 1, . . .
The second generator yields the sequence:
1, 7, 10, 5, 9, 11, 12, 6, 3, 8, 4, 2, 1, . . .
Because of the patterns evident
7.2 The total number of tones, or individual channels is:
Ws /fd = (400 MHz)/(100 Hz) = 4 106 .
The minimum number of PN bits = log 2 (4 10 6 ) = 22
where x indicates the smallest integer value not less than x. Source: [SKLA01]
7.3 Ws = 1000 fd ; Wd = 4 f
CHAPTER 7
SPREAD SPECTRUM
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7.1 The bandwidth is wider after the signal has been encoded using spread spectrum.
7.2 (1) We can gain immunity from various kinds of noise and multipath distortion. (2)
It can also be used for hid
b. For PM, s(t) = A cos(2fct + npm(t)
s1 (t) = A cos(2fct + npm 1 (t);
s2 (t) = A cos(2fct + npm 2 (t)
For the combined signal m c(t) = m1 (t) + m2 (t),
sc(t) = A cos(2fct + np[m 1 (t) + m2 (t)]), which is not a linear combination of s1 (t)
and s2 (t).
-2
6.13
slope overload distortions
DM output
1
0
6.14 s(t) = A c cos[2f ct + (t)] = 10 cos [(108 )t + 5 sin 2 (10 3 )t]
Therefore, (t) = 5 sin 2 (103 )t, and the maximum phase deviation is 5 radians. For
frequency deviation, recognize that the change in freq
and the maximum slope is A2 fa . To avoid slope overload, we require that
fs > A2 fa
or
>
2f a A
fs
Source: [COUC01]
6.12 a. A total of 28 quantization levels are possible, so the normalized step size is 2 8
= 0.003906.
b. The actual step size, in volts,
CHAPTER 6
SIGNAL ENCODING TECHNIQUES
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6.1 In differential encoding, the signal is decoded by comparing the polarity of adjacent
signal elements rather than determining the absolute value of a signal element.
6.2 A modem conve
5.12 From Equation 2.2, the ratio of transmitted power to received power is
Pt/P r = (4d/ ) 2
If we double the frequency, we halve , or if we double the distance, we double d,
so the new ratio for either of these events is:
Pt/P r2 = (8d/ ) 2
Therefore:
1
y1
p
p y1
1
x1
y12 px1 + p 2
2
2
tan =
y1 p =
1
1+
p y1 x1 y1 2 py1 + py1
x1
2
2
Because y1 = 2 px1 , this simplifies to tan = (p/y1 ).
5.7
Antenna
Isotropic
= 30 cm
Effective area (m2 )
0.007
Gain
1
= 3 mm
Effective area (m2 )
7.2 10 7
Gain
1
Infinit