Stefano Di Bella.
Driven by the process. Fascinated by the results.
Stefano Di Bella pays close attention to detail. The brand revolves around organization, productivity, and
creativity. The slogan Op
CHAPTER 8
CODING AND ERROR CONTROL
A N SWERS
NSWERS
TO
Q UESTIONS
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
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 =
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
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 le
CHAPTER 7
SPREAD SPECTRUM
A N SWERS
NSWERS
TO
Q UESTIONS
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 an
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
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 siz
CHAPTER 6
SIGNAL ENCODING TECHNIQUES
A N SWERS
NSWERS
TO
Q UESTIONS
6.1 In differential encoding, the signal is decoded by comparing the polarity of adjacent
signal elements rather than determining th
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 eit
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
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
Rebecca C.
PROJECT MANAGER
WHAT DO YOU NEED?
A COMMITTED, DETAIL-ORIENTED, MANAGER
FOR YOUR BUSINESS PROJECTS.
Thats exactly what I can offer to you and your organization, along
with a passion for ens
To: ENC 3250 Professor Brenda Maxey-Billings
From:
Date 1/18/18
Subject: Introductory Memo
Self-Introduction
Hi, my name is Devin. I am currently a student at University of North Florida. I am a Sopho
Running head: PERSONAL BRAND
1
Personal Brand
Name
Institutional Affiliation
PERSONAL BRAND
2
Mc STINGAZ
Forever committed to excellence.
Stingaz is determined and set in the production of the best qu
CHAPTER 10
CELLULAR WIRELESS NETWORKS
A N SWERS
NSWERS
TO
Q UESTIONS
10.1 Hexagon
10.2 For frequency reuse in a cellular system, the same set of frequencies are used in
multiple cells, with these cell
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
= 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 - 4
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
CHAPTER 9
SATELLITE COMMUNICATIONS
A N SWERS
NSWERS
TO
Q UESTIONS
9.1 Coverage area: global, regional, or national. The larger the area of coverage, the
more satellites must be involved in a single ne
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
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 seque