22
CHAPTER 1
INTRODUCTION
Sites
The following sites are related to topics discussed in this chapter.
o
o
www.acm.org/sigcomm/sos.html This site gives the status of varililus networking
standards.
www.ietf.org/ The Internet Engineering Task Force (IETF) home page.
RFCs
The following site lists all RFCs, including those related to IP and TCP. In future chapters we cite the RFCs pertinent to the chapter material.
o
1.6
www.ietf.org/rfc.html
KEY TERMS
Advanced Research Projects
Agency (ARPA)
American National Standards
Institute (ANSI)
American Standard Code for
Information Interchange (ASCII)
ARPANET
audio
backbone
Basic Latin
bus topology
code
Consultative Committee for
International Telegraphy
and Telephony (CCITT)
data
data communications
de facto standards
de jure standards
delay
distributed processing
Electronic Industries Association (EIA)
entity
Federal Communications Commission
(FCC)
forum
full-duplex mode, or duplex
half-duplex mode
hub
image
Institute of Electrical and Electronics
Engineers (IEEE)
International Organization for
Standardization (ISO)
International Telecommunication
Union-Telecommunication
Standards Sector (ITU-T)
Internet
Internet draft
Internet service provider (ISP)
Internet standard
internetwork or internet
local area network (LAN)
local Internet service providers
mesh topology
message
metropolitan area network (MAN)
multipoint or multidrop connection
national Internet service provider
network
SECTION 1.7
network access points (NAPs)
node
performance
physical topology
point-to-point connection
protocol
receiver
regional ISP
reliability
Request for Comment (RFC)
ROB
ring topology
security
semantics
1.7
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
SUMMARY
23
sender
simplex mode
star topology
syntax
telecommunication
throughput
timing
Transmission Control Protocol!
Internetworking Protocol (TCPIIP)
transmission medium
Unicode
video
wide area network (WAN)
YCM
SUMMARY
Data communications are the transfer of data from one device to another via some
form of transmission medium.
A data communications system must transmit data to the correct destination in an
accurate and timely manner.
The five components that make up a data communications system are the message,
sender, receiver, medium, and protocol.
Text, numbers, images, audio, and video are different forms of information.
Data flow between two devices can occur in one of three ways: simplex, half-duplex,
or full-duplex.
A network is a set of communication devices connected by media links.
In a point-to-point connection, two and only two devices are connected by a
dedicated link. In a multipoint connection, three or more devices share a link.
Topology refers to the physical or logical arrangement of a network. Devices may
be arranged in a mesh, star, bus, or ring topology.
A network can be categorized as a local area network or a wide area network.
A LAN is a data communication system within a building, plant, or campus, or
between nearby buildings.
A WAN is a data communication system spanning states, countries, or the whole
world.
An internet is a network of networks.
The Internet is a collection of many separate networks.
There are local, regional, national, and international Internet service providers.
A protocol is a set of rules that govern data communication; the key elements of
a protocol are syntax, semantics, and timing.
24
CHAPTER 1
INTRODUCTION
o
o
o
o
Standards are necessary to ensure that products from different manufacturers can
work together as expected.
The ISO, ITD-T, ANSI, IEEE, and EIA are some of the organizations involved
in standards creation.
Forums are special-interest groups that quickly evaluate and standardize new
technologies.
A Request for Comment is an idea or concept that is a precursor to an Internet
standard.
1.8
PRACTICE SET
Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1 I.
12.
13.
Identify the five components of a data communications system.
What are the advantages of distributed processing?
What are the three criteria necessary for an effective and efficient network?
What are the advantages of a multipoint connection over a point-to-point
connection?
What are the two types of line configuration?
Categorize the four basic topologies in terms of line configuration.
What is the difference between half-duplex and full-duplex transmission modes?
Name the four basic network topologies, and cite an advantage of each type.
For n devices in a network, what is the number of cable links required for a mesh,
ring, bus, and star topology?
What are some of the factors that determine whether a communication system is a
LAN or WAN?
What is an internet? What is the Internet?
Why are protocols needed?
Why are standards needed?
Exercises
14. What is the maximum number of characters or symbols that can be represented by
Unicode?
15. A color image uses 16 bits to represent a pixel. What is the maximum number of
different colors that can be represented?
16. Assume six devices are arranged in a mesh topology. How many cables are needed?
How many ports are needed for each device?
17. For each of the following four networks, discuss the consequences if a connection fails.
a. Five devices arranged in a mesh topology
b. Five devices arranged in a star topology (not counting the hub)
c. Five devices arranged in a bus topology
d. Five devices arranged in a ring topology
SECTION 1.8
PRACTICE SET
25
18. You have two computers connected by an Ethernet hub at home. Is this a LAN, a
MAN, or a WAN? Explain your reason.
19. In the ring topology in Figure 1.8, what happens if one of the stations is unplugged?
20. In the bus topology in Figure 1.7, what happens if one ofthe stations is unplugged?
21. Draw a hybrid topology with a star backbone and three ring networks.
22. Draw a hybrid topology with a ring backbone and two bus networks.
23. Performance is inversely related to delay. When you use the Internet, which of the
following applications are more sensitive to delay?
a. Sending an e-mail
b. Copying a file
c. Surfing the Internet
24. When a party makes a local telephone call to another party, is this a point-to-point
or multipoint connection? Explain your answer.
25. Compare the telephone network and the Internet. What are the similarities? What
are the differences?
Research Activities
26.
27.
28.
29.
Using the site \\iww.cne.gmu.edu/modules/network/osi.html, discuss the OSI model.
Using the site www.ansi.org, discuss ANSI's activities.
Using the site www.ieee.org, discuss IEEE's activities.
Using the site www.ietf.org/, discuss the different types of RFCs.
SECTION 2. 7 KEY TERMS
51
Books
Network models are discussed in Section 1.3 of [Tan03], Chapter 2 of [For06], Chapter 2
of [Sta04], Sections 2.2 and 2.3 of [GW04], Section 1.3 of [PD03], and Section 1.7 of
[KR05]. A good discussion about addresses can be found in Section 1.7 of [Ste94].
Sites
The following site is related to topics discussed in this chapter.
o
www.osi.org!
Information about OS1.
RFCs
The following site lists all RFCs, including those related to IP and port addresses.
o
2.7
www.ietLorg/rfc.html
KEY TERMS
access control
Address Resolution Protocol (ARP)
application layer
best-effort delivery
bits
connection control
data link layer
encoding
error
error control
flow control
frame
header
hop-to-hop delivery
host-to-host protocol
interface
Internet Control Message Protocol
(ICMP)
Internet Group Message Protocol (IGMP)
logical addressing
mail service
network layer
node-to-node delivery
open system
Open Systems Interconnection (OSI)
model
peer-to-peer process
physical addressing
physical layer
port address
presentation layer
process-to-process delivery
Reverse Address Resolution Protocol
(RARP)
routing
segmentation
session layer
source-to-destination delivery
Stream Control Transmission Protocol
(SCTP)
synchronization point
TCPIIP protocol suite
trailer
Transmission Control Protocol (TCP)
transmission rate
transport layer
transport level protocols
User Datagram Protocol (UDP)
52
CHAPTER 2
NETWORK MODELS
2.8
o
U
o
o
D
o
o
o
[J
D
U
o
o
U
U
o
L,J
U
o
SUMMARY
The International Standards Organization created a model called the Open Systems
Interconnection, which allows diverse systems to communicate.
The seven-layer OSI model provides guidelines for the development of universally
compatible networking protocols.
The physical, data link, and network layers are the network support layers.
The session, presentation, and application layers are the user support layers.
The transport layer links the network support layers and the user support layers.
The physical layer coordinates the functions required to transmit a bit stream over
a physical medium.
The data link layer is responsible for delivering data units from one station to the
next without errors.
The network layer is responsible for the source-to-destination delivery of a packet
across multiple network links.
The transport layer is responsible for the process-to-process delivery of the entire
message.
The session layer establishes, maintains, and synchronizes the interactions between
communicating devices.
The presentation layer ensures interoperability between communicating devices
through transformation of data into a mutually agreed upon format.
The application layer enables the users to access the network.
TCP/IP is a five-layer hierarchical protocol suite developed before the OSI model.
The TCP/IP application layer is equivalent to the combined session, presentation,
and application layers of the OSI model.
Four levels of addresses are used in an internet following the TCP/IP protocols: physical (link) addresses, logical (IP) addresses, port addresses, and specific addresses.
The physical address, also known as the link address, is the address of a node as
defined by its LAN or WAN.
The IP address uniquely defines a host on the Internet.
The port address identifies a process on a host.
A specific address is a user-friendly address.
2.9
PRACTICE SET
Review Questions
I. List the layers of the Internet model.
2. Which layers in the Internet model are the network support layers?
3. Which layer in the Internet model is the user support layer?
4. What is the difference between network layer delivery and transport layer delivery?
SECTION 2.9
PRACTICE SET
53
5. What is a peer-to-peer process?
6. How does information get passed from one layer to the next in the Internet
model?
7. What are headers and trailers, and how do they get added and removed?
X. What are the concerns of the physical layer in the Internet model?
9. What are the responsibilities of the data link layer in the Internet model?
10. What are the responsibilities of the network layer in the Internet model?
II. What are the responsibilities of the transport layer in the Internet model?
12. What is the difference between a port address, a logical address, and a physical
address?
13. Name some services provided by the application layer in the Internet model.
14. How do the layers of the Internet model correlate to the layers of the OSI model?
Exercises
15. How are OSI and ISO related to each other?
16. Match the following to one or more layers of the OSI model:
a. Route determination
b. Flow control
c. Interface to transmission media
d. Provides access for the end user
I 7. Match the following to one or more layers of the OSI model:
a. Reliable process-to-process message delivery
b. Route selection
c. Defines frames
d. Provides user services such as e-mail and file transfer
e. Transmission of bit stream across physical medium
\ 8. Match the following to one or more layers of the OSl model:
a. Communicates directly with user's application program
b. Error correction and retransmission
c. Mechanical, electrical, and functional interface
d. Responsibility for carrying frames between adjacent nodes
I 9. Match the following to one or more layers of the OSI model:
a. Format and code conversion services
b. Establishes, manages, and terminates sessions
c. Ensures reliable transmission of data
d. Log-in and log-out procedures
e. Provides independence from differences in data representation
20. In Figure 2.22, computer A sends a message to computer D via LANl, router Rl,
and LAN2. Show the contents of the packets and frames at the network and data
link layer for each hop interface.
54
CHAPTER 2
NETWORK MODELS
Figure 2.22 Exercise 20
Rl
Sender
8/42
LAN 1
LAN 2
Sender
21. In Figure 2.22, assume that the communication is between a process running at
computer A with port address i and a process running at computer D with port
address j. Show the contents of packets and frames at the network, data link, and
transport layer for each hop.
22. Suppose a computer sends a frame to another computer on a bus topology LAN.
The physical destination address of the frame is corrupted during the transmission.
What happens to the frame? How can the sender be informed about the situation?
23. Suppose a computer sends a packet at the network layer to another computer
somewhere in the Internet. The logical destination address of the packet is corrupted. What happens to the packet? How can the source computer be informed of
the situation?
24. Suppose a computer sends a packet at the transport layer to another computer
somewhere in the Internet. There is no process with the destination port address
running at the destination computer. What will happen?
25. If the data link layer can detect errors between hops, why do you think we need
another checking mechanism at the transport layer?
Research Activities
26. Give some advantages and disadvantages of combining the session, presentation,
and application layer in the OSI model into one single application layer in the
Internet model.
27. Dialog control and synchronization are two responsibilities of the session layer in
the OSI model. Which layer do you think is responsible for these duties in the
Internet model? Explain your answer.
28. Translation, encryption, and compression are some of the duties of the presentation
layer in the OSI model. Which layer do you think is responsible for these duties in
the Internet model? Explain your answer.
29. There are several transport layer models proposed in the OSI model. Find all of
them. Explain the differences between them.
30. There are several network layer models proposed in the OSI model. Find all of
them. Explain the differences between them.
94
CHAPTER 3
DATA AND SIGNALS
Jitter
Another performance issue that is related to delay is jitter. We can roughly say that jitter
is a problem if different packets of data encounter different delays and the application
using the data at the receiver site is time-sensitive (audio and video data, for example).
If the delay for the first packet is 20 ms, for the second is 45 ms, and for the third is
40 ms, then the real-time application that uses the packets endures jitter. We discuss jitter
in greater detail in Chapter 29.
3.7
RECOMMENDED READING
For more details about subjects discussed in this chapter, we recommend the following
books. The items in brackets [...] refer to the reference list at the end of the text.
Books
Data and signals are elegantly discussed in Chapters 1 to 6 of [Pea92]. [CouOl] gives
an excellent coverage about signals in Chapter 2. More advanced materials can be
found in [Ber96]. [Hsu03] gives a good mathematical approach to signaling. Complete
coverage of Fourier Analysis can be found in [Spi74]. Data and signals are discussed in
Chapter 3 of [Sta04] and Section 2.1 of [Tan03].
3.8
KEY TERMS
analog
analog data
analog signal
attenuation
bandpass channel
bandwidth
baseband transmission
bit rate
bits per second (bps)
broadband transmission
composite signal
cycle
decibel (dB)
digital
digital data
digital signal
distortion
Fourier analysis
frequency
frequency-domain
fundamental frequency
harmonic
Hertz (Hz)
jitter
low-pass channel
noise
nonperiodic signal
Nyquist bit rate
peak amplitude
period
periodic signal
phase
processing delay
propagation speed
SECTION 3.9
propagation time
queuing time
Shannon capacity
signal
signal-to-noise ratio (SNR)
3.9
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
SUMMARY
95
sine wave
throughput
time-domain
transmission time
wavelength
SUMMARY
Data must be transformed to electromagnetic signals to be transmitted.
Data can be analog or digital. Analog data are continuous and take continuous
values. Digital data have discrete states and take discrete values.
Signals can be analog or digital. Analog signals can have an infinite number of
values in a range; digital ,signals can have only a limited number of values.
In data communications, we commonly use periodic analog signals and nonperiodic digital signals.
Frequency and period are the inverse of each other.
Frequency is the rate of change with respect to time.
Phase describes the position of the waveform relative to time O.
A complete sine wave in the time domain can be represented by one single spike in
the frequency domain.
A single-frequency sine wave is not useful in data communications; we need to
send a composite signal, a signal made of many simple sine waves.
According to Fourier analysis, any composite signal is a combination of simple
sine waves with different frequencies, amplitudes, and phases.
The bandwidth of a composite signal is the difference between the highest and the
lowest frequencies contained in that signal.
A digital signal is a composite analog signal with an infinite bandwidth.
Baseband transmission of a digital signal that preserves the shape of the digital
signal is possible only if we have a low-pass channel with an infinite or very wide
bandwidth.
If the available channel is a bandpass channel, we cannot send a digital signal
directly to the channel; we need to convert the digital signal to an analog signal
before transmission.
For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rate. For a noisy channel, we need to use the Shannon capacity to find the
maximum bit rate.
Attenuation, distortion, and noise can impair a signal.
Attenuation is the loss of a signal's energy due to the resistance of the medium.
Distortion is the alteration of a signal due to the differing propagation speeds of
each of the frequencies that make up a signal.
Noise is the external energy that corrupts a signal.
The bandwidth-delay product defines the number of bits that can fill the link.
96
CHAPTER 3
DATA AND SIGNALS
3.10
PRACTICE SET
Review Questions
1. What is the relationship between period and frequency?
2. What does the amplitude of a signal measure? What does the frequency of a signal
measure? What does the phase of a signal measure?
3. How can a composite signal be decomposed into its individual frequencies?
4. Name three types of transmission impairment.
5. Distinguish between baseband transmission and broadband transmission.
6. Distinguish between a low-pass channel and a band-pass channel.
7. What does the Nyquist theorem have to do with communications?
8. What does the Shannon capacity have to do with communications?
9. Why do optical signals used in fiber optic cables have a very short wave length?
10. Can we say if a signal is periodic or nonperiodic by just looking at its frequency
domain plot? How?
11. Is the frequency domain plot of a voice signal discrete or continuous?
12. Is the frequency domain plot of an alarm system discrete or continuous?
13. We send a voice signal from a microphone to a recorder. Is this baseband or broadband transmission?
14. We send a digital signal from one station on a LAN to another station. Is this baseband or broadband transmission?
15. We modulate several voice signals and send them through the air. Is this baseband
or broadband transmission?
Exercises
16. Given the frequencies listed below, calculate the corresponding periods.
a. 24Hz
b. 8 MHz
c. 140 KHz
17. Given the following periods, calculate the corresponding frequencies.
a. 5 s
b. 12 Jls
c. 220 ns
18. What is the phase shift for the foIlowing?
a. A sine wave with the maximum amplitude at time zero
b. A sine wave with maximum amplitude after 1/4 cycle
c. A sine wave with zero amplitude after 3/4 cycle and increasing
19. What is the bandwidth of a signal that can be decomposed into five sine waves
with frequencies at 0, 20, 50, 100, and 200 Hz? All peak amplitudes are the same.
Draw the bandwidth.
SECTION 3.10
PRACTICE SET
97
20. A periodic composite signal with a bandwidth of 2000 Hz is composed of two sine
waves. The first one has a frequency of 100 Hz with a maximum amplitude of 20 V;
the second one has a maximum amplitude of 5 V. Draw the bandwidth.
21. Which signal has a wider bandwidth, a sine wave with a frequency of 100 Hz or a
sine wave with a frequency of 200 Hz?
22. What is the bit rate for each of the following signals?
a. A signal in which 1 bit lasts 0.001 s
b. A signal in which 1 bit lasts 2 ms
c. A signal in which 10 bits last 20 J-ls
23. A device is sending out data at the rate of 1000 bps.
a. How long does it take to send out 10 bits?
b. How long does it take to send out a single character (8 bits)?
c. How long does it take to send a file of 100,000 characters?
24. What is the bit rate for the signal in Figure 3.34?
Figure 3.34 Exercise 24
16 ns
---t=:i--J~-...;..---+-.....;.--.;....-d ... ~
TI~
1
25. What is the frequency of the signal in Figure 3.35?
Figure 3.35 Exercise 25
4ms
I,
.1
1
I
V\ f\ f\ f\ f\ f\ f\ f\ : ...
\TV V VVV V~
Time
26. What is the bandwidth of the composite signal shown in Figure 3.36.
Figure 3.36 Exercise 26
Frequency
5
5
5
5
5
I
9S
CHAPTER 3
DATA AND SIGNALS
27. A periodic composite signal contains frequencies from 10 to 30 KHz, each with an
amplitude of 10 V. Draw the frequency spectrum.
2K. A non-periodic composite signal contains frequencies from 10 to 30 KHz. The
peak amplitude is 10 V for the lowest and the highest signals and is 30 V for the
20-KHz signal. Assuming that the amplitudes change gradually from the minimum
to the maximum, draw the frequency spectrum.
20. A TV channel has a bandwidth of 6 MHz. If we send a digital signal using one
channel, what are the data rates if we use one harmonic, three harmonics, and five
harmonics?
30. A signal travels from point A to point B. At point A, the signal power is 100 W. At
point B, the power is 90 W. What is the attenuation in decibels?
31. The attenuation of a signal is -10 dB. What is the final signal power if it was originally 5 W?
32. A signal has passed through three cascaded amplifiers, each with a 4 dB gain.
What is the total gain? How much is the signal amplified?
33. If the bandwidth of the channel is 5 Kbps, how long does it take to send a frame of
100,000 bits out of this device?
3cf. The light of the sun takes approximately eight minutes to reach the earth. What is
the distance between the sun and the earth?
35. A signal has a wavelength of 1 11m in air. How far can the front of the wave travel
during 1000 periods?
36. A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is
the maximum data rate supported by this line?
37. We measure the performance of a telephone line (4 KHz of bandwidth). When the
signal is 10 V, the noise is 5 mV. What is the maximum data rate supported by this
telephone line?
3X. A file contains 2 million bytes. How long does it take to download this file using a
56-Kbps channel? 1-Mbps channel?
39. A computer monitor has a resolution of 1200 by 1000 pixels. If each pixel uses
1024 colors, how many bits are needed to send the complete contents of a screen?
40. A signal with 200 milliwatts power passes through 10 devices, each with an average
noise of 2 microwatts. What is the SNR? What is the SNRdB ?
4 I. If the peak voltage value of a signal is 20 times the peak voltage value of the noise,
what is the SNR? What is the SNR dB ?
42. What is the theoretical capacity of a channel in each of the following cases:
a. Bandwidth: 20 KHz
SNRdB =40
b. Bandwidth: 200 KHz SNRdB =4
c. Bandwidth: 1 MHz
SNRdB =20
43. We need to upgrade a channel to a higher bandwidth. Answer the following
questions:
a. How is the rate improved if we double the bandwidth?
b. How is the rate improved if we double the SNR?
SECTION 3.10
PRACTICE SET
99
44. We have a channel with 4 KHz bandwidth. If we want to send data at 100 Kbps,
what is the minimum SNRdB ? What is SNR?
45. What is the transmission time of a packet sent by a station if the length of the
packet is 1 million bytes and the bandwidth of the channel is 200 Kbps?
46. What is the length of a bit in a channel with a propagation speed of 2 x 108 mls if
the channel bandwidth is
a. 1 Mbps?
h. 10 Mbps?
c. 100 Mbps?
-1- 7. How many bits can fit on a link with a 2 ms delay if the bandwidth of the link is
a. 1 Mbps?
h. 10 Mbps?
c. 100 Mbps?
-1-X. What is the total delay (latency) for a frame of size 5 million bits that is being sent
on a link with 10 routers each having a queuing time of 2 Ils and a processing time
of 1 Ils. The length of the link is 2000 Km. The speed of light inside the link is 2 x
108 mls. The link has a bandwidth of 5 Mbps. Which component of the total delay
is dominant? Which one is negligible?
CHAPTER 1
Introduction
Solutions to Review Questions and Exercises
Review Questions
1. The five components of a data communication system are the sender, receiver,
transmission medium, message, and protocol.
2. The advantages of distributed processing are security, access to distributed databases, collaborative processing, and faster problem solving.
3. The three criteria are performance, reliability, and security.
4. Advantages of a multipoint over a point-to-point configuration (type of connection) include ease of installation and low cost.
5. Line configurations (or types of connections) are point-to-point and multipoint.
6. We can divide line configuration in two broad categories:
a. Point-to-point: mesh, star, and ring.
b. Multipoint: bus
7. In half-duplex transmission, only one entity can send at a time; in a full-duplex
transmission, both entities can send at the same time.
8. We give an advantage for each of four network topologies:
a. Mesh: secure
b. Bus: easy installation
c. Star: robust
d. Ring: easy fault isolation
9. The number of cables for each type of network is:
a. Mesh: n (n – 1) / 2
b. Star: n
c. Ring: n – 1
d. Bus: one backbone and n drop lines
10. The general factors are size, distances (covered by the network), structure, and
ownership.
1
2
11. An internet is an interconnection of networks. The Internet is the name of a specific worldwide network
12. A protocol defines what is communicated, in what way and when. This provides
accurate and timely transfer of information between different devices on a network.
13. Standards are needed to create and maintain an open and competitive market for
manufacturers, to coordinate protocol rules, and thus guarantee compatibility of
data communication technologies.
Exercises
14. Unicode uses 32 bits to represent a symbol or a character. We can define 232 different symbols or characters.
15. With 16 bits, we can represent up to 216 different colors.
16.
a. Cable links: n (n – 1) / 2 = (6 × 5) / 2 = 15
b. Number of ports: (n – 1) = 5 ports needed per device
17.
a. Mesh topology: If one connection fails, the other connections will still be working.
b. Star topology: The other devices will still be able to send data through the hub;
there will be no access to the device which has the failed connection to the hub.
c. Bus Topology: All transmission stops if the failure is in the bus. If the drop-line
fails, only the corresponding device cannot operate.
d. Ring Topology: The failed connection may disable the whole network unless it
is a dual ring or there is a by-pass mechanism.
18. This is a LAN. The Ethernet hub creates a LAN as we will see in Chapter 13.
19. Theoretically, in a ring topology, unplugging one station, interrupts the ring. However, most ring networks use a mechanism that bypasses the station; the ring can
continue its operation.
20. In a bus topology, no station is in the path of the signal. Unplugging a station has
no effect on the operation of the rest of the network.
21. See Figure 1.1
22. See Figure 1.2.
23.
a. E-mail is not an interactive application. Even if it is delivered immediately, it
may stay in the mail-box of the receiver for a while. It is not sensitive to delay.
b. We normally do not expect a file to be copied immediately. It is not very sensitive to delay.
c. Surfing the Internet is the an application very sensitive to delay. We except to
get access to the site we are searching.
24. In this case, the communication is only between a caller and the callee. A dedicated line is established between them. The connection is point-to-point.
3
Figure 1.1 Solution to Exercise 21
Hub
Station
Repeater
Station
Station
Station
Station
Repeat er
Repeat er
Station
Station
Station
Station
Figure 1.2 Solution to Exercise 22
Station
Station
Repeater
Repeater
Station
Station
25. The telephone network was originally designed for voice communication; the
Internet was originally designed for data communication. The two networks are
similar in the fact that both are made of interconnections of small networks. The
telephone network, as we will see in future chapters, is mostly a circuit-switched
network; the Internet is mostly a packet-switched network.
4
Sol-02.fm Page 1 Saturday, January 21, 2006 10:27 AM
CHAPTER 2
Network Models
Solutions to Review Questions and Exercises
Review Questions
1. The Internet model, as discussed in this chapter, include physical, data link, network, transport, and application layers.
2. The network support layers are the physical, data link, and network layers.
3. The application layer supports the user.
4. The transport layer is responsible for process-to-process delivery of the entire
message, whereas the network layer oversees host-to-host delivery of individual
packets.
5. Peer-to-peer processes are processes on two or more devices communicating at a
same layer
6. Each layer calls upon the services of the layer just below it using interfaces
between each pair of adjacent layers.
7. Headers and trailers are control data added at the beginning and the end of each
data unit at each layer of the sender and removed at the corresponding layers of the
receiver. They provide source and destination addresses, synchronization points,
information for error detection, etc.
8. The physical layer is responsible for transmitting a bit stream over a physical
medium. It is concerned with
a. physical characteristics of the media
b. representation of bits
c. type of encoding
d. synchronization of bits
e. transmission rate and mode
f. the way devices are connected with each other and to the links
9. The data link layer is responsible for
a. framing data bits
b. providing the physical addresses of the sender/receiver
c. data rate control
1
Sol-02.fm Page 2 Saturday, January 21, 2006 10:27 AM
2
10.
11.
12.
13.
14.
d. detection and correction of damaged and lost frames
The network layer is concerned with delivery of a packet across multiple networks; therefore its responsibilities include
a. providing host-to-host addressing
b. routing
The transport layer oversees the process-to-process delivery of the entire message.
It is responsible for
a. dividing the message into manageable segments
b. reassembling it at the destination
c. flow and error control
The physical address is the local address of a node; it is used by the data link layer
to deliver data from one node to another within the same network. The logical
address defines the sender and receiver at the network layer and is used to deliver
messages across multiple networks. The port address (service-point) identifies the
application process on the station.
The application layer services include file transfer, remote access, shared database management, and mail services.
The application, presentation, and session layers of the OSI model are represented
by the application layer in the Internet model. The lowest four layers of OSI correspond to the Internet model layers.
Exercises
15. The International Standards Organization, or the International Organization of
Standards, (ISO) is a multinational body dedicated to worldwide agreement on
international standards. An ISO standard that covers all aspects of network communications is the Open Systems Interconnection (OSI) model.
16.
a. Route determination: network layer
b. Flow control: data link and transport layers
c. Interface to transmission media: physical layer
d. Access for the end user: application layer
17.
a. Reliable process-to-process delivery: transport layer
b. Route selection: network layer
c. Defining frames: data link layer
d. Providing user services: application layer
e. Transmission of bits across the medium: physical layer
18.
a. Communication with user’s application program: application layer
b. Error correction and retransmission: data link and transport layers
c. Mechanical, electrical, and functional interface: physical layer
Sol-02.fm Page 3 Saturday, January 21, 2006 10:27 AM
3
d. Responsibility for carrying frames between adjacent nodes: data link layer
19.
a. Format and code conversion services: presentation layer
b. Establishing, managing, and terminating sessions: session layer
c. Ensuring reliable transmission of data: data link and transport layers
d. Log-in and log-out procedures: session layer
e. Providing independence from different data representation: presentation layer
20. See Figure 2.1.
Figure 2.1 Solution to Exercise 20
A/40
LAN1
LAN2
R1
Sender
B/42
D/80
C/82
Sender
80 82 A D Data T2
42 40 A D Data T2
21. See Figure 2.2.
Figure 2.2 Solution to Exercise 21
LAN1
A/40
LAN2
R1
Sender
B/42
D/80
C/82
Sender
42 40 A D i
j
Data
T2
80 82 A D i
j
Data
T2
22. If the corrupted destination address does not match any station address in the network, the packet is lost. If the corrupted destination address matches one of the stations, the frame is delivered to the wrong station. In this case, however, the error
detection mechanism, available in most data link protocols, will find the error and
discard the frame. In both cases, the source will somehow be informed using one
of the data link control mechanisms discussed in Chapter 11.
23. Before using the destination address in an intermediate or the destination node, the
packet goes through error checking that may help the node find the corruption
(with a high probability) and discard the packet. Normally the upper layer protocol
will inform the source to resend the packet.
Sol-02.fm Page 4 Saturday, January 21, 2006 10:27 AM
4
24. Most protocols issue a special error message that is sent back to the source in this
case.
25. The errors between the nodes can be detected by the data link layer control, but the
error at the node (between input port and output port) of the node cannot be
detected by the data link layer.
CHAPTER 3
Data and Signals
Solutions to Review Questions and Exercises
Review Questions
1. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T.
2. The amplitude of a signal measures the value of the signal at any point. The frequency of a signal refers to the number of periods in one second. The phase
describes the position of the waveform relative to time zero.
3. Using Fourier analysis. Fourier series gives the frequency domain of a periodic
signal; Fourier analysis gives the frequency domain of a nonperiodic signal.
4. Three types of transmission impairment are attenuation, distortion, and noise.
5. Baseband transmission means sending a digital or an analog signal without modulation using a low-pass channel. Broadband transmission means modulating a
digital or an analog signal using a band-pass channel.
6. A low-pass channel has a bandwidth starting from zero; a band-pass channel has a
bandwidth that does not start from zero.
7. The Nyquist theorem defines the maximum bit rate of a noiseless channel.
8. The Shannon capacity determines the theoretical maximum bit rate of a noisy
channel.
9. Optical signals have very high frequencies. A high frequency means a short wave
length because the wave length is inversely proportional to the frequency (λ = v/f),
where v is the propagation speed in the media.
10. A signal is periodic if its frequency domain plot is discrete; a signal is nonperiodic if its frequency domain plot is continuous.
11. The frequency domain of a voice signal is normally continuous because voice is a
nonperiodic signal.
12. An alarm system is normally periodic. Its frequency domain plot is therefore discrete.
13. This is baseband transmission because no modulation is involved.
14. This is baseband transmission because no modulation is involved.
15. This is broadband transmission because it involves modulation.
1
2
Exercises
16.
a. T = 1 / f = 1 / (24 Hz) = 0.0417 s = 41.7 × 10–3 s = 41.7 ms
b. T = 1 / f = 1 / (8 MHz) = 0.000000125 = 0.125 × 10–6 s = 0.125 μs
c. T = 1 / f = 1 / (140 KHz) = 0.00000714 s = 7.14 × 10–6 s = 7.14 μs
17.
a. f = 1 / T = 1 / (5 s) = 0.2 Hz
b. f = 1 / T = 1 / (12 μs) =83333 Hz = 83.333 × 103 Hz = 83.333 KHz
c. f = 1 / T = 1 / (220 ns) = 4550000 Hz = 4.55× 106 Hz = 4.55 MHz
18.
a. 90 degrees (π/2 radian)
b. 0 degrees (0 radian)
c. 90 degrees (π/2 radian)
19. See Figure 3.1
Figure 3.1 Solution to Exercise 19
Frequency domain
0
20
100
50
200
Bandwidth = 200 − 0 = 200
20. We know the lowest frequency, 100. We know the bandwidth is 2000. The highest
frequency must be 100 + 2000 = 2100 Hz. See Figure 3.2
Figure 3.2 Solution to Exercise 20
20
Frequency domain
5
100
2100
Bandwidth = 2100 − 100 = 2000
21. Each signal is a simple signal in this case. The bandwidth of a simple signal is
zero. So the bandwidth of both signals are the same.
22.
a. bit rate = 1/ (bit duration) = 1 / (0.001 s) = 1000 bps = 1 Kbps
b. bit rate = 1/ (bit duration) = 1 / (2 ms) = 500 bps
3
c. bit rate = 1/(bit duration) = 1 / (20 μs/10) = 1 / (2 μs) = 500 Kbps
23.
24.
25.
26.
27.
a. (10 / 1000) s = 0.01 s
b. (8 / 1000) s = 0. 008 s = 8 ms
c. ((100,000 × 8) / 1000) s = 800 s
There are 8 bits in 16 ns. Bit rate is 8 / (16 × 10−9) = 0.5 × 10−9 = 500 Mbps
The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz
The bandwidth is 5 × 5 = 25 Hz.
The signal is periodic, so the frequency domain is made of discrete frequencies. as
shown in Figure 3.3.
Figure 3.3 Solution to Exercise 27
Amplitude
10 volts
...
10
KHz
Frequency
30
KHz
28. The signal is nonperiodic, so the frequency domain is made of a continuous spectrum of frequencies as shown in Figure 3.4.
Figure 3.4 Solution to Exercise 28
30 volts
Amplitude
10 volts
10 volts
Frequency
10
KHz
29.
20
KHz
30
KHz
Using the first harmonic, data rate = 2 × 6 MHz = 12 Mbps
Using three harmonics, data rate = (2 × 6 MHz) /3 = 4 Mbps
Using five harmonics, data rate = (2 × 6 MHz) /5 = 2.4 Mbps
30. dB = 10 log10 (90 / 100) = –0.46 dB
31. –10 = 10 log10 (P2 / 5) → log10 (P2 / 5) = −1 → (P2 / 5) = 10−1 → P2 = 0.5 W
32. The total gain is 3 × 4 = 12 dB. The signal is amplified by a factor 101.2 = 15.85.
4
33.
34.
35.
36.
100,000 bits / 5 Kbps = 20 s
480 s × 300,000 km/s = 144,000,000 km
1 μm × 1000 = 1000 μm = 1 mm
We have
4,000 log2 (1 + 1,000) ≈ 40 Kbps
37. We have
4,000 log2 (1 + 10 / 0.005) = 43,866 bps
38. The file contains 2,000,000 × 8 = 16,000,000 bits. With a 56-Kbps channel, it takes
16,000,000/56,000 = 289 s. With a 1-Mbps channel, it takes 16 s.
39. To represent 1024 colors, we need log21024 = 10 (see Appendix C) bits. The total
number of bits are, therefore,
1200 × 1000 × 10 = 12,000,000 bits
40. We have
SNR = (200 mW) / (10 × 2 × μW) = 10,000
We then have
SNRdB = 10 log10 SNR = 40
41. We have
SNR= (signal power)/(noise power).
However, power is proportional to the square of voltage. This means we have
SNR = [(signal voltage)2] / [(noise voltage)2] =
[(signal voltage) / (noise voltage)]2 = 202 = 400
We then have
SNRdB = 10 log10 SNR ≈ 26.02
42. We can approximately calculate the capacity as
a. C = B × (SNRdB /3) = 20 KHz × (40 /3) = 267 Kbps
b. C = B × (SNRdB /3) = 200 KHz × (4 /3) = 267 Kbps
c. C = B × (SNRdB /3) = 1 MHz × (20 /3) = 6.67 Mbps
43.
a. The data rate is doubled (C2 = 2 × C1).
b. When the SNR is doubled, the data rate increases slightly. We can say that,
approximately, (C2 = C1 + 1).
44. We can use the approximate formula
C = B × (SNRdB /3) or SNRdB = (3 × C) /B
We can say that the minimum
SNRdB = 3 × 100 Kbps / 4 KHz = 75
5
This means that the minimum
SNR = 10 SNRdB/10 = 107.5 ≈ 31,622,776
45. We have
transmission time = (packet length)/(bandwidth) =
(8,000,000 bits) / (200,000 bps) = 40 s
46. We have
(bit length) = (propagation speed) × (bit duration)
The bit duration is the inverse of the bandwidth.
a. Bit length = (2 ×108 m) × [(1 / (1 Mbps)] = 200 m. This means a bit occupies
200 meters on a transmission medium.
b. Bit length = (2 ×108 m) × [(1 / (10 Mbps)] = 20 m. This means a bit occupies 20
meters on a transmission medium.
c. Bit length = (2 ×108 m) × [(1 / (100 Mbps)] = 2 m. This means a bit occupies 2
meters on a transmission medium.
47.
a. Number of bits = bandwidth × delay = 1 Mbps × 2 ms = 2000 bits
b. Number of bits = bandwidth × delay = 10 Mbps × 2 ms = 20,000 bits
c. Number of bits = bandwidth × delay = 100 Mbps × 2 ms = 200,000 bits
48. We have
Latency = processing time + queuing time +
transmission time + propagation time
Processing time = 10 × 1 μs = 10 μs = 0.000010 s
Queuing time = 10 × 2 μs = 20 μs = 0.000020 s
Transmission time = 5,000,000 / (5 Mbps) = 1 s
Propagation time = (2000 Km) / (2 × 108) = 0.01 s
Latency = 0.000010 + 0.000020 + 1 + 0.01 = 1.01000030 s
The transmission time is dominant here because the packet size is huge.
6
SECTION 4.5
KEY TERMS
135
The advantage of synchronous transmission is speed. With no extra bits or gaps to
introduce at the sending end and remove at the receiving end, and, by extension, with
fewer bits to move across the link, synchronous transmission is faster than asynchronous transmission. For this reason, it is more useful for high-speed applications such as
the transmission of data from one computer to another. Byte synchronization is accomplished in the data link layer.
We need to emphasize one point here. Although there is no gap between characters
in synchronous serial transmission, there may be uneven gaps between frames.
Isochronous
In real-time audio and video, in which uneven delays between frames are not acceptable, synchronous transmission fails. For example, TV images are broadcast at the rate
of 30 images per second; they must be viewed at the same rate. If each image is sent
by using one or more frames, there should be no delays between frames. For this type
of application, synchronization between characters is not enough; the entire stream of
bits must be synchronized. The isochronous transmission guarantees that the data
arrive at a fixed rate.
4.4
RECOMMENDED READING
For more details about subjects discussed in this chapter, we recommend the following
books. The items in brackets [...] refer to the reference list at the end of the text.
Books
Digital to digital conversion is discussed in Chapter 7 of [Pea92], Chapter 3 of
[CouOl], and Section 5.1 of [Sta04]. Sampling is discussed in Chapters 15, 16, 17, and
18 of [Pea92], Chapter 3 of [CouO!], and Section 5.3 of [Sta04]. [Hsu03] gives a good
mathematical approach to modulation and sampling. More advanced materials can be
found in [Ber96].
4.5
KEY TERMS
adaptive delta modulation
alternate mark inversion (AMI)
analog-to-digital conversion
asynchronous transmission
baseline
baseline wandering
baud rate
biphase
bipolar
bipolar with 8-zero substitution (B8ZS)
bit rate
block coding
companding and expanding
data element
data rate
DC component
delta modulation (DM)
differential Manchester
digital-to-digital conversion
digitization
136
CHAPTER 4
DIGITAL TRANSMISSION
eight binary/ten binary (8B/lOB)
eight-binary, six-ternary (8B6T)
four binary/five binary (4B/5B)
four dimensional, five-level pulse
amplitude modulation (4D-PAM5)
high-density bipolar 3-zero (HDB3)
isochronous transmission
line coding
Manchester
modulation rate
multilevel binary
multiline transmission, 3 level (MLT-3)
pulse amplitude modulation (PAM)
pulse code modulation (PCM)
pulse rate
quantization
quantization error
return to zero (RZ)
sampling
sampling rate
scrambling
self-synchronizing
serial transmission
signal element
nonreturn to zero (NRZ)
signal rate
nonreturn to zero, invert (NRZ-I)
nonreturn to zero, level (NRZ-L)
Nyquist theorem
parallel transmission
start bit
stop bit
synchronous transmission
transmission mode
polar
pseudoternary
two-binary, one quaternary (2B I Q)
unipolar
4.6
o
o
o
o
o
o
o
o
SUMMARY
Digital-to-digital conversion involves three techniques: line coding, block coding,
and scrambling.
Line coding is the process of converting digital data to a digital signal.
We can roughly divide line coding schemes into five broad categories: unipolar,
polar, bipolar, multilevel, and multitransition.
Block coding provides redundancy to ensure synchronization and inherent error
detection. Block coding is normally referred to as mB/nB coding; it replaces each
m-bit group with an n-bit group.
Scrambling provides synchronization without increasing the number of bits. Two
common scrambling techniques are B8ZS and HDB3.
The most common technique to change an analog signal to digital data (digitization) is called pulse code modulation (PCM).
The first step in PCM is sampling. The analog signal is sampled every Ts s, where Ts
is the sample interval or period. The inverse of the sampling interval is called the
sampling rate or sampling frequency and denoted by fs, where fs = lITs. There are
three sampling methods-ideal, natural, and flat-top.
According to the Nyquist theorem, to reproduce the original analog signal, one
necessary condition is that the sampling rate be at least twice the highest frequency
in the original signal.
SECTION 4.7
o
o
o
o
o
4.7
PRACTICE SET
137
Other sampling techniques have been developed to reduce the complexity of PCM.
The simplest is delta modulation. PCM finds the value of the signal amplitude for
each sample; DM finds the change from the previous sample.
While there is only one way to send parallel data, there are three subclasses of
serial transmission: asynchronous, synchronous, and isochronous.
In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or
more stop bits (1 s) at the end of each byte.
In synchronous transmission, we send bits one after another without start or stop
bits or gaps. It is the responsibility of the receiver to group the bits.
The isochronous mode provides synchronized for the entire stream of bits must. In
other words, it guarantees that the data arrive at a fixed rate.
PRACTICE SET
Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
List three techniques of digital-to-digital conversion.
Distinguish between a signal element and a data element.
Distinguish between data rate and signal rate.
Define baseline wandering and its effect on digital transmission.
Define a DC component and its effect on digital transmission.
Define the characteristics of a self-synchronizing signal.
List five line coding schemes discussed in this book.
Define block coding and give its purpose.
Define scrambling and give its purpose.
Compare and contrast PCM and DM.
What are the differences between parallel and serial transmission?
List three different techniques in serial transmission and explain the differences.
Exercises
13. Calculate the value of the signal rate for each case in Figure 4.2 if the data rate is
1 Mbps and c = 1/2.
14. In a digital transmission, the sender clock is 0.2 percent faster than the receiver clock.
How many extra bits per second does the sender send if the data rate is 1 Mbps?
15. Draw the graph of the NRZ-L scheme using each of the following data streams,
assuming that the last signa11evel has been positive. From the graphs, guess the
bandwidth for this scheme using the average number of changes in the signal level.
Compare your guess with the corresp.onding entry in Table 4.1.
a. 00000000
b. 11111111
c. 01010101
d. 00110011
138
CHAPTER 4
DIGITAL TRANSMISSION
16.
17.
18.
19.
Repeat Exercise 15 for the NRZ-I scheme.
Repeat Exercise 15 for the Manchester scheme.
Repeat Exercise 15 for the differential Manchester scheme.
Repeat Exercise 15 for the 2B 1Q scheme, but use the following data streams.
a. 0000000000000000
b. 1111111111111111
c. 0101010101010101
d. 0011001100110011
20. Repeat Exercise 15 for the MLT-3 scheme, but use the following data streams.
a. 00000000
b. 11111111
c. 01010101
d. 00011000
21. Find the 8-bit data stream for each case depicted in Figure 4.36.
Figure 4.36 Exercise 21
t
Time
a. NRZ-I
Time
b. differential Manchester
Time
c.AMI
22. An NRZ-I signal has a data rate of 100 Kbps. Using Figure 4.6, calculate the value
of the normalized energy (P) for frequencies at 0 Hz, 50 KHz, and 100 KHz.
23. A Manchester signal has a data rate of 100 Kbps. Using Figure 4.8, calculate the
value of the normalized energy (P) for frequencies at 0 Hz, 50 KHz, 100 KHz.
SECTION 4. 7
PRACTICE SET
139
24. The input stream to a 4B/5B block encoder is 0100 0000 0000 0000 0000 OOOI.
Answer the following questions:
a. What is the output stream?
b. What is the length of the longest consecutive sequence of Os in the input?
c. What is the length of the longest consecutive sequence of Os in the output?
25. How many invalid (unused) code sequences can we have in 5B/6B encoding? How
many in 3B/4B encoding?
26. What is the result of scrambling the sequence 11100000000000 using one of the
following scrambling techniques? Assume that the last non-zero signal level has
been positive.
a. B8ZS
b. HDB3 (The number of nonzero pules is odd after the last substitution)
27. What is the Nyquist sampling rate for each of the following signals?
a. A low-pass signal with bandwidth of 200 KHz?
b. A band-pass signal with bandwidth of 200 KHz if the lowest frequency is
100 KHz?
28. We have sampled a low-pass signal with a bandwidth of 200 KHz using 1024 levels
of quantization.
a. Calculate the bit rate of the digitized signal.
b. Calculate the SNRdB for this signal.
c. Calculate the PCM bandwidth of this signal.
29. What is the maximum data rate of a channel with a bandwidth of 200 KHz if we
use four levels of digital signaling.
30. An analog signal has a bandwidth of 20 KHz. If we sample this signal and send it
through a 30 Kbps channel what is the SNRdB ?
31. We have a baseband channel with a I-MHz bandwidth. What is the data rate for
this channel if we use one of the following line coding schemes?
a. NRZ-L
b. Manchester
c. MLT-3
d. 2B1Q
32. We want to transmit 1000 characters with each character encoded as 8 bits.
a. Find the number of transmitted bits for synchronous transmission.
b. Find the number of transmitted bits for asynchronous transmission.
c. Find the redundancy percent in each case.
SECTION 5.5
SUMMARY
157
gives a good mathematical approach to all materials discussed in this chapter. More
advanced materials can be found in [Ber96].
5.4
KEY TERMS
amplitude modulation (AM)
amplitude shift keying (ASK)
analog-to-analog conversion
carrier signal
constellation diagram
digi tal-to-analog conversion
5.5
o
o
o
o
o
o
o
o
o
o
o
frequency modulation (PM)
frequency shift keying (FSK)
phase modulation (PM)
phase shift keying (PSK)
quadrature amplitude modulation
(QAM)
SUMMARY
Digital-to-analog conversion is the process of changing one of the characteristics
of an analog signal based on the information in the digital data.
Digital-to-analog conversion can be accomplished in several ways: amplitude shift
keying (ASK), frequency shift keying (FSK), and phase shift keying (PSK).
Quadrature amplitude modulation (QAM) combines ASK and PSK.
In amplitude shift keying, the amplitude of the carrier signal is varied to create signal
elements. Both frequency and phase remain constant while the amplitude changes.
In frequency shift keying, the frequency of the carrier signal is varied to represent
data. The frequency of the modulated signal is constant for the duration of one signal element, but changes for the next signal element if the data element changes.
Both peak amplitude and phase remain constant for all signal elements.
In phase shift keying, the phase of the carrier is varied to represent two or more different signal elements. Both peak amplitude and frequency remain constant as the
phase changes.
A constellation diagram shows us the amplitude and phase of a signal element,
particularly when we are using two carriers (one in-phase and one quadrature).
Quadrature amplitude modulation (QAM) is a combination of ASK and PSK.
QAM uses two carriers, one in-phase and the other quadrature, with different
amplitude levels for each carrier.
Analog-to-analog conversion is the representation of analog information by an
analog signal. Conversion is needed if the medium is bandpass in nature or if only
a bandpass bandwidth is available to us.
Analog-to-analog conversion can be accomplished in three ways: amplitude modulation (AM), frequency modulation (FM), and phase modulation (PM).
In AM transmission, the carrier signal is modulated so that its amplitude varies with the
changing amplitudes of the modulating signal. The frequency and phase of the carrier
remain the same; only the amplitude changes to follow variations in the information.
In PM transmission, the frequency of the carrier signal is modulated to follow the
changing voltage level (amplitude) of the modulating signal. The peak amplitude
158
CHAPTER 5
ANALOG TRANSMISSION
o
and phase of the carrier signal remain constant, but as the amplitude of the information signal changes, the frequency of the carrier changes correspondingly.
In PM transmission, the phase of the carrier signal is modulated to follow the
changing voltage level (amplitude) of the modulating signal. The peak amplitude
and frequency of the carrier signal remain constant, but as the amplitude of the
information signal changes, the phase of the carrier changes correspondingly.
5.6
PRACTICE SET
Review Questions
1. Define analog transmission.
2. Define carrier signal and its role in analog transmission.
3. Define digital-to-analog conversion.
4. Which characteristics of an analog signal are changed to represent the digital signal
in each of the following digital-to-analog conversion?
a. ASK
b. FSK
c. PSK
d. QAM
5. Which of the four digital-to-analog conversion techniques (ASK, FSK, PSK or
QAM) is the most susceptible to noise? Defend your answer.
6. Define constellation diagram and its role in analog transmission.
7. What are the two components of a signal when the signal is represented on a con- .
stellation diagram? Which component is shown on the horizontal axis? Which is
shown on the vertical axis?
8. Define analog-to-analog conversion?
9. Which characteristics of an analog signal are changed to represent the lowpass analog
signal in each of the following analog-to-analog conversions?
a. AM
b. FM
c. PM
] 0. Which of the three analog-to-analog conversion techniques (AM, FM, or PM) is
the most susceptible to noise? Defend your answer.
Exercises
11. Calculate the baud rate for the given bit rate and type of modulation.
a. 2000 bps, FSK
b. 4000 bps, ASK
c. 6000 bps, QPSK
d. 36,000 bps, 64-QAM
SECTION 5.6
PRACTICE SET
159
12. Calculate the bit rate for the given baud rate and type of modulation.
a. 1000 baud, FSK
b. 1000 baud, ASK
c. 1000 baud, BPSK
d. 1000 baud, 16-QAM
13. What is the number of bits per baud for the following techniques?
a. ASK with four different amplitudes
b. FSK with 8 different frequencies
c. PSK with four different phases
d. QAM with a constellation of 128 points.
14. Draw the constellation diagram for the following:
a. ASK, with peak amplitude values of 1 and 3
b. BPSK, with a peak amplitude value of 2
c. QPSK, with a peak amplitude value of 3
d. 8-QAM with two different peak amplitude values, I and 3, and four different
phases.
15. Draw the constellation diagram for the following cases. Find the peak amplitude
value for each case and define the type of modulation (ASK, FSK, PSK, or QAM).
The numbers in parentheses define the values of I and Q respectively.
a. Two points at (2, 0) and (3, 0).
b. Two points at (3, 0) and (-3, 0).
c. Four points at (2, 2), (-2, 2), (-2, -2), and (2, -2).
d. Two points at (0 , 2) and (0, -2).
16. How many bits per baud can we send in each of the following cases if the signal
constellation has one of the following number of points?
a. 2
b. 4
c. 16
d. 1024
17. What is the required bandwidth for the following cases if we need to send 4000 bps?
Let d = 1.
a. ASK
b. FSK with 2~f = 4 KHz
c. QPSK
d. 16-QAM
18. The telephone line has 4 KHz bandwidth. What is the maximum number of bits we
can send using each of the following techniques? Let d = O.
a. ASK
b. QPSK
c. 16-QAM
d.64-QAM
160
CHAPTER 5
ANALOG TRANSMISSION
19. A corporation has a medium with a I-MHz bandwidth (lowpass). The corporation
needs to create 10 separate independent channels each capable of sending at least
10 Mbps. The company has decided to use QAM technology. What is the minimum number of bits per baud for each channel? What is the number of points in
the constellation diagram for each channel? Let d = O.
20. A cable company uses one of the cable TV channels (with a bandwidth of 6 MHz)
to provide digital communication for each resident. What is the available data rate
for each resident if the company uses a 64-QAM technique?
21. Find the bandwidth for the following situations if we need to modulate a 5-KHz
voice.
a. AM
=5)
PM (set ~ = 1)
b. PM (set ~
c.
22. Find the total number of channels in the corresponding band allocated by FCC.
a. AM
b. FM
CHAPTER 4
Digital Transmission
Solutions to Review Questions and Exercises
Review Questions
1. The three different techniques described in this chapter are line coding, block coding, and scrambling.
2. A data element is the smallest entity that can represent a piece of information (a
bit). A signal element is the shortest unit of a digital signal. Data elements are
what we need to send; signal elements are what we can send. Data elements are
being carried; signal elements are the carriers.
3. The data rate defines the number of data elements (bits) sent in 1s. The unit is bits
per second (bps). The signal rate is the number of signal elements sent in 1s. The
unit is the baud.
4. In decoding a digital signal, the incoming signal power is evaluated against the
baseline (a running average of the received signal power). A long string of 0s or 1s
can cause baseline wandering (a drift in the baseline) and make it difficult for the
receiver to decode correctly.
5. When the voltage level in a digital signal is constant for a while, the spectrum creates very low frequencies, called DC components, that present problems for a system that cannot pass low frequencies.
6. A self-synchronizing digital signal includes timing information in the data being
transmitted. This can be achieved if there are transitions in the signal that alert the
receiver to the beginning, middle, or end of the pulse.
7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransition coding.
8. Block coding provides redundancy to ensure synchronization and to provide inherent error detecting. In general, block coding changes a block of m bits into a block
of n bits, where n is larger than m.
9. Scrambling, as discussed in this chapter, is a technique that substitutes long zerolevel pulses with a combination of other levels without increasing the number of
bits.
1
2
10. Both PCM and DM use sampling to convert an analog signal to a digital signal.
PCM finds the value of the signal amplitude for each sample; DM finds the change
between two consecutive samples.
11. In parallel transmission we send data several bits at a time. In serial transmission
we send data one bit at a time.
12. We mentioned synchronous, asynchronous, and isochronous. In both synchronous and asynchronous transmissions, a bit stream is divided into independent
frames. In synchronous transmission, the bytes inside each frame are synchronized; in asynchronous transmission, the bytes inside each frame are also independent. In isochronous transmission, there is no independency at all. All bits in the
whole stream must be synchronized.
Exercises
13. We use the formula s = c × N × (1/r) for each case. We let c = 1/2.
a. r = 1
→ s = (1/2) × (1 Mbps) × 1/1
= 500 kbaud
b. r = 1/2 → s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaud
c. r = 2
→ s = (1/2) × (1 Mbps) × 1/2
= 250 Kbaud
d. r = 4/3 → s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud
14. The number of bits is calculated as (0.2 /100) × (1 Mbps) = 2000 bits
15. See Figure 4.1. Bandwidth is proportional to (3/8)N which is within the range in
Table 4.1 (B = 0 to N) for the NRZ-L scheme.
Figure 4.1 Solution to Exercise 15
Average Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8
B
(3 / 8) N
Case a
Case c
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
0
0
1
1
0
0
1
1
Case b
Case d
16. See Figure 4.2. Bandwidth is proportional to (4.25/8)N which is within the range
in Table 4.1 (B = 0 to N) for the NRZ-I scheme.
17. See Figure 4.3. Bandwidth is proportional to (12.5 / 8) N which is within the range
in Table 4.1 (B = N to B = 2N) for the Manchester scheme.
18. See Figure 4.4. B is proportional to (12/8) N which is within the range in Table 4.1
(B = N to 2N) for the differential Manchester scheme.
3
Figure 4.2 Solution to Exercise 16
Average Number of Changes = (0 + 9 + 4 + 4) / 4 = 4.25 for N = 8
B
(4.25 / 8) N
Case a
Case c
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
0
0
1
1
0
0
1
1
Case b
Figure 4.3
Case d
Solution to Exercise 17
Average Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8
B
(12.5 / 8) N
Case a
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
Case b
Figure 4.4
Case c
0
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
Case d
Solution to Exercise 18
Average Number of Changes = (16 + 8 + 12 + 12) / 4 = 12 for N = 8
B
(12 / 8) N
Case a
Case c
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
0
0
1
1
0
0
1
1
Case b
Case d
4
19. See Figure 4.5. B is proportional to (5.25 / 16) N which is inside range in Table 4.1
(B = 0 to N/2) for 2B/1Q.
Figure 4.5 Solution to Exercise 19
Average Number of Changes = (0 + 7 + 7 + 7) / 4 = 5.25 for N = 16
B
(5.25 / 8) N
Case a
00
+3
+1
−1
00
00
00
00
00
00
00
Case c
01
+3
+1
−1
10
01
10
01
10
01
10
11
00
11
00
11
00
11
−3
−3
11
11
11
11
11
11
11
11
+3
+1
−1
00
+3
+1
−1
−3
−3
Case b
Case d
20. See Figure 4.6. B is proportional to (5.25/8) × N which is inside the range in Table
4.1 (B = 0 to N/2) for MLT-3.
Figure 4.6 Solution to Exercise 20
Average Number of Changes = (0 + 7 + 4 + 3) / 4 = 4.5 for N = 8
B
(4.5 / 8) N
Case a
Case c
0
0
0
0
0
0
0
0
+V
+V
−V
1
0
1
0
1
0
1
0
0
0
1
1
0
0
0
−V
1
1
1
1
1
1
1
1
+V
+V
−V
−V
Case b
0
Case d
21. The data stream can be found as
a. NRZ-I: 10011001.
b. Differential Manchester: 11000100.
c. AMI: 01110001.
22. The data rate is 100 Kbps. For each case, we first need to calculate the value f / N.
We then use Figure 4.6 in the text to find P (energy per Hz). All calculations are
approximations.
5
a. f /N = 0/100
=0
→
P = 1.0
b. f /N = 50/100 = 1/2 →
P = 0.5
c. f /N = 100/100 = 1
→
P = 0.0
d. f /N = 150/100 = 1.5
→
P = 0.2
23. The data rate is 100 Kbps. For each case, we first need to calculate the value f/N.
We then use Figure 4.8 in the text to find P (energy per Hz). All calculations are
approximations.
a. f /N = 0/100
=0
→
P = 0.0
b. f /N = 50/100 = 1/2 →
P = 0.3
c. f /N = 100/100 = 1
→
P = 0.4
d. f /N = 150/100 = 1.5
→
P = 0.0
24.
a. The output stream is 01010 11110 11110 11110 11110 01001.
b. The maximum length of consecutive 0s in the input stream is 21.
c. The maximum length of consecutive 0s in the output stream is 2.
25. In 5B/6B, we have 25 = 32 data sequences and 26 = 64 code sequences. The number
of unused code sequences is 64 − 32 = 32. In 3B/4B, we have 2 3 = 8 data
sequences and 24 = 16 code sequences. The number of unused code sequences is
16 − 8 = 8.
26. See Figure 4.7. Since we specified that the last non-zero signal is positive, the first
bit in our sequence is positive.
Figure 4.7 Solution to Exercise 26
a. B8ZS
1
1
1
0
0
0
0
0
0
B
0
V
1
1
1
0
0
0
0
0
0
0
0
0
0
0
V
B
0
B
0
0
0
V
V
b. HDB3
27.
a. In a low-pass signal, the minimum frequency 0. Therefore, we have
fmax = 0 + 200 = 200 KHz. → fs = 2 × 200,000 = 400,000 samples/s
6
b. In a bandpass signal, the maximum frequency is equal to the minimum frequency plus the bandwidth. Therefore, we have
fmax = 100 + 200 = 300 KHz. → fs = 2 × 300,000 = 600,000 samples /s
28.
a. In a lowpass signal, the minimum frequency is 0. Therefore, we can say
fmax = 0 + 200 = 200 KHz →
fs = 2 × 200,000 = 400,000 samples/s
The number of bits per sample and the bit rate are
N = 400 KHz × 10 = 4 Mbps
nb = log21024 = 10 bits/sample
b. The value of nb = 10. We can easily calculate the value of SNRdB
SNRdB = 6.02 × nb + 1.76 = 61.96
c. The value of nb = 10. The minimum bandwidth can be calculated as
BPCM = nb × Banalog = 10 × 200 KHz = 2 MHz
29. The maximum data rate can be calculated as
Nmax = 2 × B × nb = 2 × 200 KHz × log24 = 800 kbps
30. We can first calculate the sampling rate (fs) and the number of bits per sample (nb)
fmax = 0 + 4 = 4 KHz
→
fs = 2 × 4 = 8000 sample/s
We then calculate the number of bits per sample.
→ nb = 30000 / 8000 = 3.75
We need to use the next integer nb = 4. The value of SNRdB is
SNRdB = 6.02 × nb + 1.72 = 25.8
31. We can calculate the data rate for each scheme:
a. NRZ
b. Manchester
c. MLT-3
d. 2B1Q
→
→
→
→
N=2
N=1
N=3
N=4
×B=2
×B=1
×B=3
×B=4
× 1 MHz = 2 Mbps
× 1 MHz = 1 Mbps
× 1 MHz = 3 Mbps
× 1 MHz = 4 Mbps
32.
a. For synchronous transmission, we have 1000 × 8 = 8000 bits.
b. For asynchronous transmission, we have 1000 × 10 = 10000 bits. Note that we
assume only one stop bit and one start bit. Some systems send more start bits.
c. For case a, the redundancy is 0%. For case b, we send 2000 extra for 8000
required bits. The redundancy is 25%.
CHAPTER 5
Analog Transmission
Solutions to Review Questions and Exercises
Review Questions
1. Normally, analog transmission refers to the transmission of analog signals using a
band-pass channel. Baseband digital or analog signals are converted to a complex
analog signal with a range of frequencies suitable for the channel.
2. A carrier is a single-frequency signal that has one of its characteristics (amplitude,
frequency, or phase) changed to represent the baseband signal.
3. The process of changing one of the characteristics of an analog signal based on the
information in digital data is called digital-to-analog conversion. It is also called
modulation of a digital signal. The baseband digital signal representing the digital
data modulates the carrier to create a broadband analog signal.
4.
a. ASK changes the amplitude of the carrier.
b. FSK changes the frequency of the carrier.
c. PSK changes the phase of the carrier.
d. QAM changes both the amplitude and the phase of the carrier.
5. We can say that the most susceptible technique is ASK because the amplitude is
more affected by noise than the phase or frequency.
6. A constellation diagram can help us define the amplitude and phase of a signal
element, particularly when we are using two carriers. The diagram is useful when
we are dealing with multilevel ASK, PSK, or QAM. In a constellation diagram, a
signal element type is represented as a dot. The bit or combination of bits it can
carry is often written next to it.The diagram has two axes. The horizontal X axis is
related to the in-phase carrier; the vertical Y axis is related to the quadrature carrier.
7. The two components of a signal are called I and Q. The I component, called inphase, is shown on the horizontal axis; the Q component, called quadrature, is
shown on the vertical axis.
8. The process of changing one of the characteristics of an analog signal to represent
the instantaneous amplitude of a baseband signal is called analog-to-analog con-
1
2
version. It is also called the modulation of an analog signal; the baseband analog
signal modulates the carrier to create a broadband analog signal.
9.
a. AM changes the amplitude of the carrier
b. FM changes the frequency of the carrier
c. PM changes the phase of the carrier
10. We can say that the most susceptible technique is AM because the amplitude is
more affected by noise than the phase or frequency.
Exercises
11. We use the formula S = (1/r) × N, but first we need to calculate the value of r for
each case.
a. r = log22
b. r = log22
c. r = log24
d. r = log264
=1
=1
=2
=6
→
→
→
→
S = (1/1) × (2000 bps)
S = (1/1) × (4000 bps)
S = (1/2) × (6000 bps)
S = (1/6) × (36,000 bps)
= 2000 baud
= 4000 baud
= 3000 baud
= 6000 baud
12. We use the formula N = r × S, but first we need to calculate the value of r for each
case.
a. r = log22
b. r = log22
c. r = log22
d. r = log216
=1
=1
=1
=4
→
→
→
→
N = (1) × (1000 bps)
N = (1) × (1000 bps)
N = (1) × (1000 bps)
N = (4) × (1000 bps)
= 1000 bps
= 1000 bps
= 1000 bps
= 4000 bps
13. We use the formula r = log2L to calculate the value of r for each case.
a. log24
b. log28
c. log24
d. log2128
=2
=3
=2
=7
14. See Figure 5.1.
a. We have two signal elements with peak amplitudes 1 and 3. The phase of both
signal elements are the same, which we assume to be 0 degrees.
b. We have two signal elements with the same peak amplitude of 2. However,
there must be 180 degrees difference between the two phases. We assume one
phase to be 0 and the other 180 degrees.
c. We have four signal elements with the same peak amplitude of 3. However,
there must be 90 degrees difference between each phase. We assume the first
phase to be at 45, the second at 135, the third at 225, and the fourth at 315
degrees. Note that this is one out of many configurations. The phases can be at
3
Figure 5.1 Solution to Exercise 14
a. ASK
b. BPSK
Q
Q
I
1
I
–2
3
2
Q
Q
3
3
3
3
1
1
1
1
I
3
I
3
3
c. QPSK
3
d. 8-QAM
0, 90, 180, and 270. As long as the differences are 90 degrees, the solution is
satisfactory.
d. We have four phases, which we select to be the same as the previous case. For
each phase, however, we have two amplitudes, 1 and 3 as shown in the figure.
Note that this is one out of many configurations. The phases can be at 0, 90,
180, and 270. As long as the differences are 90 degrees, the solution is satisfactory.
15. See Figure 5.2
Figure 5.2 Solution to Exercise 15
a.
b.
Q
Q
I
2
I
–3
3
3
Q
Q
2
2
–2
2
I
I
–2
–2
c.
d.
a. This is ASK. There are two peak amplitudes both with the same phase (0
degrees). The values of the peak amplitudes are A1 = 2 (the distance between
4
the first dot and the origin) and A2= 3 (the distance between the second dot and
the origin).
b. This is BPSK, There is only one peak amplitude (3). The distance between each
dot and the origin is 3. However, we have two phases, 0 and 180 degrees.
c. This can be either QPSK (one amplitude, four phases) or 4-QAM (one amplitude and four phases). The amplitude is the distance between a point and the
origin, which is (22 + 22)1/2 = 2.83.
d. This is also BPSK. The peak amplitude is 2, but this time the phases are 90 and
270 degrees.
16. The number of points define the number of levels, L. The number of bits per baud
is the value of r. Therefore, we use the formula r = log2L for each case.
a. log22
b. log24
c. log216
d. log21024
=1
=2
=4
= 10
17. We use the formula B = (1 + d) × (1/r) × N, but first we need to calculate the
value of r for each case.
a. r = 1
b. r = 1
c. r = 2
d. r = 4
→
→
→
→
B= (1 + 1) × (1/1) × (4000 bps)
B = (1 + 1) × (1/1) × (4000 bps) + 4 KHz
B = (1 + 1) × (1/2) × (4000 bps)
B = (1 + 1) × (1/4) × (4000 bps)
= 8000 Hz
= 8000 Hz
= 2000 Hz
= 1000 Hz
18. We use the formula N = [1/(1 + d)] × r × B, but first we need to calculate the
value of r for each case.
a. r = log22 = 1
b. r = log24=2
c. r = log216= 4
d. r = log264= 6
→
→
→
→
N= [1/(1 + 0)] × 1
N = [1/(1 + 0)] × 2
N = [1/(1 + 0)] × 4
N = [1/(1 + 0)] × 6
× (4 KHz)
× (4 KHz)
× (4 KHz)
× (4 KHz)
= 4 kbps
= 8 kbps
= 16 kbps
= 24 kbps
19.
First, we calculate the bandwidth for each channel = (1 MHz) / 10 = 100 KHz. We
then find the value of r for each channel:
B = (1 + d) × (1/r) × (N) → r = N / B
→ r = (1 Mbps/100 KHz) = 10
We can then calculate the number of levels: L = 2r = 210 = 1024. This means that
that we need a 1024-QAM technique to achieve this data rate.
20. We can use the formula: N = [1/(1 + d)] × r × B = 1 × 6 × 6 MHz = 36 Mbps
21.
a. BAM = 2 × B = 2 × 5
= 10 KHz
5
b. BFM = 2 × (1 + β) × B = 2 × (1 + 5) × 5
c. BPM = 2 × (1 + β) × B = 2 × (1 + 1) × 5
= 60 KHz
= 20 KHz
22. We calculate the number of channels, not the number of coexisting stations.
a. n = (1700 - 530) KHz / 10 KHz
b. n = (108 - 88) MHz / 200 KHz
= 117
= 100
208
CHAPTER 7
TRANSMISSION MEDIA
Applications
The infrared band, almost 400 THz, has an excellent potential for data transmission.
Such a wide bandwidth can be used to transmit digital data with a very high data rate.
The Infrared Data Association (IrDA), an association for sponsoring the use of infrared
waves, has established standards for using these signals for communication between
devices such as keyboards, mice, PCs, and printers. For example, some manufacturers
provide a special port called the IrDA port that allows a wireless keyboard to communicate with a PC. The standard originally defined a data rate of 75 kbps for a distance
up to 8 m. The recent standard defines a data rate of 4 Mbps.
Infrared signals defined by IrDA transmit through line of sight; the IrDA port on
the keyboard needs to point to the PC for transmission to occur.
Infrared signals can be used for short-range communication
in a closed area using line-of-sight propagation.
7.3
RECOMMENDED READING
For more details about subjects discussed in this chapter, we recommend the following
books. The items in brackets [...] refer to the reference list at the end of the text.
Books
Transmission media is discussed in Section 3.8 of [GW04], Chapter 4 of [Sta04], Section 2.2 and 2.3 of [Tan03]. [SSS05] gives a full coverage of transmission media.
7.4
KEY TERMS
angle of incidence
Bayone-Neil-Concelman (BNC)
connector
cladding
IrDA port
line-of-sight propagation
microwave
MT-RJ
coaxial cable
multimode graded-index fiber
core
multimode step-index fiber
omnidirectional antenna
optical fiber
parabolic dish antenna
critical angle
electromagnetic spectrum
fiber-optic cable
gauge
Radio Government (RG) number
ground propagation
radio wave
guided media
horn antenna
infrared wave
reflection
refraction
RJ45
SECTION 7.6
shielded twisted-pair (STP)
single-mode fiber
sky propagation
straight-tip (ST) connector
subscriber channel (SC) connector
transmission medium
7.5
PRACTICE SET
209
twisted-pair cable
unguided medium
unidirectional antenna
unshielded twisted-pair
(UTP)
wireless communication
SUMMARY
o
Transmission media lie below the physical layer.
D A guided medium provides a physical conduit from one device to another. Twistedpair cable, coaxial cable, and optical fiber are the most popular types of guided
media.
D Twisted-pair cable consists of two insulated copper wires twisted together. Twistedpair cable is used for voice and data communications.
D Coaxial cable consists of a central conductor and a shield. Coaxial cable can carry
signals of higher frequency ranges than twisted-pair cable. Coaxial cable is used in
cable TV networks and traditional Ethernet LANs.
Fiber-optic cables are composed of a glass or plastic inner core surrounded by
cladding, all encased in an outside jacket. Fiber-optic cables carry data signals in
the form of light. The signal is propagated along the inner core by reflection. Fiberoptic transmission is becoming increasingly popular due to its noise resistance, low
attenuation, and high-bandwidth capabilities. Fiber-optic cable is used in backbone
networks, cable TV networks, and Fast Ethernet networks.
D Unguided media (free space) transport electromagnetic waves without the use of a
physical conductor.
Wireless data are transmitted through ground propagation, sky propagation, and lineof-sight propagation.Wireless waves can be classified as radio waves, microwaves, or
infrared waves. Radio waves are omnidirectional; microwaves are unidirectional.
Microwaves are used for cellular phone, satellite, and wireless LAN communications.
D Infrared waves are used for short-range communications such as those between a
PC and a peripheral device. It can also be used for indoor LANs.
o
o
7.6
PRACTICE SET
Review Questions
1.
2.
3.
4.
5.
What is the position of the transmission media in the OSI or the Internet model?
Name the two major categories of transmission media.
How do guided media differ from unguided media?
What are the three major classes of guided media?
What is the significance of the twisting in twisted-pair cable?
210
CHAPTER 7
6.
7.
8.
9.
10.
TRANSMISSION MEDIA
What is refraction? What is reflection?
What is the purpose of cladding in an optical fiber?
Name the advantages of optical fiber over twisted-pair and coaxial cable.
How does sky propagation differ from line-of-sight propagation?
What is the difference between omnidirectional waves and unidirectional waves?
Exercises
11. Using Figure 7.6, tabulate the attenuation (in dB) of a 18-gauge UTP for the indicated
frequencies and distances.
Table 7.5 Attenuation/or I8-gauge UTP
Distance
dB at 1 KHz
dRat 10KHz
dB at 100 KHz
1 Krn
lOKm
15 Krn
20Km
12. Use the result of Exercise 11 to infer that the bandwidth of a UTP cable decreases
with an increase in distance.
13. If the power at the beginning of a 1 KIn 18-gauge UTP is 200 mw, what is the
power at the end for frequencies 1 KHz, 10 KHz, and 100 KHz? Use the result of
Exercise 11.
14. Using Figure 7.9, tabulate the attenuation (in dB) of a 2.6/9.5 mm coaxial cable for
the indicated frequencies and distances.
Table 7.6 Attenuation/or 2.6/9.5 mm coaxial cable
Distance
dB at 1 KHz
dB at 10 KHz
dB at 100KHz
1 Km
lOKrn
15Km
20Km
15. Use the result of Exercise 14 to infer that the bandwidth of a coaxial cable
decreases with the increase in distance.
16. If the power at the beginning of a 1 KIn 2.6/9.5 mm coaxial cable is 200 mw, what
is the power at the end for frequencies 1 KHz, 10KHz, and 100 KHz? Use the
result of Exercise 14.
17. Calculate the bandwidth of the light for the following wavelength ranges (assume a
propagation speed of 2 x 108 m):
a. 1000 to 1200 nm
b. 1000 to 1400 nm
SECTION 7.6
PRACTICE SET
211
18. The horizontal axes in Figure 7.6 and 7.9 represent frequencies. The horizontal
axis in Figure 7.16 represents wavelength. Can you explain the reason? lfthe propagation speed in an optical fiber is 2 x 108 ill, can you change the units in the horizontal axis to frequency? Should the vertical-axis units be changed too? Should the
curve be changed too?
19. Using Figure 7.16, tabulate the attenuation (in dB) of an optical fiber for the indicated
wavelength and distances.
Table 7.7 Attenuation for optical fiber
Distance
dB at 800 nm
dB at 1000 nm
dB at 1200nm
1 Km
lOKm
15Km
20Km
20. A light signal is travelling through a fiber. What is the delay in the signal if the
length of the fiber-optic cable is 10 m, 100 m, and 1 Km (assume a propagation
speed of 2 x 108 ill)?
21. A beam oflight moves from one medium to another medium with less density. The
critical angle is 60°. Do we have refraction or reflection for each of the following
incident angles? Show the bending of the light ray in each case.
a. 40°
b. 60°
c. 80
0
236
CHAPTER 8
SWITCHING
multistage switch
output port
packet-switched network
routing processor
setup phase
space-division switching
switch
switching
switching fabric
8.7
o
o
o
o
o
o
o
o
table lookup
teardown phase
time-division switching
time-slot interchange (TSI)
time-space-time (TST)
switch
trap
virtual-circuit identifier (VCI)
virtual-circuit network
SUMMARY
A switched network consists of a series of interlinked nodes, called switches. Traditionally' three methods of switching have been important: circuit switching, packet
switching, and message switching.
We can divide today's networks into three broad categories: circuit-switched networks,
packet-switched networks, and message-switched. Packet-switched networks can also
be divided into two subcategories: virtual-circuit networks and datagram networks
A circuit-switched network is made of a set of switches connected by physical links,
in which each link is divided into n channels. Circuit switching takes place at the
physical layer. In circuit switching, the resources need to be reserved during the
setup phase; the resources remain dedicated for the entire duration of data transfer
phase until the teardown phase.
In packet switching, there is no resource allocation for a packet. This means that
there is no reserved bandwidth on the links, and there is no scheduled processing
time for each packet. Resources are allocated on demand.
In a datagram network, each packet is treated independently of all others. Packets in
this approach are referred to as datagrams. There are no setup or teardown phases.
A virtual-circuit network is a cross between a circuit-switched network and a datagram network. It has some characteristics of both.
Circuit switching uses either of two technologies: the space-division switch or the
time-division switch.
A switch in a packet-switched network has a different structure from a switch used in
a circuit-switched network.We can say that a packet switch has four types of components: input ports, output ports, a routing processor, and switching fabric.
8.8
PRACTICE SET
Review Questions
I. Describe the need for switching and define a switch.
2. List the three traditional switching methods. What are the most common today?
SECTION 8.8
PRACTICE SET
237
3. What are the two approaches to packet-switching?
4. Compare and contrast a circuit-switched network and a packet-switched network.
5. What is the role of the address field in a packet traveling through a datagram
network?
6. What is the role of the address field in a packet traveling through a virtual-circuit
network?
7. Compare space-division and time-division switches.
8. What is TSI and its role in a time-division switching?
9. Define blocking in a switched network.
10. List four major components of a packet switch and their functions.
Exercises
11. A path in a digital circuit-switched network has a data rate of I Mbps. The exchange
of 1000 bits is required for the setup and teardown phases. The distance between
two parties is 5000 km. Answer the following questions if the propagataion speed is
2 X 108 m:
a. What is the total delay if 1000 bits of data are exchanged during the data transfer
phase?
b. What is the total delay if 100,000 bits of data are exchanged during the data
transfer phase?
c. What is the total delay if 1,000,000 bits of data are exchanged during the data
transfer phase?
d. Find the delay per 1000 bits of data for each of the above cases and compare
them. What can you infer?
12. Five equal-size datagrams belonging to the same message leave for the destination one after another. However, they travel through different paths as shown in
Table 8.1.
Table 8.1
Exercise 12
Datagram
Path Length
Visited Switches
1
3200Km
1,3,5
2
11,700 Km
1,2,5
3
12,200 Km
1,2,3,5
4
10,200 Km
1,4,5
5
10,700 Km
1,4,3,5
We assume that the delay for each switch (including waiting and processing) is 3,
10, 20, 7, and 20 ms respectively. Assuming that the propagation speed is 2 x 108 m,
find the order the datagrams arrive at the destination and the delay for each. Ignore
any other delays in transmission.
238
CHAPTER 8
SWITCHING
13. Transmission of information in any network involves end-to-end addressing and
sometimes local addressing (such as YCI). Table 8.2 shows the types of networks
and the addressing mechanism used in each of them.
Table 8.2 Exercise 13
Network
Setup
Circuit-switched
End-ta-end
Teardown
End-ta-end
End-ta-end
Datagram
Virtual-circuit
Data Transfer
End-to-end
Local
End-to-end
Answer the following questions:
a. Why does a circuit-switched network need end-to-end addressing during the setup
and teardown phases? Why are no addresses needed during the data transfer phase
for this type of network?
b. Why does a datagram network need only end-to-end addressing during the data
transfer phase, but no addressing during the setup and teardown phases?
c. Why does a virtual-circuit network need addresses during all three phases?
14. We mentioned that two types of networks, datagram and virtual-circuit, need a
routing or switching table to find the output port from which the information
belonging to a destination should be sent out, but a circuit-switched network has
no need for such a table. Give the reason for this difference.
15. An entry in the switching table of a virtual-circuit network is normally created
during the setup phase and deleted during the teardown phase. In other words, the
entries in this type of network reflect the current connections, the activity in the
network. In contrast, the entries in a routing table of a datagram network do not
depend on the current connections; they show the configuration of the network and
how any packet should be routed to a final destination. The entries may remain
the same even if there is no activity in the network. The routing tables, however, are
updated if there are changes in the network. Can you explain the reason for these
two different characteristics? Can we say that a virtual-circuit is a connectionoriented network and a datagram network is a connectionLess network because of the
above characteristics?
16. The minimum number of columns in a datagram network is two; the minimum number of columns in a virtual-circuit network is four. Can you explain the reason? Is the
difference related to the type of addresses carried in the packets of each network?
17. Figure 8.27 shows a switch (router) in a datagram network.
Find the output port for packets with the following destination addresses:
Packet 1: 7176
Packet 2: 1233
Packet 3: 8766
Packet 4: 9144
18. Figure 8.28 shows a switch in a virtual circuit network.
SECTION 8.8
Figure 8.27
PRACTICE SET
239
Exercise 17
Destination
address
Output
port
1233
3
2
1
4
2
1456
3255
4470
7176
8766
9144
4
2
3
3
2
Figure 8.28 Exercise 18
Incoming
Outgoing
Port
VCI
Port
VCI
1
2
2
3
3
14
71
92
58
78
56
3
22
41
45
43
70
11
4
4
1
2
2
3
4
2
Find the output port and the output VCI for packets with the following input port
and input VCI addresses:
Packet 1: 3, 78
Packet 2: 2, 92
Packet 3: 4, 56
Packet 4: 2, 71
19. Answer the following questions:
a. Can a routing table in a datagram network have two entries with the same destination address? Explain.
b. Can a switching table in a virtual-circuit network have two entries with the same
input port number? With the same output port number? With the same incoming
VCls? With the same outgoing VCls? With the same incoming values (port, VCI)?
With the same outgoing values (port, VCI)?
20. It is obvious that a router or a switch needs to do searching to find information in
the corresponding table. The searching in a routing table for a datagram network is
based on the destination address; the searching in a switching table in a virtualcircuit network is based on the combination of incoming port and incoming VCI.
Explain the reason and define how these tables must be ordered (sorted) based on
these values.
2]. Consider an n X k crossbar switch with n inputs and k outputs.
a. Can we say that switch acts as a multiplexer if n > k?
b. Can we say that switch acts as a demultiplexer if n < k?
240
CHAPTER 8
SWITCHING
22. We need a three-stage space-division switch with N = 100. We use 10 crossbars at
the first and third stages and 4 crossbars at the middle stage.
a. Draw the configuration diagram.
b. Calculate the total number of crosspoints.
c. Find the possible number of simultaneous connections.
d. Find the possible number of simultaneous connections if we use one single crossbar (100 x 100).
e. Find the blocking factor, the ratio of the number of connections in c and in d.
23. Repeat Exercise 22 if we use 6 crossbars at the middle stage.
24. Redesign the configuration of Exercise 22 using the Clos criteria.
25. We need to have a space-division switch with 1000 inputs and outputs. What is the
total number of crosspoints in each of the following cases?
a. Using one single crossbar.
b. Using a multi-stage switch based on the Clos criteria
26. We need a three-stage time-space-time switch with N = 100. We use 10 TSIs at the
first and third stages and 4 crossbars at the middle stage.
a. Draw the configuration diagram.
b. Calculate the total number of crosspoints.
c. Calculate the total number of memory locations we need for the TSIs.
CHAPTER 7
Transmission Media
Solutions to Review Questions and Exercises
Review Questions
1. The transmission media is located beneath the physical layer and controlled by
the physical layer.
2. The two major categories are guided and unguided media.
3. Guided media have physical boundaries, while unguided media are unbounded.
4. The three major categories of guided media are twisted-pair, coaxial, and fiberoptic cables.
5. Twisting ensures that both wires are equally, but inversely, affected by external
influences such as noise.
6. Refraction and reflection are two phenomena that occur when a beam of light
travels into a less dense medium. When the angle of incidence is less than the critical angle, refraction occurs. The beam crosses the interface into the less dense
medium. When the angle of incidence is greater than the critical angle, reflection
occurs. The beam changes direction at the interface and goes back into the more
dense medium.
7. The inner core of an optical fiber is surrounded by cladding. The core is denser
than the cladding, so a light beam traveling through the core is reflected at the
boundary between the core and the cladding if the incident angle is more than the
critical angle.
8. We can mention three advantages of optical fiber cable over twisted-pair and coaxial cables: noise resistance, less signal attenuation, and higher bandwidth.
9. In sky propagation radio waves radiate upward into the ionosphere and are then
reflected back to earth. In line-of-sight propagation signals are transmitted in a
straight line from antenna to antenna.
10. Omnidirectional waves are propagated in all directions; unidirectional waves are
propagated in one direction.
1
2
Exercises
11. See Table 7.1 (the values are approximate).
Table 7.1 Solution to Exercise 11
Distance
dB at 1 KHz
dB at 10 KHz
dB at 100 KHz
1 Km
−3
−5
−7
10 Km
−30
−50
−70
15 Km
−45
−75
−105
20 Km
−60
−100
−140
12. As the Table 7.1 shows, for a specific maximum value of attenuation, the highest
frequency decreases with distance. If we consider the bandwidth to start from zero,
we can say that the bandwidth decreases with distance. For example, if we can tolerate a maximum attenuation of 50 dB (loss), then we can give the following listing of distance versus bandwidth.
Distance
1 Km
10 Km
15 Km
20 Km
Bandwidth
100 KHz
50 KHz
1 KHz
0 KHz
13. We can use Table 7.1 to find the power for different frequencies:
1 KHz
10 KHz
100 KHz
P2 = P1 ×10−3/10
P2 = P1 ×10−5/10
P2 = P1 ×10−7/10
dB = −3
dB = −5
dB = −7
= 100.23 mw
= 63.25 mw
= 39.90 mw
The table shows that the power is reduced 5 times, which may not be acceptable
for some applications.
14. See Table 7.2 (the values are approximate).
Table 7.2 Solution to Exercise 14
Distance
dB at 1 KHz
dB at 10 KHz
dB at 100 KHz
1 Km
−3
−7
−20
10 Km
−30
−70
−200
15 Km
−45
−105
−300
20 Km
−60
−140
−400
15. As Table 7.2 shows, for a specific maximum value of attenuation, the highest frequency decreases with distance. If we consider the bandwidth to start from zero,
we can say that the bandwidth decreases with distance. For example, if we can tol-
3
erate a maximum attenuation of 50 dB (loss), then we can give the following listing of distance versus bandwidth.
Distance
1 Km
10 Km
15 Km
20 Km
Bandwidth
100 KHz
1 KHz
1 KHz
0 KHz
16. We can use Table 7.2 to find the power for different frequencies:
1 KHz
10 KHz
100 KHz
dB = −3
dB = −7
dB = −20
P2 = P1 ×10−3/10
P2 = P1 ×10−7/10
P2 = P1 ×10−20/10
= 100.23 mw
= 39.90 mw
= 2.00 mw
The table shows that power is decreased 100 times for 100 KHz, which is unacceptable for most applications.
17. We can use the formula f = c / λ to find the corresponding frequency for each wave
length as shown below (c is the speed of propagation):
a. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1200 × 10−9] = 33 THz
b. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1400 × 10−9] = 57 THz
18.
a. The wave length is the inverse of the frequency if the propagation speed is
fixed (based on the formula λ = c / f). This means all three figures represent the
same thing.
b. We can change the wave length to frequency. For example, the value 1000 nm
can be written as 200 THz.
c. The vertical-axis units may not change because they represent dB/km.
d. The curve must be flipped horizontally.
19. See Table 7.3 (The values are approximate).
Table 7.3 Solution to Exercise 19
Distance
dB at 800 nm
dB at 1000 nm
dB at 1200 nm
1 Km
−3
−1.1
−0.5
10 Km
−30
−11
−5
15 Km
−45
−16.5
−7.5
20 Km
−60
−22
−10
20. The delay = distance / (propagation speed). Therefore, we have:
a. Delay = 10/(2 × 108) = 0.05 ms
b. Delay = 100/(2 × 108) = 0.5 ms
c. Delay = 1000/(2 × 108) = 5 ms
4
21. See Figure 7.1.
Figure 7.1 Solution to Exercise 21
Refraction
a. 40 degrees
Critical angle = 60
Critical angle
Refraction
b. 60 degrees
Critical angle = 60
Critical angle
Reflection
c. 80 degrees
Critical angle = 60
Critical angle
a. The incident angle (40 degrees) is smaller than the critical angle (60 degrees).
We have refraction.The light ray enters into the less dense medium.
b. The incident angle (60 degrees) is the same as the critical angle (60 degrees).
We have refraction. The light ray travels along the interface.
c. The incident angle (80 degrees) is greater than the critical angle (60 degrees).
We have reflection. The light ray returns back to the more dense medium.
CHAPTER 8
Switching
Solutions to Review Questions and Exercises
Review Questions
1. Switching provides a practical solution to the problem of connecting multiple
devices in a network. It is more practical than using a bus topology; it is more efficient than using a star topology and a central hub. Switches are devices capable of
creating temporary connections between two or more devices linked to the switch.
2. The three traditional switching methods are circuit switching, packet switching,
and message switching. The most common today are circuit switching and packet
switching.
3. There are two approaches to packet switching: datagram approach and virtualcircuit approach.
4. In a circuit-switched network, data are not packetized; data flow is somehow a
continuation of bits that travel the same channel during the data transfer phase. In a
packet-switched network data are packetized; each packet is somehow an independent entity with its local or global addressing information.
5. The address field defines the end-to-end (source to destination) addressing.
6. The address field defines the virtual circuit number (local) addressing.
7. In a space-division switch, the path from one device to another is spatially separate
from other paths. The inputs and the outputs are connected using a grid of electronic microswitches. In a time-division switch, the inputs are divided in time
using TDM. A control unit sends the input to the correct output device.
8. TSI (time-slot interchange) is the most popular technology in a time-division
switch. It used random access memory (RAM) with several memory locations.
The RAM fills up with incoming data from time slots in the order received. Slots
are then sent out in an order based on the decisions of a control unit.
9. In multistage switching, blocking refers to times when one input cannot be connected to an output because there is no path available between them—all the possible intermediate switches are occupied. One solution to blocking is to increase the
number of intermediate switches based on the Clos criteria.
1
2
10. A packet switch has four components: input ports, output ports, the routing processor, and the switching fabric. An input port performs the physical and data link
functions of the packet switch. The output port performs the same functions as the
input port, but in the reverse order. The routing processor performs the function of
table lookup in the network layer. The switching fabric is responsible for moving
the packet from the input queue to the output queue.
Exercises
11. We assume that the setup phase is a two-way communication and the teardown
phase is a one-way communication. These two phases are common for all three
cases. The delay for these two phases can be calculated as three propagation delays
and three transmission delays or
3 [(5000 km)/ (2 ×108 m/s)]+ 3 [(1000 bits/1 Mbps)] = 75 ms + 3 ms = 78 ms
We assume that the data transfer is in one direction; the total delay is then
delay for setup and teardown + propagation delay + transmission delay
a.
b.
c.
d.
78 + 25 + 1 = 104 ms
78 + 25 + 100 = 203 ms
78 + 25 + 1000 = 1103 ms
In case a, we have 104 ms. In case b we have 203/100 = 2.03 ms. In case c, we
have 1103/1000 = 1.101 ms. The ratio for case c is the smallest because we use
one setup and teardown phase to send more data.
12. We assume that the transmission time is negligible in this case. This means that we
suppose all datagrams start at time 0. The arrival timed are calculated as:
First:
Second:
Third:
Fourth:
Fifth:
(3200 Km) / (2 × 108 m/s)
(11700 Km) / (2 × 108 m/s)
(12200 Km) / (2 × 108 m/s)
(10200 Km) / (2 × 108 m/s)
(10700 Km) / (2 × 108 m/s)
+ (3 + 20 + 20)
+ (3 + 10 + 20)
+ (3 + 10+ 20 + 20)
+ (3 + 7 + 20)
+ (3 + 7 + 20 + 20)
=
=
=
=
=
59.0 ms
91.5 ms
114.0 ms
81.0 ms
103.5 ms
The order of arrival is: 3 → 5 → 2 → 4 → 1
13.
a. In a circuit-switched network, end-to-end addressing is needed during the setup
and teardown phase to create a connection for the whole data transfer phase.
After the connection is made, the data flow travels through the already-reserved
resources. The switches remain connected for the entire duration of the data
transfer; there is no need for further addressing.
b. In a datagram network, each packet is independent. The routing of a packet is
done for each individual packet. Each packet, therefore, needs to carry an endto-end address. There is no setup and teardown phases in a datagram network
(connectionless transmission). The entries in the routing table are somehow
permanent and made by other processes such as routing protocols.
3
c. In a virtual-circuit network, there is a need for end-to-end addressing during
the setup and teardown phases to make the corresponding entry in the switching
table. The entry is made for each request for connection. During the data transfer phase, each packet needs to carry a virtual-circuit identifier to show which
virtual-circuit that particular packet follows.
14. A datagram or virtual-circuit network handles packetized data. For each packet,
the switch needs to consult its table to find the output port in the case of a datagram
network, and to find the combination of the output port and the virtual circuit identifier in the case of a virtual-circuit network. In a circuit-switched network, data
are not packetized; no routing information is carried with the data. The whole path
is established during the setup phase.
15. In circuit-switched and virtual-circuit networks, we are dealing with connections.
A connection needs to be made before the data transfer can take place. In the case
of a circuit-switched network, a physical connection is established during the setup
phase and the is broken during the teardown phase. In the case of a virtual-circuit
network, a virtual connection is made during setup and is broken during the teardown phase; the connection is virtual, because it is an entry in the table. These two
types of networks are considered connection-oriented. In the case of a datagram
network no connection is made. Any time a switch in this type of network receives
a packet, it consults its table for routing information. This type of network is considered a connectionless network.
16. The switching or routing in a datagram network is based on the final destination
address, which is global. The minimum number of entries is two; one for the final
destination and one for the output port. Here the input port, from which the packet
has arrived is irrelevant. The switching or routing in a virtual-circuit network is
based on the virtual circuit identifier, which has a local jurisdiction. This means
that two different input or output ports may use the same virtual circuit number.
Therefore, four pieces of information are required: input port, input virtual circuit
number, output port, and output virtual circuit number.
17.
Packet 1: 2
Packet 2: 3
Packet 3: 3
Packet 4: 2
18.
Packet 1: 2, 70
Packet 2: 1, 45
Packet 3: 3, 11
Packet 4: 4, 41
19.
a. In a datagram network, the destination addresses are unique. They cannot be
duplicated in the routing table.
b. In a virtual-circuit network, the VCIs are local. A VCI is unique only in relationship to a port. In other words, the (port, VCI) combination is unique. This
means that we can have two entries with the same input or output ports. We can
4
have two entries with the same VCIs. However, we cannot have two entries
with the same (port, VCI) pair.
20. When a packet arrives at a router in a datagram network, the only information in
the packet that can help the router in its routing is the destination address of the
packet. The table then is sorted to make the searching faster. Today’s routers use
some sophisticated searching techniques. When a packet arrives at a switch in a
virtual-circuit network, the pair (input port, input VCI) can uniquely determined
how the packet is to be routed; the pair is the only two pieces of information in the
packet that is used for routing. The table in the virtual-circuit switch is sorted
based on the this pair. However, since the number of port numbers is normally
much smaller than the number of virtual circuits assigned to each port, sorting is
done in two steps: first according to the input port number and second according to
the input VCI.
21.
a. If n > k, an n × k crossbar is like a multiplexer that combines n inputs into k outputs. However, we need to know that a regular multiplexer discussed in Chapter
6 is n × 1.
b. If n < k, an n × k crossbar is like a demultiplexer that divides n inputs into k outputs. However, we need to know that a regular demultiplexer discussed in
Chapter 6 is 1 × n.
22.
a. See Figure 8.1.
Figure 8.1 Solution to Exercise 22 Part a
…
4 × 10
Stage 2
n = 10
…
…
…
…
…
…
Stage 1
10 × 10
n = 10 N = 100
…
10 × 4
n = 10
…
n = 10
4 × 10
4 × 10
…
10 × 4
10 × 10
10
Crossbars
…
N = 100 n = 10
10 × 4
4
Crossbars
…
n = 10
…
10
Crossbars
Stage 3
b. The total number of crosspoints are
Number of crosspoints = 10 (10 × 4) + 4 (10 × 10) + 10 (4 × 10) = 1200
c. Only four simultaneous connections are possible for each crossbar at the first
stage. This means that the total number of simultaneous connections is 40.
d. If we use one crossbar (100 × 100), all input lines can have a connection at the
same time, which means 100 simultaneous connections.
e. The blocking factor is 40/100 or 40 percent.
5
23.
a. See Figure 8.2.
Figure 8.2 Solution to Exercise 23 Part a
…
6 × 10
Stage 1
Stage 2
n = 10 N = 100
n = 10
…
…
…
…
…
…
10 × 6
n = 10
…
6 × 10
…
n = 10
10 × 10
6 × 10
…
10 × 6
10 × 10
10
Crossbars
…
N = 100 n = 10
10 × 6
6
Crossbars
…
n = 10
…
10
Crossbars
Stage 3
b. The total number of crosspoints are
Number of crosspoints = 10 (10 × 6) + 6 (10 × 10) + 10 (6 × 10) = 1800
c. Only six simultaneous connections are possible for each crossbar at the first
stage. This means that the total number of simultaneous connections is 60.
d. If we use one crossbar (100 × 100), all input lines can have a connection at the
same time, which means 100 simultaneous connections.
e. The blocking factor is 60/100 or 60 percent.
24. According to Clos, n = (N/2)1/2 = 7.07. We can choose n = 8. The number of crossbars in the first stage can be 13 (to have similar crossbars). Some of the input lines
can be left unused. We then have k = 2n − 1 = 15. Figure 8.3 shows the configuration.
Figure 8.3 Solution to Exercise 24 Part a
…
15 × 8
Stage 2
We can calculate the total number of crosspoints as
n=8
…
…
…
…
…
…
Stage 1
13 × 13
n=8
…
8 × 15
n=8
…
n=8
15 × 8
15 × 8
…
8 × 15
13 × 13
13
Crossbars
…
N = 104 n = 8
8 × 15
15
Crossbars
…
n=8
…
13
Crossbars
Stage 3
N = 104
6
13 (8 × 15) + 15 (13 × 13) + 13 (15 × 8) = 5655
The number of crosspoints is still much less than the case with one crossbar
(10,000). We can see that there is no blocking involved because each 8 input line
has 15 intermediate crossbars. The total number of crosspoints here is a little
greater than the minimum number of crosspoints according to Clos using the formula 4N[(2N)1/2 − 1], which is 5257.
25.
a. Total crosspoints = N2 = 10002 = 1,000,000
b. Total crosspoints ≥ 4Ν[(2Ν)1/2 −1] ≥ 174,886. With less than 200,000 crosspoints we can design a three-stage switch. We can use n = (N/2)1/2 =23 and
choose k = 45. The total number of crosspoints is 178,200.
26. We give two solutions.
a. We first solve the problem using only crossbars and then we replace the crossbars at the first and the last stage with TSIs. Figure 8.1 shows the solution using
only crossbars. We can replace the crossbar at the first and third stages with
TSIs as shown in Figure 8.4. The total number of crosspoints is 400 and the
total number of memory locations is 200. Each TSI at the first stage needs one
TDM multiplexer and one TDM demultiplexer. The multiplexer is 10 × 1, but
the demultiplexer is 1 × 4. In other words, the input frame has 10 slots and the
output frame has only 4 slots. The data in the first slot of all input TSIs are
directed to the first switch, the output in the second slot are directed to the second switch, and so on.
Figure 8.4 First solution to Exercise 26
10 TSIs
10 TSIs
4
Crossbars
10 × 10
10 × 10
10 × 10
10 × 10
Stage 1
Stage 2
Stage 3
7
b. We can see the inefficiency in the first solution. Since the slots are separated in
time, only one of the switches at the middle stage is active at each moment. This
means that, instead of 4 crossbars, we could have used only one with the same
result. Figure 8.5 shows the new design. In this case we still need 200 memory
locations but only 100 crosspoints.
Figure 8.5 Second solution to Exercise 26
10 TSIs
10 TSIs
1
Crossbars
10 × 10
Stage 1
Stage 2
Stage 3
SECTION 9. 7
4.
5.
6.
7.
KEY TERMS
261
CMs and CMTSs for the minislots used for timesharing of the upstream channels.
We will learn about this timesharing when we discuss contention protocols in
Chapter 12.
The CM sends a packet to the ISP, asking for the Internet address.
The CM and CMTS then exchange some packets to establish security parameters,
which are needed for a public network such as cable TV.
The CM sends its unique identifier to the CMTS.
Upstream communication can start in the allocated upstream channel; the CM can
contend for the minislots to send data.
Downstream Communication
In the downstream direction, the communication is much simpler. There is no contention because there is only one sender. The CMTS sends the packet with the address of
the receiving eM, using the allocated downstream channel.
9.6
RECOMMENDED READING
For more details about subjects discussed in this chapter, we recommend the following
books. The items in brackets [...] refer to the reference list at the end of the text.
Books
[CouOl] gives an interesting discussion about telephone systems, DSL technology,
and CATV in Chapter 8. [Tan03] discusses telephone systems and DSL technology in
Section 2.5 and CATV in Section 2.7. [GW04] discusses telephone systems in Section 1.1.1 and standard modems in Section 3.7.3. A complete coverage of residential
broadband (DSL and CATV) can be found in [Max99].
9.7
KEY TERMS
56Kmodem
800 service
900 service
ADSL Lite
ADSLmodem
analog leased service
analog switched service
asymmetric DSL (ADSL)
cable modem (CM)
cable modem transmission system
(CMTS)
cable TV network
common carrier
community antenna TV (CATV)
competitive local exchange carrier
(CLEC)
Data Over Cable System Interface
Specification (DOCSIS)
demodulator
digital data service (DDS)
digital service
digital subscriber line (DSL)
digital subscriber line access multiplexer
(DSLAM)
discrete multitone technique (DMT)
distribution hub
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CHAPTER 9
USING TELEPHONE AND CABLE NETWORKS FOR DATA TRANSMISSION
downloading
downstream data band
end office
fiber node
head end
high-bit-rate DSL (HDSL)
hybrid fiber-coaxial (HFC) network
in-band signaling
incumbent local exchange carrier
(ILEC)
interexchange carrier (IXC)
ISDN user port (ISUP)
local access transport area (LATA)
local exchange carrier (LEC)
local loop
long distance company
message transport port (MTP) level
modem
modulator
out-of-band signaling
plain old telephone system (POTS)
point of presence (POP)
ranging
regional cable head (RCH)
regional office
9.8
o
o
o
o
server control point (SCP)
signal point (SP)
signal transport port (STP)
signaling connection control point
(SCep)
Signaling System Seven (SS7)
switched/56 service
switching office
symmetric DSL (SDSL)
tandem office
telephone user port (TUP)
transaction capabilities application port
(TCAP)
trunk
uploading
upstream data band
Y.32
Y.32bis
Y.34bis
Y.90
Y.92
very-high-bit-rate DSL (VDSL)
video band
V-series
wide-area telephone service (WATS)
SUMMARY
The telephone, which is referred to as the plain old telephone system (POTS),
was originally an analog system. During the last decade, the telephone network
has undergone many technical changes. The network is now digital as well as
analog.
The telephone network is made of three major components: local loops, trunks,
and switching offices. It has several levels of switching offices such as end offices,
tandem offices, and regional offices.
The United States is divided into many local access transport areas (LATAs). The
services offered inside a LATA are called intra-LATA services. The carrier that
handles these services is called a local exchange carrier (LEC). The services
between LATAs are handled by interexchange carriers (lXCs).
In in-band signaling, the same circuit is used for both signaling and data. In out-ofband signaling, a portion of the bandwidth is used for signaling and another portion
SECTION 9.9
o
o
o
o
o
o
PRACTICE SET
263
for data. The protocol that is used for signaling in the telephone network is called
Signaling System Seven (SS7).
Telephone companies provide two types of services: analog and digital. We can
categorize analog services as either analog switched services or analog leased services. The two most common digital services are switched/56 service and digital
data service (DDS).
Data transfer using the telephone local loop was traditionally done using a dial-up
modem. The term modem is a composite word that refers to the two functional
entities that make up the device: a signal modulator and a signal demodulator.
Most popular modems available are based on the V-series standards. The V.32 modem
has a data rate of 9600 bps. The V32bis modem supports 14,400-bps transmission.
V90 modems, called 56K modems, with a downloading rate of 56 kbps and uploading rate of 33.6 kbps are very common. The standard above V90 is called V92. These
modems can adjust their speed, and if the noise allows, they can upload data at the rate
of 48 kbps.
Telephone companies developed another technology, digital subscriber line (DSL), to
provide higher-speed access to the Internet. DSL technology is a set of technologies,
each differing in the first letter (ADSL, VDSL, HDSL, and SDSL. ADSL provides
higher speed in the downstream direction than in the upstream direction. The high-bitrate digital subscriber line (HDSL) was designed as an alternative to the T-l line
(1.544 Mbps). The symmetric digital subscriber line (SDSL) is a one twisted-pair version of HDSL. The very high-bit-rate digital subscriber line (VDSL) is an alternative
approach that is similar to ADSL.
Community antenna TV (CATV) was originally designed to provide video services
for the community. The traditional cable TV system used coaxial cable end to end.
The second generation of cable networks is called a hybrid fiber-coaxial (HFC)
network. The network uses a combination of fiber-optic and coaxial cable.
Cable companies are now competing with telephone companies for the residential
customer who wants high-speed access to the Internet. To use a cable network for
data transmission, we need two key devices: a cable modem (CM) and a cable modem
transmission system (CMTS).
9.9
PRACTICE SET
Review Questions
1.
2.
3.
4.
What are the three major components of a telephone network?
Give some hierarchical switching levels of a telephone network.
What is LATA? What are intra-LATA and inter-LATA services?
Describe the SS7 service and its relation to the telephone network.
S. What are the two major services provided by telephone companies in the United
States?
6. What is dial-up modem technology? List some of the common modem standards
discussed in this chapter and give their data rates.
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CHAPTER 9
USING TELEPHONE AND CABLE NETWORKS FOR DATA TRANSMISSION
7. What is DSL technology? What are the services provided by the telephone companies
using DSL? Distinguish between a DSL modem and a DSLAM.
8. Compare and contrast a traditional cable network with a hybrid fiber-coaxial network.
9. How is data transfer achieved using CATV channels?
10. Distinguish between CM and CMTS.
Exercises
11. Using the discussion of circuit-switching in Chapter 8, explain why this type of
switching was chosen for telephone networks.
12. In Chapter 8, we discussed the three communication phases involved in a circuitswitched network. Match these phases with the phases in a telephone call between
two parties.
13. In Chapter 8, we learned that a circuit-switched network needs end-to-end addressing
during the setup and teardown phases. Define end-to-end addressing in a telephone
network when two parties communicate.
14. When we have an overseas telephone conversation, we sometimes experience a
delay. Can you explain the reason?
15. Draw a barchart to compare the different downloading data rates of common modems.
16. Draw a barchart to compare the different downloading data rates of common DSL
technology implementations (use minimum data rates).
17. Calculate the minimum time required to download one million bytes of information
using each of the following technologies:
a. V32 modem
b. V32bis modem
c. V90 modem
18. Repeat Exercise 17 using different DSL implementations (consider the minimum
rates).
19. Repeat Exercise 17 using a cable modem (consider the minimum rates).
20. What type of topology is used when customers in an area use DSL modems for
data transfer purposes? Explain.
21. What type of topology is used when customers in an area use cable modems for
data transfer purposes? Explain.
CHAPTER 9
Using Telephone and Cable Networks
Solutions to Review Questions and Exercises
Review Questions
1. The telephone network is made of three major components: local loops, trunks,
and switching offices.
2. The telephone network has several levels of switching offices such as end offices,
tandem offices, and regional offices.
3. A LATA is a small or large metropolitan area that according to the divestiture of
1984 was under the control of a single telephone-service provider. The services
offered by the common carriers inside a LATA are called intra-LATA services. The
services between LATAs are handled by interexchange carriers (IXCs). These carriers, sometimes called long-distance companies, provide communication services
between two customers in different LATAs.
4. Signaling System Seven (SS7) is the protocol used to provide signaling services in
the telephone network. It is very similar to the five-layer Internet model.
5. Telephone companies provide two types of services: analog and digital.
6. Dial-up modems use part of the bandwidth of the local loop to transfer data. The
latest dial-up modems use the V-series standards such as V.32 and V.32bis (9600
bps), V.34bis (28,800 or 33,600 bps), V.90 (56 kbps for downloading and 33.6
kbps for uploading), and V.92. (56 kbps for downloading and 48 kbps for uploading.
7. Telephone companies developed digital subscriber line (DSL) technology to provide higher-speed access to the Internet. DSL technology is a set of technologies,
each differing in the first letter (ADSL, VDSL, HDSL, and SDSL). The set is often
referred to as xDSL, where x can be replaced by A, V, H, or S. DSL uses a device
called ADSL modem at the customer site. It uses a device called a digital subscriber line access multiplexer (DSLAM) at the telephone company site.
8. The traditional cable networks use only coaxial cables to distribute video information to the customers. The hybrid fiber-coaxial (HFC) networks use a combination of fiber-optic and coaxial cable to do so.
1
2
9. To provide Internet access, the cable company has divided the available bandwidth
of the coaxial cable into three bands: video, downstream data, and upstream data.
The downstream-only video band occupies frequencies from 54 to 550 MHz. The
downstream data occupies the upper band, from 550 to 750 MHz. The upstream
data occupies the lower band, from 5 to 42 MHz.
10. The cable modem (CM) is installed on the subscriber premises. The cable modem
transmission system (CMTS) is installed inside the distribution hub by the cable
company. It receives data from the Internet and passes them to the combiner,
which sends them to the subscriber. The CMTS also receives data from the subscriber and passes them to the Internet.
Exercises
11. Packet-switched networks are well suited for carrying data in packets. The end-toend addressing or local addressing (VCI) occupies a field in each packet. Telephone networks were designed to carry voice, which was not packetized. A circuit-switched network, which dedicates resources for the whole duration of the
conversation, is more suitable for this type of communication.
12. The setup phase can be matched to the dialing process. After the callee responds,
the data transfer phase (here voice transfer phase) starts. When any of the parties
hangs up, the data transfer is terminated and the teardown phase starts. It takes a
while before all resources are released.
13. In a telephone network, the telephone numbers of the caller and callee are serving
as source and destination addresses. These are used only during the setup (dialing)
and teardown (hanging up) phases.
14. The delay can be attributed to the fact that some telephone companies use satellite
networks for overseas communication. In these case, the signals need to travel several thousands miles (earth station to satellite and satellite to earth station).
15. See Figure 9.1.
Figure 9.1 Solution to Exercise 15
60 kbps
56 kbps
50 kbps
40 kbps
30 kbps
20 kbps
14.4 kbps
10 kbps
9600 bps
V.32
V.32bis
V.90
3
16. See Figure 9.2.
Figure 9.2 Solution to Exercise 16
55 Mbps
54 Mbps
25 Mbps
24 Mbps
6 Mbps
5 Mbps
4 Mbps
3 Mbps
2 Mbps
1 Mbps
ADSL
Lower
ADSL Lite
HDSL
SDSL
VDSL
Higher
17.
→
→
→
a. V.32
b. V.32bis
c. V.90
Time = (1,000,000 × 8) /9600
Time = (1,000,000 × 8) / 14400
Time = (1,000,000 × 8) / 56000
≈ 834 s
≈ 556 s
≈ 143 s
18.
a. ADSL
b. ADSL Lite
c. HDSL
d. SDSL
e. VDSL
→
→
→
→
→
Time = (1,000,000 × 8) / 1,500,000
Time = (1,000,000 × 8) / 1,500,000
Time = (1,000,000 × 8) / 1,500,000
Time = (1,000,000 × 8) / 768,000
Time = (1,000,000 × 8) / 25,000,000
≈ 5.3 s
≈ 5.3 s
≈ 5.3 s
≈ 10.42 s
≈ 0.32 s
19. We can calculate time based on the assumption of 10 Mbps data rate:
Time = (1,000,000 × 8) / 10,000,000 ≈ 0.8 seconds
20. The DSL technology is based on star topology with the hub at the telephone office.
The local loop connects each customer to the end office. This means that there is
no sharing; the allocated bandwidth for each customer is not shared with neighbors. The data rate does not depend on how many people in the area are transferring data at the same time.
4
21. The cable modem technology is based on the bus (or rather tree) topology. The
cable is distributed in the area and customers have to share the available bandwidth. This means if all neighbors try to transfer data, the effective data rate will be
decreased.