Ch1 v1

1.1
1-1 DATA COMMUNICATIONS1-1 DATA COMMUNICATIONS
The termThe term telecommunicationtelecommunication means communication at ameans communication at a
distance. The worddistance. The word datadata refers to information presentedrefers to information presented
in whatever form is agreed upon by the parties creatingin whatever form is agreed upon by the parties creating
and using the data.and using the data. Data communicationsData communications are theare the
exchange of data between two devices via some form ofexchange of data between two devices via some form of
transmission medium such as a wire cable.transmission medium such as a wire cable.
 Components of a data communications system
 Data Flow
Topics discussed in this section:Topics discussed in this section:

1.2
Figure 1.1 Components of a data communication system

1.3
Figure 1.2 Data flow (simplex, half-duplex, and full-duplex)

1.4
1-2 NETWORKS1-2 NETWORKS
AA networknetwork is a set of devices (often referred to asis a set of devices (often referred to as nodesnodes))
connected by communicationconnected by communication linkslinks. A node can be a. A node can be a
computer, printer, or any other device capable of sendingcomputer, printer, or any other device capable of sending
and/or receiving data generated by other nodes on theand/or receiving data generated by other nodes on the
network. A link can be a cable, air, optical fiber, or anynetwork. A link can be a cable, air, optical fiber, or any
medium which can transport a signal carryingmedium which can transport a signal carrying
information.information.
 Network Criteria
 Physical Structures
 Categories of Networks

1.5
Network Criteria
 Performance
 Depends on Network Elements
 Measured in terms of Delay and Throughput
 Reliability
 Failure rate of network components
 Measured in terms of availability/robustness
 Security
 Data protection against corruption/loss of data due to:
 Errors
 Malicious users

1.6
Physical Structures
 Type of Connection
 Point to Point - single transmitter and receiver
 Multipoint - multiple recipients of single transmission
 Physical Topology
 Connection of devices
 Type of transmission - unicast, mulitcast, broadcast

1.7
Figure 1.3 Types of connections: point-to-point and multipoint

1.8
Figure 1.4 Categories of topology

1.9
Figure 1.5 A fully connected mesh topology (five devices)

1.10
Figure 1.6 A star topology connecting four stations

1.11
Figure 1.7 A bus topology connecting three stations

1.12
Figure 1.8 A ring topology connecting six stations

1.13
Figure 1.9 A hybrid topology: a star backbone with three bus networks

1.14
Categories of Networks
 Local Area Networks (LANs)
 Short distances
 Designed to provide local interconnectivity
 Wide Area Networks (WANs)
 Long distances
 Provide connectivity over large areas
 Metropolitan Area Networks (MANs)
 Provide connectivity over areas such as a city, a campus

1.15
Figure 1.10 An isolated LAN connecting 12 computers to a hub in a closet

1.16
Figure 1.11 WANs: a switched WAN and a point-to-point WAN

1.17
Figure 1.12 A heterogeneous network made of four WANs and two LANs

McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Multiplexing
1.18

6.19
Bandwidth utilization is the wise use of
available bandwidth to achieve
specific goals.
Efficiency can be achieved by
multiplexing; i.e., sharing of the
bandwidth between multiple users.
Note

6.20
MULTIPLEXINGMULTIPLEXING
Whenever the bandwidth of a medium linking twoWhenever the bandwidth of a medium linking two
devices is greater than the bandwidth needs of thedevices is greater than the bandwidth needs of the
devices, the link can be shared. Multiplexing is the setdevices, the link can be shared. Multiplexing is the set
of techniques that allows the (simultaneous)of techniques that allows the (simultaneous)
transmission of multiple signals across a single datatransmission of multiple signals across a single data
link. As data and telecommunications use increases,link. As data and telecommunications use increases,
so does traffic.so does traffic.
 Frequency-Division Multiplexing
 Wavelength-Division Multiplexing
 Synchronous Time-Division Multiplexing
 Statistical Time-Division Multiplexing

6.21
Dividing a link into channels

6.22
Categories of multiplexing

6.23
Frequency-division multiplexing (FDM)

6.24
FDM is an analog multiplexing technique
that combines analog signals.
It uses the concept of modulation
Note

6.27
FDM demultiplexing example

6.28
Assume that a voice channel occupies a bandwidth of 4
kHz. We need to combine three voice channels into a link
with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the
configuration, using the frequency domain. Assume there
are no guard bands.
Solution
We shift (modulate) each of the three voice channels to a
different bandwidth, as shown in Figure 6.6. We use the
20- to 24-kHz bandwidth for the first channel, the 24- to
28-kHz bandwidth for the second channel, and the 28- to
32-kHz bandwidth for the third one. Then we combine
them as shown in Figure 6.6.
Example 1

6.30
Five channels, each with a 100-kHz bandwidth, are to be
multiplexed together. What is the minimum bandwidth of
the link if there is a need for a guard band of 10 kHz
between the channels to prevent interference?
Solution
For five channels, we need at least four guard bands.
This means that the required bandwidth is at least
5 × 100 + 4 × 10 = 540 kHz,
as shown in Figure 6.7.
Example 2

6.32
Figure 9 Analog hierarchy

6.33
Figure 10 Wavelength-division multiplexing (WDM)

6.34
WDM is an analog multiplexing
technique to combine optical signals.
Note

6.35
Figure 11 Prisms in wavelength-division multiplexing and demultiplexing

6.36
Figure 12 Time Division Multiplexing (TDM)

6.37
TDM is a digital multiplexing technique
for combining several low-rate digital
channels into one high-rate one.
Note

6.38
Figure 13 Synchronous time-division multiplexing

6.39
In synchronous TDM, the data rate
of the link is n times faster, and the unit
duration is n times shorter.
Note

6.40
In Figure 6.13, the data rate for each one of the 3 input
connection is 1 kbps. If 1 bit at a time is multiplexed (a
unit is 1 bit), what is the duration of (a) each input slot,
(b) each output slot, and (c) each frame?
Solution
We can answer the questions as follows:
a. The data rate of each input connection is 1 kbps. This
means that the bit duration is 1/1000 s or 1 ms. The
duration of the input time slot is 1 ms (same as bit
duration).
Example 6.5

6.41
b. The duration of each output time slot is one-third of
the input time slot. This means that the duration of the
output time slot is 1/3 ms.
c. Each frame carries three output time slots. So the
duration of a frame is 3 × 1/3 ms, or 1 ms.
Note: The duration of a frame is the same as the duration
of an input unit.
Example 6.5 (continued)

6.43
Four 1-kbps connections are multiplexed together. A unit
is 1 bit. Find (a) the duration of 1 bit before multiplexing,
(b) the transmission rate of the link, (c) the duration of a
time slot, and (d) the duration of a frame.
Solution
We can answer the questions as follows:
a. The duration of 1 bit before multiplexing is 1 / 1 kbps,
or 0.001 s (1 ms).
b. The rate of the link is 4 times the rate of a connection,
or 4 kbps.
Example 7

6.44
c. The duration of each time slot is one-fourth of the
duration of each bit before multiplexing, or 1/4 ms or
250 s. Note that we can also calculate this from theμ
data rate of the link, 4 kbps. The bit duration is the
inverse of the data rate, or 1/4 kbps or 250 s.μ
d. The duration of a frame is always the same as the
duration of a unit before multiplexing, or 1 ms. We
can also calculate this in another way. Each frame in
this case has four time slots. So the duration of a
frame is 4 times 250 s, or 1 ms.μ

2.45
OSI MODEL
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2.46
2-1 LAYERED TASKS2-1 LAYERED TASKS
We use the concept ofWe use the concept of layerslayers in our daily life. Asin our daily life. As
an example, let us consider two friends whoan example, let us consider two friends who
communicate through postal mail. The process ofcommunicate through postal mail. The process of
sending a letter to a friend would be complex ifsending a letter to a friend would be complex if
there were no services available from the postthere were no services available from the post
office.office.
Sender, Receiver, and Carrier
Hierarchy

2.47
2.1 Tasks involved in sending a letter

2.48
2-2 THE OSI MODEL2-2 THE OSI MODEL
Established in 1947, the International StandardsEstablished in 1947, the International Standards
Organization (Organization (ISOISO) is a multinational body) is a multinational body
dedicated to worldwide agreement ondedicated to worldwide agreement on
international standards. An ISO standard thatinternational standards. An ISO standard that
covers all aspects of network communications iscovers all aspects of network communications is
the Open Systems Interconnection (the Open Systems Interconnection (OSIOSI) model. It) model. It
was first introduced in the late 1970s.was first introduced in the late 1970s.
Layered Architecture
Peer-to-Peer Processes
Encapsulation

2.49
ISO is the organization.
OSI is the model.
Note

2.50
Figure 2.2 Seven layers of the OSI model

2.51
Figure 2.3 The interaction between layers in the OSI model

2.52
Figure 2.4 An exchange using the OSI model

2.53
2-3 LAYERS IN THE OSI MODEL2-3 LAYERS IN THE OSI MODEL
In this section we briefly describe the functionsIn this section we briefly describe the functions
of each layer in the OSI model.of each layer in the OSI model.
Physical Layer
Data Link Layer
Network Layer
Transport Layer
Session Layer
Presentation Layer
Application Layer

2.54
Figure 2.5 Physical layer

2.55
The physical layer is responsible for movements of
individual bits from one hop (node) to the next.
Note

2.56
Figure 2.6 Data link layer

2.57
The data link layer is responsible for moving
frames from one hop (node) to the next.
Note

2.58
Figure 2.7 Hop-to-hop delivery

2.60
The network layer is responsible for the
delivery of individual packets from
the source host to the destination host.
Note

2.61
Figure 2.9 Source-to-destination delivery

2.62
Figure 2.10 Transport layer

2.63
The transport layer is responsible for the delivery
of a message from one process to another.
Note

2.64
Figure 2.11 Reliable process-to-process delivery of a message

2.65
Figure 2.12 Session layer

2.66
The session layer is responsible for dialog
control and synchronization.
Note

2.67
Figure 2.13 Presentation layer

2.68
The presentation layer is responsible for translation,
compression, and encryption.
Note

2.69
Figure 2.14 Application layer

2.70
The application layer is responsible for
providing services to the user.
Note

2.71
Figure 2.15 Summary of layers

2.72
2-4 TCP/IP PROTOCOL SUITE2-4 TCP/IP PROTOCOL SUITE
The layers in theThe layers in the TCP/IP protocol suiteTCP/IP protocol suite do notdo not
exactly match those in the OSI model. Theexactly match those in the OSI model. The
original TCP/IP protocol suite was defined asoriginal TCP/IP protocol suite was defined as
having four layers:having four layers: host-to-networkhost-to-network,, internetinternet,,
transporttransport, and, and applicationapplication. However, when TCP/IP. However, when TCP/IP
is compared to OSI, we can say that the TCP/IPis compared to OSI, we can say that the TCP/IP
protocol suite is made of five layers:protocol suite is made of five layers: physicalphysical,,
data linkdata link,, networknetwork,, transporttransport, and, and applicationapplication..
Physical and Data Link Layers
Network Layer
Transport Layer
Application Layer

2.73
Figure 2.16 TCP/IP and OSI model

2.74
2-5 ADDRESSING2-5 ADDRESSING
Four levels of addresses are used in an internetFour levels of addresses are used in an internet
employing the TCP/IP protocols:employing the TCP/IP protocols: physicalphysical,, logicallogical,,
portport, and, and specificspecific..
Physical Addresses
Logical Addresses
Port Addresses
Specific Addresses

2.75
Figure 2.17 Addresses in TCP/IP

2.76
Figure 2.18 Relationship of layers and addresses in TCP/IP

2.77
In Figure 2.19 a node with physical address 10
sends a frame to a node with physical address
87. The two nodes are connected by a link (bus
topology LAN). As the figure shows, the
computer with physical address 10 is the sender,
and the computer with physical address 87 is the
receiver.
Example 2.1

2.78
Figure 2.19 Physical addresses

2.79
Most local-area networks use a 48-bit (6-byte)
physical address written as 12 hexadecimal
digits; every byte (2 hexadecimal digits) is
separated by a colon, as shown below:
Example 2.2
07:01:02:01:2C:4B
A 6-byte (12 hexadecimal digits) physical
address.

2.80
Figure 2.20 shows a part of an internet with two
routers connecting three LANs. Each device
(computer or router) has a pair of addresses
(logical and physical) for each connection. In this
case, each computer is connected to only one
link and therefore has only one pair of
addresses. Each router, however, is connected
to three networks (only two are shown in the
figure). So each router has three pairs of
addresses, one for each connection.
Example 2.3

2.82
Figure 2.21 shows two computers
communicating via the Internet. The sending
computer is running three processes at this time
with port addresses a, b, and c. The receiving
computer is running two processes at this time
with port addresses j and k. Process a in the
sending computer needs to communicate with
process j in the receiving computer. Note that
although physical addresses change from hop to
hop, logical and port addresses remain the same
from the source to destination.
Example 2.4

2.83
Figure 2.21 Port addresses

2.84
The physical addresses will change from hop to hop,
but the logical addresses usually remain the same.
Note

2.85
Example 2.5
A port address is a 16-bit address represented
by one decimal number as shown.
753
A 16-bit port address represented
as one single number.

3.86
3-5 DATA RATE LIMITS3-5 DATA RATE LIMITS
A very important consideration in dataA very important consideration in data
communications is how fast we can sendcommunications is how fast we can send
data, in bits per second, over a channel.data, in bits per second, over a channel.
Data rate depends on three factors:Data rate depends on three factors:
1.1. The bandwidth availableThe bandwidth available
2.2. The level of the signals we useThe level of the signals we use
33. The quality of the channel (the level of. The quality of the channel (the level of
noise)noise)
Noiseless Channel: Nyquist Bit Rate
 Noisy Channel: Shannon Capacity
 Using Both Limits

3.87
Increasing the levels of a signal
increases the probability of an error
occurring, in other words it reduces the
reliability of the system. Why??
Note

3.88
Capacity of a System
 The bit rate of a system increases with an
increase in the number of signal levels we
use to denote a symbol.
 A symbol can consist of a single bit or “n” bits.
 The number of signal levels = 2n
.
 As the number of levels goes up, the spacing
between level decreases -> increasing the
probability of an error occurring in the
presence of transmission impairments.

3.89
Nyquist Theorem
 Nyquist gives the upper bound for the bit rate
of a transmission system by calculating the bit
rate directly from the number of bits in a
symbol (or signal levels) and the bandwidth of
the system (assuming 2 symbols/per cycle
and first harmonic).
 Nyquist theorem states that for a noiseless
channel:
C = 2 B log22n
C= capacity in bps
B = bandwidth in Hz

3.90
Does the Nyquist theorem bit rate agree
with the intuitive bit rate described in
baseband transmission?
Solution
They match when we have only two levels.
We said, in baseband transmission, the bit
rate is 2 times the bandwidth if we use
only the first harmonic in the worst case.
However, the Nyquist formula is more
general than what we derived intuitively; it
Example 3.33

3.91
Consider a noiseless channel with a
bandwidth of 3000 Hz transmitting a signal
with two signal levels. The maximum bit
rate can be calculated as
Example 3.34

3.92
Consider the same noiseless channel
transmitting a signal with four signal levels
(for each level, we send 2 bits). The
maximum bit rate can be calculated as
Example 3.35

3.93
We need to send 265 kbps over a
noiseless channel with a bandwidth of 20
kHz. How many signal levels do we need?
Solution
We can use the Nyquist formula as shown:
Example 3.36
Since this result is not a power of 2, we
need to either increase the number of
levels or reduce the bit rate. If we have 128
levels, the bit rate is 280 kbps. If we have

3.94
Shannon’s Theorem
 Shannon’s theorem gives the capacity
of a system in the presence of noise.
C = B log2(1 + SNR)

3.95
Consider an extremely noisy channel in
which the value of the signal-to-noise ratio
is almost zero. In other words, the noise is
so strong that the signal is faint. For this
channel the capacity C is calculated as
Example 3.37
This means that the capacity of this
channel is zero regardless of the
bandwidth. In other words, we cannot
receive any data through this channel.

3.96
We can calculate the theoretical highest bit
rate of a regular telephone line. A
telephone line normally has a bandwidth of
3000. The signal-to-noise ratio is usually
3162. For this channel the capacity is
calculated as
Example 3.38
This means that the highest bit rate for a
telephone line is 34.860 kbps. If we want to
send data faster than this, we can either
increase the bandwidth of the line or
improve the signal-to-noise ratio.

3.97
The signal-to-noise ratio is often given in
decibels. Assume that SNRdB = 36 and the
channel bandwidth is 2 MHz. The
theoretical channel capacity can be
calculated as
Example 3.39

3.98
For practical purposes, when the SNR is
very high, we can assume that SNR + 1 is
almost the same as SNR. In these cases,
the theoretical channel capacity can be
simplified to
Example 3.40
For example, we can calculate the
theoretical capacity of the previous
example as

3.99
We have a channel with a 1-MHz
bandwidth. The SNR for this channel is 63.
What are the appropriate bit rate and
signal level?
Solution
First, we use the Shannon formula to find
the upper limit.
Example 3.41

3.100
The Shannon formula gives us 6 Mbps, the
upper limit. For better performance we
choose something lower, 4 Mbps, for
example. Then we use the Nyquist formula
to find the number of signal levels.

3.101
The Shannon capacity gives us the
upper limit; the Nyquist formula tells us
how many signal levels we need.
Note

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