data commucation

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@ Ashek Mahmud Khan; Dept. of CSE (JUST); 01725-402592

01725-402592

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Data Communication (Chapter:01)

Data: Data is the smallest part of the information. The word data refers to information

presented in whatever form is agreed upon by the parties creating and using the data.

Information: Information is processed data.

Q. 2013-1(a): what do you mean by data Communication? Write three

fundamental characteristics of data communications system.

Data Communication: Data communications are the exchange of data between two devices

via some form of transmission medium such as a wire cable.

The effectiveness of a data communications system depends on four fundamental

characteristics: (i) Delivery (ii) Accuracy (iii) Timeliness and (iv) Jitter.

(i) Delivery: The system must deliver data to the correct destination. Data must be received

by the intended device or user and only by that device or user.

(ii) Accuracy: The system must deliver the data accurately. Data that have been altered in

transmission and left uncorrected are unusable.

(iii) Timeliness: The system must deliver data in a timely manner. Data delivered late are

useless.

(iv) Jitter: Jitter refers to the variation in the packet arrival time. It is the uneven delay in the

delivery of audio or video packets.

Q. 2012-1(a): Describe basic components of data communication.

Components: A data communications system has five components.

(i) Message (ii) Sender (iii) Receiver (iv) Transmission medium (v) Protocol

(i) Message: The message is the information (data) to be communicated. Popular forms of

information include text, numbers, pictures, audio, and video.

(ii) Sender: The sender is the device that sends the data message. It can be a computer, work

station, telephone handset, video camera, and soon.

(iii) Receiver: The receiver is the device that receives the message. It can be a computer,

work station, telephone handset, television, and soon.

(iv) Transmission medium: The transmission medium is the physical path by which a

message travels from sender to receiver. Some examples of transmission media include

twisted-pair wire, coaxial cable, fiber-optic cable, and radio waves.

Ashek Mahmud Khan

Dept. of CSE (JUST)

01725-402592

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(v) Protocol: A protocol is a set of rules that govern data communications. It represents an

agreement between the communicating devices. Without a protocol, two devices may be

connected but not communicating, just as a person speaking

Data Representation: Information today comes in different forms such as text, numbers,

images, audio, and video.

Text: In data communications, text is represented as a bit pattern, a sequence of bits (Os or

1s). Different sets of bit patterns have been designed to represent text symbols. Each set is

called a code, and the process of representing symbols is called coding.

Numbers: Numbers are also represented by bit patterns. However, a code such as ASCII is

not used to represent numbers; the number is directly converted to a binary number to

simplify mathematical operations.

Images: Images are also represented by bit patterns. In its simplest form, an image is

composed of a matrix of pixels (picture elements), where each pixel is a small dot. The size

of the pixel depends on the resolution.

Audio: Audio refers to the recording or broad casting of sound or music. Audio is by nature

different from text, numbers, or images. It is continuous, not discrete.

Video: Video refers to the recording or broadcasting of a picture or movie. Video can either

be produced as a continuous entity (e.g., by a TV camera), or it can be a combination of

images, each a discrete entity, arranged to convey the idea of motion.

Data Flow (Data Transmission Mode): The way in which data is transmitted from one

place to another is called data transmission mode. It is also called data the transmission

mode. It is indicates the direction of flow of information. Sometimes, data transmission

modes are called directional modes.

Q. 2013-6(b): Define following transmission links:

Difference types of data transmission modes are as follows:

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(i) Simplex mode (ii) Half-duplex mode (iii) Full-duplex mode

(i) Simplex mode: In simplex mode, data can flow in only one direction. In this mode, a

sender can only send data and cannot receive it. Similarly, a receiver can only receive data

but cannot send it. Data sent from computer to printer is an example of simplex mode.

(ii) Half-duplex mode: In half-duplex mode, data can flow in both directions but only in one

direction at a time. In this mode, data is sent and received alternatively. It is like a one-lane

bridge where two-way traffic must give way in order to cross the other.

(iii) Full-duplex mode: In full duplex-mode, data can flow in both directions at the same

time. It is the fastest directional mode of data communication. The telephone communication

system is an example of full-duplex communication mode.

Q. 2012-1(a)/(08-09): Compare Simplex mode, Half-duplex mode and Full-duplex mode.

Simplex mode Half-duplex mode Full-duplex mode (i) In simplex mode, data can

flow in only one direction. (i) In half-duplex mode, data

can flow in both directions

but only in one direction at a

time.

(i) In full duplex-mode, data

can flow in both directions at

the same time.

(ii) In this mode, a sender

can only send data and

cannot receive it.

(ii) In this mode, data is sent

and received alternatively. (ii) It is the fastest directional

mode of data

communication. (iii) Data sent from computer

to printer is an example of

simplex mode.

(iii) For example, a walkie-

talkie is a half-duplex device

because only one party can

talk at a time.

(iii) The telephone

communication system is an

example of full-duplex

communication mode.

http://www.webopedia.com/TERM/D/device.html

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Networks: A network is a set of devices (often referred to as nodes) connected by

communication links. A node can be a computer, printer, or any other device capable of

sending and/or receiving data generated by other nodes on the network.

Distributed Processing: Most networks use distributed processing, in which a task is

divided among multiple computers. Instead of one single large machine being responsible for

all aspects of a process, separate computers(usually a personal computer or workstation)

handle a subset.

Network Criteria: A network must be able to meet a certain number of criteria. The most

important of these are performance, reliability, and security.

Performance: Performance can be measured in many ways, including transit time and

response time. Transit time is the amount of time required for a message to travel from one

device to another. Response time is the elapsed time between an inquiry and a response.

Performance depends on some factors:

(i) the number of users

(ii) the type of transmission medium

(iii) the type of transmission medium

(iv) the efficiency of the software.

Reliability: In addition to accuracy of delivery, network reliability is measured by the

frequency of failure, the time it takes a link to recover from a failure, and the network's

robustness in a catastrophe.

Security: Network security issues include protecting data from unauthorized access,

protecting data from damage and development, and implementing policies and procedures

for recovery from breaches and data losses.

Type of Connection:

There are two possible types of connections: (i) point-to-point and (ii) multipoint.

Point-to-Point: A point-to-point connection provides a dedicated link between two

devices. The entire capacity of the link is reserved for transmission between those two

devices.

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Multipoint: A multipoint (also called multi-drop) connection is one in which more than

two specific devices share a single link.

Fig. Multipoint

Q. 2012-1(b): Discuss the advantages of a multipoint connection over a point to

point connection.

Point to point connection is limited to two devices, where else more than two devices

share a single link in multipoint connection. Multipoint connection can be used for

fail-over and reliability.

Q. 2012-2(b): Define topology? Explain mesh & star topology?

Physical Topology: The term physical topology refers to the way in which a network is

laid out physically. Two or more devices connect to a link; two or more links form a

topology. There are four basic topologies possible: (i) mesh (ii) star (iii) bus and (iv) ring

Mesh Topology: In a mesh topology, every device has a dedicated point-to-point link to

every other device. The term dedicated means that the link carries traffic only between the

two devices it connects.

Fig. Mesh topology

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Star Topology: In a star topology, each device has a dedicated point-to-point link only to

a central controller, usually called a hub. The devices are not directly linked to one another.

Unlike a mesh topology, a star topology does not allow direct traffic between devices.

Fig. Star topology

Bus topology: A bus topology, on the other hand, is multipoint. One long cable acts as a

back bone to link all the devices in a network. Nodes are connected to the bus cable by drop

lines and taps. A drop line is a connection running between the device and the main cable.

Fig. Bus topology

Ring Topology: In a ring topology, each device has a dedicated point-to-point connection

with only the two devices on either side of it. A signal is passed along the ring in one

direction, from device to device, until it reaches its destination. Each device in the ring

incorporates a repeater.

Fig. Ring topology

Hybrid Topology: A network can be hybrid. Hybrid topology is an integration of two or

more different topologies to form a resultant topology. For example, if there is a ring

topology in one office department while a bus topology in another department, connecting

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these two will result in Hybrid topology. Star-Ring and Star-Bus networks are most common

examples of hybrid network.

Fig. Hybrid topology

Advantages of Hybrid Network Topology:

(i) Reliable (ii) Scalable (iii) Flexible (iv) Effective

Disadvantages of Hybrid Topology:

(i) Complexity of Design (ii) Costly Hub (iii) Costly Infrastructure

Q. 2012-1(b): Compare mesh topology with star topology.

Star topology Mesh topology

(i) In a star topology, each device has a

dedicated point-to-point link only to a

central controller, usually called a hub.

(i) In a mesh topology, every device has a

dedicated point-to-point link to every other

device. (ii) The devices are not directly linked to

one another. (ii) The devices are directly linked to one

another. (iii) A star topology does not allow direct

traffic between devices.

(iii) The term dedicated means that the link

carries traffic only between the two devices

it connects.

(iv) Star topology has minimal line cost. (iv) While mesh topology is the most

expensive network.

(v) Star topology communication is not very

fast between any two nodes

(v) While in mesh topology communication

is very fast between any two nodes.

Categories of Networks: Today when we speak of networks, we are generally referring to

two primary categories: (i) local area networks and (ii) wide-area networks.

(i) Local area networks (LAN): A local area network (LAN) is a computer network that

interconnects computers within a limited area such as a home, school, computer laboratory,

office building, using network media. LAN size is limited to a few kilometers.

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(ii) Wide area networks (WAN): A wide-area network (WAN) is a network that covers a

broad area. A wide area network (WAN) provides long distance transmission of data, image,

audio, and video information overlarge geographic areas that may comprise a country, a

continent, or even the whole world.

Metropolitan Area Networks (MAN): A metropolitan area network (MAN) is a network

with a size between a LAN and a WAN. It normally covers the area inside a town or a city.

A good example of a MAN is the part of the telephone Company network that can provide a

high-speed DSL line to the customer. Another example is the cable TV network.

Q. CT-1(d): Define Protocols and write the key elements of its.

Q. 2012-2(b)/(08-09): Define protocol and standard.

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Protocol: A protocol is a set of rules that govern data communications. A protocol defines

what is communicated, how it is communicated, and when it is communicated. The key

elements of a protocol are syntax, semantics, and timing.

Syntax: The term syntax refers to the structure or format of the data, meaning the order in

which they are presented.

Semantics: The word semantics refers to the meaning of each section of bits.

Timing: The term timing refers to two characteristics: when data should be sent and how

fast they can be sent. For example, if a sender produces data at 100Mbps but the receiver can

process data at only 1Mbps.

Standards: Standards are essential in creating and maintaining an open and competitive

market for equipment manufacturers and in guaranteeing national and international

interoperability of data and telecommunications technology and processes. Standards

provide guidelines to manufacturers, vendors, government agencies.

Data communication standards fall in to two categories: de facto (meaning "by fact" or "by

convention") and de jure (meaning "by law" or "by regulation").

De facto: Standards that have not been approved by an organized body but have been

adopted as standards through widespread use are de facto standards.

De jure: Those standards that have been legislated by an officially recognized body are de

jure standards.

Standards Organizations: Standards are developed through the cooperation of standards

creation committees, forums, and government regulatory agencies.

Standards Creation Committees:

International Organization for Standardization (ISO): The ISO is a multinational body

whose member ship is drawn mainly from the standards creation committees of various

governments throughout the world.

International Telecommunication Union-Telecommunication Standards Sector (ITU-T).

American National Standards Institute (ANSI): Despite its name, the American National

Standards Institute is a completely private, nonprofit corporation not affiliated with the U.S.

federal government.

Institute of Electrical and Electronics Engineers (IEEE): The Institute of Electrical and

Electronics Engineers is the largest professional engineering society in the world.

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Chapter:02 (Network Models)

The OSI Model: The OSI model is a layered framework for the design of network systems

that allows communication between all types of computer systems. It consists of seven

separate but related layers, each of which defines a part of the process of moving

information across a network.

Peer-to-Peer Processes: At the physical layer, communication is direct

Physical Layer: The physical layer is responsible for movements of individual bits from

one hop (node) to the next.

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The physical layer is also concerned with the following:

(i) Physical characteristics of interfaces and medium. The physical layer defines the

characteristics of the interface between the devices and the transmission medium. It also

defines the type of transmission medium.

(ii) Representation of bits. The physical layer data consists of a stream of bits (sequence

of Os or 1s) with no interpretation.

(iii) Data rate. The transmission rate the number of bits sent each second is also defined by

the physical layer.

(iv) Synchronization of bits. The sender and receiver not only must use the same bit rate

but also must be synchronized at the bit level.

(v) Line configuration. The physical layer is concerned with the connection of devices to

the media. In a point-to-point configuration, two devices are connected through a dedicated

link.

(vi) Physical topology. The physical topology define show devices are connected to make

a network.

(vii) Transmission mode. The physical layer also defines the direction of transmission

between two devices: simplex, half-duplex, or full-duplex.

Q. 2012-4(a): What are the responsibilities of data link layer in the internet model?

Data Link Layer: The data link layer is responsible for moving frames from one hop

(node) to the next. The data link layer transforms the physical layer, a raw transmission

facility, to a reliable link. It makes the physical layer appear error-free to the upper layer

(network layer). The data link layer is responsible for moving frames from one hop (node) to

the next. Other responsibilities of the data link layer include the following:

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(i) Framing. The data link layer divides the stream of bits received from the network layer

into manage able data units called frames.

(ii) Physical addressing. (iii) Flow control. (iv) Error control (v) Access control.

Q. 2012-1(c): Elucidate the responsibilities of the network layer of OSI model.

Network Layer: The network layer is responsible for the source-to-destination delivery of

a packet, possibly across multiple networks (links). The network layer is responsible for the

delivery of individual packets from the source host to the destination host.

(i) Logical addressing. The physical addressing implemented by the data link layer

handles the addressing problem locally.

(ii) Routing. When independent networks or links are connected to create internetworks

(network of networks) or a large network, the connecting devices route or switch the packets

to their final destination.

Transport Layer: The transport layer is responsible for process-to-process delivery of the

entire message. A process is an application program running on a host. The transport layer is

responsible for the delivery of a message from one process to another.

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(i) Service-point addressing. (ii) Segmentation and reassembly. (iii) Connection control. (iv)

Flow control. (v) Error control.

Presentation Layer: The presentation layer is concerned with the syntax and semantics of

the information exchanged between two systems. The presentation layer is responsible for

translation, compression, and encryption.

Addressing: Four levels of addresses are used in an internet employing the TCP/IP

protocols: physical (link) addresses, logical (IP) addresses, port addresses, and specific

addresses.

TCP/IP Protocol: The original TCP/IP protocol suite was defined as having four layers:

host-to-network, internet, transport, and application.

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TCP/IP and OSI model:

TCP/IP is a hierarchical protocol made up of interactive modules, each of which provides a

specific functionality; however, the modules are not necessarily interdependent. The term

hierarchical means that each upper-level protocol is supported by one or more lower-level

protocols.

Network Layer: At the network layer (or, more accurately, the internet work layer),

TCP/IP supports the Internet working Protocol. IP, in turn, uses four supporting protocols:

ARP, RARP, ICMP, and IGMP.

Internetworking Protocol (IP): The Internetworking Protocol (IP) is the transmission

mechanism used by the TCP/IP protocols.

Address Resolution Protocol: The Address Resolution Protocol (ARP) is used to

associate logical address with a physical address.

Reverse Address Resolution Protocol: The Reverse Address Resolution Protocol

(RARP) allows a host to discover its Internet address when it knows only its physical

address.

Internet Control Message Protocol: The Internet Control Message Protocol (ICMP) is a

mechanism used by hosts and gateways to send notification of datagram problems back to

the sender.

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Internet Group Message Protocol: The Internet Group Message Protocol (IGMP) is

used to facilitate the simultaneous transmission of a message to a group of recipients.

User Datagram Protocol: The User Datagram Protocol (UDP) is the simpler of the two

standard TCP/IP transport protocols.

Transmission Control Protocol: The Transmission Control Protocol (TCP) provides full

transport-layer services to applications. TCP is a reliable stream transport protocol.

Physical Addresses: The physical address, also known as the link address, is the address

of a node as defined by its LAN or WAN. It is included in the frame used by the data link

layer. It is the lowest level address. The physical addresses have authority over the network

(LAN or WAN).

Logical Addresses: Logical addresses are necessary for universal communications that are

independent of underlying physical networks. Physical addresses are not adequate in an

internet work environment where different networks can have different address formats. A

universal addressing system is needed in which each host can be identified uniquely,

regardless of the underlying physical network.

Port Addresses: In the TCP/IP architecture, the label assigned to a process is called a port

address. A port address in TCP/IP is 16 bits in length. In the TCP/IP architecture, the label

assigned to a process is called a port address. A port address in TCP/IP is 16 bits in length.

Specific Addresses: Some applications have user friendly addresses that are designed for

that specific address. Examples include the e-mail address (for example,

[email protected]).

“Ashek Mahmud Khan”

Chapter: 03 (Data and Signals)

Analog Data: The term analog data refers to information that is continuous. Analog data is

data that is represented in a physical way. Analog data, such as the sounds made by a human

voice, take on continuous values.

Digital Data: Digital data refers to information that has discrete states. For example, data

are stored in computer memory in the form of Os and 1s. They can be converted to a digital

signal or modulated into an analog signal for transmission across a medium.

Q. 2013-1(d): Define continuous and discrete time signal.

Continuous time signal: A Continuous signal or Continuous time signal is a varying

mailto:[email protected]

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quantity(a signal).

Discrete time signal: A discrete signal or discrete time signal is a time series consisting of

a sequence of quantities.

Analog and Digital Signals: 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.

Q. 2013-2(a): Describe Periodic and Non periodic Signals.

Periodic and Non periodic Signals: Both analog and digital signals can take one of two

forms: periodic or non periodic.

A periodic signal completes a pattern with in a measurable time frame, called a period, and

repeats that pattern over subsequent identical periods. The completion of one full pattern is

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called a cycle.

A non periodic signal changes without exhibiting a pattern or cycle that repeats over time.

Both analog and digital signals can be periodic or non periodic. In data communications, we

commonly use periodic analog signals and non periodic digital signals.

Sine Wave: The sine wave is the most fundamental form of a periodic analog signal.

Periodic analog signals: Periodic analog signals can be classified as simple or composite.

A simple periodic analog signal, a sine wave, cannot be decomposed into simpler signals. A

composite periodic analog signal is composed of multiple sine waves.

Period and Frequency: Period refers to the amount of time, in seconds, a signal needs to

complete1cycle. Frequency refers to the number of periods in 1s. Note that period and

frequency are just one characteristic defined in two ways. Period is the inverse of frequency,

and frequency is the inverse of period, as the following formulas show.

Example 3.3: The power we use at home has a frequency of 60Hz (50 Hz in Europe).The

period of this sine wave can be determined as follows:

This means that the period of the power for our lights at home is 0.0116s, or 16.6ms. Our

eyes are not sensitive enough to distinguish these rapid changes in amplitude.

ms

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Example 3.4 : Express a period of 100ms in microseconds.

Solution: From Table 3.1 we find the equivalents of 1ms (l ms is 10-3s) and 1s (1s is 106

micros). We make the following substitutions:

Example 3.5: The period of a signal is 100ms.What is its frequency in kilo hertz?

Solution: First we change 100ms to seconds, and then we calculate the frequency from the

period (1Hz=10-3kHz).

Q. 2013-1(c): Define Phase difference? A sine wave is offset (1/6) one-sixth of a

cycle with respect to time 0. What is its phase in degrees and radians?

Phase difference:

Solution: We know that 1complete cycle is 360°. Therefore, (1/6) one-sixth cycle is:

1

6× 360° = 60°; 60° = 60 ×

2𝜋

360=

𝜋

3 rad= 1.046 rad

Example 3.11: A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz.

What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the

same amplitude.

Solution: Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth.

Then

The spectrum contains all integer frequencies. We show this by a series of spikes –

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Q. 2013-2(c): Define the following terminology (i) Wave length (ii) Time and

Frequency Domains (iii) Bandwidth

(i) Wave length: Wave length binds the period or the frequency of a simple sine wave to

the propagation speed of the medium. Wave length can be calculated if one is given the

propagation speed (the speed of light) and the period of the signal. However, since period

and frequency are related to each other, if we represent wavelength by 𝜆, propagation speed

by c (speed of light), and frequency by 𝑓,we get

Wave length = propagation speed × period = Propagation speed

frequency

𝜆 =𝑐

𝑓

(ii) Time and Frequency Domains: A sine wave is comprehensively defined by its

amplitude, frequency, and phase. We have been showing a sine wave by using what is called

a time-domain plot. The time-domain plot shows changes in signal amplitude with respect to

time.

(iii) Bandwidth: The band width of a composite signal is the difference between the

highest and the lowest frequencies contained in that signal. The band-width is

normally a difference between two numbers. For example, if a composite signal

contains frequencies between 1000 and 5000, its bandwidth is 5000- 1000, or 4000.

There are two types of band-width- (i) Low pass band-width (ii) High pass band-width

Q. 2013-2(c): Describe digital signal as a composite analog signal.

Composite Signals: Composite Signal is a signal that is composed of other signals. The

constituent signals originate separately and join to form the Composite signal.

A composite signal can be periodic or non periodic. A periodic composite signal can be

decomposed into a series of simple sine waves with discrete frequencies that have integer

values (1, 2, 3, and soon).

A non periodic composite signal can be decomposed into a combination of an infinite

number of simple sine waves with continuous frequencies, frequencies that have real values.

Digital Signals: In addition to being represented by an analog signal, information can also

be represented by a digital signal. For example, a 1 can be encoded as a positive voltage and

a 0 as zero voltage. A digital signal can have more than two levels.

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Bit Rate: Term-bit rate (instead of frequency) is used to describe digital signals. The bit

rate is the number of bits sent in 1s, expressed in bits per second (bps).

Example 3.18 Assume we need to download text documents at the rate of 100 pages per

minute. What is the required bit rate of the channel?

Solution: A page is an average of 24 lines with 80 characters in each line. If we assume that

one character requires 8 bits, the bit rate is

100×24×80×8 = 1,536,000bps = 1.536Mbps

Example 3.19: A digitized voice channel, as we will see in Chapter 4, is made by digitizing

a 4-kHz band width analog voice signal. We need to sample the signal at twice the highest

frequency (two samples per hertz).We assume that each sample requires 8 bits. What is the

required bit rate?

Solution: The bit rate can be calculated as

2×4000×8 = 64,000bps = 64kbps

Example 3.20: What is the bit rate for high-definition TV (HDTV)?

Solution: HDTV uses digital signals to broadcast high quality video signals. The HDTV

Screen is normally a ratio of 16: 9 (in contrast to 4:3 for regular TV), which means the

screen is wider .There are 1920 by1080 pixels per screen, and the screen is renewed 30 times

per second. Twenty-four bits represents one color pixel. We can calculate the bit rate as

1920×1080×30×24 = 1,492,992,000 or 1.5Gbps

The TV stations reduce this rate to 20 to 40Mbps through compression.

Bit Length: The bit length is the distance one bit occupies on the transmission medium.

Bit length = propagation speed×bit duration

Baseband Transmission: Baseband transmission means sending a digital signal over a

channel without changing the digital signal to an analog signal.

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Q. 2012-4(d): Name and Explain three types of transmission impairment.

Transmission impairment: Three causes of impairment are attenuation, distortion, and

noise.

Attenuation: Attenuation means a loss of energy. When a signal, simple or composite,

travels through a medium, it loses some of its energy in overcoming the resistance of the

medium.

Distortion: Distortion means that the signal changes its form or shape. Distortion can occur

in a composite signal made of different frequencies.

Q. 2013-2(a): Mention different types of noises in data transmission.

Noise is another cause of impairment. Several types of noise, such as thermal noise, induced

noise, cross talk, and impulse noise, may corrupt the signal.

(i) Thermal: (white) caused by electrons random movement in the medium (extra

superimposed signals)

(ii) Induced: Caused by appliances acting as a sending antenna with medium as receiving

one.

(iii) Crosstalk: Mutual affect between wires acting as sending/ receiving antenna

(iv) Impulse: Caused by a power lines, lightning shaped as a “spike” (high frequency on

very short time)

Q. 2012-2(c): Example 3.26: Suppose a signal travels through a transmission medium and

its power is reduced to one-half.

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This means that P2 = 1

2P1 In this case, the attenuation (loss of power) can be calculated as

10log10𝑝2

𝑝1=10log10

0.5p1

p1 = 10log10 (0.5) =10× (-0.3) = -3dB

A loss of 3dB (-3dB) is equivalent to losing one-half the power.

Signal-to-Noise Ratio (SNR): The signal-to-noise ratio is defined as

SNR= average signal power

average noise power

SNR is the ratio of two powers, it is often described in decibel units, SNRdB, defined

as

SNRdB = l0log10SNR

Example 3.31: The power of a signal is 10mW and the power of the noise is 1𝜇𝑊; what

are the values of SNR and SNRdB?

Solution: The values of SNR and SNRdB can be calculated as follows:

SNR=10,000𝜇𝑊

1𝜇𝑊= 10,000

SNRdB= 10logl0 10,000 = 10logl0104

=40

Example 3.32: The values of SNR and SNRdB for a noise less channel are

SNR= signal power

0= ∞

SNRdB=10logl0∞ = ∞

We can never achieve this ratio in real life; it is an ideal.

Data Rate Limits: A very important consideration in data communications is how

fast we can send data, in bits per second over a channel. Data rate depends on three

factors:

1. The band width available

2. The level of the signals we use

3. The quality of the channel (the level of noise)

Noise less Channel: Nyquist Bit Rate

For a noise less channel, the Nyquist bit rate formula defines the theoretical maximum bit

rate

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Bit Rate=2×bandwidth× log2 L

In this formula, band width is the bandwidth of the channel, L is the number of signal levels

used to represent data, and Bit Rate is the bit rate in bits per second.

Example 3.34: Consider a noise less channel with a bandwidth of 3000Hz transmitting a

signal with two signal levels. The maximum bit rate can be calculated as

Bit Rate=2 ×3000×log2 2=6000bps

Example 3.35: Consider the same (Example 3.34) noise less channel transmitting a signal

with four signal levels (for each level, we send 2 bits). The maximum bit rate can be

calculated as

Bit Rate=2 ×3000×log2 4=12,000bps

Example 3.36: We need to send 265 kbps over a noise less channel with a bandwidth of 20

kHz. How many signal levels do we need?

Solution: We can use the formula as shown:

265,000=2×20,000×log2 L

log2 L=6.625 L=26.625

=98.7 levels

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 have128 levels, the bit rate is 280kbps.If we have 64 levels, the bit

rate is 240kbps.

Determine the theoretical highest data rate for a noisy channel:

Capacity=band-width×log2 (1+SNR)

Example 3.37: 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

C= B log2 (1 +SNR) = B×1og2 (l+0) = B log2 1= B×0 = 0

This means that the capacity of this channel is zero regardless of the band-width. In other

words, we cannot receive any data through this channel.

Example 3.38: We can calculate the theoretical highest bit rate of a regular telephone line.

A telephone line normally has a bandwidth of 3000Hz (300 to 3300Hz) assigned for data

communications. The signal-to-noise ratio is usually 3162. For this channel the capacity is

calculated as

25


C= B log2 (1 +SNR) = 3000 log2 (l+3162)= 3000 log2 3163 (By Calculator)

=3000×11.62=34,860bps

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 band-width of the line or improve the signal-

to-noise ratio.

Band width: One characteristic that measures network performance is band width.

Throughput: The through put is a measure of how fast we can actually send data through a

network. Although, at first glance, band-width in bits per second and throughput seem the

same, they are different.

Q. 2012-2(e)/ (08-09): Example 3.41: 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.

C= B log2 (1 +SNR) = 106 log2 (1 +63) = 10

6 log2 64 = 6Mbps

The Shannon formula gives us 6Mbps, the upper limit. For better performance we choose

something lower, 4Mbps, for example. Then we use the Nyquist formula to find the number

of signal levels.

4Mbps= 2 × 1MHz× log2L → 2log2L=4 → log2L=2 → L=22 → L=4

Latency (Delay): The latency or delay defines how long it takes for an entire message to

completely arrive at the destination from the time the first bit is sent out from the source. We

can say that latency is made of four components: propagation time, transmission time,

queuing time and processing delay.

Latency=propagation time + transmission time+ queuing time+ processing delay

Propagation Time: Propagation time measures the time required for a bit to travel from

the source to the destination. The propagation time is calculated by dividing the distance by

the propagation speed

Example 3.45: What is the propagation time if the distance between the two points is

12,000km? Assume the propagation speed to be 2.4×108 m/s.

Solution: We can calculate the propagation time as-

Propagation time = 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆

𝑷𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒐𝒏 𝒔𝒑𝒆𝒆𝒅 =

12,000×1000

2.4×108 = 50ms

Transmission time:

26


Transmission time = 𝑴𝒆𝒔𝒔𝒂𝒈𝒆 𝑺𝒊𝒛𝒆

𝑩𝒂𝒏𝒅𝒘𝒊𝒅𝒕𝒉

Example 3.46: What are the propagation time and the transmission time for a 2.5-kbyte

message (an e-mail) if the bandwidth of the net work is 1Gbps? Assume that the distance

between the sender and the receiver is 12,000 km and that light travels at 2.4x108 m/s.

Solution: We can calculate the propagation and transmission time as-


𝑷𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒐𝒏 𝒔𝒑𝒆𝒆𝒅=

12,000×1000

2.4×108 = 50ms


𝑩𝒂𝒏𝒅𝒘𝒊𝒅𝒕𝒉=

2500 ×8

109= 20 × 10−6ms

Example 3.47: What are the propagation time and the transmission time for a 5-Mbyte

message (an image) if the bandwidth of the network is 1Mbps? Assume that the distance

between the sender and the receiver is 12,000km and that light travels at 2.4x108 m/s.

Solution: We can calculate the propagation and transmission times as-


𝑷𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒐𝒏 𝒔𝒑𝒆𝒆𝒅=

12,000×1000

2.4×108 = 50ms


𝑩𝒂𝒏𝒅𝒘𝒊𝒅𝒕𝒉=

5,000,000×8

106= 40ms

Jitter: Another performance issue that is related to delay is jitter.

Q. 2012-3(b)/09: Ex-19: What is the bandwidth of a signal that can be decomposed into

five sine waves with frequencies at 0, 20, 50,100, and 200Hz? All peak amplitudes are the

same. Draw the bandwidth.

Solution: Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth.

Then

B= fh - fl = 200-0 = 200Hz

27


Digital Transmission (Chapter:04)

Q.2013-3(a): With a necessary diagram explain line coding and decoding.

Line coding: Line coding is the process of converting digital data to digital signals. We

assume that data, in the form of text, numbers, graphical images, audio or video, are stored

in computer memory as sequences of bits. Line coding converts a sequence of bits to a

digital signal.

Signal Element versus Data Element:

Data Rate versus Signal Rate: The data rate defines the number of data elements (bits)

sent in Is. The unit is bits per-second (bps).The signal rate is the number of signal elements

sent in Is. The unit is the baud. There are several common terminologies used in the

literature. The data rate is sometimes called the bit rate; the signal rate is sometimes called

28


the pulse rate, the modulation rate, or the baud rate. We now need to consider the

relationship between data rate and signal rate (bit rate and baud rate).

S=c×N×1

𝑟 baud

Where N is the data rate (bps); c is the case factor, which varies for each case; S is the

number of signal elements; and r is the previously defined factor.

Data Rate versus Signal Rate:

Data Rate Signal Rate

(i) The data rate defines the number of

data elements (bits) sent in Is.

(i) The signal rate is the number of

signal elements sent in Is.

(ii) The unit is bits per-second (bps). (ii) The unit is the baud.

(iii) The data rate is sometimes called the

bit rate.

(iii) The signal rate is sometimes called

the pulse rate, the modulation rate, or

the baud rate.

Example: 4.1 A signal is carrying data in which one data element is encoded as one signal

element (r=1). If the bit rate is 100kbps, what is the average value of the baud rate if c is

between 0 and l?

Solution: We assume that the average value of c is 1

2. The baud rate is then

S=c×N×1

𝑟 =

1

2 × 100,000 ×

1

1 = 50,000 = 50kbaud

Band width: Although the actual band width of a digital signal is infinite, the effective

bandwidth is finite. The minimum bandwidth can be given as-

Bmin = c×N×1

𝑟

We can solve for the maximum data rate if the bandwidth of the channel is given.

Nmax = 1

𝑐×B×r

Example 4.2 The maximum data rate of a channel (see Chapter 3) is Nmax=2× B× log2L

(defined by the Nyquist formula). Does this agree with the previous formula for Nmax?

Solution: A signal with L levels actually can carry log2L bits per level. If each level

corresponds to one signal element and we assume the average case(c= 1

2), then we have

Nmax = 1

𝑐× B×r = 2×B× log2L

29


Baseline Wandering: In decoding a digital signal, the receiver calculates a running average

of the received signal power. This average is called the baseline.

DC Components: When the voltage level in a digital signal is constant for a while, the

spectrum creates very low frequencies (results of Fourier analysis). These frequencies

around zero, called DC (direct-current) components.

Self-synchronization: To correctly interpret the signals received from the sender, the

receiver's bit intervals must correspond exactly to the sender's bit intervals. If the receiver

clock is faster or slower, the bit intervals are not matched and the receiver might misinterpret

the signals.

Line Coding Schemes:

Unipolar Scheme: In a unipolar scheme, all the signal levels are on one side of the time

axis, either above or below.

NRZ (Non-Return-to-Zero) Traditionally, a unipolar scheme was designed as a non-return-

to-zero (NRZ) scheme in which the positive voltage defines bit I and the zero voltage

defines bit O.

30


Polar Schemes: In polar schemes, the voltages are on the both sides of the time axis. For

example, the voltage level for 0 can be positive and the voltage level for I can be negative.

Non-Return-to-Zero (NRZ) In polar NRZ encoding, we use two levels of voltage amplitude.

We can have two versions of polar NRZ: NRZ-L and NRZ-I, as shown in Figure-

Example4.4: A system is using NRZ-I to transfer 10-Mbps data. What are the average

signal rate and minimum band width?

Solution: The average signal rate is S=N/2=500k baud. The minimum band width for this

average baud rate is Bmin= S= 500kHz.

Return to Zero (RZ): The main problem with NRZ encoding occurs when the sender and

receiver clocks are not synchronized. The receiver does not know when one bit has ended

and the next bit is starting. One solution is the return-to-zero (RZ) scheme, which uses three

values: positive, negative, and zero.

Bipolar Schemes: In bipolar encoding (sometimes called multi level binary), there are three

voltage levels: positive, negative, and zero. The voltage level for one data element is at zero,

while the voltage level for the other element alternates between positive and negative.

AMI and Pseudo ternary Figure shows two variations of bipolar encoding: AMI and pseudo

ternary. A common bipolar encoding scheme is called bipolar alternate mark inversion

(AMI). In the term alternate mark inversion, the word mark comes from telegraphy and

31


means 1. So AMI means alternate 1 inversion. A neutral zero voltage represents binary 0.

Binary 1s are represented by alternating positive and negative voltages. A variation of AMI

encoding is called pseudo ternary in which the 1 bit is encoded as a zero voltage and the 0

bit is encoded as alternating positive and negative voltages.

Q.2012-3(c)/09: For the bit stream 01001110, sketch the waveforms for each of the

following technique (i) NZR-L (ii) RZ (iii) Manchester (iv) Differential Manchester.

(i) NZR-L: 1 Negative voltage; 0 Positive voltages.

NZR- I: 1 existence of a signal transition at the beginning of the bit time (either a

low-to-high or a high-to-low transition); 0 no signal transition at the beginning of the bit

time

(ii) RZ: The Return to Zero (RZ) scheme uses three voltage values. +, 0, -. Each symbol has

a transition in the middle. Either from high to zero or from low to zero.

32


(iii) Manchester: With Manchester, there is always a transition in the middle of a bit period.

** Next bit 1 = Not Inverse. 0 = 1=

** Next bit 0 = Inverse.

(iv) Differential Manchester:

1 absence of transition at the beginning of the bit interval

0 presence of transition at the beginning of the bit interval

Block Coding: Block coding is normally referred to as mB/nB coding; it replaces each m-

bit group with an n-bit group. Block coding normally involves three steps: division,

substitution, and combination.

33


Scrambling: Biphase schemes that are suitable for dedicated links between stations in a

LAN are not suitable for long-distance communication because of their wide bandwidth

requirement. The combination of block coding and NRZ line coding is not suitable for long-

distance encoding either, because of the DC component. Two common scrambling

techniques are B8ZS and HDB3.

B8ZS: Bipolar with S-zero substitution (BSZS) is commonly used in North America. In this

technique, eight consecutive zero-level voltages are replaced by the sequence OOOVBOVB.

The V in the sequence denotes violation; this is a nonzero voltage that breaks an AMI rule of

encoding (opposite polarity from the previous).The B in the sequence denotes bipolm; which

means a nonzero level voltage in accordance with the AMI rule.

HDB3: High-density bipolar 3-zero (HDB3) is commonly used outside of North America.

In this technique, which is more conservative than B8ZS, four consecutive zero-level

voltages are replaced with a sequence of OOOV or BOOV.

34


Analog signal to digital Conversion: Analog-to-digital conversion is an electronic

process in which a continuously variable (analog) signal is changed, without altering its

essential content, into a multi-level (digital) signal. In this section we describe two

techniques, pulse code modulation and delta modulation.

Q.2013-3(b)/2012-5(a): Explain pulse code modulation (PCM) for converting analog

signal to digital data.

Pulse Code Modulation (PCM): The most common technique to change an analog signal

to digital data (digitization) is called pulse code modulation (PCM).

PCM consists of three steps to digitize an analog signal: (i) Sampling (ii) Quantization (iii) Binary encoding

Q. 2012-4(a): Define Sampling and quantization.

Sampling: The first step in PCM is sampling.

Analog signal is sampled every TS secs.

Ts is referred to as the sampling interval.

fs = 1/Ts is called the sampling rate or sampling frequency.

There are 3 sampling methods:

◦Ideal – an impulse at each sampling instant

◦Natural – a pulse of short width with varying amplitude

◦Flat-top – sample and hold, like natural but with single amplitude value

The sampling process is sometimes referred to as pulse amplitude modulation (PAM).

http://searchcio-midmarket.techtarget.com/definition/analog

http://searchcio-midmarket.techtarget.com/definition/digital

35


Quantization: The result of sampling is a series of pulses with amplitude values between

the maximum and minimum amplitudes of the signal. The set of amplitudes can be infinite

with non integral values between the two limits. These values cannot be used in the

encoding process. The following are the steps in quantization:

1. We assume that the original analog signal has instantaneous amplitudes between Vmin and

Vmax

2. We divide the range into L zones, each of height ∆ (delta).

∆=𝑉𝑚𝑎𝑥 −𝑉𝑚𝑖𝑛

𝐿

3. We assign quantized values of 0 to L- I to the midpoint of each zone.

4. We approximate the value of the sample amplitude to the quantized values.

Binary Encoding: The last step in PCM is encoding. After each sample is quantized and

the number of bits per sample is decided, each sample can be changed to an l lb-bit code

word. The bit rate can be found from the formula

Bit rate= sampling rate × number of bits per sample = fs×nb

Delta Modulation (DM): PCM is a very complex technique. Other techniques have been

developed to reduce the complexity of PCM. The simplest is delta modulation.

36


Delta modulation components:

Delta demodulation components:

Data transmission and modes: The transmission of binary data across a link can be

accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with

each clock tick. There are three sub classes of serial transmission: asynchronous,

synchronous, and isochronous.

http://sharmamonika95.blog.com/files/2012/08/delta-modulation-components.gif

http://sharmamonika95.blog.com/files/2012/08/delta-demodulation-components.gif

37


Q. 2012-5(d): Define Parallel and Serial Transmission.

Parallel transmission: When referring to a data transmission, parallel transmission refers

to the process of sending two or more packets of information that are relayed over two or

more lines at the same time.

Serial Transmission: In serial transmission one bit follows another, so we need only one

communication channel rather than n to transmit data between two communicating devices.

Serial transmission occurs in one of three ways: asynchronous, synchronous, and

isochronous.

Q. 2012/09-4(c): What are the advantage and disadvantages of parallel transmission?

Advantages of Parallel Data Transmission:

Fastest form of transmission -able to send multiple bits simultaneously

Doesn’t require high frequency of operation

Disadvantages of Parallel Data Transmission:

Requires separate lines for each bit of a word

Costly to run long distances due to multiple wires

Suffers from electromagnetic interference

Cable lengths more limited than a serial cable

http://www.computerhope.com/jargon/p/packet.htm

38


Asynchronous Transmission: Asynchronous transmission is so named because the

timing of a signal is unimportant. Instead, information is received and translated by agreed

upon patterns. In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or

more stop bits (Is) at the end of each byte. There may be a gap between each byte.

Asynchronous data transmission is a transmission which data transmit character by

character transmission sender to data receiver.

Synchronous Transmission: In synchronous transmission, the bit stream is combined

into longer "frames," which may contain multiple bytes. 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.

Isochronous transmission: In real-time audio and video, in which uneven delays between

frames are not accept-able, synchronous transmission fails. Isochronous transmission is

designed to provide steady bit flow for multimedia applications

Q. 2012-2(b): Compare Synchronous transmission and Asynchronous transmission

Synchronous transmission Asynchronous transmission

(i) supports high data transfer rate (i) slower data transfer rate

(ii) needs clock signal between the sender

and the receiver

(ii) does not need clock signal between the

sender and the receiver

(iii) Hardware is more expensive (iii) Cheap, because asynchronous

transmission requires less hardware

(iv) Synchronous common is used in circuit

switched networks.

(iv) Asynchronous common is used in packet

switched networks.

(iv) It is based on synchronization between

the sender and the receiver.

(v) This communication doesn't need any

synchronization between the sender and the

receiver.

39


Chapter: 05 (Analog Transmission)

Digital-to-analog conversion: Digital-to-analog conversion is the process of changing one

of the characteristics of an analog signal based on the information in digital data.

There are three Types of digital-to-analog conversion: (i) Amplitude shift keying (ASK), (ii)

Frequency shift keying (FSK) and (iii) Phase shift keying (PSK).

Data Rate versus Signal Rate: We can define the data rate (bit rate) and the signal rate

(baud rate) as we did for digital transmission. The relationship between them is

S=N×1

𝑟 baud

Where N is the data rate (bps) and r is the number of data elements carried in one signal

element. Bit rate is the number of bits per second. Baud rate is the number of signal elements

per second.

Q. 2012-7(a): Example 5.1 An analog signal carries 4 bits per signal element. If 1000 signal

elements are sent per second, find the bit rate.

Solution: In this case, r = 4, S= 1000, and N is unknown. We can find the value of N from

S=N×1

𝑟 or, N= S× 𝑟=1000×4=4000 bps

Example 5.2 An analog signal has a bit rate of 8000bps and a baud rate of 1000baud. How

many data elements are carried by each signal element? How many signal elements do we

need?

Solution: In this example, S=1000, N=8000, and r and L are unknown. We find first the

value of r and the n the value of L.

S=N×1

𝑟 or, r = ×

N

𝑆 =

8000

1000= 8bits/baud

r = log2L or, L=2r =2

8= 256

Q. 2012-3(b): Describe the ASK, FSK and PSK.

40


Q. 2012-7(b): Amplitude Shift Keying: ASK is implemented by changing the amplitude of

a carrier signal to reflect amplitude levels in the digital signal. For example: a digital “1”

could not affect the signal, where as a digital “0” would, by making it zero.

Binary ASK (BASK): The bandwidth B of ASK is proportional to the signal rate S.

B = (1+d) S where “d” is due to modulation and filtering, lies between 0 and 1.

Figure 5.3 Binary amplitude shift keying

Implementation: If digital data are presented as a unipolar NRZ digital signal with a high

voltage of 1V and a low voltage of 0V, the implementation can achieved by multiplying the

NRZ digital signal by the carrier signal coming from an oscillator. When the amplitude of

the NRZ signal is 1, the amplitude of the carrier frequency is held; when the amplitude of the

NRZ signal is 0, the amplitude of the carrier frequency is zero.

Figure 5.4 Implementation of binary ASK

Example 5.3: We have an available bandwidth of 100 kHz which spans from 200 to 300

kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK

with d = 1?

Solution: The middle of the bandwidth is located at 250 kHz. This means that our carrier

frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate

(with d = 1 and r = 1).

41


Frequency Shift Keying: The digital data stream changes the frequency of the carrier

signal, fc. For example, a “1” could be represented by f1=fc +f, and a “0” could be

represented by f2= fc-f. / In this case two binary values are represented carrier frequency.

Binary frequency shift keying: One way to think about binary FSK (or BFSK) is to

consider two carrier frequencies. In Figure 5.6, we have selected two carrier frequencies, f1

and f2. We use the first carrier if the data element is 0; we use the second if the data element

is 1.

Band width of FSK: If the difference between the two frequencies (f1 and f2) is 2f, then

the required BW B will be: B = (1+d)×S +2f

Example 5.5 We have an available bandwidth of 100 kHz which spans from 200 to 300

kHz. What should be the carrier frequency and the bit rate if we modulated our data by using

FSK with d = 1?

Solution: This problem is similar to Example 5.3, but we are modulating by using FSK. The

midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means

Phase Shift Keying: 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. The bandwidth requirement, B is: B = (1+d) ×S

Q. 2012-3(c): Quadrature PSK (QPSK): To increase the bit rate, we can code 2 or more

bits onto one signal element.

42


In QPSK, we parallelize the bit stream so that every two incoming bits are split up and PSK

a carrier frequency. One carrier frequency is phase shifted 90o from the other - in quadrature.

The two PSKed signals are then added to produce one of 4 signal elements. L = 4 here.

Constellation Diagrams: A constellation diagram helps us to define the amplitude and

phase of a signal when we are using two carriers, one in quadrature of the other. The X-axis

represents the in-phase carrier and the Y-axis represents quadrature carrier.

Q. 2012-7(d): 4-QAM: Quadrature amplitude modulation is a combination of ASK and

PSK so that a maximum contrast between each signal unit (bit, di bit, tri bit, and so on) is

achieved.

We can have x variations in phase and y variations of amplitude

x • y possible variation (greater data rates)

43


The minimum bandwidth required for QAM transmission is the same as that required for

ASK and PSK transmission. QAM has the same advantages as PSK over ASK.

Example 5.7: Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value

of d = 0.

Solution: For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the

signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6

MHz.

Example 5.8: Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK

signals.

Solution: Figure 5.13 shows the three constellation diagrams-

Figure 5.13 Three constellation diagrams

Chapter: 06 (Bandwidth Utilization: Multiplexing and Spreading)

Bandwidth Utilization: Bandwidth utilization is the wise use of available bandwidth to

achieve specific goals.

Efficiency can be achieved by multiplexing; privacy and anti-jamming can be achieved by

spreading.

44


Multiplexing: Whenever the bandwidth of a medium linking two devices is greater than the

bandwidth needs of the devices, the link can be shared. Multiplexing is the set of techniques

that allows the simultaneous transmission of multiple signals across a single data link. As

data and telecommunications use increases, so does traffic.

Figure 6.1: Dividing a link into channels

There are three basic multiplexing techniques: frequency-division multiplexing, Wave

length-division multiplexing, and time-division multiplexing. The first two are techniques

designed for analog signals, the third, for digital signals.

Q. 2013-4(a)/2012-8(b): Explain Frequency-Division Multiplexing (FDM) and its

process.

Frequency-Division Multiplexing: Frequency-division multiplexing (FDM) is an analog

technique that can be applied when the bandwidth of a link (in hertz) is greater than the

combined bandwidths of the signals to be transmitted.

FDM is an analog multiplexing technique that combines analog signals.

Q. 2012-8(a): Multiplexing Process: Figure 6.4 is a conceptual illustration of the

multiplexing process. Each source generates a signal of a similar frequency range. Inside the

multiplexer, these similar signals modulates different carrier frequencies (f1, f2, and f3). The

45


resulting modulated signals are then combined into a single composite signal that is sent out

over a media link that has enough bandwidth to accommodate it.

Figure 6.4 FDM process

De multiplexing Process: The de multiplexer uses a series of filters to decompose the

multiplexed signal into its constituent component signals. The individual signals are then

passed to a demodulator that separates them from their carriers and passes them to the output

lines. Figure 6.5 is a conceptual illustration of de multiplexing process.

Figure 6.5 FDM de multiplexing example

Example 6.2: 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

46


Figure 6.7

Example 6.3: Four data channels (digital), each transmitting at 1 Mbps, use a satellite

channel of 1 MHz. Design an appropriate configuration, using FDM.

Solution: The satellite channel is analog. We divide it into four channels, each channel

having a 250-kHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4

bits is modulated to 1 Hz. One solution is 16-QAM modulation. Figure 6.8 shows one

possible configuration.

Figure 6.8

Wavelength Division Multiplexing: Wave length-division multiplexing (WDM) is

designed to use the high-data-rate capability of fiber-optic cable. The optical fiber data rate

is higher than the data rate of metallic transmission cable. Using a fiber-optic cable for one

single line wastes the available bandwidth. Multiplexing allows us to combine several lines

into one.

Although WDM technology is very complex, the basic idea is very simple. We want to

combine multiple light sources into one single light at the multiplexer and do the reverse at

the de multiplexer. The combining and splitting of light sources are easily handled by a

prism.

47


Figure 6.11 Prisms in wavelength-division multiplexing and de multiplexing

Q. 2012 (09) -7(b): Explain Time-Division Multiplexing.

Time-Division Multiplexing: Time division multiplexing (TDM) is a digital process that

allows several connections to share the high band width of a link Instead of sharing a portion

of the bandwidth as in FDM, time is shared.

TDM is a digital multiplexing technique for combining several low-rate channels into one

high-rate one.

Q. 2012-8(c): Compare between FDM and WDM

Frequency-Division Multiplexing (FDM):

• Each logical channel is transmitted on a separate frequency.

• Television and radio uses FDM to broadcast many channels over the same media.

• Filters separate the multiplexed signal back into its constituent component signals.

Wavelength Division Multiplexing (WDM):

• Theoretically identical to Frequency Division Multiplexing.

• Used in optical systems while FDM is used in electrical systems.

• Requires more spacing between channels.

Example 6.5: In Figure 6.13, the data rate for each input connection is 3 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:

48


(a) The data rate of each input connection is 1 kbps. This means that the bit duration is

1/1000 s or 1ms. The duration of the input time slot is 1 ms (same as bit duration).

(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 1/3ms.

(c) Each frame carries three output time slots. So the duration of a frame is 3×1/3 ms, or

1ms.The duration of a frame is the same as the duration of an input unit.

Multilevel multiplexing: Multilevel multiplexing is a technique used when the data rate of

an input line is a multiple of others. For example, in Figure 6.19, we have two inputs of

20kbps and three inputs of 40kbps. The first two input lines can be multiplexed together to

provide a data rate equal to the last three. A second level of multiplexing can create an

output of 160kbps.

Pulse Stuffing: Sometimes the bit rates of sources are not multiple integers of each other.

Therefore, neither of the above two techniques can be applied. One solution is to make the

highest input data rate the dominant data rate and then add dummy bits to the input lines

with lower rates. This will increase their rates. This technique is called pulse stuffing, bit

padding, or bit stuffing.

Example 6.11: Two channels, one with a bit rate of 100 kbps and another with a bit rate of

200 kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is

the frame duration? What is the bit rate of the link?

Solution: We can allocate one slot to the first channel and two slots to the second channel.

Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1

bit from the first channel. The bit rate is 100,000 frames/s × 3 bits per frame, or 300 kbps.

49


Chapter: 07 (Transmission Media)

Transmission Media: A transmission medium can be broadly defined as anything that can

carry information from a source to a destination.

In data communications the definition of the information and the transmission medium is

more specific. The transmission medium is usually free space, metallic cable, or fiber-optic

cable. The information is usually a signal that is the result of a conversion of data from

another form.

In telecommunications, transmission media can be divided in to two broad categories:

guided and unguided. Guided media include twisted-pair cable, coaxial cable, and fiber-optic

cable. Unguided medium is free space.

Guided media: Guided media, which are those that provide a conduit from one device to

another, include twisted-pair cable, coaxial cable, and fiber-optic cable.

Twisted-pair cable: A twisted pair consists of two conductors (normally copper), each with

its own plastic insulation, twisted together, as shown in Figure7.3.

Figure 7.3 Twisted-pair cable

One of the wires is used to carry signals to the receiver, and the other is used only as

aground reference. The receiver uses the difference between the two.

In addition to the signal sent by the sender on one of the wires, interference (noise) and cross

talk may affect both wires and create unwanted signals.

50


Q. 2013-8(c): Describe Unshielded Versus Shielded Twisted-Pair Cable.

The most common twisted-pair cable used in communications is referred to as unshielded

twisted-pair (UTP). IBM has also produced a version of twisted-pair cable for its use called

shielded twisted-pair (STP). STP cable has a metal foil or braided-mesh covering that

encases each pair of insulated conductors. Figure shows the difference between UTP and

STP.

Coaxial cable: Coaxial cable (or coax) carries signals of higher frequency ranges than those

in twisted-pair cable, in part because the two media are constructed quite differently. Instead

of having two wires, coax has a central core conductor of solid or stranded wire enclosed in

an insulating sheath, which is, in turn, encased in an outer conductor of metal foil, braid, or a

combination of the two.

Coaxial Cable Standards: Coaxial cables are categorized by their radio government (RG)

ratings.

Q. 2012 (09) - 1(c): Describe communication system using optical fiber with a proper

diagram.

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Fiber-Optic Cable: A fiber-optic cable is made of glass or plastic and transmits signals in

the form of light. To understand optical fiber, we first need to explore several aspects of the

nature of light.

Figure 7.10 Fiber optics: Bending of light ray

Optical fibers use reflection to guide light through a channel. A glass or plastic core is

surrounded by a cladding of less dense glass or plastic. The difference in density of the two

materials must be such that a beam of light moving through the core is reflected off the

cladding instead of being refracted into it. See Figure7.11.

Figure 7.11 Optical fiber

Q. 2012-4(c)/4(e)-09: Write the advantages and disadvantages of optical fiber.

Advantages:

Less signal degradation: The loss of signal in optical fiber is less than in copper wire.

Bandwidth: Fiber optic cables have a much greater bandwidth than metal cables.

Low Power Loss: An optical fiber offers low power loss.

Interference: Fiber optic cables are immune to electromagnetic interference.

Weight: Fiber optic cables are much thinner and lighter than metal wires.

Safety: Since the fiber is a dielectric, it does not present a spark hazard.

Security: Optical fibers are difficult to tap.

Flexibility: An optical fiber has greater tensile strength than copper or steel fibers of the

same diameter.

Cost: The raw materials for glass are plentiful, unlike copper.

Disadvantages:

Limited Application: Can only be used on ground, cannot leave the ground or be associated

with the mobile communication.

Cost: Cables are expensive to install but last longer than copper cables.

Transmission: Transmission on optical fiber requires repeating at distance intervals.

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Fragility: The optical fibers are easily broken

Protection: Optical fibers require more protection around the cable compared to copper.

Unguided media: Unguided media (free space) transport electromagnetic waves without the

use of a physical conductor.

Propagation Modes: Current technology supports two modes (multi mode and single mode)

for propagating light along optical channels, each requiring fiber with different physical

characteristics. Multi-mode can be implemented in two forms: step-index or graded-index.

Multimode: Multi mode is so named because multiple beams from a light source move

through the core in different paths.

In multi mode step-index fiber, the density of the core remains constant from the center to

the edges. A beam of light moves through this constant density in a straight line until it

reaches the interface of the core and the cladding.

A second type of fiber, called multi mode graded-index fiber, decreases this distortion of

the signal through the cable.

Single-Mode: Single-mode uses step-index fiber and a highly focused source of light that

limits beams to a small range of angles, all close to the horizontal.

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Radio waves: Radio waves are omnidirectional. Radio waves are used for multicast

communications, such as radio and television, and paging systems.

Microwaves: Microwaves are unidirectional. Microwaves are used for unicast communication

such as cellular telephones, satellite networks, and wireless LANs.

Q. 2012-4(d)-09: Distinguish between micro-wave and radio wave.

Radio wave Micro-wave

1. Radio wave is omnidirectional. 1. microwaves are unidirectional

2. The frequency of radio waves can take values from

300 GHz to 3 kHz

2. Microwaves are defined to have frequencies

ranging from 300 GHz to only 300MHz.

3. Radio waves in general have long distance

communication capabilities

3. Microwaves do not have these abilities

4. Radio waves are mostly used in the

communication field

4. Microwaves are used in industries and astronomy

5. Microwaves are a type of radio waves with short

frequencies

Infrared: 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.

Satellite Communication: In a communication context, a satellite is a specialized wireless

receiver/transmitter that is launched by a rocket and placed in orbit around the earth.

There are hundreds of satellites currently in operation. They are used for such diverse

purposes as weather forecasting, television broadcast, internet communications, and the

Global Positioning System.

Q. 2012-4(b): How do guided media differ from unguided media?

Guided media it use only local area networks (twisted pair, coaxial, & Optical Fiber Cable).

But unguided media are used worldwide networks (satellite communication, radio

communication).

54


Chapter: 08 (Switching)

Q. 2013-5(a): Describe a block diagram of switched network.

Switching Techniques: In large networks there might be multiple paths linking sender and

receiver. A switched network consists of a series of inter linked nodes, called switches.

Switches are devices capable of creating temporary connections between two or more

devices linked to the switch.

The end systems (communicating devices) are labeled A, B, C, D, and so on, and the switches

are labeled I, II, III, IV, and V. Each switch is connected to multiple links.

There are four typical switching techniques available for digital traffic.

Circuit Switching

Packet Switching

Message Switching

Cell Switching

Q. 2013-5(b): Mention with diagram the taxonomy of switch network. Briefly explain

the circuit switched network.

Fig: Taxonomy of switched networks

Q. 2012-7(a): Define circuit switching and packet switching. Compare them.

Circuit switched network /Circuit switching: A circuit-switched network consists of a set

of switches connected by physical links. A connection between two stations is a dedicated

path made of one or more links.

55


However, each connection uses only one dedicated channel on each link. Each link is

normally divided into n channels by using FDM or TDM.

A circuit-switched network is made of a set of switches connected by physical links, in

which each link is divided into n channels.

Packet switching: A switch used 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

components: input ports, output ports, the routing processor, and the switching fabric.

Compare: Circuit-switched networks and packet-switched networks have traditionally

occupied different spaces within corporations. Circuit-switched networks were used for

phone calls and packet-switched networks handled data. But because of the reach of phone

lines and the efficiency and low cost of data networks, the two technologies have shared

chores for years.

Packet-switched networks move data in separate, small blocks -- packets -- based on the

destination address in each packet. When received, packets are reassembled in the proper

sequence to make up the message. Circuit-switched networks require dedicated point-to-

point connections during calls.

Three Phases: The actual communication in a circuit-switched network requires three

phases: connection setup, data transfer, and connection tear down.

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Data Transfer Phase: After the establishment of the dedicated circuit (channels), the two

parties can transfer data.

Tear down Phase: When one of the parties needs to disconnect, a signal is sent to each

switch to release the resources.

Circuit-Switched Technology in Telephone Networks: Switching at the physical layer in

the traditional telephone network uses the circuit-switching approach.

Data gram network: In data communications, we need to send messages from one end

system to another. If the message is going to pass through a packet-switched network, it

needs to be divided into packets of fixed or variable size. The size of the packet is

determined by the network and the governing protocol. /In packet switching, an

independent data unit.

Q. 2013-5(c): Explain efficiency and delay of datagram network.

Efficiency: The efficiency of a datagram network is better than that of a circuit-switched

network; resources are allocated only when there are packets to be transferred. If a source

sends a packet and there is a delay of a few minutes before another packet can be sent, the

resources can be reallocated during these minutes for other packets from other sources.

Delay: There may be greater delay in a datagram network than in a virtual-circuit network.

Although there are no setup and teardown phases, each packet may experience a wait at a

switch before it is forwarded. In addition, since not all packets in a message necessarily

travel through the same switches, the delay is not uniform for the packets of a message.

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The packet travels through two switches. There are three transmission times (3T), three

propagation delays (slopes 3't of the lines), and two waiting times (WI + w2)' We ignore the

processing time in each switch. The total delay is

Total delay =3T + 3t +WI +W2

Q. 2012-7(c): Define Virtual-circuit.

Virtual-circuit: A virtual-circuit network is a cross between a circuit-switched network and

a datagram network. It has some characteristics of both.

Q. 2012-8(d)/09: Describe the diagram approach of packet switching with its

advantages and disadvantage.

Advantages:

» Security

» Bandwidth used to full potential

» Devices of different speeds can communicate

» Not affected by line failure (redirects signal)

» Availability – no waiting for a direct connection to become available

» During a crisis or disaster, when the public telephone network might stop working, e-

mails and texts can still be sent via packet switching

58


Disadvantages:

» Under heavy use there can be a delay

» Data packets can get lost or become corrupted

» Protocols are needed for a reliable transfer

» Not so good for some types data streams (e.g. real-time video streams can lose frames

due to the way packets arrive out of sequence)

Q. 2012-6(a): What are the limiting factors of a crossbar switch? How dose multi stage

switch alleviated this problem?

Crossbar Switch: A crossbar switch connects n inputs to m outputs in a grid, using

electronic micro switches (transistors) at each cross point (see Figure 8.17). The major

limitation of this design is the number of cross points required. To connect n inputs tom

outputs using a crossbar switch requires n × m cross points. For example, to connect 1000

inputs to 1000 outputs requires a switch with 1,000,000 cross points. A crossbar with this

number of cross points is impractical. Such a switch is also inefficient because statistics

show that, in practice, fewer than 25percent of the cross points are in use at any given time.

The rest are idle.

Multi stage Switch: The solution to the limitations of the crossbar switch is the multistage

switch, which combines cross bar switches in several (normally three) stages, as shown in

Figure8.18.In a single cross bar switch, only one row or column (one path) is active for any

connection. So we need N×N cross points. If we can allow multiple paths inside the switch,

we can decrease the number of cross points. Each cross point in the middle stage can be

accessed by multiple cross points in the first or third stage.

59


Figure 8.18 Multistage switch

Chapter: 10 (Error Detection and Correction)

Errors Detected and corrected: Data can be corrupted during transmission. Some

applications require that errors be detected and corrected.

Types of Errors:

1. Single-Bit Error

2. Burst Error

Single-Bit Error: In a single-bit error, only 1 bit in the data unit has changed.

Burst Error: A burst error means that 2 or more bits in the data unit have changed.

Detection versus Correction: The correction of errors is more difficult than the detection.

In error detection, we are looking only to see if any error has occurred. The answer is a

simple yes or no. A single-bit error is the same for us as a burst error.

In error correction, we need to know the exact number of bits that are corrupted and more

importantly, their location in the message. The number of the errors and the size of the

message are important factors.

Redundancy: The central concept in detecting or correcting errors is redundancy. To be

able to detect or correct errors, we need to send some extra bits with our data. These

redundant bits are added by the sender and removed by the receiver.

60


Coding: Redundancy is achieved through various coding schemes. The sender adds

redundant bits through a process that creates a relationship between the redundant bits and

the actual data bits. We can divide coding schemes into two broad categories:

1. Block coding

2. Convolution coding.

In coding, we need to use modulo-2 arithmetic. Operations in this arithmetic are very simple;

addition and subtraction give the same results. We use the XOR (exclusive OR) operation

for both addition and subtraction.

Q. 2013-7 (b): Explain code words and data words in blocking coding.

In modulo-N arithmetic, we use only the integers in the range 0 to N - 1, inclusive. In block

coding, we divide our message into blocks, each of k bits, called data words. We add r

redundant bits to each block to make the length n = k + r. The resulting n-bit blocks are

called code words. For the moment, it is important to know that we have a set of data words,

each of size k, and a set of code words, each of size of n. With k bits, we can create a

combination of 2k data words; with n bits, we can create a combination of 2𝑛 codewords.

Since n > k, the number of possible code words is larger than the number of possible data-

words. The block coding process is one-to-one; the same data word is always encoded as the

same codeword. This means that we have 2𝑛 - 2𝑘 code words that are not used. We call these

code words invalid or illegal.

61


Error detection: Error detection means to decide whether the received data is correct or not

without having a copy of the original message. In block coding, errors be detected by using

the following two conditions:

a. The receiver has (or can find) a list of valid code words.

b. The original codeword has changed to an invalid one.

Example 10.2: Let us assume that k =2 and n =3. Table 10.1 shows the list of data words

and code words.

Table 10.1 A code for error detection (Example 10.2)

Data words Code words

00 000

01 011

10 101

11 110

00 Even =0; 11 Even in 1=0; 111 odd in 1=1; remember this rules

Hamming Distance: The Hamming distance between two words is the number of

differences between corresponding bits.

Example 10.4: Let us find the Hamming distance between two pairs of words.

1. The Hamming distance d(000, 011) is 2 because 000⊕011 is 011 (two Is).

2. The Hamming distance d(10101, 11110) is 3 because 10101 ⊕11110 is 01011 (three Is).

62


Minimum Hamming Distance: The minimum Hamming distance is the smallest Hamming

distance between all possible pairs in a set of words.

Example 10.5: Find the minimum Hamming distance of the coding scheme in Table 10.1.

Solution: We first find all Hamming distances.

d(000, 011) =2 d(000,101)=2 d(000,110)=2 d(011,101)=2

d(011,110)=2 d(101,110)=2

The 𝑑𝑚𝑖𝑛 in this case is 2.

Example 10.6: Find the minimum Hamming distance of the coding scheme in Table 10.2.

Solution: We first find all the Hamming distances.

d(00000, 01011) = 3 d(00000, 10101) =3 d(00000, 11110) = 4

d(01011, 10101) =4 d(0l011, 11110) = 3 d(10101, 11110) =3

The 𝑑𝑚𝑖𝑛 in this case is 3.

To guarantee the detection of up to s errors in all cases, the minimum Hamming distance

in a block code must be dmin =S + 1.

Example 10.7: The minimum Hamming distance for our first code scheme (Table 10.1) is 2.

This code guarantees detection of only a single error. For example, if the third codeword (l0

1) is sent and one error occurs, the received codeword does not match any valid codeword. If

two errors occur, however, the received codeword may match a valid codeword and the

errors are not detected.

Linear block code: Al most all block codes used today belong to a subset called linear

block codes. The use of nonlinear block codes for error detection and correction is not as

wide spread because their structure makes theoretical analysis and implementation difficult.

In a linear block code, the exclusive OR (XOR) of any two valid code words creates another

valid codeword.

A simple parity-check code is a single-bit error-detecting code in which n =k + 1 with

𝑑𝑚𝑖𝑛 =2.

Table 10.3 Simple parity-check code C(5, 4)

0000 00000 1000 10001

0001 00011 1001 10010

0010 00101 1010 10100

0011 00110 1011 10111

0100 01001 1100 11000

0101 01010 1101 11011

0110 01100 1110 11101

0111 01111 1111 11110

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Figure 10.11 Two-dimensional parity-check code

64


Encoder: Let us take a closer look at the encoder. The encoder takes the data word and

augments it with n- k number of as. It then divides the augmented data word by the divisor,

as shown in Figure:

Decoder: The code word can change during transmission. The decoder does the same

division process as the encoder. The remainder of the division is the syndrome. If the

syndrome is all Os, there is no error; the data word is separated from the received code word

and accepted. Otherwise, everything is discarded.

65


Q. 2012-5(c)-09: Describe the operation of CRC.

Cyclic Redundancy Check: We can create cyclic codes to correct errors. However, the

theoretical background required is beyond the scope of this book. In this section, we simply

discuss a category of cyclic codes called the cyclic redundancy check (CRC) that is used in

networks such as LANs and WANs.

Table 10.6 shows an example of a CRC code. We can see both the linear and cyclic

properties of this code.

Table 10.6: A CRC code with C (7, 4)

Figure 10.14: CRC encoder and decoder

In the encoder, the data word has k bits (4 here); the code word has n bits (7 here). The size

of the data word is augmented by adding n- k (3 here) Os to the right-hand side of the word.

Then-bit result is fed into the generator. The generator uses a divisor of size n-k+1 (4 here),

predefined and agreed upon. The generator divides the augmented data word by the divisor

(modulo-2division). The quotient of the division is discarded; the remainder is appended to

the data word to create the codeword.

66


The decoder receives the possibly corrupted code word. A copy of all n bits is fed to the

checker which is a replica of the generator. The remainder produced by the checker is a

syndrome of n- k (3 here) bits.

A copy of a 4-bit data word is fed into the generator that creates three parity checks 𝑟1, 𝑟2,

and 𝑟3 as shown below:

𝑟0, =𝑎2, +𝑎1 +𝑎0 modulo-2

𝑟1 =𝑎3+ 𝑎2+ 𝑎1 modulo-2

𝑟2=𝑎1 +𝑎0+𝑎3 modulo-2

67


Check sum: The last error detection method we discuss here is called the check sum. The

check sum is used in the Internet by several protocols although not at the data link layer.

However, we briefly discuss it here to complete our discussion on error checking.

Example 10.18: Suppose our data is a list of five 4-bit numbers that we want to send to a

destination. In addition to sending these numbers, we send the sum of the numbers. For

example, if the set of numbers is (7, 11, 12, 0, 6), we send (7, 11, 12, 0, 6, 36), where 36 is

the sum of the original numbers. The receiver adds the five numbers and compares the result

with the sum. If the two are the same, the receiver assumes no error, accepts the five

numbers, and discards the sum. Otherwise, there is an error somewhere and the data are not

accepted.

Example 10.19: We can make the job of the receiver easier if we send the negative

(complement) of the sum, called the check sum. In this case, we send (7, 11, 12, 0, 6, -36).

The receiver can add all the numbers received (including the check sum). If the result is 0, it

assumes no error; otherwise, there is an error.

Example10.20: How can we represent the number 21 in one's complement arithmetic using

only four bits?

Solution: The number 21in binary is 10101(it needs five bits). We can wrap the left most bit

and add it to the four right most bits. We have (0101 +1) = 0110 or 6.

Example 10.21: How can we represent the number-6 in one's complement arithmetic using

only four bits?

Solution: In one's complement arithmetic, the negative or complement of a number is found

by inverting all bits. Positive 6 is 0110; negative 6 is 1001. If we consider only unsigned

numbers, this is 9. In other words, the complement of 6 is 9. Another way to find the

complement of a number in one's complements arithmetic is to subtract the number from

2n-1(16- 1 in this case).

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Data transparency: The ability to send any bit pattern as data without it being mistaken for

control bits.

• FDMA- (Frequency division multiple access), DQPSK modulation scheme, 48.6kbps bit rate

• TDMA- (Time division multiple access), DQPSK modulation scheme, 48.6kbps bit

rate

• CDMA- (Code-Division Multiple Access), QPSK/(Offset) OQPSK modulation

scheme, 1.2288Mbps bit rate

Chapter-11 (Data Link Control)

Data link control: Data link control functions include framing, flow and error control, and

software-implemented protocols that provide smooth and reliable transmission of frames

between nodes.

Framing: Framing in the data link layer separates a message from one source to a

destination, or from other messages to other destinations, by adding a sender address and a

destination address.

(i) Fixed-Size Framing

(ii) Variable-Size Framing

Fixed-Size Framing: Frames can be of fixed or variable size. In fixed-size framing, there is

no need for defining the boundaries of the frames; the size itself can be used as a delimiter.

Variable-Size framing: In variable-size framing, we need a way to define the end of the

frame and the beginning of the next.

Protocols: Now let us see how the data link layer can combine framing, flow control, and

error control to achieve the delivery of data from one node to another. The protocols are

normally implemented in software by using one of the common programming languages.

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Figure 11.5 Taxonomy of protocols discussed in this chapter

data commucation

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