Computer Networks Module I

COMPUTER NETWORKS
Ajit K Nayak, Ph.D.
Department of Computer Science &Information Technology,
School of Computer Science and Engineering, ITER, SOA
University.
Lecture Notes
Module I

Computer Networking / Module I / AKN / 2
Out Line of Module I
 Overview of Data Communications and Networking
 Physical Layer
 Digital Transmission
 Analog Transmission
 Multiplexing
 Transmission Media
 Circuit switching and Telephone Network
Readings: “Data Communications and Networking” Behrouz
A Forouzan, Chapter 1 - Chapter 7

Overview of Data Communications
and Networking
Lecture I
• Data Communication
• Networks & Internet
• Protocols & Standards
• Layered Tasks
• Internet Model
• OSI Model

Data Communication
 Sharing of information is “Data Communication”
 Sharing can be local (face to face)
 Remote (over a distance)
 “Data” refers to facts, concepts and / or
instructions
 In the context of computers, data represented in the
form of 0‟s and 1‟s
 “Data Communication” is “Exchange of data
between two/more devices via a transmission
medium.

Characteristics of Data Communication
 Delivery: system must deliver data to correct
destination
 Accuracy: Accurate data should be delivered
 Timeliness: Data delivered late are useless

Components of Data Communication
 Message: It is the Information (data) to be
communicated (shared) with others
 Sender: The device that sends the message
 Receiver: The device that receives the message
 Medium: Physical path by which a message travels
from sender to receiver
 Protocol: A set of rules that governs the data
communication

Direction of Data Flow
 Communication can be simplex, Half-duplex, or
full-duplex.
 Simplex: communication is
unidirectional
 Half-duplex: bi-directional
but not at the same time
 Full-duplex: bi-directional
and simultaneously.
Any real life
examples?

Networks & Distributed processing
 Interconnection of „Intelligent devices‟ is called a
„computer network‟
 In „Distributed processing‟ a task is divided and
submitted among multiple computers using network
 Network Criteria: to design an effective and efficient
network the most important criteria are
 „Performance‟ depends on
 No of users: large no of users may slow down the „response time‟
due to heavy traffic
 Type of transmission medium: defines the speed at which the data
can travel (speed of light is the upper bound)
 Hardware: A high-speed computer with greater storage provides
better performance
 Software: efficient mechanisms to transform raw data into
transmittable signal, to route the signals, to ensure error-free
delivery etc.

Network Criteria
 Reliability depends on
 Frequency of failure: all networks fail occasionally
 Recovery time: how long does it takes to restore
the service
 Catastrophe: networks should be protected from
fire, earthquake, theft, etc.
 Security depends on
 Unauthorized access should be prevented
 Should be protected from viruses, spywares,
adwares, malwares etc.

Physical Structure
 It refers to the way two or more devices are
attached to a link
 Point-to-Point: provides a dedicated link
between two devices. i.e. entire capacity
of the link is reserved for transmission
between those two devices
 Multi-point: In this configuration more
than two devices share the same link
 If several devices can use the link
simultaneously then called „spatially
shared connection‟
 If devices take turns then it is a time-shared connection
(temporally)

Topology
 Topology of a network is the geometric
representation of the links and nodes of a
physical network.
ETC.

Mesh Topology
 Every device has a dedicated point-to-
point link to every other device
 A fully connected mesh network has
n(n-1)/2 links
 Every device required to have at least
n-1 I/O ports
 Eliminates traffic problem as links are
not shared
 It is robust as breaking one link couldn't defunct the network
completely
 Privacy/security is maintained
 Installation and reconfiguration is difficult due to complicated
connections
 Expensive in terms of cost and space
 Not Difficult to add/remove a device

Star Topology
 Each computer has a point-point
link only to a central controller
called the HUB
 HUB acts as an exchange to send
data from one device to another
 Less expensive than mesh
 It is robust as one link failure causes that device to go out of
the network and it does not affect others
 Easy fault finding
 when one device sending data to another device, all other
devices have to be idle
 however, a switch in place of hub can eliminate this problem

Bus Topology
 Multi-point
 One long cable acts as a
backbone to link all the
devices
 There is a limit on the no of
drop lines (tapes) as in each
tape some energy is lost
 Installation is easy
 It uses less cabling than star or mesh
 difficult reconnection and fault finding
 Adding new device may require modification/replacement of the
backbone otherwise the performance will be degraded
 Fault in bus stops all transmission, the damaged area reflects
signal back in the direction of origin, creating noise in both
directions

Ring Topology
 Point-to-point
 Each device is linked only
to its immediate
neighbours
 To add or remove a
device requires moving
two connections only
 Each device in the ring incorporates a repeater to regenerate a
signal before passing to neighbour.
 Easy to install and reconfiguration
 Maximum ring length and no of devices are fixed
 failure of one device causes network failure if not bypassed
 unidirectional data traffic

Category of networks
 The networks may be categorized according to
its size, ownership, distance it covers and its
physical architecture.

Local Area Network(LAN)
 LAN is a privately owned
networks within a single
building or campus
 Size is restricted? (10m-1KM)
 Common LAN topologies are
bus, ring, star
 Speed is high (100Mbps – 1 Gbps)
 These are designed to share resources (hardware/software)
between personal computers or workstations
 the size is restricted as the H/w will not work correctly over
wires that exceed the bound as electrical signal becomes weaker
over distance due to resistance.
 Also the delay increases as the distance, but LANs are designed
for specific delays?

Figure 1.13 LAN (Continued)
Example: LAN of an organisation

Metropolitan Area Network(MAN)
 MAN is designed to extend
over an entire city
 It may be either
private(cable TV, Bank
ATMs), or public
(Telephone)
 May be a single network like cable TV or may be a means of
connecting a number of LANs into a larger network so that the
resources may be shared
 It forms the basic long distance connection in a large network &
technologies that provide high speed digital access to individual
homes & business
 Also sometimes called the access network, as it provides access
to various services, say cable TV, Internet etc.

Metropolitan Area Network(MAN)
 It utilizes public, leased or private communication devices
 The end systems are connected to subnets, which are intelligent
entities and contains communication channels and routers
 A WAN wholly owned by a single company is called an
„enterprise network „
 speed is less than LANs
 WAN provides long distance
transmission of data, voice,
image, and video information
over large geographical
areas that may comprise a
country, a continent or even
the whole world

A metropolitan area network based on cable TV.

The Internet
 It is a specific world wide network (i.e. A network of networks)
that interconnects millions of computing devices throughout the
world
 Computing devices include
 PCs, UNIX based workstations, servers(?)
 PDAs, TVs, Mobile computers, automobiles, Toaters, …
 End systems are connected either directly by „communication
links‟ or indirectly by intermediate switching devices called
„switches/Routers‟
 Communication links include
 Coaxial cable, copper wire, fiber optics, radio spectrum
 Different communication links can transmit data at different
speeds. The link transmission rate is called „bandwidth‟
 Switches/Routers receives a chunk of information (called a
packet) and forwards it towards destination

Internet Today
 It is difficult to give an accurate representation of the Internet
as it is continuously changing
 It is represented in form of hierarchy of Service providers
 International Service Providers
 That connect nations together
 National Service Providers
 Are backbone networks created and maintained by specialized companies
like SprintLink, PSINet, etc
 Theses networks are connected by complex switching stations called
Network Access Points (NAPs)
 Regional Service Providers
 Are smaller ISPs that are connected to one or more NSPs
 Local Service Providers
 Provide direct service to end users, may be connected to regional ISPs or
directly to NSPs

Internet today
History of Internet
- read yourself
(page 15, sec 1.3)

Services provided by Internet
 The www including browsing & internet commence
 E-mail including attachment
 Instant messages
 Peer-to-peer file sharing
 VOIP
 Online Games
 Tele Conferencing
 Video-on-demand
 Remote Login (SSH client, Telnet) etc…
 Remote file transfer
 . . .

Protocol !!!
 What is a Protocol?
 What does a protocol do?
 How would you recognize a protocol if you
met one?
A Human Analogy
 What you do when you want to ask some one
for the time of day?

Protocol
 First you offer a
greeting (Hi )
 The typical response to
a Hi is a returned Hi
 This response is an
indication that you can
proceed and ask for the
time
 And the conversation
continues . . .

Protocol
 But what happens when a different response comes to
the initial Hi like
 Don’t bother me! OR
 I don’t speak English OR
 Some unprintable reply! OR
 No response at all !!!
 Then human protocol would be not to ask for the time
of day
 In our human protocol, there are specific messages
we send, and specific actions we take in response to
the received reply messages

Protocol
 If people run different protocols! Say
 If one person has manners and other does not
 If one understands concept of time other does not
 Then protocols do not interoperate and no
useful work can be accomplished.
 The same is true in networking – It takes two
(or more) communicating entities running the
same protocol in order to accomplish a task
 But the exception is that the entities
exchanging messages and taking action are
Hardware and/or Software components of
some device

A Network Protocol
 Visiting a Web site
 Type in the URL in Web
browser
 First your computer will
send a connection request
message to the Web Server
 Web Server will respond by
returning a connection reply
message
 Your computer then sends
the name of the web page
 Finally the server returns
the page to you.

Defining A Protocol
A Protocol defines the format and the order
of messages exchanged between two or
more communicating entities, as well as the
actions taken on the transmission and/or
receipt of a message of other event.
. . . J. F. Kurose

Protocols contd.
 A protocol defines what is communicated, How
it is communicated, when it is communicated
 The key elements of a protocol are
 Syntax: refers to structure or format of data, i.e. the
order in which they are presented
Example: a date
 Semantics: refers to structure meaning of each
section
 Timing: refers to two characteristics. i. When data
should be sent. ii. How fast they can be sent
 Depends on link availability, and speed of receiver
day Yearmonth
8 8 16

Standards
 The standard provides a model for development that
makes it possible for a product to work regardless of
the individual manufacturer
 Example: A steering wheel of a car from one make may not
feet into other make
 Standards are essential in creating and maintaining an
open and competitive market and guarantees
international inter-operability
 Two categories of standards
 De Facto: that have just happened without any formal plan
 De Jure: are formal, legal standards adopted by some
authorized or officially recognized body

Standards Organizations
 Standards Creation Committees
 International Standards Organization (ISO)
 International Telecommunications Union-Telecommunication standards (ITU-T)
 American National Standards Institute (ANSI)
 Institute of Electrical and Electronics Engineers (IEEE)
 Electronic Industries Association (EIA)
 Forums
 The forums work with universities and users to test, evaluate and the conclusion is
presented to standard bodies to standardize new technologies
 Regulatory Agencies
 Govt. agencies responsible for protecting the public interest.
 Internet Standards
 Internet draft is a working document with no official status and a 6 month life
time.
 If recommended by IETF then a draft may be published as a Request for
Comment (RFC)

Layered Tasks
 The service that we expect from a Computer Network are much more
complex than just sending a signal from one device to another.
 To solve a complex problem we apply the strategy “Divide and Rule”. i.e. the
main problem is divided into some small tasks/ levels of reduced complexity
and then handled individually.
 In other words Each level is responsible to solve a more focused problem of
the original problem is a called layer in network terminology.
 Each layer observes a different level of abstraction and performs some well
defined functions.
 Each layer uses the service of the layer below below it and each layer
provides service to its upper layer.
 There exists an interface between each pair of adjacent layers that defines
the information and services a layer must provide to the adjacent layer.

Example: Sending a letter

Example: The philosopher-translator-secretary architecture.

The Internet model
 The layered protocol stack that is used in
practice is a five ordered layer Internet
model, also called TCP/IP protocol suite
 The responsibility of each layer is well
defined and focused
 Each end user device engaged in communication must have these layers in it
(in form of HW or SW)
 An intermediate device may not have all the layers but at least first three
layers
 Layer x on one device communicates with layer x of other device.
 The processes on each machine that communicate at a given layer are called
peer-to-peer processes.

Peer-to-peer processes

An exchange using the Internet model

Physical layer
 The responsibility of physical layer is to coordinate the functions
required to transmit a bit stream over a physical medium
 The duties are
 Defines the characteristics of the interface between devices and
transmission medium
 Type of transmission medium, topology, etc…
 Representation of bits
 Encoding, voltage level, duration etc…
 Data rate
 Synchronization of bits
 Sender‟s and receiver‟s clock shynchronization

Data link layer
 is responsible for transmitting frames from
one node to the next
 The duties are
 Framing
 Stream of bits received from upper layer is divided into manageable
data units(?) called frame
 Physical addressing
 Adds the address of sender and receiver in the header
 Flow control
 This mechanism helps to prevents overflow at receiving side
 Error control
 Mechanism to detect/correct errors in transmission
 Access Control
 Which device has the control over the link at a given time

Datalink layer contd.
 Physical addressing and hop-hop delivery can
be done in one network only
 If the message is to be passed across the
network then network layer functionality is
required.

Network Layer
 The network layer is responsible for the delivery of packets
from the original source to the final destination possibly
across multiple networks.
 The Duties are
 Logical addressing
 It adds logical addresses into the packet header
 Routing
 Forwarding the packet towards the destination

Source-to-Destination

An Example
sending from a node with network
address A and physical address 10 to a
node with a network address P and
physical address 95
Because the two devices are located on
different networks, we cannot use
physical addresses only;as the physical
addresses only have local jurisdiction.
What we need here are universal
addresses that can pass through the
LAN boundaries. The network (logical)
addresses have this characteristic.

Transport layer
 The transport layer is responsible for delivery of a message
from one process to another.
 The Duties
 Port addressing
 Actual transmission occurs from a specific process on one device to a
process of another.
 Port address (an integer) defines the process/application in a device
 Segmentation and reassembly
 Message received from application layer is divided in to transmittable
segments containing sequence nos

Transport layer contd.
 Connection control
 Two types of connection service is allowed
 Connection oriented: establish the connection, use the connection, release
the connection. (guarantee of delivery)
 Example: telephone
 Connection less: each message carries the destination address and routed
through the system
 Example: postal service
 Flow Control
 Responsible for end-to-end flow control as well as
intermediate flow control (congestion)
 Error Control
 End-to-end error control

Application layer
 The application layer is responsible for providing
services to the user.
 It provides user interfaces and support services
such as email, remote file transfer, remote logins
etc…

Summary of duties

OSI model
 Session Layer is the network dialog controller, It
establishes maintains and synchronizes the interaction
between communicating systems
 Duties are
 Dialog control
 Synchronization at data level
 Presentation layer is concerned with syntax and
semantics of the information exchanged between two
systems
 Duties are
 Translation: converting to bit streams
 Encryption: to ensure privacy
 Compression: increases virtual BW

The Physical Layer
Lecture II
• Signals
• Digital Transmission
• Analog Transmission
• Multiplexing
• Transmission Media

Position of the physical layer

Signals
 Information is transmitted in the form of
electromagnetic signals
 Signals are of two types
 Analog Signal is a continuous signal in which the signal
intensity varies smoothly over time
 Digital Signal is a discrete signal in which the signal intensity
maintains a constant level for some period and then changes
to another constant level.
 Analog Data: human voice, Digital data: data stored in a
computer

Periodic / Aperiodic Signals
Periodic Signal: A signal completes a pattern within a measurable
time frame (period)
The completion of one full pattern is called a cycle. The
period is constant for any given periodic signal
Aperiodic Signal: Changes without exhibiting a pattern
In data communication, we commonly use periodic and analog
signals and aperiodic digital signals
Aperiodic Signal
Periodic Signal

Analog Signals
 The sine wave is the most fundamental form of a
periodic signal
 Represented as s(t)=Asin(2ft+)
 Characterstics
 Amplitude: intensity of signal at any given time
 Frequency: no of cycles/periods in one second, measured in
Hz
 Frequency = 1/Period
 Phase: describes the position of the waveform relative to
time zero
 A complete cycle is 360o = 2
 Wavelength:The distance a signal can travel in one period
  = c/f, c: speed of light

Amplitude Period and frequency

Time and frequency domains
A signal can also be represented in frequency domain

Composite signals
 A single-frequency sine wave is not useful in
data communications; we need to change one
or more of its characteristics to make it useful.
 When we change one or more characteristics
of a single-frequency signal, it becomes a
composite signal made of many frequencies.
 A composite signal is composed of multiple
sine waves called harmonics

Example : A Square wave
 According to Fourier analysis, this signal can be
decomposed in to a series of sine waves i.e.
 f is called fundamental frequency
 3f is third harmonic, and 5f 5th harmonic
 To recreate the complete square wave requires
all the odd harmonics upto infinity
...])5(2sin[
5
4
])3(2sin[
3
4
2sin
4
)(  tf
A
tf
A
ft
A
ts 






Three harmonics

Frequency spectrum
The Signal using the
frequency domain and
containing all its
components is called the
frequency spectrum of
that signal
 The range of frequencies that a medium can pass is called its Bandwidth
 The bandwidth is a property of a medium: It is the difference between
the highest and the lowest frequencies that the medium can satisfactorily
pass.

Example
A signal has a spectrum with frequencies between 1000 and
2000 Hz (bandwidth of 1000 Hz). A medium can pass
frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz).
Can this signal faithfully pass through this medium?
Solution
The answer is definitely no. Although the signal can have the
same bandwidth (1000 Hz), the range does not overlap. The
medium can only pass the frequencies between 3000 and 4000
Hz; the signal is totally lost.

Digital Signals
 Digital signals can be better described by two terms
 Bit interval: time required to send a single bit
 Bit rate: number of bit intervals in one second
 A digital signal is a composite signal having an infinite
number of frequencies i.e. infinite bandwidth
 The digital BW is bits per sec (bps)

Analog vs Digital
• Channels or links are of two types
• low-pass: lower limit is zero and
upper limit is any frequency ()
• band-pass: has a band width with
frequencies f1and f2
 A digital signal theoretically needs a BW between o and 
 if the upper limit will be relaxed than digital transmission can use a low-pass
channel
 An analog signal has a narrower BW with frequencies f1and f2
 Also BW of analog signal can be shifted, i.e. f1and f2 can be shifted to f3 and
f4
Analog signal can use a band-pass channel

Data rate limits
 Data rate depends on
 The BW available
 The levels of signal that can be used
 The quality of channel (i.e. the level of noise)
 Nyquist Bit rate: noise less channel
 Bit rate= 2  BW  lg L
 For a noise less channel the nyquist bit rate defines the
theoretical maximum bit rate
 BW: band width of channel, L: no of signal levels used to
represent data
 Shannon Capacity: noisy channel
 Capacity = BW  lg (1+SNR)
 The signal-to-noise ratio is the statistical ratio of power of
the signal to the power of the noise

Signal to noise ratio
 SNR=Avg. Signal Power/Avg. Noise Power
 SNRdb = 10 log10 SNR
 Example:
 SNRdb=36, BW=2MHz, Find C
 SNR=10SNRdb/10
 C = B log2 (1+SNR) = 24Mbps

Example
We have a channel with a 1 MHz bandwidth. The SNR for this
channel is 63; what is the appropriate bit rate and signal level?
Solution
C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
6 Mbps = 2  1 MHz  log2 L  L = 8
First, we use the Shannon formula to find our upper limit.

Signal
 Data rate depends on
 The BW available
 The levels of signal that can be used
 The quality of channel (i.e. the level of noise)
 Nyquist Bit rate: noise less channel
 Bit rate= 2  BW  lg L
 For a noise less channel the nyquist bit rate defines the
theoretical maximum bit rate
 BW: band width of channel, L: no of signal levels used to
represent data
 Shannon Capacity: noisy channel
 Capacity = BW  lg (1+SNR)
 The signal-to-noise ratio is the statistical ratio of power of
the signal to the power of the noise

Transmission Impairment
 In practice the signal sent at sending end using
a transmission medium is not exactly same at
receiving end due to some impairments
 Attenuation: loss of energy
 Decibel: is the unit to measure the relative strength
of two signals
 dB = 10 log (P1/P2)
 It is negative if attenuated and +ve if amplified

Distortion
 Signal changes its forms at the receiving end
 It is normally happens in case of composite
signals
 As each signal component has its own
propagation speed thus received out of phase

Noise
 Several types of noise such as
 thermal noise: random motion of electrons in a wire
 induced noise: sources such as motors and elecrical
appliances
 cross talk: effect of one wire over the other
 impulse noise: is a spike may corrupt the original
signal that comes from power lines and lightning

More terminologies
 Throughput: number of
bits passed per second at a
given point
 Propagation Delay: the
time required for a bit to
travel from one point to
another
 Wavelength: is the
distance a signal can travel in
 = c / f

Digital Transmission
Line coding
Block Coding
Sampling
Transmission Mode

What is Line Coding?
 Is the process of converting binary data (a
sequence of bits) to a digital signal

Signal Level versus Data Level
 No of values allowed in a signal
 No of values used to represent data

DC Component
 A component having zero frequency
 Can‟t be passed through a transformer
 Energy consumed is useless

Pulse Rate versus Bit Rate
 No of pulses per second
 Minimum amount of time required to transmit a symbol
 No of Bits per second
 If a pulse carries one bit then pulse rate and bit rate are
same
Example
A signal has two data levels with a pulse duration of
1 ms. We calculate the pulse rate and bit rate as
follows:Pulse Rate = 1/ 10-3= 1000 pulses/s
Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps

 No Synchronization: if receivers clock is faster
 A Signal that includes timing information along
with data is called a self-synchronizing signal
 i.e. transitions in the signal alerts the receiver to
reset the clock
Self Synchronization

Example
In a digital transmission, the receiver clock is
0.1 percent faster than the sender clock. How
many extra bits per second does the receiver
receive if the data rate is 1 Kbps? How many
if the data rate is 1 Mbps?
Solution
At 1 Kbps:
1000 bits sent 1001 bits received1 extra bps
At 1 Mbps:
1,000,000 bits sent 1,001,000 bits received1000 extra bps

Line Coding Schemes

Unipolar encoding uses only one voltage
level.
Note:
UniPolar Encoding

Unipolar Encoding
 One is coded as +ve voltage
 Zero is coded as –ve voltage

Polar encoding uses two voltage levels
(positive and negative).
Note:
Polar Encoding

Polar Encoding
 Avarage voltage level is decreased
 DC component problem is avoided
 Four Important type of polar encoding are:
There are many others also!

In NRZ-L the level of the signal is
dependent upon the state of the bit.
Note:
NRZ-L Encoding

In NRZ-I the signal is inverted if a 1 is
encountered.
Note:
NRZ-I Encoding

NRZ Encoding
 Loss of synchronization incase of continuous
ones or zeros

RZ uses three values i.e. +ve, zero & -ve
Signal change occurs during each bit
Note:
RZ Encoding

RZ Encoding
 A +ve voltage means 1 and –ve voltage means
zero.
 But signal returns to zero at mid of the bit
interval

RZ is a good encoded digital signal that contain
a provision for synchronization.
But it requires two signal changes to encode 1
bit  more bandwidth!
Note:
RZ Encoding

In Manchester encoding, the transition at
the middle of the bit is used for both
synchronization and bit representation.
Note:
Manchester Encoding

Manchester Encoding
 It achieves the synchronization but with two levels of
amplitude
 Datarate(R) = 1/tb , tb: bit duration in seconds
 Modulation rate (D) = R/b, b: no of bits per signal
element

In differential Manchester encoding, the
transition at the middle of the bit is used
only for synchronization.
The bit representation is defined by the
inversion or noninversion at the
beginning of the bit.
Note:
Diff-Manchester Encoding

Diff-Manchester Encoding
 Manchester Encoding used for 802.3 base band
– CSMA/CD Lans
 Diff-Manchester is used foe 802.5 token ring
LAn

In bipolar encoding, we use three levels:
positive, zero,
and negative.
Note:
Bipolar Encoding

Bipolar Encoding

2B1Q Encoding
 Two Binary One Quaternary
 Each pulse represents 2 bits
-1
-3

MLT-3 Encoding
 Multi transmission, three level (MLT-3)
 The signal transition from one level to the next
at the beginning of a 1 bit

 To ensure synchronization some
redundant bits may be introduced
Steps in Transformation
 Division
 Substitution
 Line Coding
Block Coding

Block Coding

Substitution

4B/5B Encoding
 Each 4-bit 'nibble' of received data has an extra
5th bit added.
 If input data is dealt with in 4-bit nibbles there
are 24 = 16 different bit patterns. With 5-bit
'packets' there are 25 = 32 different bit
patterns.
 As a result, the 5-bit patterns can always have
two '1's in them even if the data is all '0's a
translation.
 This enables clock synchronizations required
for reliable data transfer.

Data Code Data Code
0000 11110 1000 10010
0001 01001 1001 10011
0010 10100 1010 10110
0011 10101 1011 10111
0100 01010 1100 11010
0101 01011 1101 11011
0110 01110 1110 11100
0111 01111 1111 11101
4B/5B encoding

Example 8B/6T
 sends 8 data bits as six ternary (one of three voltage
levels i.e. +, 0, -) signals.
 Each bit block of 8-bit group with a six symbol code
 i.e. 8 bit  28 & six symbol 36 possibilities
 i.e. the carrier just needs to be running at 3/4 of the
speed of the data rate.
 Helps to maintain synchronization and error checking

Pulse Amplitude Modulation
 Generates a series of pulses by sampling a
given analog signal
 Sampling is measuring amplitude in equal intervals

Pulse amplitude modulation has some
applications, but it is not used by itself in
data communication. However, it is the
first step in another very popular
conversion method called
pulse code modulation.
Note:
PAM

PCM: Quantization
 It is a method of assigning integral values in a
specific range to sampled instances

Binary encoding
 Each quantized value is translated into a 7bit
binary equivalent.
 The eighth bit indicates the sign

Line coding
 The binary digits are transformed to a digital
signal by using one of the line coding
techniques.

Analog to PCM Digital Code

According to the Nyquist theorem, the
sampling rate must be at least 2 times the
highest frequency.
Note:
Sampling rate
 Accuracy of reproduction depend on the no of
samples taken
 What should be the sampling rate?

Nyquist Theorem

Example
What sampling rate is needed for a signal with a
bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
Solution
The sampling rate must be twice the highest frequency in the
signal:
Sampling rate = 2 x (11,000) = 22,000 samples/s

Example
A signal is sampled. Each sample requires at least
12 levels of precision (+0 to +5 and -0 to -5). How
many bits should be sent for each sample?
Solution
We need 4 bits; 1 bit for the sign and 3 bits for the value.
A 3-bit value can represent 23 = 8 levels (000 to 111), which is
more than what we need.
A 2-bit value is not enough since 22 = 4.
A 4-bit value is too much because 24 = 16.

Example
We want to digitize the human voice. What is the bit
rate, assuming 8 bits per sample?
Solution
The human voice normally contains frequencies from 0 to
4000 Hz.
Sampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample
= 8000 x 8 = 64,000 bps = 64 Kbps

Transmission mode

 Information is organized into group of bits
 All bits of one group are transmitted with each clock tick
from one device to other
 More speed
 Cost is high restricted to short distance
Parallel Transmission

Serial Transmission
 One bit follows another using same line
 Reduced cost (by a factor n)
 Parallel/serial converter required
 May used for large distance

In asynchronous transmission, we send 1
start bit (0) at the beginning and 1 or
more stop bits (1s) at the end of each byte.
There may be a gap between each byte.
Note:
Asynchronous Transmission
 Serial transmission occurs in one of the two
ways

 Insertion of extra bits & a gap makes it slower
 But cheap and effective
 Suitable for low speed communication like KB to
computer. i.e. typing is done one character at a time
and unpredictable gap between characters.

Asynchronous here means “asynchronous
at the byte level,” but the bits are still
synchronized; their durations are the
same.
Note:
 When receiver detects a start bit, it starts a timer and
begins counting
 After receiving a stop bit it ignores all pulses till next
start bit arrives and resets the timer

In synchronous transmission,
we send bits one after another without
start/stop bits or gaps.
It is the responsibility of the receiver to
group the bits.
Note:
Synchronous Transmission

Synchronous Transmission
 More speed
 Synchronization is necessary
 Accuracy is completely dependent on the
ability of the receiving device to keep an
accurate count of the bits as they come in
 Byte synchronization is done in datalink layer

Modulation of Digital Data
Digital-to-Analog Conversion
Amplitude Shift Keying (ASK)
Frequency Shift Keying (FSK)
Phase Shift Keying (PSK)
Quadrature Amplitude Modulation
Bit/Baud Comparison
Analog Transmission

Digital to analog modulation
It is Needed if the transmission line is analog but the data
produced is binary.
Example: sending data from a computer via a public access
telephone line

Bit rate is the number of bits per second. Baud
rate is the number of signal units per second.
Baud rate is less than or equal to the bit rate.
Note:
Bit rate / Baud rate
The sending device produces a signal that acts as a basis
of information signal called carrier signal or carrier
frequency
The digital information is then modulates the carrier signal
by modifying one or more of its characteristics.

Example
An analog signal carries 4 bits in each signal unit. If
1000 signal units are sent per second, find the baud
rate and the bit rate
Solution
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps
Example
The bit rate of a signal is 3000. If each signal
unit carries 6 bits, what is the baud rate?
Solution
Baud rate = 3000 / 6 = 500 baud/s

Amplitude Shift Keying
• The intensity of the signal is
varied to represent binary one
or zero
• ASK is highly susceptible to
noise interference, i.e a zero
may be changed to 1 or vice
versa
• If one of the bit values is represented by no
voltage then it is called on/off keying (OOK). It
results in reduction of energy transmitted.
• ASK modulated signal contains many simple
frequencies
• band width is given by BW=(1+d) Nbaud
• Where Nbaud is the baud rate and d is a factor
of modulation with minimum value=0

Example
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw
the full-duplex ASK diagram of the system. Find the carriers
and the bandwidths in each direction. Assume there is no
gap between the bands in the two directions.
Solution
For full-duplex ASK, the bandwidth for each direction is
BW = 10000 / 2 = 5000 Hz
The carrier frequencies can be chosen at the middle of each band
fc (forward) = 1000 + 5000/2 = 3500 Hz
fc (backward) = 11000 – 5000/2 = 8500 Hz

Frequency Shift Keying
 Frequency of carrier signal
varies to represent a binary
1 or 0
 Effect of noise is less,
receiving device ignores
spikes but more Bandwidth
is required
 Although there are two
carrier frequencies, the
process of modulation
produces a composite signal
 Bandwidth = fc1 – fc0 + Nbaud

Example
Find the maximum bit rates for an FSK signal if the
bandwidth of the medium is 12,000 Hz and the
difference between the two carriers is 2000 Hz.
Transmission is in full-duplex mode.
Solution
Because the transmission is full duplex, only 6000 Hz is
allocated for each direction.
BW = baud rate + fc1 - fc0
Baud rate = BW - (fc1 - fc0 ) = 6000 - 2000 = 4000
But because the baud rate is the same as the bit rate, the bit
rate is 4000 bps.

Phase Shift Keying
 Phase of carrier signal varies
to represent a binary 1
(180o)or 0 (0o) also called 2-
PSK or binary PSK
 Avoids problems of noise and
bandwidth
 Can be represented in a
constallation diagram or
phase-state diagram
 BW=same as of ASK
 More variations in phase may
be added to represent more
than one bit

Other variations of PSK
4-PSK / Q-PSK, 2 bits per baud
8-PSK, 3 bits per baud
i. The bit rate increases as
compared to baud rate
ii. But needs sophisticated
devices to distinguish small
difference in phase

QAM is a combination of ASK and PSK
so that a maximum contrast between each
signal unit (bit, dibit, tribit, and so on) is
achieved.
Note:
Quadrature Amplitude Modulation

4-QAM & 8-QAM Constellation

16-QAM constellations
QAM is less susceptible to noise than ASK?
Bandwidth required for QAM is same as PSK and ASK

Bit/Baud Comparison

A telephone line has a bandwidth of almost 2400 Hz for data
transmission.
Modem Standards
Modem stands for modulator/demodulator.

Modulation/Demodulation
 A modulator creates a band-pass
signal from binary data.
 A demodulator recovers the
binary data from the modulated
signal

V series modems
V.32 constellation & BW
• published by ITU-T
• it uses a technique called trellis coded modulation I.e. QAM plus
one redundant bit
• 32 QAM with a baud rate of 2400 and datarate is
2400*4=9600kbps (1 bit redundant)

V.32bis constellation & BW
 Uses 128-QAM (7 bits/ baud
with 1 bit for error control)
 datarate (2400*6)=14400 bps
V.90
 Asymetric modems, i.e. downloading speed is 56 kbps and
uploading speed is 33.6 kbps
 This is possible if one party is using digital signaling
V.92
 can adjust their speed I.e. if noise allows than it can upload at a
rate of 48 Kbps
 Additional features like modem can interrupt internet connection
for a incoming phone call etc.

Traditional modems
56 K Modems
• Sampled, digitized and
at telephone comp
• The quantization noise
introduced thus data rate is
limited according to
shannon capacity i.e. 33.6k
• signal not affected by
quantization noise and not
limited by shannon capacity
• sampling is done at a rate
of 8000 samples/sec with 8 bits
per sample.
• One bit is used for control thus
speed becomes 8000*7=56 kbps

Modulation of Analog Signals
Methods:
Amplitude Modulation (AM)
Frequency Modulation (FM)
Phase Modulation (PM)
• Representation of analog information by an analog signal
• i.e. shifting the center frequency of baseband signal up to
the radio carrier
• It is needed because
• To reduce Antenna length (length  1/f)
• helps in frequency division multiplexing
• To support medium characteristics

Amplitude modulation
• The carrier signal is modulated so that
its amplitude varies with the changing
amplitude of modulating signal
• Phase and frequency remains the same
• The modulating signal becomes an
envelope to the carrier
• The bandwidth of an AM signal is twice
the bandwidth of the modulating signal
• BWt = 2  BWm
• BWt is total bandwidth
• BWm is bandwidth of modulating signal

Frequency modulation
• The carrier signal is modulated so
that its frequency varies with the
changing amplitude of modulating
signal
• Phase and peak amplitde remains
the same
•The bandwidth of an AM signal is
ten times the bandwidth of the
modulating signal
• BWt = 10  BWm
• BWt is total bandwidth
• BWm is bandwidth of modulating
signal

The Physical Layer contd.
Lecture III
• Multiplexing
• Transmission Media
• Switching

Multiplexing
 It is not practical to have a separate line for each other device
we want to communicate
 Therefore, it is better to share communication medium
 The technique used to share a link by more than one device is
called multiplexing
 Multiplexing needs that the BW of the link should be greater
than the total individual BW of the devices connected.
 In a multiplexed system one link may contain more than one
channel

Categories of multiplexing

Frequency Division Multiplexing
 FDM is an analog
multiplexing technique
that combines signals
 Signals generated by
each device modulate
different carrier
frequencies
 These modulated
signals are combined to
form a composite
signal
 Demultiplexer uses a
series of filters to
decompose the signal
into its component
signals

FDM
• Carrier frequencies are separated by sufficient BW to
accommodate modulated signal
•These BW ranges are channels through which the various
signal travel
• Channels must be separated by strips of unused BWs
(called Guard Bands) to prevent signals from overlapping
• Carrier frequencies must not interfere with the original
signals
f
t

Example 1
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 without the use of
guard bands.
Solution
Shift (modulate)
each of the three
voice channels to
a different
bandwidth, as
shown in Figure

Example
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 x 100 + 4 x 10
= 540 KHz
as shown in Figure

Example
Four data channels (digital), each transmitting at 1 Mbps, use
a satellite channel of 1 MHz. Design an appropriate
configuration using FDMSolution
• 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 are modulated to 1 Hz.
• One solution is 16-
QAM modulation.
• Figure shows one
possible configuration.

Analog hierarchy

Wave Division Multiplexing
 Very narrow bands of light
from different sources are
combined to make a wider
band of light
 A prism is used to bend a beam of light based on the angle of
incidence and frequency and acts like a multiplexer
 Another prism may be used to reverse the process and acts like
a demultiplexer

Time division Multiplexing
 Each shared connection occupies a portion of time but
uses full BW f
t
 The data flow of each connection is
divided into units
 For n input connections, a frame is
organised into a minimum of n units
 Each slot
carrying one
unit from each
section
 Data rate of the
link has to be n
times the data
rate of one unit

Time division Multiplexing contd.
 If the data rate of a link is 3 times the data rate of a
connection
 then the duration of a unit on a connection will be 3
times that of a time slot

Example
Four 1-Kbps connections are multiplexed together. A unit is
1 bit. Find (1) the duration of 1 bit before multiplexing, (2)
the transmission rate of the link, (3) the duration of a time
slot, and (4) the duration of a frame?
Solution
1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms).
2. The rate of the link is 4 times the rate of connection, i.e. 4
Kbps.
3. The duration of each time slot is 1/4 th of the bit duration
before multiplexing i.e. 1/4 ms or 250 ms.
or inverse of data rate i.e. 1/4 Kbps = 250 ms.
4. The duration of a frame is same as duration of each unit,
i.e. 1 ms.
or 4 times the bit duration i.e. 4 * 250 ms = 1ms

Example
Four channels are multiplexed using TDM. If each channel
sends 100 bytes/s and we multiplex 1 byte per channel,
show the frame traveling on the link, the size of the frame,
the duration of a frame, the frame rate, and the bit rate for
the link.
Solution

Example
A multiplexer combines four 100-Kbps channels using a time
slot of 2 bits. Show the output with four arbitrary inputs.
What is the frame rate? What is the frame duration? What is
the bit rate? What is the bit duration?
Solution

Synchronization
• Synchronization between multiplexer and demultiplexer is
important otherwise a bit of one channel may be received by
other channel
• To avoid this one or more synchronization bits may be added
called Framing bits

Example
We have four sources, each
creating 250 characters per
second. If the interleaved unit
is a character and 1
synchronizing bit is added to
each frame, find
(1) the data rate of each source,
(2) the duration of each character
in each source,
(3) the frame rate,
(4) the duration of each frame,
(5) the number of bits in each
frame, and
Solution
1. The data rate of each source
is 2508=2000 bps
2. The duration of a character
is 1/250 s, or 4 ms.
3. The link needs to send 250
frames per second.
4. The duration of each frame
is 1/250 s, or 4 ms.
5. Each frame is 4 x 8 + 1 = 33
bits.
6. The data rate of the link is
250 x 33, or 8250 bps.

Bit Padding
 If one or more devices are faster than other
devices than faster devices are given more time
slots than others
 e.g. we can accommodate a device 5 times
faster than others by giving time slots as 5:1
 When speeds are not integer multiples of each
other then bit padding is used
 In bit padding the multiplexer adds extra bits to
device‟s source stream to force the speed
relationships as integer multiples

Example 9
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 frame duration is 1/100,000 s, or 10 ms.
The bit rate is 100,000 frames/s x 3 bits/frame, or 300 Kbps.

DS hierarchy
Telephone companies implement TDM through hierarchy of
digital signals called Digital Signal service

T-1 line for multiplexing telephone lines
o Digital Signal services are implemented by T Lines (T-1 to T-4)
o T Lines are digital lines designed for transmission of digital
data, audio or video

T-1 frame structure
• The frame used on a T-1 line is usually 193 bits divided into 24
slots of 8 bits each plus 1 extra bit for synchronization (24*8 + 1)
• If a T-1 line carries 8000 frames then data rate = 193*8000 =
1.544 Kbps

E Line
Rate
(Mbps)
Voice
Channels
E-1 2.048 30
E-2 8.448 120
E-3 34.368 480
E-4 139.264 1920
• Europeans use E Lines in place T Lines. Both are conceptually
same only capacity differs

Multiplexing and inverse multiplexing
• Inverse multiplexing takes data from high speed line and breaks
it into portions that can be sent across several lower speed lines
• If an organisation wants to send data, audio and video, each
requires a different bandwidth
• using an agreement called Bandwidth on Demand
• The organisation can use any of the channel whenever and
however it needs them

Transmission Media
 Signals in the form of electromagnetic energy is
propagated through transmission media from
one device to another device
 A selected portion of electromagnetic spectrum
are currently usable for telecommunication like
Power, radio waves, infrared, visible light, ultra-
violate, and X, gamma and cosmic rays etc.

Classes of transmission media

Guided Media
 Provides a conduit from one device to another,
includes
 Twisted-Pair Cable
 Consists of two conductors, each with its own plastic
insulation, twisted together
 Due to twists, the noise interference and crosstalk affects
both wires equally thus cancels each other
 i.e. no of twists per unit length determines the quality of the
cable; more twists mean better quality

Unshielded vs Shielded Twisted-Pair Cable
 STP has a metal foil or braided-mesh covering that
encases each pair of insulated conductor
 Metal casing improves mechanical strength, prevents
penetration of noise or cross talk but is bulkier and more
expensive
 STP is produced by IBM and seldom used else where.
 EIA developed standards for UTP in 7 categories

Categories of Unshielded Twisted-Pair cables
Category Bandwidth Data Rate Digital/Analog Use
1 very low < 100 kbps Analog Telephone
2 < 2 MHz 2 Mbps Analog/digital T-1 lines
3 16 MHz 10 Mbps Digital LANs
7 (draft) 600 MHz 600 Mbps Digital LANs

UTP Contd.
 RJ-45 (Registered-Jack)is
used for 4-pair UTP cable
 UTP can pass a wide range
of frequencies
 Performance is measured
as attenuation versus
frequency and distance
 Attenuation is measured as
decibels per mile and is
increased sharply after
100KHz

Coaxial Cable
 It can carry higher frequency ranges
than UTP
 The outer metallic wrapping serves
both as a shield against noise and
as the second conductor
 These cables are categorized by
their radio government (RG) ratings
 These are categorized according to
gauge of wire, thickness and type of
insulation, construction of the shield
and size of type of outer casing
Category
Impedan
ce
Use
RG-59 75 W Cable TV
RG-58 50 W
Thin
Ethernet
RG-11 50 W
Thick
Ethernet

Coaxial Cable contd.
 BNC connectors are
used(Bayone-Neill-Concelman)
 BNC connector is used to connect
end of the cable to a device
 BNC-T is used in ethernet
 BNC terminator is used at the end
of the cable
 Attenuation is much higher than
the UTP
 Frequent use of repeaters is
needed to avoid attenuation

Fiber-Optic cables
 Transmits signals in the form of visible light
 It uses the refraction property of light for transmission
 i.e. light travels in a straight line in an uniform
medium and changes the direction when passes from
one medium to another having different density
Core: glass or plastic, cladding: covering with less dense glass or plastic

Propagation modes
Current technology allows two modes of propagating
light along optical channels
Multimode: multiple beams
Single mode: single focused beam

Mechanism
 Distortion is less as compared to step-index as distance traveled is less
and received time variation is less
 Single Mode:
 Uses focused source of light and step-index fiber having small diameter
 Propagation of beams is almost horizontal
 Multimode step index:
 The density of core remains constant
from core center to edges.
 Light moves in a straight line and
reflects back from edge
 Distortion is more as various rays
received at different times
 Multimode graded index:
 The density of core varies (decreases)
from core center to edges.
 Light undergoes a series of refraction

Fiber Optics contd. Type Core
Clad
ding
Mode
50/125 50 125
Multimode,
graded-index
62.5/125 62.5 125
Multimode,
graded-index
100/125 100 125
Multimode,
graded-index
7/125 7 125 Single-mode
 Optical fibers are defined by
the ratio of their diameter of
their core to cladding
 Cable composition
 Outer jacket is made of either
PVC or teflon
 Inside the jacket are Kevlar
strands to strengthen the cable
 Below the Kevlar another plastic
coating is there
 The fiber is at the center of the
cable, and it consists of cladding
and the core

Fiber Optics contd.
 It uses three different types of
connectors
 Subscriber channel(SC) connector
used in cable TV with a push/pull
locking system
 Straight Tip (ST) connector is used
for connecting cable to networking
devices with a bayonet locking
system
 MT-RJ is a new connector with same
size as RJ-45
 Attenuation is flatter than TP and
coax thus less no of repeaters are
needed to transmit(10 times less)

Advantages and Disadvantages
Adavntages
 Higher Bandwidth
 BW is not limited by medium but by signal generation and reception
 Less Signal Attenuation
 Can run 50 KM without regeneration
 No electromagnetic interference
 Resistance to corrosive materials
 Light weight
 Tapping is difficult
Disadvantages
 Installation and Maintenance
 Unidirectional (two fibers needed to make it bi-directional)
 Cost

Unguided Media
 It transports electromagnetic waves without using a
physical conductor called Wireless Communication
 Unguided signals can travel from source to destination
in several ways

Radio and microwaves of Electromagnetic spectrum is divided into 8 ranges
Band Range Propagation Application
VLF 3–30 KHz Ground Long-range radio navigation
LF 30–300 KHz Ground
Radio beacons and
navigational locators
MF 300 KHz–3 MHz Sky AM radio
HF 3–30 MHz Sky
Citizens band (CB),
ship/aircraft communication
VHF 30–300 MHz
Sky and
line-of-sight
VHF TV,
FM radio
UHF 300 MHz–3 GHz Line-of-sight
UHF TV, cellular phones,
paging, satellite
SHF 3–30 GHz Line-of-sight Satellite communication
EHF 30–300 GHz Line-of-sight Long-range radio navigation

Wireless transmission waves
 Wireless transmission is broadly divided into three groups
 Radio Wave: Between 3KHz to 1GHz, omni directional, can travel long
distance thus making suitable for log-distance broadcasting like AM
radio, FM radio, TV, cordless phones etc.
 Low and medium frequencies can penetrate walls, uses omni directional
antennas, high interference
 Microwave: Ranging from 1 and 300GHz, unidirectional, low interference
uses unidirectional antennas with line-of-Sight (LOS) propagation
 Very high frequency microwaves cannot penetrate walls, used for long
distance transmission, cellular phones, wireless LANs, two types:
terrestrial microwave and satellite microwave
 Infrared: frequencies from 300GHz to 400THz, can be used for very
short range communication, cannot penetrate walls, confined to one
room only(remote control of TV), no licensing required
 May be used to communicate between devices such as keyboards, mice,
PCs, printers, handset, PDAs etc.

Antennas
Unidirectional Antenna
Radiation and reception of electromagnetic waves
Coupling of wires to space for radio transmission
It works as an adapter between a guided and unguided
media

Switching
 To connect multiple devices over a distance we
adopt a method called switching
 Switches are hardware and/or software devices
capable of creating temporary connections as
per requirements
 A switched network consists of a series of
interlinked switches
 Switching Methods
 Circuit switching
 Packet switching

Circuit Switching
 It creates a direct physical connection between two
devices i.e. it establishes a physical circuit before
transmission
 It uses a device
with n I/P s and
m O/Ps
 Circuit Switching Techniques
 Space Division Switches
 Crossbar switch, multistage switch
 Time division switches
 Time Slot Interchange, TDM Bus

Crossbar switch
 It connects n I/Ps and m O/Ps in a grid
 Each cross point consists of a electronic switch
 The order of switch required is huge O(nm)
 It is impractical because of the size of the crossbar
 It is also inefficient because in practice 25% of the
switches are used at a given time

Multistage switch
 Uses crossbar switches in several stages
 The design of multistage switch depends on the no of stages
and the no of switches required in each stage
 Number of outputs in one stage=number of switches in the
next stage
 The number of cross points required is much less than a
crossbar switch
 The reduction in the number of cross points results in blocking.
i.e. one input is blocked to connect to a output due to
unavailability of a path

Time Division Switches
 It uses time division multiplexing to achieve switching
 Time Slot Interchange(TSI)
 It changes ordering of slots based on desired connections
 It consists of RAM with several memory location
 Size of each location is same as size of time slot
 TSI fills up incoming data inorder of reception
 Slots are sent out in an order based on the decission of
control unit

TDM Bus
 In this case the I/P and
O/P are connected to a
high speed bus through
input output gates
 Each input gate is closed
during the time slots and
only one output gate is
closed.
 The controlling unit
decided which switches
are to be closed

TDM Bus
 Space division switches have no delay and time division
switches requires cross points
 Combining both technologies will result in switches that are
optimised both in physically (no of components) and temporally
(delay)
 It can be designed as TST, TSST, STTS, etc.

Telephone Network
 Telephone network is made of three major components: local
loops, trunks, and switching offices
 Local loop: that connects the subscriber telephone to the nearest
end office or local central office
 Trunk: transmission media that handle the communication
between offices, normally handles hundreds or thousands of
connections through multiplexing
 Switching Office: A switch connects several local loops or trunks
and allows a connection between different subscribers.

Making a Connection
 Accessing the switching station at the end offices is
accomplished through dialing
 In case of rotary dialing a digital signal is sent to the end office
 In case of touch-tone technique two analog signals are sent to
the end office, depending on the row and column of the switch
position.
 e.g. for 8, the signals 852Hz and 1336Hz are sent

Voice communication used analog signals
in the past, but is now moving to digital
signals. On the other hand, dialing started
with digital signals (rotary) and is now
moving to analog signals (touch-tone).
Note:

Packet Switching
 Circuit switching are best suited for voice
communication, as data communication are bursty in
nature i.e. data transmitted in blocks with gaps
between them
 A circuit switched link assumes a single data rate for
both devices
 In Circuit switching all transmissions are equal, priority
base communication is not allowed
 In Packet switching data transmitted in discrete units
called packets
 There are two approaches for packet switching
 Datagram approach, and Virtual Circuit approach

Datagram Approach
 In this approach each packet treated independently called
datagrams
 Each datagram contains appropriate information about the
destinations and the network carries the datagrams towards
destination
 Datagrams may reach at destination out of order
 The links joining each pair of nodes may contain multiple
channels. Each of these channels is capable of carrying
datagrams from several sources or from a single source

Virtual Circuit Approach
 In this approach the relationship between all packets belonging
to a message is preserved
 A single route is chosen between sender and receiver at the
beginning of session
 All packets now travel one after another along the same route
 It is implemented in two formats
 Switched Virtual Circuit (SVC), and Permanent Virtual Circuit (PVC)
 Switched Virtual Circuit
 A Virtual Circuit is created whenever it is needed (e.g. TCP‟s three way
handshake) and exists for the duration of the specific exchange
 Each time a device makes a connection to another device, the route may
be same or may differ in response to varying network conditions
 Permanent Virtual Circuit
 The same virtual circuit is provided between two users on a contineous
basis. The circuit is dedicated to specific users without making a
connection establishment or release

A Comparison for data traffic
 A circuit switch connection creates a physical path between two
points where as a virtual circuit creates a route between two
points
 The Network resources (link and switches) that make a path
are dedicated but that make a route can be shared by other
connections
 The line efficiency is greater in Packet switching as a single link
can be shared by many packets over time
 A packet switching network can perform data-rate conversion.
i.e. two stations having different data rates can exchange
packets but it is not possible in circuit switching
 In a typical user/host data connection, much of the time line is
idle thus making circuit switching inefficient
 When traffic becomes heavy on a circuit switching network,
some calls are blocked, but in packet switching network

Effect of Packet Size
 Virtual circuit from x to y
 a and b are intermediate switches
 Message of size 40 octets
 Packet header 3 octets (control
information)
 Case I: entire message sent as one
packet
 Case II: entire message sent as
two packets
 Case III: entire message sent as
five packets
 Case IV: entire message sent as
ten packets

Packet Size contd.
 Case I
 packet is first transmitted from X to a. when the entire packet is received
by a, it can then be transmitted to b.
 Ignoring switching time, total transmission time is 433=129 octet time
 Case II
 Node a can begin transmitting the first packet as soon it has arrived
from X, without waiting for the second packet. Overlapping in
transmission time!
 Total transmission time is 234=92 octet time
 Case III
 packets are transmitted still faster due to more number of overlapping
 Case IV
 Time is increased as fixed header becomes an overhead. i.e. 3 10=30
octets of header information for 40 octets of data!

One more comparison
 Performance
 Propagation delay
 Time it takes a signal to
propagate from one
node to another
 Transmission Time
 Time it takes for a
transmitter to push a
block of data to the
medium
 Propagation delay
 Time it takes for a node
to perform the
necessary processing as
it switches data

Circuit Switching Datagram Virtual-Circuit
Dedicated transmission path No dedicated path No dedicated path
Continuous transmission of data Transmission of packet Transmission of packet
Fast enough for interactive Fast enough for interactive Fast enough for interactive
Messages are not stored Packets may be stored until
transmitted
Packets may be stored until
delivered
The path is established for
entire conversation
Route established for each
packet
Route established for entire
conversation
Call set-up delay, transmission
delay
Packet transmission delay Call setup delay, packet
transmission delay
Busy signal if called party busy Sender may be notified if
packet not delivered
Sender notified of connection
denial
Overload may block call setup;
no delay for established calls
Overload increases packet delay Overload may block call set-up;
increases packet delay
Usually no speed or code
conversion
Speed and code conversion Speed and code conversion
Fixed Bandwidth Dynamic use of bandwidth Dynamic use of bandwidth
No overhead bits after call
setup
Overhead bits in each
packet
Overhead bits in each
packet

Network Performance
 Throughput
 Is a measure of the actual transmission of data in a network per unit
time.
 Latency
 Propagation time + Transmission Time + Queuing Time + Processing Delay
 Propagation Time = Distance/Propagation speed
 Transmission Time= Message size/Bandwidth
 Bandwidth Delay Product
 BDP defines the number of bits that can fill the link

End of Module I

Computer Networks Module I

More Related Content

What's hot

Similar to Computer Networks Module I

More from Ajit Nayak

Recently uploaded

Computer Networks Module I