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 requestmessage to the Web Server

Web Server will respond by returning a connection replymessage

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 threelayers

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 SignalPeriodic Signal


Analog Signals

The sine wave is the most fundamental form of a periodic signal

Represented as s(t)=Asin(2ft+)

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

42sin

4)( tf

Atf

Aft

Ats


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 received1 extra bps

At 1 Mbps:

1,000,000 bits sent 1,001,000 bits received1000 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


Asynchronous Transmission 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:

Asynchronous Transmission 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 modemsV.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 andat telephone comp

• The quantization noise introduced thus data rate islimited 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 bitsper 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

(6) the data rate of the link.

Solution

1. The data rate of each source

is 2508=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 LineRate

(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

CategoryImpedan

ceUse

RG-59 75 W Cable TV

RG-58 50 WThin

Ethernet

RG-11 50 WThick

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 CoreClad

dingMode

50/125 50 125Multimode,

graded-index

62.5/125 62.5 125Multimode,

graded-index

100/125 100 125Multimode,

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 DisadvantagesAdavntages

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 GroundRadio beacons and

navigational locators

MF 300 KHz–3 MHz Sky AM radio

HF 3–30 MHz SkyCitizens band (CB),

ship/aircraft communication

VHF 30–300 MHzSky and

line-of-sight

VHF TV,

FM radio

UHF 300 MHz–3 GHz Line-of-sightUHF 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(nm)

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 433=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 234=92 octet time

Case III packets are transmitted still faster due to more number of overlapping

Total transmission time is 117=77 octet time

Case IV Total transmission time is 712=84 octet time

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