Computer Network Module 1

COMPUTER NETWORKS (BCSE 3306) Last Updated 14th Jan 07

Lecture Notes Module I Ajit K Nayak [email protected] Department of Computer Science Engineering & Application

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

Text: “Data Communications and Networking” Third Edition, Behrouz A Forcuzan, Tata Mc Graw-Hill. Chapter 1 - Chapter 7 Computer Networking / Module I / AKN / 2

Lecture I Overview of Data Communications and Networking • Data Communication • Networks & Internet • Protocols & Standards • Layered Tasks • Internet Model • OSI Model Computer Networking / Module I / AKN / 3

Data Communication „

Sharing of information is “Data Communication” „ „

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Sharing can be local (face to face) Remote (over a distance)

“Data” refers to facts, concepts and / or instructions „

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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. Computer Networking / Module I / AKN / 4

Characteristics of Data Communication „

Delivery: system must deliver data to correct destination

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Accuracy: Accurate data should be delivered

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Timeliness: Data delivered late are useless

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Components of Data Communication

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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 Computer Networking / Module I / AKN / 6

Direction of Data Flow „

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Communication can be simplex, Half-duplex, or full-duplex. Simplex: communication is unidirectional

Any real life examples?

Half-duplex: bi-directional but not at the same time Full-duplex: bi-directional and simultaneously. Computer Networking / Module I / AKN / 7

Networks & Distributed processing „

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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 „

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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 Computer Networking / Module I / AKN / 8 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. „

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Security depends on Unauthorized access should be prevented „ Should be protected from viruses, spywares, adwares, malwares etc. „

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Physical Structure „

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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) Computer Networking / Module I / AKN / 10

Topology „

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

ETC.

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Mesh Topology Every device has a dedicated point-topoint 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 Computer Networking / Module I / AKN / 12 „ Not Difficult to add/remove a device „

Star Topology „

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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 „

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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 Computer Networking / Module I / AKN / 14 „

Ring Topology „ „

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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 „

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Category of networks „

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

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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? „

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Figure 1.13

LAN (Continued)

Example: LAN of an organisation Computer Networking / Module I / AKN / 18

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. „

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Wide Area Network(WAN) „

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

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 „

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A metropolitan area network based on cable TV.

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The Internet „

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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 „ „

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End systems are connected either directly by ‘communication links’ or indirectly by intermediate switching devices called ‘switches/Routers’ Communication links include „

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PCs, UNIX based workstations, servers(?) PDAs, TVs, Mobile computers, automobiles, Toaters, …

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 towardsComputer destination Networking / Module I / AKN / 22

Internet Today „

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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 „

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National Service Providers „

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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 „

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That connect nations together

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 Computer Networking / Module I / AKN / 23

Internet today

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

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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 ... Computer Networking / Module I / AKN / 25

Protocol !!! „ „ „

What is a Protocol? What does a protocol do? How would you recognize a protocol if you met one?

A Human Analogy ¾

What you do when you want to ask some one for the time of day?

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Protocol „

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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 . . . Computer Networking / Module I / AKN / 27

Protocol „

But what happens when a different response comes to the initial Hi like „ „ „ „

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Don’t bother me! I don’t speak English Some unprintable reply! No response at all !!!

OR OR OR

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 Computer Networking / Module I / AKN / 28

Protocol „

If people run different protocols! Say „ „

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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 Computer Networking / Module I / AKN / 29 some device

A Network Protocol „ „

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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. Computer Networking / Module I / AKN / 30

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

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Protocols contd. „

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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 day month Year Example: a date 8 8 16 „ 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 „

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Depends on link availability, and speed of receiver Computer Networking / Module I / AKN / 32

Standards „

The standard provides a model for development that makes it possible for a product to work regardless of the individual manufacturer „

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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 Computer Networking / Module I / AKN / 33

Standards Organizations „

Standards Creation Committees „ „ „ „ „

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Forums „

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The forums work with universities and users to test, evaluate and the conclusion is presented to standard bodies to standardize new technologies

Regulatory Agencies „

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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)

Govt. agencies responsible for protecting the public interest.

Internet Standards „

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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) Computer Networking / Module I / AKN / 34

Layered Tasks „

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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 Computer Networking / Module I / AKN / 35 the adjacent layer.

Example:

Sending a letter

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Example:

The philosopher-translator-secretary architecture.

Location A I like rabbits

Location B

Message

Philosopher

J'aime bien les lapins

3

2

1

3

L: Dutch Ik vind konijnen leuk

Fax #--L: Dutch Ik vind konijnen leuk

Information for the remote translator

Information for the remote secretary

Translator

Secretary

L: Dutch Ik vind konijnen leuk

Fax #--L: Dutch Ik vind konijnen leuk

2

1

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The Internet model „

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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 Computer Networking / Module I / AKN / 38 layer are called peer-to-peer processes.

Peer-to-peer processes

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An exchange using the Internet model

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Physical layer „

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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 „

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Representation of bits „

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Type of transmission medium, topology, etc… Encoding, voltage level, duration etc…

Data rate Synchronization of bits „

Sender’s and receiver’s clock shynchronization

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Data link layer „

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is responsible for transmitting frames from one node to the next The duties are „

Framing „

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Physical addressing „

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This mechanism helps to prevents overflow at receiving side

Error control „

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Adds the address of sender and receiver in the header

Flow control „

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Stream of bits received from upper layer is divided into manageable data units(?) called frame

Mechanism to detect/correct errors in transmission

Access Control „

Which device has the control over the link at a given time Computer Networking / Module I / AKN / 42

Datalink layer contd. „

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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. Computer Networking / Module I / AKN / 43

Network Layer „

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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 „

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It adds logical addresses into the packet header

Routing „

Forwarding the packet towards the destination

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Source-to-Destination

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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. Computer Networking / Module I / AKN / 46

Transport layer „

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The transport layer is responsible for delivery of a message from one process to another. The Duties „

Port addressing „

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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

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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) „

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Connection less: each message carries the destination address and routed through the system „

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Example: telephone

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 Computer Networking / Module I / AKN / 48

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…

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Summary of duties

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OSI model „

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Session Layer is the network dialog controller, It establishes maintains and synchronizes the interaction between communicating systems Duties are „ „

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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 Computer Networking / Module I / AKN / 51

Lecture II The Physical Layer • Signals • Digital Transmission • Analog Transmission • Multiplexing • Transmission Media Computer Networking / Module I / AKN / 52

Position of the physical layer

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Signals „

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Information is transmitted in the form of electromagnetic signals Signals are of two types „

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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

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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

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Analog Signals

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The sine wave is the most fundamental form of a periodic signal Represented as s(t)=Asin(2πft+Φ)

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Characterstics

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Amplitude: intensity of signal at any given time Frequency: no of cycles/periods in one second, measured in Hz „

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Frequency = 1/Period

Phase: describes the position of the waveform relative to time zero „

A complete cycle is 360o = 2π Computer Networking / Module I / AKN / 56

Amplitude Period and frequency

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Time and frequency domains

A signal can also be represented in frequency domain

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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 „

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Example : A Square wave

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According to Fourier analysis, this signal can be decomposed in to a series of sine waves i.e. 4A

4A 4A s (t ) = sin 2πft + sin[ 2π (3 f )t ] + sin[ 2π (5 f )t ] + ... π 3π 5π „ „ „

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 Computer Networking / Module I / AKN / 60

Three harmonics

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Frequency spectrum

The Signal using the frequency domain and containing all its components is called the frequency spectrum of that signal ƒ The range of frequencies that a medium can pass is called its Bandwidth ƒ 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.

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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. Computer Networking / Module I / AKN / 63

Digital Signals „

Digital signals can be better described by two terms „ „

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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)

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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

ƒ A digital signal theoretically needs a BW between o and ∞ ƒ if the upper limit will be relaxed than digital transmission can use a low-pass channel

ƒ An analog signal has a narrower BW with frequencies f1and f2 ƒ Also BW of analog signal can be shifted, i.e. f1and f2 can be shifted to f3 and f4 ƒ Analog signal can use a band-pass channel Computer Networking / Module I / AKN / 65

Data rate limits „

Data rate depends on „ „ „

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Nyquist Bit rate: noise less channel „ „

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The BW available The levels of signal that can be used The quality of channel (i.e. the level of noise) 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 Computer the signal to the power of the noise Networking / Module I / AKN / 66

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 First, we use the Shannon formula to find our upper limit. 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. 4 Mbps = 2 × 1 MHz × log2 L Î L = 4 Computer Networking / Module I / AKN / 67

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 „

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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 Computer Networking / Module I / AKN / 68

Distortion „ „

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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

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Noise „

Several types of noise such as „ „

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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

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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 λ=c/f Computer Networking / Module I / AKN / 71

Digital Transmission Line coding Block Coding Sampling Transmission Mode

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What is Line Coding? „

Is the process of converting binary data (a sequence of bits) to a digital signal

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Signal Level versus Data Level „ „

No of values allowed in a signal No of values used to represent data

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DC Component „

A component having zero frequency „ „

Can’t be passed through a transformer Energy consumed is useless

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Pulse Rate versus Bit Rate „

No of pulses per second „

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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 Computer Networking / Module I / AKN / 76

Self Synchronization „

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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 Computer Networking / Module I / AKN / 77

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 Î1001 bits receivedÎ1 extra bps At 1 Mbps: 1,000,000 bits sent Î1,001,000 bits receivedÎ1000 extra bps Computer Networking / Module I / AKN / 78

Line Coding Schemes

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

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Unipolar Encoding „ „

One is coded as +ve voltage Zero is coded as –ve voltage

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Polar Encoding

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

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Polar Encoding „ „ „

Avarage voltage level is decreased DC component problem is avoided Four Important type of polar encoding are:

There are many others also! Computer Networking / Module I / AKN / 83

NRZ-L Encoding

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

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NRZ-I Encoding

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

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NRZ Encoding

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Loss of synchronization incase of continuous ones or zeros Computer Networking / Module I / AKN / 86

RZ Encoding

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

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RZ Encoding

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A +ve voltage means 1 and –ve voltage means zero. But signal returns to zero at mid of the bit interval Computer Networking / Module I / AKN / 88

RZ Encoding Note: 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!

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

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Manchester Encoding

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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 Computer Networking / Module I / AKN / 91

Diff-Manchester Encoding Note: 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. Computer Networking / Module I / AKN / 92

Diff-Manchester Encoding

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Manchester Encoding used for 802.3 base band – CSMA/CD Lans Diff-Manchester is used foe 802.5 token ring LAn Computer Networking / Module I / AKN / 93

Bipolar Encoding

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

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Bipolar Encoding

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2B1Q Encoding „ „

Two Binary One Quaternary Each pulse represents 2 bits

-1 -3

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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

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Block Coding ™ To ensure synchronization some redundant bits may be introduced Steps in Transformation ¾ Division ¾ Substitution ¾ Line Coding Computer Networking / Module I / AKN / 98

Block Coding

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Substitution

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4B/5B Encoding „

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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. Computer Networking / Module I / AKN / 101

4B/5B encoding 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

Computer Networking / Module I / AKN / 102

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

Computer Networking / Module I / AKN / 103

Pulse Amplitude Modulation „

„

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

Computer Networking / Module I / AKN / 104

PAM Note: 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. Computer Networking / Module I / AKN / 105

PCM: Quantization „

It is a method of assigning integral values in a specific range to sampled instances

Computer Networking / Module I / AKN / 106

Binary encoding „

„

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

Computer Networking / Module I / AKN / 107

Line coding „

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

Computer Networking / Module I / AKN / 108

Analog to PCM Digital Code

Computer Networking / Module I / AKN / 109

Sampling rate „

„

Accuracy of reproduction depend on the no of samples taken What should be the sampling rate?

Note: According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency. Computer Networking / Module I / AKN / 110

Nyquist Theorem

Computer Networking / Module I / AKN / 111

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 Computer Networking / Module I / AKN / 112

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. Computer Networking / Module I / AKN / 113

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 Computer Networking / Module I / AKN / 114

Transmission mode

Computer Networking / Module I / AKN / 115

Parallel Transmission „ „

„ „

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 Computer Networking / Module I / AKN / 116

Serial Transmission „

„ „ „

One bit follows another using same line

Reduced cost (by a factor n) Parallel/serial converter required May used for large distance Computer Networking / Module I / AKN / 117

Asynchronous Transmission „

Serial transmission occurs in one of the two ways

Note: 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. Computer Networking / Module I / AKN / 118

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. Computer Networking / Module I / AKN / 119

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

Note: Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same. Computer Networking / Module I / AKN / 120

Synchronous Transmission Note: 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. Computer Networking / Module I / AKN / 121

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 Computer Networking / Module I / AKN / 122

Modulation of Digital Data Analog Transmission Digital-to-Analog Conversion Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation Bit/Baud Comparison Computer Networking / Module I / AKN / 123

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

Computer Networking / Module I / AKN / 124

Bit rate / Baud rate Note: 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. 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. Computer Networking / Module I / AKN / 125

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

Computer Networking / Module I / AKN / 126

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

Computer Networking / Module I / AKN / 127

Example Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the fullduplex 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 Computer Networking / Module I / AKN / 128

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 Computer Networking / Module I / AKN / 129

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. Computer Networking / Module I / AKN / 130

Phase Shift Keying „

„

„

„ „

Phase of carrier signal varies to represent a binary 1 (180o)or 0 (0o) also called 2PSK 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 Computer Networking / Module I / AKN / 131

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 Computer Networking / Module I / AKN / 132

Quadrature Amplitude Modulation

Note: 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.

Computer Networking / Module I / AKN / 133

4-QAM & 8-QAM Constellation

Computer Networking / Module I / AKN / 134

16-QAM constellations

QAM is less susceptible to noise than ASK? Bandwidth required for QAM is same as PSK and ASK Computer Networking / Module I / AKN / 135

Bit/Baud Comparison

Computer Networking / Module I / AKN / 136

Modem Standards Modem stands for modulator/demodulator.

A telephone line has a bandwidth of almost 2400 Hz for data transmission. Computer Networking / Module I / AKN / 137

Modulation/Demodulation

‰ A modulator creates a band-pass signal from binary data. ‰ A demodulator recovers the binary data from the modulated signal

Computer Networking / Module I / AKN / 138

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)

Computer Networking / Module I / AKN / 139

V.32bis constellation & BW ƒ Uses 128-QAM (7 bits/ baud with 1 bit for error control) ƒ datarate (2400*6)=14400 bps

V.90 ƒ Asymetric modems, i.e. downloading speed is 56 kbps and uploading speed is 33.6 kbps ƒ This is possible if one party is using digital signaling

V.92 ƒ can adjust their speed I.e. if noise allows than it can upload at a rate of 48 Kbps ƒ Additional features like modem can interrupt internet connection for a incoming phone call etc. Computer Networking / Module I / AKN / 140

Traditional modems

• Sampled, digitized and at telephone comp • The quantization noise introduced thus data rate is limited according to shannon capacity i.e. 33.6k

56 K Modems • 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

Computer Networking / Module I / AKN / 141

Modulation of Analog Signals • 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 α 1/f) • helps in frequency division multiplexing • To support medium characteristics

Methods: Amplitude Modulation (AM) Frequency Modulation (FM) Phase Modulation (PM) Computer Networking / Module I / AKN / 142

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

Computer Networking / Module I / AKN / 143

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

Computer Networking / Module I / AKN / 144

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

Computer Networking / Module I / AKN / 145

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

Computer Networking / Module I / AKN / 146

Categories of multiplexing

Computer Networking / Module I / AKN / 147

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

Computer Networking / Module I / AKN / 148

FDM f

t

• 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 Computer Networking / Module I / AKN / 149

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 Computer Networking / Module I / AKN / 150

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

Computer Networking / Module I / AKN / 151

Example Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM Solution • The satellite channel is analog. We divide it into four channels, each channel having a 250-KHz bandwidth. • Each digital channel of 1 Mbps is modulated such that each 4 bits are modulated to 1 Hz. • One solution is 16QAM modulation. • Figure shows one possible configuration.

Computer Networking / Module I / AKN / 152

Analog hierarchy

Computer Networking / Module I / AKN / 153

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

Computer Networking / Module I / AKN / 154

Time division Multiplexing „

„

„

„

„

Each shared connection occupies a portion of time but uses full BW

f

The data flow of each connection is divided into units For n input connections, a frame is t 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 Computer Networking / Module I / AKN / 155

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

Computer Networking / Module I / AKN / 156

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 µs. 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 Computer Networking / Module I / AKN / 157

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

Computer Networking / Module I / AKN / 158

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

Computer Networking / Module I / AKN / 159

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

Computer Networking / Module I / AKN / 160

Example

Solution

We have four sources, each creating 250 1. The data rate of each source characters per second. If the is 250×8=2000 bps interleaved unit is a character and 1 2. The duration of a character synchronizing bit is added to each is 1/250 s, or 4 ms. frame, find 3. The link needs to send 250 (1) the data rate of each source, frames per second. (2) the duration of each character in each 4. The duration of each frame is 1/250 s, or 4 ms. source, 5. Each frame is 4 x 8 + 1 = 33 (3) the frame rate, bits. 6. The data rate of the link is (4) the duration of each frame, 250 x 33, or 8250 bps. (5) the number of bits in each frame, and (6) the data rate of the link. Computer Networking / Module I / AKN / 161

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 Computer Networking / Module I / AKN / 162

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. Computer Networking / Module I / AKN / 163

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

Computer Networking / Module I / AKN / 164

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

Computer Networking / Module I / AKN / 165

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

Computer Networking / Module I / AKN / 166

• Europeans use E Lines in place T Lines. Both are conceptually same only capacity differs

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

Computer Networking / Module I / AKN / 167

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

Computer Networking / Module I / AKN / 168

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, ultraviolate, and X, gamma and cosmic rays etc.

Computer Networking / Module I / AKN / 169

Classes of transmission media

Computer Networking / Module I / AKN / 170

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 Computer Networking / Module I / AKN / 171

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 Computer Networking / Module I / AKN / 172

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

4

20 MHz

20 Mbps

Digital

LANs

5

100 MHz

100 Mbps

Digital

LANs

6

200 MHz

200 Mbps

Digital

LANs

7 (draft)

600 MHz

600 Mbps

Digital

LANs

Computer Networking / Module I / AKN / 173

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 Computer Networking / Module I / AKN / 174

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 Ω

Cable TV

RG-58

50 Ω

Thin Ethernet

RG-11

50 Ω

Thick Ethernet

Computer Networking / Module I / AKN / 175

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

Computer Networking / Module I / AKN / 176

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 Computer Networking / Module I / AKN / 177

Propagation modes Current technology allows two modes of propagating light along optical channels

Multimode: multiple beams Single mode: single focused beam Computer Networking / Module I / AKN / 178

Mechanism „

Multimode step index: „

„

„

„

Multimode graded 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 The density of core varies (decreases) from core center to edges. Light undergoes a series of refraction 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

Computer Networking / Module I / AKN / 179

Fiber Optics contd. „

„

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

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

Single-mode

7/125

Computer Networking / Module I / AKN / 180

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) Computer Networking / Module I / AKN / 181

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 „

Computer Networking / Module I / AKN / 182

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

Computer Networking / Module I / AKN / 183

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

Computer Networking / Module I / AKN / 184

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. Computer Networking / Module I / AKN / 185

Antennas ‰Radiation ‰Coupling

and reception of electromagnetic waves

of wires to space for radio transmission

‰It

works as an adapter between a guided and unguided media Unidirectional Antenna

Computer Networking / Module I / AKN / 186

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 Computer Networking / Module I / AKN / 187

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 Computer Networking / Module I / AKN / 188

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 Computer Networking / Module I / AKN / 189

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 Computer Networking / Module I / AKN / 190

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 Computer Networking / Module I / AKN / 191

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 Computer Networking / Module I / AKN / 192

TDM Bus „

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„

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.

Computer Networking / Module I / AKN / 193

Telephone Network

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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. Computer Networking / Module I / AKN / 194

Making a Connection „

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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

Computer Networking / Module I / AKN / 195

Note: 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 (touchtone).

Computer Networking / Module I / AKN / 196

Packet Switching „

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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 Computer Networking / Module I / AKN / 197

Datagram Approach

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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 Computer Networking / Module I / AKN / 198

Virtual Circuit Approach „

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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 „

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Switched Virtual Circuit „

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Switched Virtual Circuit (SVC), and Permanent Virtual Circuit (PVC) 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 Computer Networking / Module I / AKN / 199

A Comparison for data traffic „

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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 Computer Networking / Module I / AKN / 200

Effect of Packet Size

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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 Computer Networking / Module I / AKN / 201

Packet Size contd. „

Case I „

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Case II „

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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 „ „

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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

packets are transmitted still faster due to more number of overlapping Total transmission time is 11×7=77 octet time

Case IV „ „

Total transmission time is 7×12=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! Computer Networking / Module I / AKN / 202

One more comparison „

Performance „

Propagation delay „

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Transmission Time „

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Time it takes a signal to propagate from one node to another 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 Computer Networking / Module I / AKN / 203

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 Overhead bits in each packet packet Computer Networking / Module I / AKN / 204

End of Module I

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