module i final

COMPUTER NETWORKS

T C [email protected]

Department of Computer Science Engineering & Application

Lecture NotesModule I

Computer Networking / Module I / TCM / 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

Text: “Data Communications and Networking” 4th Edition, Behrouz A Forcuzan, Tata Mc Graw-Hill.Chapter 1 - Chapter 7


Overview of Data Communications Overview of Data Communications and Networkingand Networking

Lecture ILecture I

• Data CommunicationData Communication• Networks & InternetNetworks & Internet• Protocols & StandardsProtocols & Standards• Layered TasksLayered Tasks• Internet ModelInternet Model• OSI ModelOSI 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?


Computer Networks Interconnection of ‘Intelligent devices’ is called

a ‘computer 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

Security depends on Unauthorized access should be prevented Should be protected from viruses, spy

wares, 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 at same time.


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 ( nC2

)

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 disturb the

network completely Privacy/security is maintained Installation and reconfiguration is difficult due to complicated connections Expensive in terms of cost and space Very 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 has 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.


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.


Wide Area Network(WAN)

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.


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 presentedExample: 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 Yearmonth8 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 two/three layers

Layer x on one device communicates with layer x of other device.

The processes on each machine that communicate at a given layer are called peer-to-peer processes.


Peer-to-peer processes


An exchange using the Internet model


Physical layer The responsibility of physical layer is to coordinate the

functions required to transmit a bit stream over a physical medium

The duties are Defines the characteristics of the interface between devices and

transmission medium Type of transmission medium, topology, etc…

Representation of bits Encoding, voltage level, duration etc…

Data rate Synchronization of bits

Sender’s and receiver’s clock shynchronization


Data link layer

is responsible for transmitting frames from one node to the next

The duties are Framing

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

Physical addressing Adds the address of sender and receiver in the header

Flow control This mechanism helps to prevents overflow at receiving side

Error control Mechanism to detect/correct errors in transmission

Access Control Which device has the control over the link at a given time


Datalink layer contd. Physical addressing and hop-hop delivery

can be done in one network only

If the message is to be passed across the network then network layer functionality is required.


Network Layer The network layer is responsible for the delivery of packets

from the original source to the final destination possibly across multiple networks.

The Duties are Logical addressing

It adds logical addresses into the packet header Routing

Forwarding the packet towards the destination


Source-to-Destination


An Example

sending from a node with network address A and physical address 10 to a node with a network address P and physical address 95

Because the two devices are located on different networks, we cannot use physical addresses only;as the physical addresses only have local jurisdiction.

What we need here are universal addresses that can pass through the LAN boundaries. The network (logical) addresses have this characteristic.


Transport layer The transport layer is responsible for delivery of a message

from one process to another. The Duties

Port addressing Actual transmission occurs from a specific process on one device to a

process of another. Port address (an integer) defines the process/application in a device

Segmentation and reassembly Message received from application layer is divided in to transmittable

segments containing sequence nos


Transport layer contd. Connection control

Two types of connection service is allowed Connection oriented: establish the connection, use the connection, release

the connection. (guarantee of delivery) Example: telephone

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

Example: postal service

Flow Control Responsible for end-to-end flow control as well as

intermediate flow control (congestion) Error Control

End-to-end error control


Application layer

The application layer is responsible for providing services to the user. It provides user interfaces and support

services such as email, remote file transfer, remote logins etc…


Summary of duties


OSI model

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

Duties are Dialog control Synchronization at data level

Presentation layer is concerned with syntax and semantics of the information exchanged between two systems

Duties are Translation: converting to bit streams Encryption: to ensure privacy Compression: increases virtual BW


The Physical LayerThe Physical Layer

Lecture IILecture II

• SignalsSignals• Digital TransmissionDigital Transmission• Analog TransmissionAnalog Transmission• MultiplexingMultiplexing• Transmission MediaTransmission 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+) Characteristics

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


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 When we change one or more characteristics of a single-frequency signal, it becomes a of a single-frequency signal, it becomes a composite signal made of many frequencies.composite signal made of many frequencies.

A composite signal is composed of multiple A composite signal is composed of multiple sine waves called sine waves called harmonicsharmonics


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 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?

SolutionSolution

The answer is definitely no. Although the signal can have the The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and 4000 medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.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


ExampleExample

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?

SolutionSolution

C = B logC = B log22 (1 + SNR) = 10 (1 + SNR) = 1066 log log22 (1 + 63) = 10 (1 + 63) = 1066 log log22 (64) = 6 Mbps (64) = 6 Mbps

Then we use the Nyquist formula to find the number of signal levels.

4 Mbps = 2 4 Mbps = 2 1 MHz 1 MHz log log22 LL L = 4 L = 4

First, we use the Shannon formula to find our upper limit.First, we use the Shannon formula to find our upper limit.


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 (P2/P1) 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


Signal-to-Noise-Ratio SNR=avg.signal power avg.noise power


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 codingBlock CodingSamplingTransmission 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/ 10Pulse Rate = 1/ 10-3-3= 1000 pulses/s= 1000 pulses/s

Bit Rate = Pulse Rate x logBit Rate = Pulse Rate x log22 L = 1000 x log L = 1000 x log22 2 = 1000 bps 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


ExampleExample

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?

SolutionSolution

At 1 Kbps:1000 bits sent 1001 bits received1 extra bpsAt 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:Note:

UniPolar Encoding


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


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

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

Note:Note:

NRZ-L Encoding


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

Note:Note:

NRZ-I Encoding


NRZ Encoding

Loss of synchronization incase of continuous ones or zeros


RZ uses three values i.e. +ve, zero & -veRZ uses three values i.e. +ve, zero & -ve

Signal change occurs during each bitSignal change occurs during each bit

Note: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 RZ is a good encoded digital signal that contain a provision for synchronization.a provision for synchronization.

But it requires two signal changes to encode 1 But it requires two signal changes to encode 1 bit bit moremore bandwidth! bandwidth!

Note:Note:

RZ Encoding


In Manchester encoding, the transition at In Manchester encoding, the transition at the middle of the bit is used for both the middle of the bit is used for both

synchronization and bit representation.synchronization and bit representation.

Note: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 In differential Manchester encoding, the transition at the middle of the bit is used transition at the middle of the bit is used

only for synchronization. only for synchronization. The bit representation is defined by the The bit representation is defined by the

inversion or noninversion at the inversion or noninversion at the beginning of the bit.beginning of the bit.

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

Note:Note:

Bipolar Encoding


Bipolar Encoding AMI(alternate mark inversion)


Multilevel scheme2B1Q 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 ‘pattern' of received data has

an extra 5th bit . If input data is dealt with in 4-bit patterns

there are 24 = 16 different bit patterns. With 5-bit ‘pattern' 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 1111011110 1000 1001010010

0001 0100101001 1001 1001110011

0010 1010010100 1010 1011010110

0011 1010110101 1011 1011110111

0100 0101001010 1100 1101011010

0101 0101101011 1101 1101111011

0110 0111001110 1110 1110011100

0111 0111101111 1111 1110111101

4B/5B encoding4B/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


Analog-to-Digital-data conversion Pulse Code Modulation

Generates a series of pulses by sampling a given analog signal

Sampling is measuring amplitude in equal intervals


Pulse amplitude modulation has some Pulse amplitude modulation has some applications, but it is not used by itself in applications, but it is not used by itself in data communication. However, it is the data communication. However, it is the

first step in another very popular first step in another very popular conversion method called conversion method called

pulse code modulation.pulse code modulation.

Note: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 According to the Nyquist theorem, the sampling rate must be at least 2 times the sampling rate must be at least 2 times the

highest frequency.highest frequency.

Note:Note:

Sampling rate Accuracy of reproduction depend on the

no of samples taken What should be the sampling rate?


Nyquist Theorem


ExampleExample

What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)?

SolutionSolution

The sampling rate must be twice the highest frequency in the signal:

Sampling rate = 2 x (11,000) = 22,000 samples/sSampling rate = 2 x (11,000) = 22,000 samples/s


ExampleExample

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?

SolutionSolution

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.


ExampleExample

We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?

SolutionSolution

The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/sSampling rate = 4000 x 2 = 8000 samples/s

Bit rate = sampling rate x number of bits per sample Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps= 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 In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or start bit (0) at the beginning and 1 or

more stop bits (1s) at the end of each byte. more stop bits (1s) at the end of each byte. There may be a gap between each byte.There may be a gap between each byte.

Note: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 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, In synchronous transmission, we send bits one after another without we send bits one after another without

start/stop bits or gaps. start/stop bits or gaps. It is the responsibility of the receiver to It is the responsibility of the receiver to

group the bits.group the bits.

Note: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 ConversionAmplitude Shift Keying (ASK)Frequency Shift Keying (FSK)Phase Shift Keying (PSK)Quadrature Amplitude ModulationBit/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: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 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 rateSolutionSolution

Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s)Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps

Example Example

The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is the baud rate?SolutionSolution

Baud rate = 3000 / 6 = 500 baud/sBaud 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


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 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.

SolutionSolution

Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = baud rate + fc1 BW = baud rate + fc1 fc0 fc0 Baud rate = BW Baud rate = BW (fc1 (fc1 fc0 ) = 6000 fc0 ) = 6000 2000 = 4000 2000 = 4000But 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: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


Modulation of Analog SignalsModulation of Analog Signals

Methods:Amplitude Modulation (AM)Frequency Modulation (FM)Phase Modulation (PM)

• Representation of analog information by an analog signal


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.The Physical Layer contd.

Lecture IIILecture III

• MultiplexingMultiplexing• Transmission MediaTransmission Media• SwitchingSwitching


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 1Example 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.

SolutionSolution

Shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure


Example 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?

SolutionSolution

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


ExampleExample

Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM

SolutionSolution• 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.


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


ExampleExample

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?

SolutionSolution

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 s.

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


ExampleExampleFour 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.

SolutionSolution


Example 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?

SolutionSolution


Example 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.

SolutionSolution

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.


Transmission Media Signals in the form of electromagnetic

energy is propagated through transmission media from one device to another device


Classes of transmission media


Guided Media Provides a specific path from one device to

another, includes Twisted-Pair Cable

Consists of two conductors(normally copper), each with its own plastic insulation, twisted together

Due to twists, the noise interference and crosstalk affects both wires equally thus cancels each other


Unshielded vs Shielded Twisted-Pair Cable

STP has a metal foil covering each pair of insulated conductor

Metal casing improves mechanical strength, prevents of noise or cross talk but it is more expensive

STP is produced by IBM and seldom used else where.


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


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


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

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


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.

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.


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[Datagram,Virtual-circuit] Message switching


Circuit Switching It consists of a set of switches connected by physical links. A connection

between two stations is dedicated path made of one or more links. It creates a direct physical connection between two devices i.e. it

establishes a physical circuit before transmission

Circuit Switching Techniques Space Division Switches

Crossbar switch, multistage switch Time division switches

Time Slot Interchange


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 TSI fills up incoming data inorder of reception Slots are sent out in an order based on the decission

of control unit


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


Comparison 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


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


End of Module I

module i final

Documents

exchange of data

information data

raw data

accurate data

distance data

s data communication

highspeed computer

line of module