PES SCHOOL OF ENGINEERINGHosur Road, BANGALORE - 560 100

Deptt. Of CSE

1ST INTERNAL TEST (V Sem CSE/ISE)

SUB: 06CS55 COMPUTERNETWORKS-I MAXIMUM MARKS: 50 DURATION: 1HR 30MIN Faculty: Ms Swapna

Date: 02 Sept 2008

ANSWER ANY FIVE. ALL QUESTIONS CARRY EQUAL MARKS

1. a)Explain UDP,TCP,SCTP protocols governing transport layer in TCP/IP architecture. 6marks

ANSWER:

b) What do you understand by the term layering? Why do we require Standards? 4marks

ANSWER:

Layering is defined as:-

Decomposition or modularisation of overall communication functionin to a set of layers. Layering helps in simplifying network design,network management& product development. It provides flexibilityto modify and develop network.

Standards are essential 1) In creating and maintaining an open and competitive market for

equipment manufacturers and in guaranteeing national and international interoperatabilty of data and telecommunicationstechnology.

2) They provide guidelines in manufacturers,vendors,governmentagencies & other service providers to ensure the kind of interconnectivity necessary in today’s market place and in internationalcommunications.

2. a) Define data communications. Describe different components of data communication systems with a block diagram. 6marks

ANSWER:

Data Communications are the exchange of data between two devicesvia some form of transmission medium such as wire cable. For datacommunications to occur, the communicating devices must be part of communication system.

A data communication system has five components (see fig above)1. Message: Is the information(data) to be communicated. Popular

form of information include text,numbers,pictures,audio,and video.2. Sender: Is the device that sends the message. It can be a computer,

workstation, telephone handset etc.3. Receiver :Is the device that receives the message. It can be a

computer,workstation, telephone handset etc.

4. Transmission medium :Is the physical path by which a message travels from sender to receiver. Examples include twisted-pair wire, coaxial cable, fiber optic cable, and radiowaves.

5. Protocol: Is a set of rules that govern data communications. It represents an agreement between the communicating devices.

b) Name the four basic network topologies and cite an advantage of each type. 4marks

Mesh topologyAdvantage: 1)Use of dedicated links guarantees that each connectionthat each connection can carry its own data load, thus eliminating the traffic problems that occur when links must be shared by multiple devices.

2)It is robust. 3)Privacy or Security.

Disadvantage:1)Amount of cabling and number of i/o ports required.2)Every device connected to other device, installation and reconnection Are difficult.3)Hardware required to connect each link(I/O Ports and cable)can be Prohibitively expensive.

Star topology Advantage:

1)Less expensive than Mesh topology2)Easy to install and reconfigure because it needs only one link and one i/o port to connect it to any number of other devices.3)Robustness 4)If one link is fails, only that link is affected. All the other links remain active, this factor leads to easy fault identification and fault isolation.Disadvantage:1) Is the dependency of the whole topology on one single point, the hub.

If the hub goes down, the whole system is dead.2) Though it requires less cabling than a mesh,it requires more cabling

than other topologies.

Bus topology Advantage : 1) Ease of installation 2)Less cabling: Backbone cable can be laid along the most efficient path Then connected to the nodes by drop lines of various lengths. Disadvantage: 1)Difficulty in reconnection and fault isolation. 2) A fault or break in the bus cable stops all transmission.

Ring topology Advantage: 1)Easy to install or reconfigure. 2)Fault isolation is simplified. Disadvantage: Unidirectional traffic: A break in the ring can disable the entire network.

3. a) What are the differences between a port address, a logical address and a physical address? 6marks ANSWER: Physical address: 1)It is also known as the link address. 2)It is address of a node as defined by its LAN or WAN.

3)It is included in the frame used by the datalink layer. 4)It is lowest-level of address. It will change from hop to hop. 5)Size and format of these addresses vary depending on the network.

6)Example: Ethernet uses a 6-byte(48-bit)physical address that is Imprinted on the network interface card(NIC)

Logical address: 1)A Universal addressing system is needed in which each host can be Identified uniquely regardless of the underlying physical network. Logical addresses are designed for this purpose. 2)Logical address in the internet is currently a 32-bit address that can Uniquely define a host connected to the internet.Logical addresses do not change they usually remain the same. 3)Example:IP address which uniquely defines the host on the Internet.

Port address: 1)It identifies a process on a host. 2)It addresses each application/process like browser client to browser Server(http protocol);email to email(SMTP);terminal to host(TELNET). 3)Example: Computer A can communicate with computer B by using FTP And computer A can communicate with computer C by using TELNET. Port addresses is the method used to label different processes. 4)A Port address in TCP/IP is 16bits in length.

b) What are the functions of data link layer, transport layer, session layer and presentation layer? 4marks ANSWER Data Link layer: Responsible for moving frames from one hop(node) to the next.Other responsibilties:Framing,Physical addressing,flow control,error

control and access control.

Transport layer: Responsible for process to process delivery of the entire Message. Other responsibilties:Service-point addressing,Segmentation and reassembly connection control,flow control and error control.

Session layer:Dialog control and Synchronization

Presentation layer: It is concerned with the syntax and semantics of the Information exchanged between the two systems.Translation,Encryption, Compression.

4. List and explain any five line coding techniques and represent the sequence 101011100 using the techniques. 10marks

5. Explain the PCM technique of changing analog signal to digital signalwith neat diagrams of PCM encoder and PCM decoder. 10marks

Answer:

PCM encoder:

PCM encoder has three processes1) Sampling: Analog signal is sampled every Ts.Ts is sampling interval or

period. Inverse of the sampling interval is called sampling rate or sampling frequency(fs)Three types of sampling methods1)Ideal sampling, pulses from analog signal are sampled.2)In natural sampling, a high speed switch is turned on for only small period of time when the sampling occurs. The result is a sequence of samples that retains the shape of the analog signal.3)Sample and hold method creates flat top samples by using a circuit.

After sampling result is still an analog signal with non-integral values.Sampling rate must be atleast 2 times the highest frequency ,not the bandwidth and bandwidth should be limited.

2) Quantization: Assuming original analog signal has instantaneous amplitudes between Vmax and Vmin.We divide range into L zones, each of height delta.Δ=Vmax-Vmin/ L.We assign quantized values of 0 to L-1 to the midpoint of each zone.We approximate the value of the sample amplitude to quantized values.In quantization input values are the real values and output values are approximated values.

3)Encoding: After each sample is quantized and the number of bits per sample is decided ,each sample can be changed to an nb-bit log code word.nb=logL.Bit rate=sampling rateX number of bits per sample.

PCM decoderIt is required for recovery of the original signal.It uses circuitry to convert the code words into a pulse that holds the amplitude until the next pulse.After the Staircase signal is completed ,it is passed through a low pass filter to smooth the staircase signal into an analog signal.The filter has the same cutoff freqas the original signal at the sender.If the signal has been sampled at the nyquist

sampling rate and if there is enough quantization levels,the original signal will be recreated.Maximum and minimum values of the original signal can be achieved by using amplification.Bandwidth of the digital signal is Nb times greater than analog signal.Maximum data rate =2XB.WXlog L

6. a)Explain broadband transmission with neat diagrams 6marks

Explanation :Refer notes

ANSWER:

b)Define a DC component and its effect on digital transmission. 4marks

DC Component or Direct current component are frequencies around zero.In Digital transmission when a voltage level in a digital signal is constant for awhile,the spectrum creates very low frequencies.These frequencies around zero,called DC components which present problems for a system that cannotpass low frequencies or a system that uses electrical coupling(Transformer)

Example:A Telephone line cannot pass frequencies below 200hz.Also a long Distance link may use one or more transformers to isolate different parts of the line electrically.For these systems,we need a scheme with no Dc

component.

7. List the three techniques in serial transmission and explain the transmission in detail. 10marks

Serial Transmission:

Asynchronous Transmission:

Synchronous transmission

Isochronous transmission:Refer notes