Search in the document preview
Data Communication & Computer Networks Solution Assignment 1
Q1: Explain the major classification of Computer Networks? Ans: In its simplest form, data communication takes place between two devices that are directly connected by some form of point-to-point transmission medium. Often, however, it is impractical for two devices to be directly, point-to-point connected. This is so for one (or both) of the following contingencies: The devices are very far apart. It would be inordinately expensive, for example, to string a dedicated link between two devices thousands of miles apart. There is a set of devices, each of which may require a link to many of the others at various times. Examples are all of the telephones in the world and all of the terminals and computers owned by a single organization. Except for the case of a very few devices, it is impractical to provide a dedicated wire between each pair of devices. The solution to this problem is to attach each device to a communications network. The two major categories into which communications networks are traditionally classified: wide-area networks (WANs) and local-area networks (LANs). The distinction between the two, both in terms of technology and application, has become somewhat blurred in recent years, but it remains a useful way of organizing the discussion. Wide-Area Networks Wide-area networks have been traditionally been considered to be those that cover a large geographical area, require the crossing of public right-of-ways, and rely at least in part on circuits provided by a common carrier. Typically, a WAN consists of a number of interconnected switching nodes. A transmission from any one device is routed through these internal nodes to the specified destination device. These nodes (including the boundary nodes) are not concerned with the content of the data; rather, their purpose is to provide a switching facility that will move the data from node to node until they reach their destination. Traditionally, WANs have been implemented using one of two technologies: circuit switching and packet switching. More recently, frame relay and ATM networks have assumed major roles. Circuit Switching In a circuit-switched network, a dedicated communications path is established between two stations through the nodes of the network. That path is a connected sequence of physical links between nodes. On each link, a logical channel is dedicated to the connection. Data generated by the source station are transmitted along the dedicated path as rapidly as possible. At each node, incoming data are routed or switched to the appropriate outgoing channel without delay. The most common example of circuit switching is the telephone network. Packet Switching
A quite different approach is used in a packet-switched network. In this case, it is not necessary to dedicate transmission capacity along a path through the network. Rather, data are sent out in a sequence of small chunks, called packets. Each packet is passed through the network from node to node along some path leading from source to destination. At each node, the entire packet is received, stored briefly, and then transmitted to the next node. Packet-switched networks are commonly used for terminal- to-computer and computer-to-computer communications. Frame Relay Packet switching was developed at a time when digital long-distance transmission facilities exhibited a relatively high error rate compared to today's facilities. As a result, there is a considerable amount of overhead built into packet-switched schemes to compensate for errors. The overhead includes additional bits added to each packet to introduce redundancy and additional processing at the end stations and the intermediate switching nodes to detect and recover from errors. With modern high-speed telecommunications systems, this overhead is unnecessary and counterproductive. It is unnecessary because the rate of errors has been dramatically lowered and any remaining errors can easily be caught in the end systems by logic that operates above the level of the packet-switching logic; it is counterproductive because the overhead involved soaks up a significant fraction of the high capacity provided by the network. Frame relay was developed to take advantage of these high data rates and low error rates. Whereas the original packet-switching networks were designed with a data rate to the end user of about 64 kbps, frame relay networks are designed to operate efficiently at user data rates of up to 2 Mbps. The key to achieving these high data rates is to strip out most of the overhead involved with error control. ATM Asynchronous transfer mode (ATM), sometimes referred to as cell relay, is a culmination of all of the developments in circuit switching and packet switching over the past 25 years. ATM can be viewed as an evolution from frame relay. The most obvious difference between frame relay and ATM is that frame relay uses variable-length packets, called frames, and ATM uses fixed-length packets, called cells. As with frame relay, ATM provides little overhead for error control, depending on the inherent reliability of the transmission system and on higher layers of logic in the end systems to catch and correct errors. By using a fixed-packet length, the processing overhead is reduced even further for ATM compared to frame relay. The result is that ATM is designed to work in the range of 10s and 100s of Mbps, compared to the 2-Mbps target of frame relay. ATM can also be viewed as an evolution from circuit switching. With circuit switching, only fixed-data-rate circuits are available to the end system. ATM allows the definition of multiple virtual channels with data rates that are dynamically defined at the time the virtual channel is created. By using full, fixed-size cells, ATM is so efficient that it can offer a constant-data-rate channel even though it is using a packet-switching technique. Thus, ATM extends circuit switching to allow multiple channels with the data rate on each channel dynamically set on demand. Local Area Networks
As with wide-area networks, a local-area network is a communications network that interconnects a variety of devices and provides a means for information exchange among those devices. There are several key distinctions between LANs and WANs: 1. The scope of the LAN is small, typically a single building or a cluster of buildings. This difference in geographic scope leads to different technical solutions, as we shall see. 2. It is usually the case that the LAN is owned by the same organization that owns the attached devices. For WANs, this is less often the case, or at least a significant fraction of the network assets are not owned. This has two implications. First, care must be taken in the choice of LAN, as there may be a substantial capital investment (compared to dial-up or leased charges for wide area networks) for both purchase and maintenance. Second, the network management responsibility for a local network falls solely on the user. 3. The internal data rates of LANs are typically much greater than those of wide area networks. Traditionally, LANs make use of a broadcast network approach rather than a switching approach. With a broadcast communication network, there are no intermediate switching nodes. At each station, there is a transmitter-receive that communicates over a medium shared by other stations. A transmission from any one station is broadcast to and received by all other stations. A simple example of this is a CB radio system, in which all users tuned to the same channel may communicate. More recently, examples of switched LANs have appeared. The two most prominent examples are ATM LANs, which simply use an ATM network in a local area, and Fiber Channel. We will examine these LANs, as well as the more common broadcast LAN’s. Q 2: Explain structure and Function of OSI layers? Ans: The OSI, or Open System Interconnection, model defines a networking framework for implementing protocols in seven layers. Control is passed from one layer to the next, starting at the application layer in one station, and proceeding to the bottom layer, over the channel to the next station and back up the hierarchy. Application (Layer 7):- This layer supports application and end-user processes. Communication partners are identified, quality of service is identified, user authentication and privacy are considered, and any constraints on data syntax are identified. Everything at this layer is application- specific. This layer provides application services for file transfers, e-mail, and other network software services. Telnet and FTP are applications that exist entirely in the application level. Tiered application architectures are part of this layer. Presentation (Layer 6):- This layer provides independence from differences in data representation (e.g., encryption) by translating from application to network format, and vice versa. The presentation layer works to transform data into the form that the application layer can accept. This layer formats and encrypts data to be sent across a network, providing freedom from compatibility problems. It is sometimes called the syntax layer. Session (Layer 5):-
This layer establishes, manages and terminates connections between applications. The session layer sets up, coordinates, and terminates conversations, exchanges, and dialogues between the applications at each end. It deals with session and connection coordination. Transport (Layer 4):- This layer provides transparent transfer of data between end systems, or hosts, and is responsible for end-to-end error recovery and flow control. It ensures complete data transfer. Network (Layer 3):- This layer provides switching and routing technologies, creating logical paths, known as virtual circuits, for transmitting data from node to node. Routing and forwarding are functions of this layer, as well as addressing, internetworking, error handling, congestion control and packet sequencing. Data Link (Layer 2):- At this layer, data packets are encoded and decoded into bits. It furnishes transmission protocol knowledge and management and handles errors in the physical layer, flow control and frame synchronization. The data link layer is divided into two sublayers: The Media Access Control (MAC) layer and the Logical Link Control (LLC) layer. The MAC sub layer controls how a computer on the network gains access to the data and permission to transmit it. The LLC layer controls frame synchronization, flow control and error checking. Physical (Layer 1):- This layer conveys the bit stream - electrical impulse, light or radio signal -- through the network at the electrical and mechanical level. It provides the hardware means of sending and receiving data on a carrier, including defining cables, cards and physical aspects. Fast Ethernet, RS232, and ATM are protocols with physical layer components.
Q3: Define Channel Capacity and Explain Shannon capacity formula? Ans: The maximum rate at which data can be transmitted over a given communication path, or channel, under given conditions is called Channel Capacity. Consider the relationship between data rate and error rate. If the data rate is increased, then more bits are affected by a given pattern noise so at a given noise level, the higher the data rate, the higher the error rate. It means that a greater signal strength would improve the ability to receive data correctly in the presence of noise. Thus Signal to noise ratio is the ratio of the power of the signal to the power contained in the noise. It is shown in decibels as: SNRdb = 10 log10 (signal power/noise power) This SNR ratio is used in Shannon capacity formula i.e. C = B log2 (1+SNR) Where C is the capacity of the channel in bits per second and B is the bandwidth of the channel in Hertz. From equation it can be seen that data rate can be increased by increasing either signal strength or bandwidth. But by increasing signal strength intermodulation noise is increased. Hence by increasing Bandwidth, SNR decreases. Q4: What does Signal propagation mean? Ans: :
Principle of energy transfer (transmission) Or signal propagation If we apply some kind of signal (e.g.: heat, voltage or current etc) at one end of a conducting material this signal is propagated to the other end of the material after a finite time of propagation delay. If we plot the signal versus time at the transmitting end the same graph is reproduced at the receiving end (assuming no distortion) after some propagation delay
Analog Signal An analog signal is continuously varying signal e.g.: • Sinusoidal signal
• Plot of room temperature versus time
Digital Signal Digital signal takes the form of pulses where we have something or nothing Absence of something – 0 OR Presence of something – 1
Q5: A periodic signal has been decomposed using Fourier Analysis to yield four sine waves of frequencies 100, 400,600 and 800 hz. What is the bandwidth of the resulting periodic signal. Ans: The resulting signal has the bandwidth: B = 800-100 = 700 Hz Q6: A digital signaling system is required at 9600 bps.
a) if a signal element encodes 4 bit word, what is the minimum required bandwidth of the channel.
b) Incase of 8-bit word Ans: (a) As C= 2B log2 M And log2 M = 4 where M=no. of levels = 16 So, C = 2B * 4 = 9600 B=1200 Hz
(b) Here, Log2 M = 8,
So C = 2B log2 M 9600=2B * 8
Or B = 600 Hz Q7: Given a channel with intended capacity of 20 Mbps, the bandwidth of the channel is 3 Mhz. Assuming white thermal noise what signal to noise ratio is requiredto achieve this capacity. Ans:
C = 20 Mbps B = 3Mbps C = B log2 (1 + SNR) 20 M = 3 M log2 (1 + SNR)
SNR = 2 (20/3) - 1 = 101 SNR db = 10 log10 (101) = 20.04db Q8: a) Suppose that a digitized TV picture is to be transmitted from a source that uses a matrix of 480 X 500 picture elements (pixels), where each pixel can take on one of 32 intensity values. Assume that 30 pictures are sent per second. (This digital source is roughly equivalent to broadcast TV standards that have been adopted). Find the source rate R (bp). Sol: intensities levels are represented by 5 bits R = (480 X 500) * 5 * 30 = 36 Mbps
b) Assume that the TV picture is to be transmitted over a channel with 4.5
MHz bandwidth and a 35-db signal to noise ratio. Find the capacity of the
channel (bps). (5)
SNR = antilog (35/10)
C = 4.5 X 10 6 * log2(1 + SNR)
= 52.3 Mbps
c) Discuss how the parameters given in part (a) could be modified to allow
transmission of color TV signals without increasing the required value for
Answer:- For RGB we would have to send 32 intensities levels for each color so instead of sending 5 bits for intensities we will send 15 bits overall for the color intensities. Instead of sending 30 pictures we will send 24 pictures and we also change resolution.