22 CHAPTER 1 INTRODUCTION Sites The following sites are related to topics discussed in this chapter. o o www.acm.org/sigcomm/sos.html This site gives the status of varililus networking standards. www.ietf.org/ The Internet Engineering Task Force (IETF) home page. RFCs The following site lists all RFCs, including those related to IP and TCP. In future chapters we cite the RFCs pertinent to the chapter material. o 1.6 www.ietf.org/rfc.html KEY TERMS Advanced Research Projects Agency (ARPA) American National Standards Institute (ANSI) American Standard Code for Information Interchange (ASCII) ARPANET audio backbone Basic Latin bus topology code Consultative Committee for International Telegraphy and Telephony (CCITT) data data communications de facto standards de jure standards delay distributed processing Electronic Industries Association (EIA) entity Federal Communications Commission (FCC) forum full-duplex mode, or duplex half-duplex mode hub image Institute of Electrical and Electronics Engineers (IEEE) International Organization for Standardization (ISO) International Telecommunication Union-Telecommunication Standards Sector (ITU-T) Internet Internet draft Internet service provider (ISP) Internet standard internetwork or internet local area network (LAN) local Internet service providers mesh topology message metropolitan area network (MAN) multipoint or multidrop connection national Internet service provider network SECTION 1.7 network access points (NAPs) node performance physical topology point-to-point connection protocol receiver regional ISP reliability Request for Comment (RFC) ROB ring topology security semantics 1.7 o o o o o o o o o o o o o o o SUMMARY 23 sender simplex mode star topology syntax telecommunication throughput timing Transmission Control Protocol! Internetworking Protocol (TCPIIP) transmission medium Unicode video wide area network (WAN) YCM SUMMARY Data communications are the transfer of data from one device to another via some form of transmission medium. A data communications system must transmit data to the correct destination in an accurate and timely manner. The five components that make up a data communications system are the message, sender, receiver, medium, and protocol. Text, numbers, images, audio, and video are different forms of information. Data flow between two devices can occur in one of three ways: simplex, half-duplex, or full-duplex. A network is a set of communication devices connected by media links. In a point-to-point connection, two and only two devices are connected by a dedicated link. In a multipoint connection, three or more devices share a link. Topology refers to the physical or logical arrangement of a network. Devices may be arranged in a mesh, star, bus, or ring topology. A network can be categorized as a local area network or a wide area network. A LAN is a data communication system within a building, plant, or campus, or between nearby buildings. A WAN is a data communication system spanning states, countries, or the whole world. An internet is a network of networks. The Internet is a collection of many separate networks. There are local, regional, national, and international Internet service providers. A protocol is a set of rules that govern data communication; the key elements of a protocol are syntax, semantics, and timing. 24 CHAPTER 1 INTRODUCTION o o o o Standards are necessary to ensure that products from different manufacturers can work together as expected. The ISO, ITD-T, ANSI, IEEE, and EIA are some of the organizations involved in standards creation. Forums are special-interest groups that quickly evaluate and standardize new technologies. A Request for Comment is an idea or concept that is a precursor to an Internet standard. 1.8 PRACTICE SET Review Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 I. 12. 13. Identify the five components of a data communications system. What are the advantages of distributed processing? What are the three criteria necessary for an effective and efficient network? What are the advantages of a multipoint connection over a point-to-point connection? What are the two types of line configuration? Categorize the four basic topologies in terms of line configuration. What is the difference between half-duplex and full-duplex transmission modes? Name the four basic network topologies, and cite an advantage of each type. For n devices in a network, what is the number of cable links required for a mesh, ring, bus, and star topology? What are some of the factors that determine whether a communication system is a LAN or WAN? What is an internet? What is the Internet? Why are protocols needed? Why are standards needed? Exercises 14. What is the maximum number of characters or symbols that can be represented by Unicode? 15. A color image uses 16 bits to represent a pixel. What is the maximum number of different colors that can be represented? 16. Assume six devices are arranged in a mesh topology. How many cables are needed? How many ports are needed for each device? 17. For each of the following four networks, discuss the consequences if a connection fails. a. Five devices arranged in a mesh topology b. Five devices arranged in a star topology (not counting the hub) c. Five devices arranged in a bus topology d. Five devices arranged in a ring topology SECTION 1.8 PRACTICE SET 25 18. You have two computers connected by an Ethernet hub at home. Is this a LAN, a MAN, or a WAN? Explain your reason. 19. In the ring topology in Figure 1.8, what happens if one of the stations is unplugged? 20. In the bus topology in Figure 1.7, what happens if one ofthe stations is unplugged? 21. Draw a hybrid topology with a star backbone and three ring networks. 22. Draw a hybrid topology with a ring backbone and two bus networks. 23. Performance is inversely related to delay. When you use the Internet, which of the following applications are more sensitive to delay? a. Sending an e-mail b. Copying a file c. Surfing the Internet 24. When a party makes a local telephone call to another party, is this a point-to-point or multipoint connection? Explain your answer. 25. Compare the telephone network and the Internet. What are the similarities? What are the differences? Research Activities 26. 27. 28. 29. Using the site \\iww.cne.gmu.edu/modules/network/osi.html, discuss the OSI model. Using the site www.ansi.org, discuss ANSI's activities. Using the site www.ieee.org, discuss IEEE's activities. Using the site www.ietf.org/, discuss the different types of RFCs. SECTION 2. 7 KEY TERMS 51 Books Network models are discussed in Section 1.3 of [Tan03], Chapter 2 of [For06], Chapter 2 of [Sta04], Sections 2.2 and 2.3 of [GW04], Section 1.3 of [PD03], and Section 1.7 of [KR05]. A good discussion about addresses can be found in Section 1.7 of [Ste94]. Sites The following site is related to topics discussed in this chapter. o www.osi.org! Information about OS1. RFCs The following site lists all RFCs, including those related to IP and port addresses. o 2.7 www.ietLorg/rfc.html KEY TERMS access control Address Resolution Protocol (ARP) application layer best-effort delivery bits connection control data link layer encoding error error control flow control frame header hop-to-hop delivery host-to-host protocol interface Internet Control Message Protocol (ICMP) Internet Group Message Protocol (IGMP) logical addressing mail service network layer node-to-node delivery open system Open Systems Interconnection (OSI) model peer-to-peer process physical addressing physical layer port address presentation layer process-to-process delivery Reverse Address Resolution Protocol (RARP) routing segmentation session layer source-to-destination delivery Stream Control Transmission Protocol (SCTP) synchronization point TCPIIP protocol suite trailer Transmission Control Protocol (TCP) transmission rate transport layer transport level protocols User Datagram Protocol (UDP) 52 CHAPTER 2 NETWORK MODELS 2.8 o U o o D o o o [J D U o o U U o L,J U o SUMMARY The International Standards Organization created a model called the Open Systems Interconnection, which allows diverse systems to communicate. The seven-layer OSI model provides guidelines for the development of universally compatible networking protocols. The physical, data link, and network layers are the network support layers. The session, presentation, and application layers are the user support layers. The transport layer links the network support layers and the user support layers. The physical layer coordinates the functions required to transmit a bit stream over a physical medium. The data link layer is responsible for delivering data units from one station to the next without errors. The network layer is responsible for the source-to-destination delivery of a packet across multiple network links. The transport layer is responsible for the process-to-process delivery of the entire message. The session layer establishes, maintains, and synchronizes the interactions between communicating devices. The presentation layer ensures interoperability between communicating devices through transformation of data into a mutually agreed upon format. The application layer enables the users to access the network. TCP/IP is a five-layer hierarchical protocol suite developed before the OSI model. The TCP/IP application layer is equivalent to the combined session, presentation, and application layers of the OSI model. Four levels of addresses are used in an internet following the TCP/IP protocols: physical (link) addresses, logical (IP) addresses, port addresses, and specific addresses. The physical address, also known as the link address, is the address of a node as defined by its LAN or WAN. The IP address uniquely defines a host on the Internet. The port address identifies a process on a host. A specific address is a user-friendly address. 2.9 PRACTICE SET Review Questions I. List the layers of the Internet model. 2. Which layers in the Internet model are the network support layers? 3. Which layer in the Internet model is the user support layer? 4. What is the difference between network layer delivery and transport layer delivery? SECTION 2.9 PRACTICE SET 53 5. What is a peer-to-peer process? 6. How does information get passed from one layer to the next in the Internet model? 7. What are headers and trailers, and how do they get added and removed? X. What are the concerns of the physical layer in the Internet model? 9. What are the responsibilities of the data link layer in the Internet model? 10. What are the responsibilities of the network layer in the Internet model? II. What are the responsibilities of the transport layer in the Internet model? 12. What is the difference between a port address, a logical address, and a physical address? 13. Name some services provided by the application layer in the Internet model. 14. How do the layers of the Internet model correlate to the layers of the OSI model? Exercises 15. How are OSI and ISO related to each other? 16. Match the following to one or more layers of the OSI model: a. Route determination b. Flow control c. Interface to transmission media d. Provides access for the end user I 7. Match the following to one or more layers of the OSI model: a. Reliable process-to-process message delivery b. Route selection c. Defines frames d. Provides user services such as e-mail and file transfer e. Transmission of bit stream across physical medium \ 8. Match the following to one or more layers of the OSl model: a. Communicates directly with user's application program b. Error correction and retransmission c. Mechanical, electrical, and functional interface d. Responsibility for carrying frames between adjacent nodes I 9. Match the following to one or more layers of the OSI model: a. Format and code conversion services b. Establishes, manages, and terminates sessions c. Ensures reliable transmission of data d. Log-in and log-out procedures e. Provides independence from differences in data representation 20. In Figure 2.22, computer A sends a message to computer D via LANl, router Rl, and LAN2. Show the contents of the packets and frames at the network and data link layer for each hop interface. 54 CHAPTER 2 NETWORK MODELS Figure 2.22 Exercise 20 Rl Sender 8/42 LAN 1 LAN 2 Sender 21. In Figure 2.22, assume that the communication is between a process running at computer A with port address i and a process running at computer D with port address j. Show the contents of packets and frames at the network, data link, and transport layer for each hop. 22. Suppose a computer sends a frame to another computer on a bus topology LAN. The physical destination address of the frame is corrupted during the transmission. What happens to the frame? How can the sender be informed about the situation? 23. Suppose a computer sends a packet at the network layer to another computer somewhere in the Internet. The logical destination address of the packet is corrupted. What happens to the packet? How can the source computer be informed of the situation? 24. Suppose a computer sends a packet at the transport layer to another computer somewhere in the Internet. There is no process with the destination port address running at the destination computer. What will happen? 25. If the data link layer can detect errors between hops, why do you think we need another checking mechanism at the transport layer? Research Activities 26. Give some advantages and disadvantages of combining the session, presentation, and application layer in the OSI model into one single application layer in the Internet model. 27. Dialog control and synchronization are two responsibilities of the session layer in the OSI model. Which layer do you think is responsible for these duties in the Internet model? Explain your answer. 28. Translation, encryption, and compression are some of the duties of the presentation layer in the OSI model. Which layer do you think is responsible for these duties in the Internet model? Explain your answer. 29. There are several transport layer models proposed in the OSI model. Find all of them. Explain the differences between them. 30. There are several network layer models proposed in the OSI model. Find all of them. Explain the differences between them. 94 CHAPTER 3 DATA AND SIGNALS Jitter Another performance issue that is related to delay is jitter. We can roughly say that jitter is a problem if different packets of data encounter different delays and the application using the data at the receiver site is time-sensitive (audio and video data, for example). If the delay for the first packet is 20 ms, for the second is 45 ms, and for the third is 40 ms, then the real-time application that uses the packets endures jitter. We discuss jitter in greater detail in Chapter 29. 3.7 RECOMMENDED READING For more details about subjects discussed in this chapter, we recommend the following books. The items in brackets [...] refer to the reference list at the end of the text. Books Data and signals are elegantly discussed in Chapters 1 to 6 of [Pea92]. [CouOl] gives an excellent coverage about signals in Chapter 2. More advanced materials can be found in [Ber96]. [Hsu03] gives a good mathematical approach to signaling. Complete coverage of Fourier Analysis can be found in [Spi74]. Data and signals are discussed in Chapter 3 of [Sta04] and Section 2.1 of [Tan03]. 3.8 KEY TERMS analog analog data analog signal attenuation bandpass channel bandwidth baseband transmission bit rate bits per second (bps) broadband transmission composite signal cycle decibel (dB) digital digital data digital signal distortion Fourier analysis frequency frequency-domain fundamental frequency harmonic Hertz (Hz) jitter low-pass channel noise nonperiodic signal Nyquist bit rate peak amplitude period periodic signal phase processing delay propagation speed SECTION 3.9 propagation time queuing time Shannon capacity signal signal-to-noise ratio (SNR) 3.9 o o o o o o o o o o o o o o o o o o o o SUMMARY 95 sine wave throughput time-domain transmission time wavelength SUMMARY Data must be transformed to electromagnetic signals to be transmitted. Data can be analog or digital. Analog data are continuous and take continuous values. Digital data have discrete states and take discrete values. Signals can be analog or digital. Analog signals can have an infinite number of values in a range; digital ,signals can have only a limited number of values. In data communications, we commonly use periodic analog signals and nonperiodic digital signals. Frequency and period are the inverse of each other. Frequency is the rate of change with respect to time. Phase describes the position of the waveform relative to time O. A complete sine wave in the time domain can be represented by one single spike in the frequency domain. A single-frequency sine wave is not useful in data communications; we need to send a composite signal, a signal made of many simple sine waves. According to Fourier analysis, any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases. The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal. A digital signal is a composite analog signal with an infinite bandwidth. Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth. If the available channel is a bandpass channel, we cannot send a digital signal directly to the channel; we need to convert the digital signal to an analog signal before transmission. For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rate. For a noisy channel, we need to use the Shannon capacity to find the maximum bit rate. Attenuation, distortion, and noise can impair a signal. Attenuation is the loss of a signal's energy due to the resistance of the medium. Distortion is the alteration of a signal due to the differing propagation speeds of each of the frequencies that make up a signal. Noise is the external energy that corrupts a signal. The bandwidth-delay product defines the number of bits that can fill the link. 96 CHAPTER 3 DATA AND SIGNALS 3.10 PRACTICE SET Review Questions 1. What is the relationship between period and frequency? 2. What does the amplitude of a signal measure? What does the frequency of a signal measure? What does the phase of a signal measure? 3. How can a composite signal be decomposed into its individual frequencies? 4. Name three types of transmission impairment. 5. Distinguish between baseband transmission and broadband transmission. 6. Distinguish between a low-pass channel and a band-pass channel. 7. What does the Nyquist theorem have to do with communications? 8. What does the Shannon capacity have to do with communications? 9. Why do optical signals used in fiber optic cables have a very short wave length? 10. Can we say if a signal is periodic or nonperiodic by just looking at its frequency domain plot? How? 11. Is the frequency domain plot of a voice signal discrete or continuous? 12. Is the frequency domain plot of an alarm system discrete or continuous? 13. We send a voice signal from a microphone to a recorder. Is this baseband or broadband transmission? 14. We send a digital signal from one station on a LAN to another station. Is this baseband or broadband transmission? 15. We modulate several voice signals and send them through the air. Is this baseband or broadband transmission? Exercises 16. Given the frequencies listed below, calculate the corresponding periods. a. 24Hz b. 8 MHz c. 140 KHz 17. Given the following periods, calculate the corresponding frequencies. a. 5 s b. 12 Jls c. 220 ns 18. What is the phase shift for the foIlowing? a. A sine wave with the maximum amplitude at time zero b. A sine wave with maximum amplitude after 1/4 cycle c. A sine wave with zero amplitude after 3/4 cycle and increasing 19. What is the bandwidth of a signal that can be decomposed into five sine waves with frequencies at 0, 20, 50, 100, and 200 Hz? All peak amplitudes are the same. Draw the bandwidth. SECTION 3.10 PRACTICE SET 97 20. A periodic composite signal with a bandwidth of 2000 Hz is composed of two sine waves. The first one has a frequency of 100 Hz with a maximum amplitude of 20 V; the second one has a maximum amplitude of 5 V. Draw the bandwidth. 21. Which signal has a wider bandwidth, a sine wave with a frequency of 100 Hz or a sine wave with a frequency of 200 Hz? 22. What is the bit rate for each of the following signals? a. A signal in which 1 bit lasts 0.001 s b. A signal in which 1 bit lasts 2 ms c. A signal in which 10 bits last 20 J-ls 23. A device is sending out data at the rate of 1000 bps. a. How long does it take to send out 10 bits? b. How long does it take to send out a single character (8 bits)? c. How long does it take to send a file of 100,000 characters? 24. What is the bit rate for the signal in Figure 3.34? Figure 3.34 Exercise 24 16 ns ---t=:i--J~-...;..---+-.....;.--.;....-d ... ~ TI~ 1 25. What is the frequency of the signal in Figure 3.35? Figure 3.35 Exercise 25 4ms I, .1 1 I V\ f\ f\ f\ f\ f\ f\ f\ : ... \TV V VVV V~ Time 26. What is the bandwidth of the composite signal shown in Figure 3.36. Figure 3.36 Exercise 26 Frequency 5 5 5 5 5 I 9S CHAPTER 3 DATA AND SIGNALS 27. A periodic composite signal contains frequencies from 10 to 30 KHz, each with an amplitude of 10 V. Draw the frequency spectrum. 2K. A non-periodic composite signal contains frequencies from 10 to 30 KHz. The peak amplitude is 10 V for the lowest and the highest signals and is 30 V for the 20-KHz signal. Assuming that the amplitudes change gradually from the minimum to the maximum, draw the frequency spectrum. 20. A TV channel has a bandwidth of 6 MHz. If we send a digital signal using one channel, what are the data rates if we use one harmonic, three harmonics, and five harmonics? 30. A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the attenuation in decibels? 31. The attenuation of a signal is -10 dB. What is the final signal power if it was originally 5 W? 32. A signal has passed through three cascaded amplifiers, each with a 4 dB gain. What is the total gain? How much is the signal amplified? 33. If the bandwidth of the channel is 5 Kbps, how long does it take to send a frame of 100,000 bits out of this device? 3cf. The light of the sun takes approximately eight minutes to reach the earth. What is the distance between the sun and the earth? 35. A signal has a wavelength of 1 11m in air. How far can the front of the wave travel during 1000 periods? 36. A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is the maximum data rate supported by this line? 37. We measure the performance of a telephone line (4 KHz of bandwidth). When the signal is 10 V, the noise is 5 mV. What is the maximum data rate supported by this telephone line? 3X. A file contains 2 million bytes. How long does it take to download this file using a 56-Kbps channel? 1-Mbps channel? 39. A computer monitor has a resolution of 1200 by 1000 pixels. If each pixel uses 1024 colors, how many bits are needed to send the complete contents of a screen? 40. A signal with 200 milliwatts power passes through 10 devices, each with an average noise of 2 microwatts. What is the SNR? What is the SNRdB ? 4 I. If the peak voltage value of a signal is 20 times the peak voltage value of the noise, what is the SNR? What is the SNR dB ? 42. What is the theoretical capacity of a channel in each of the following cases: a. Bandwidth: 20 KHz SNRdB =40 b. Bandwidth: 200 KHz SNRdB =4 c. Bandwidth: 1 MHz SNRdB =20 43. We need to upgrade a channel to a higher bandwidth. Answer the following questions: a. How is the rate improved if we double the bandwidth? b. How is the rate improved if we double the SNR? SECTION 3.10 PRACTICE SET 99 44. We have a channel with 4 KHz bandwidth. If we want to send data at 100 Kbps, what is the minimum SNRdB ? What is SNR? 45. What is the transmission time of a packet sent by a station if the length of the packet is 1 million bytes and the bandwidth of the channel is 200 Kbps? 46. What is the length of a bit in a channel with a propagation speed of 2 x 108 mls if the channel bandwidth is a. 1 Mbps? h. 10 Mbps? c. 100 Mbps? -1- 7. How many bits can fit on a link with a 2 ms delay if the bandwidth of the link is a. 1 Mbps? h. 10 Mbps? c. 100 Mbps? -1-X. What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2 Ils and a processing time of 1 Ils. The length of the link is 2000 Km. The speed of light inside the link is 2 x 108 mls. The link has a bandwidth of 5 Mbps. Which component of the total delay is dominant? Which one is negligible? CHAPTER 1 Introduction Solutions to Review Questions and Exercises Review Questions 1. The five components of a data communication system are the sender, receiver, transmission medium, message, and protocol. 2. The advantages of distributed processing are security, access to distributed databases, collaborative processing, and faster problem solving. 3. The three criteria are performance, reliability, and security. 4. Advantages of a multipoint over a point-to-point configuration (type of connection) include ease of installation and low cost. 5. Line configurations (or types of connections) are point-to-point and multipoint. 6. We can divide line configuration in two broad categories: a. Point-to-point: mesh, star, and ring. b. Multipoint: bus 7. In half-duplex transmission, only one entity can send at a time; in a full-duplex transmission, both entities can send at the same time. 8. We give an advantage for each of four network topologies: a. Mesh: secure b. Bus: easy installation c. Star: robust d. Ring: easy fault isolation 9. The number of cables for each type of network is: a. Mesh: n (n – 1) / 2 b. Star: n c. Ring: n – 1 d. Bus: one backbone and n drop lines 10. The general factors are size, distances (covered by the network), structure, and ownership. 1 2 11. An internet is an interconnection of networks. The Internet is the name of a specific worldwide network 12. A protocol defines what is communicated, in what way and when. This provides accurate and timely transfer of information between different devices on a network. 13. Standards are needed to create and maintain an open and competitive market for manufacturers, to coordinate protocol rules, and thus guarantee compatibility of data communication technologies. Exercises 14. Unicode uses 32 bits to represent a symbol or a character. We can define 232 different symbols or characters. 15. With 16 bits, we can represent up to 216 different colors. 16. a. Cable links: n (n – 1) / 2 = (6 × 5) / 2 = 15 b. Number of ports: (n – 1) = 5 ports needed per device 17. a. Mesh topology: If one connection fails, the other connections will still be working. b. Star topology: The other devices will still be able to send data through the hub; there will be no access to the device which has the failed connection to the hub. c. Bus Topology: All transmission stops if the failure is in the bus. If the drop-line fails, only the corresponding device cannot operate. d. Ring Topology: The failed connection may disable the whole network unless it is a dual ring or there is a by-pass mechanism. 18. This is a LAN. The Ethernet hub creates a LAN as we will see in Chapter 13. 19. Theoretically, in a ring topology, unplugging one station, interrupts the ring. However, most ring networks use a mechanism that bypasses the station; the ring can continue its operation. 20. In a bus topology, no station is in the path of the signal. Unplugging a station has no effect on the operation of the rest of the network. 21. See Figure 1.1 22. See Figure 1.2. 23. a. E-mail is not an interactive application. Even if it is delivered immediately, it may stay in the mail-box of the receiver for a while. It is not sensitive to delay. b. We normally do not expect a file to be copied immediately. It is not very sensitive to delay. c. Surfing the Internet is the an application very sensitive to delay. We except to get access to the site we are searching. 24. In this case, the communication is only between a caller and the callee. A dedicated line is established between them. The connection is point-to-point. 3 Figure 1.1 Solution to Exercise 21 Hub Station Repeater Station Station Station Station Repeat er Repeat er Station Station Station Station Figure 1.2 Solution to Exercise 22 Station Station Repeater Repeater Station Station 25. The telephone network was originally designed for voice communication; the Internet was originally designed for data communication. The two networks are similar in the fact that both are made of interconnections of small networks. The telephone network, as we will see in future chapters, is mostly a circuit-switched network; the Internet is mostly a packet-switched network. 4 Sol-02.fm Page 1 Saturday, January 21, 2006 10:27 AM CHAPTER 2 Network Models Solutions to Review Questions and Exercises Review Questions 1. The Internet model, as discussed in this chapter, include physical, data link, network, transport, and application layers. 2. The network support layers are the physical, data link, and network layers. 3. The application layer supports the user. 4. The transport layer is responsible for process-to-process delivery of the entire message, whereas the network layer oversees host-to-host delivery of individual packets. 5. Peer-to-peer processes are processes on two or more devices communicating at a same layer 6. Each layer calls upon the services of the layer just below it using interfaces between each pair of adjacent layers. 7. Headers and trailers are control data added at the beginning and the end of each data unit at each layer of the sender and removed at the corresponding layers of the receiver. They provide source and destination addresses, synchronization points, information for error detection, etc. 8. The physical layer is responsible for transmitting a bit stream over a physical medium. It is concerned with a. physical characteristics of the media b. representation of bits c. type of encoding d. synchronization of bits e. transmission rate and mode f. the way devices are connected with each other and to the links 9. The data link layer is responsible for a. framing data bits b. providing the physical addresses of the sender/receiver c. data rate control 1 Sol-02.fm Page 2 Saturday, January 21, 2006 10:27 AM 2 10. 11. 12. 13. 14. d. detection and correction of damaged and lost frames The network layer is concerned with delivery of a packet across multiple networks; therefore its responsibilities include a. providing host-to-host addressing b. routing The transport layer oversees the process-to-process delivery of the entire message. It is responsible for a. dividing the message into manageable segments b. reassembling it at the destination c. flow and error control The physical address is the local address of a node; it is used by the data link layer to deliver data from one node to another within the same network. The logical address defines the sender and receiver at the network layer and is used to deliver messages across multiple networks. The port address (service-point) identifies the application process on the station. The application layer services include file transfer, remote access, shared database management, and mail services. The application, presentation, and session layers of the OSI model are represented by the application layer in the Internet model. The lowest four layers of OSI correspond to the Internet model layers. Exercises 15. The International Standards Organization, or the International Organization of Standards, (ISO) is a multinational body dedicated to worldwide agreement on international standards. An ISO standard that covers all aspects of network communications is the Open Systems Interconnection (OSI) model. 16. a. Route determination: network layer b. Flow control: data link and transport layers c. Interface to transmission media: physical layer d. Access for the end user: application layer 17. a. Reliable process-to-process delivery: transport layer b. Route selection: network layer c. Defining frames: data link layer d. Providing user services: application layer e. Transmission of bits across the medium: physical layer 18. a. Communication with user’s application program: application layer b. Error correction and retransmission: data link and transport layers c. Mechanical, electrical, and functional interface: physical layer Sol-02.fm Page 3 Saturday, January 21, 2006 10:27 AM 3 d. Responsibility for carrying frames between adjacent nodes: data link layer 19. a. Format and code conversion services: presentation layer b. Establishing, managing, and terminating sessions: session layer c. Ensuring reliable transmission of data: data link and transport layers d. Log-in and log-out procedures: session layer e. Providing independence from different data representation: presentation layer 20. See Figure 2.1. Figure 2.1 Solution to Exercise 20 A/40 LAN1 LAN2 R1 Sender B/42 D/80 C/82 Sender 80 82 A D Data T2 42 40 A D Data T2 21. See Figure 2.2. Figure 2.2 Solution to Exercise 21 LAN1 A/40 LAN2 R1 Sender B/42 D/80 C/82 Sender 42 40 A D i j Data T2 80 82 A D i j Data T2 22. If the corrupted destination address does not match any station address in the network, the packet is lost. If the corrupted destination address matches one of the stations, the frame is delivered to the wrong station. In this case, however, the error detection mechanism, available in most data link protocols, will find the error and discard the frame. In both cases, the source will somehow be informed using one of the data link control mechanisms discussed in Chapter 11. 23. Before using the destination address in an intermediate or the destination node, the packet goes through error checking that may help the node find the corruption (with a high probability) and discard the packet. Normally the upper layer protocol will inform the source to resend the packet. Sol-02.fm Page 4 Saturday, January 21, 2006 10:27 AM 4 24. Most protocols issue a special error message that is sent back to the source in this case. 25. The errors between the nodes can be detected by the data link layer control, but the error at the node (between input port and output port) of the node cannot be detected by the data link layer. CHAPTER 3 Data and Signals Solutions to Review Questions and Exercises Review Questions 1. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T. 2. The amplitude of a signal measures the value of the signal at any point. The frequency of a signal refers to the number of periods in one second. The phase describes the position of the waveform relative to time zero. 3. Using Fourier analysis. Fourier series gives the frequency domain of a periodic signal; Fourier analysis gives the frequency domain of a nonperiodic signal. 4. Three types of transmission impairment are attenuation, distortion, and noise. 5. Baseband transmission means sending a digital or an analog signal without modulation using a low-pass channel. Broadband transmission means modulating a digital or an analog signal using a band-pass channel. 6. A low-pass channel has a bandwidth starting from zero; a band-pass channel has a bandwidth that does not start from zero. 7. The Nyquist theorem defines the maximum bit rate of a noiseless channel. 8. The Shannon capacity determines the theoretical maximum bit rate of a noisy channel. 9. Optical signals have very high frequencies. A high frequency means a short wave length because the wave length is inversely proportional to the frequency (λ = v/f), where v is the propagation speed in the media. 10. A signal is periodic if its frequency domain plot is discrete; a signal is nonperiodic if its frequency domain plot is continuous. 11. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal. 12. An alarm system is normally periodic. Its frequency domain plot is therefore discrete. 13. This is baseband transmission because no modulation is involved. 14. This is baseband transmission because no modulation is involved. 15. This is broadband transmission because it involves modulation. 1 2 Exercises 16. a. T = 1 / f = 1 / (24 Hz) = 0.0417 s = 41.7 × 10–3 s = 41.7 ms b. T = 1 / f = 1 / (8 MHz) = 0.000000125 = 0.125 × 10–6 s = 0.125 μs c. T = 1 / f = 1 / (140 KHz) = 0.00000714 s = 7.14 × 10–6 s = 7.14 μs 17. a. f = 1 / T = 1 / (5 s) = 0.2 Hz b. f = 1 / T = 1 / (12 μs) =83333 Hz = 83.333 × 103 Hz = 83.333 KHz c. f = 1 / T = 1 / (220 ns) = 4550000 Hz = 4.55× 106 Hz = 4.55 MHz 18. a. 90 degrees (π/2 radian) b. 0 degrees (0 radian) c. 90 degrees (π/2 radian) 19. See Figure 3.1 Figure 3.1 Solution to Exercise 19 Frequency domain 0 20 100 50 200 Bandwidth = 200 − 0 = 200 20. We know the lowest frequency, 100. We know the bandwidth is 2000. The highest frequency must be 100 + 2000 = 2100 Hz. See Figure 3.2 Figure 3.2 Solution to Exercise 20 20 Frequency domain 5 100 2100 Bandwidth = 2100 − 100 = 2000 21. Each signal is a simple signal in this case. The bandwidth of a simple signal is zero. So the bandwidth of both signals are the same. 22. a. bit rate = 1/ (bit duration) = 1 / (0.001 s) = 1000 bps = 1 Kbps b. bit rate = 1/ (bit duration) = 1 / (2 ms) = 500 bps 3 c. bit rate = 1/(bit duration) = 1 / (20 μs/10) = 1 / (2 μs) = 500 Kbps 23. 24. 25. 26. 27. a. (10 / 1000) s = 0.01 s b. (8 / 1000) s = 0. 008 s = 8 ms c. ((100,000 × 8) / 1000) s = 800 s There are 8 bits in 16 ns. Bit rate is 8 / (16 × 10−9) = 0.5 × 10−9 = 500 Mbps The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz The bandwidth is 5 × 5 = 25 Hz. The signal is periodic, so the frequency domain is made of discrete frequencies. as shown in Figure 3.3. Figure 3.3 Solution to Exercise 27 Amplitude 10 volts ... 10 KHz Frequency 30 KHz 28. The signal is nonperiodic, so the frequency domain is made of a continuous spectrum of frequencies as shown in Figure 3.4. Figure 3.4 Solution to Exercise 28 30 volts Amplitude 10 volts 10 volts Frequency 10 KHz 29. 20 KHz 30 KHz Using the first harmonic, data rate = 2 × 6 MHz = 12 Mbps Using three harmonics, data rate = (2 × 6 MHz) /3 = 4 Mbps Using five harmonics, data rate = (2 × 6 MHz) /5 = 2.4 Mbps 30. dB = 10 log10 (90 / 100) = –0.46 dB 31. –10 = 10 log10 (P2 / 5) → log10 (P2 / 5) = −1 → (P2 / 5) = 10−1 → P2 = 0.5 W 32. The total gain is 3 × 4 = 12 dB. The signal is amplified by a factor 101.2 = 15.85. 4 33. 34. 35. 36. 100,000 bits / 5 Kbps = 20 s 480 s × 300,000 km/s = 144,000,000 km 1 μm × 1000 = 1000 μm = 1 mm We have 4,000 log2 (1 + 1,000) ≈ 40 Kbps 37. We have 4,000 log2 (1 + 10 / 0.005) = 43,866 bps 38. The file contains 2,000,000 × 8 = 16,000,000 bits. With a 56-Kbps channel, it takes 16,000,000/56,000 = 289 s. With a 1-Mbps channel, it takes 16 s. 39. To represent 1024 colors, we need log21024 = 10 (see Appendix C) bits. The total number of bits are, therefore, 1200 × 1000 × 10 = 12,000,000 bits 40. We have SNR = (200 mW) / (10 × 2 × μW) = 10,000 We then have SNRdB = 10 log10 SNR = 40 41. We have SNR= (signal power)/(noise power). However, power is proportional to the square of voltage. This means we have SNR = [(signal voltage)2] / [(noise voltage)2] = [(signal voltage) / (noise voltage)]2 = 202 = 400 We then have SNRdB = 10 log10 SNR ≈ 26.02 42. We can approximately calculate the capacity as a. C = B × (SNRdB /3) = 20 KHz × (40 /3) = 267 Kbps b. C = B × (SNRdB /3) = 200 KHz × (4 /3) = 267 Kbps c. C = B × (SNRdB /3) = 1 MHz × (20 /3) = 6.67 Mbps 43. a. The data rate is doubled (C2 = 2 × C1). b. When the SNR is doubled, the data rate increases slightly. We can say that, approximately, (C2 = C1 + 1). 44. We can use the approximate formula C = B × (SNRdB /3) or SNRdB = (3 × C) /B We can say that the minimum SNRdB = 3 × 100 Kbps / 4 KHz = 75 5 This means that the minimum SNR = 10 SNRdB/10 = 107.5 ≈ 31,622,776 45. We have transmission time = (packet length)/(bandwidth) = (8,000,000 bits) / (200,000 bps) = 40 s 46. We have (bit length) = (propagation speed) × (bit duration) The bit duration is the inverse of the bandwidth. a. Bit length = (2 ×108 m) × [(1 / (1 Mbps)] = 200 m. This means a bit occupies 200 meters on a transmission medium. b. Bit length = (2 ×108 m) × [(1 / (10 Mbps)] = 20 m. This means a bit occupies 20 meters on a transmission medium. c. Bit length = (2 ×108 m) × [(1 / (100 Mbps)] = 2 m. This means a bit occupies 2 meters on a transmission medium. 47. a. Number of bits = bandwidth × delay = 1 Mbps × 2 ms = 2000 bits b. Number of bits = bandwidth × delay = 10 Mbps × 2 ms = 20,000 bits c. Number of bits = bandwidth × delay = 100 Mbps × 2 ms = 200,000 bits 48. We have Latency = processing time + queuing time + transmission time + propagation time Processing time = 10 × 1 μs = 10 μs = 0.000010 s Queuing time = 10 × 2 μs = 20 μs = 0.000020 s Transmission time = 5,000,000 / (5 Mbps) = 1 s Propagation time = (2000 Km) / (2 × 108) = 0.01 s Latency = 0.000010 + 0.000020 + 1 + 0.01 = 1.01000030 s The transmission time is dominant here because the packet size is huge. 6 SECTION 4.5 KEY TERMS 135 The advantage of synchronous transmission is speed. With no extra bits or gaps to introduce at the sending end and remove at the receiving end, and, by extension, with fewer bits to move across the link, synchronous transmission is faster than asynchronous transmission. For this reason, it is more useful for high-speed applications such as the transmission of data from one computer to another. Byte synchronization is accomplished in the data link layer. We need to emphasize one point here. Although there is no gap between characters in synchronous serial transmission, there may be uneven gaps between frames. Isochronous In real-time audio and video, in which uneven delays between frames are not acceptable, synchronous transmission fails. For example, TV images are broadcast at the rate of 30 images per second; they must be viewed at the same rate. If each image is sent by using one or more frames, there should be no delays between frames. For this type of application, synchronization between characters is not enough; the entire stream of bits must be synchronized. The isochronous transmission guarantees that the data arrive at a fixed rate. 4.4 RECOMMENDED READING For more details about subjects discussed in this chapter, we recommend the following books. The items in brackets [...] refer to the reference list at the end of the text. Books Digital to digital conversion is discussed in Chapter 7 of [Pea92], Chapter 3 of [CouOl], and Section 5.1 of [Sta04]. Sampling is discussed in Chapters 15, 16, 17, and 18 of [Pea92], Chapter 3 of [CouO!], and Section 5.3 of [Sta04]. [Hsu03] gives a good mathematical approach to modulation and sampling. More advanced materials can be found in [Ber96]. 4.5 KEY TERMS adaptive delta modulation alternate mark inversion (AMI) analog-to-digital conversion asynchronous transmission baseline baseline wandering baud rate biphase bipolar bipolar with 8-zero substitution (B8ZS) bit rate block coding companding and expanding data element data rate DC component delta modulation (DM) differential Manchester digital-to-digital conversion digitization 136 CHAPTER 4 DIGITAL TRANSMISSION eight binary/ten binary (8B/lOB) eight-binary, six-ternary (8B6T) four binary/five binary (4B/5B) four dimensional, five-level pulse amplitude modulation (4D-PAM5) high-density bipolar 3-zero (HDB3) isochronous transmission line coding Manchester modulation rate multilevel binary multiline transmission, 3 level (MLT-3) pulse amplitude modulation (PAM) pulse code modulation (PCM) pulse rate quantization quantization error return to zero (RZ) sampling sampling rate scrambling self-synchronizing serial transmission signal element nonreturn to zero (NRZ) signal rate nonreturn to zero, invert (NRZ-I) nonreturn to zero, level (NRZ-L) Nyquist theorem parallel transmission start bit stop bit synchronous transmission transmission mode polar pseudoternary two-binary, one quaternary (2B I Q) unipolar 4.6 o o o o o o o o SUMMARY Digital-to-digital conversion involves three techniques: line coding, block coding, and scrambling. Line coding is the process of converting digital data to a digital signal. We can roughly divide line coding schemes into five broad categories: unipolar, polar, bipolar, multilevel, and multitransition. Block coding provides redundancy to ensure synchronization and inherent error detection. Block coding is normally referred to as mB/nB coding; it replaces each m-bit group with an n-bit group. Scrambling provides synchronization without increasing the number of bits. Two common scrambling techniques are B8ZS and HDB3. The most common technique to change an analog signal to digital data (digitization) is called pulse code modulation (PCM). The first step in PCM is sampling. The analog signal is sampled every Ts s, where Ts is the sample interval or period. The inverse of the sampling interval is called the sampling rate or sampling frequency and denoted by fs, where fs = lITs. There are three sampling methods-ideal, natural, and flat-top. According to the Nyquist theorem, to reproduce the original analog signal, one necessary condition is that the sampling rate be at least twice the highest frequency in the original signal. SECTION 4.7 o o o o o 4.7 PRACTICE SET 137 Other sampling techniques have been developed to reduce the complexity of PCM. The simplest is delta modulation. PCM finds the value of the signal amplitude for each sample; DM finds the change from the previous sample. While there is only one way to send parallel data, there are three subclasses of serial transmission: asynchronous, synchronous, and isochronous. In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1 s) at the end of each byte. In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits. The isochronous mode provides synchronized for the entire stream of bits must. In other words, it guarantees that the data arrive at a fixed rate. PRACTICE SET Review Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. List three techniques of digital-to-digital conversion. Distinguish between a signal element and a data element. Distinguish between data rate and signal rate. Define baseline wandering and its effect on digital transmission. Define a DC component and its effect on digital transmission. Define the characteristics of a self-synchronizing signal. List five line coding schemes discussed in this book. Define block coding and give its purpose. Define scrambling and give its purpose. Compare and contrast PCM and DM. What are the differences between parallel and serial transmission? List three different techniques in serial transmission and explain the differences. Exercises 13. Calculate the value of the signal rate for each case in Figure 4.2 if the data rate is 1 Mbps and c = 1/2. 14. In a digital transmission, the sender clock is 0.2 percent faster than the receiver clock. How many extra bits per second does the sender send if the data rate is 1 Mbps? 15. Draw the graph of the NRZ-L scheme using each of the following data streams, assuming that the last signa11evel has been positive. From the graphs, guess the bandwidth for this scheme using the average number of changes in the signal level. Compare your guess with the corresp.onding entry in Table 4.1. a. 00000000 b. 11111111 c. 01010101 d. 00110011 138 CHAPTER 4 DIGITAL TRANSMISSION 16. 17. 18. 19. Repeat Exercise 15 for the NRZ-I scheme. Repeat Exercise 15 for the Manchester scheme. Repeat Exercise 15 for the differential Manchester scheme. Repeat Exercise 15 for the 2B 1Q scheme, but use the following data streams. a. 0000000000000000 b. 1111111111111111 c. 0101010101010101 d. 0011001100110011 20. Repeat Exercise 15 for the MLT-3 scheme, but use the following data streams. a. 00000000 b. 11111111 c. 01010101 d. 00011000 21. Find the 8-bit data stream for each case depicted in Figure 4.36. Figure 4.36 Exercise 21 t Time a. NRZ-I Time b. differential Manchester Time c.AMI 22. An NRZ-I signal has a data rate of 100 Kbps. Using Figure 4.6, calculate the value of the normalized energy (P) for frequencies at 0 Hz, 50 KHz, and 100 KHz. 23. A Manchester signal has a data rate of 100 Kbps. Using Figure 4.8, calculate the value of the normalized energy (P) for frequencies at 0 Hz, 50 KHz, 100 KHz. SECTION 4. 7 PRACTICE SET 139 24. The input stream to a 4B/5B block encoder is 0100 0000 0000 0000 0000 OOOI. Answer the following questions: a. What is the output stream? b. What is the length of the longest consecutive sequence of Os in the input? c. What is the length of the longest consecutive sequence of Os in the output? 25. How many invalid (unused) code sequences can we have in 5B/6B encoding? How many in 3B/4B encoding? 26. What is the result of scrambling the sequence 11100000000000 using one of the following scrambling techniques? Assume that the last non-zero signal level has been positive. a. B8ZS b. HDB3 (The number of nonzero pules is odd after the last substitution) 27. What is the Nyquist sampling rate for each of the following signals? a. A low-pass signal with bandwidth of 200 KHz? b. A band-pass signal with bandwidth of 200 KHz if the lowest frequency is 100 KHz? 28. We have sampled a low-pass signal with a bandwidth of 200 KHz using 1024 levels of quantization. a. Calculate the bit rate of the digitized signal. b. Calculate the SNRdB for this signal. c. Calculate the PCM bandwidth of this signal. 29. What is the maximum data rate of a channel with a bandwidth of 200 KHz if we use four levels of digital signaling. 30. An analog signal has a bandwidth of 20 KHz. If we sample this signal and send it through a 30 Kbps channel what is the SNRdB ? 31. We have a baseband channel with a I-MHz bandwidth. What is the data rate for this channel if we use one of the following line coding schemes? a. NRZ-L b. Manchester c. MLT-3 d. 2B1Q 32. We want to transmit 1000 characters with each character encoded as 8 bits. a. Find the number of transmitted bits for synchronous transmission. b. Find the number of transmitted bits for asynchronous transmission. c. Find the redundancy percent in each case. SECTION 5.5 SUMMARY 157 gives a good mathematical approach to all materials discussed in this chapter. More advanced materials can be found in [Ber96]. 5.4 KEY TERMS amplitude modulation (AM) amplitude shift keying (ASK) analog-to-analog conversion carrier signal constellation diagram digi tal-to-analog conversion 5.5 o o o o o o o o o o o frequency modulation (PM) frequency shift keying (FSK) phase modulation (PM) phase shift keying (PSK) quadrature amplitude modulation (QAM) SUMMARY Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the information in the digital data. Digital-to-analog conversion can be accomplished in several ways: amplitude shift keying (ASK), frequency shift keying (FSK), and phase shift keying (PSK). Quadrature amplitude modulation (QAM) combines ASK and PSK. In amplitude shift keying, the amplitude of the carrier signal is varied to create signal elements. Both frequency and phase remain constant while the amplitude changes. In frequency shift keying, the frequency of the carrier signal is varied to represent data. The frequency of the modulated signal is constant for the duration of one signal element, but changes for the next signal element if the data element changes. Both peak amplitude and phase remain constant for all signal elements. In phase shift keying, the phase of the carrier is varied to represent two or more different signal elements. Both peak amplitude and frequency remain constant as the phase changes. A constellation diagram shows us the amplitude and phase of a signal element, particularly when we are using two carriers (one in-phase and one quadrature). Quadrature amplitude modulation (QAM) is a combination of ASK and PSK. QAM uses two carriers, one in-phase and the other quadrature, with different amplitude levels for each carrier. Analog-to-analog conversion is the representation of analog information by an analog signal. Conversion is needed if the medium is bandpass in nature or if only a bandpass bandwidth is available to us. Analog-to-analog conversion can be accomplished in three ways: amplitude modulation (AM), frequency modulation (FM), and phase modulation (PM). In AM transmission, the carrier signal is modulated so that its amplitude varies with the changing amplitudes of the modulating signal. The frequency and phase of the carrier remain the same; only the amplitude changes to follow variations in the information. In PM transmission, the frequency of the carrier signal is modulated to follow the changing voltage level (amplitude) of the modulating signal. The peak amplitude 158 CHAPTER 5 ANALOG TRANSMISSION o and phase of the carrier signal remain constant, but as the amplitude of the information signal changes, the frequency of the carrier changes correspondingly. In PM transmission, the phase of the carrier signal is modulated to follow the changing voltage level (amplitude) of the modulating signal. The peak amplitude and frequency of the carrier signal remain constant, but as the amplitude of the information signal changes, the phase of the carrier changes correspondingly. 5.6 PRACTICE SET Review Questions 1. Define analog transmission. 2. Define carrier signal and its role in analog transmission. 3. Define digital-to-analog conversion. 4. Which characteristics of an analog signal are changed to represent the digital signal in each of the following digital-to-analog conversion? a. ASK b. FSK c. PSK d. QAM 5. Which of the four digital-to-analog conversion techniques (ASK, FSK, PSK or QAM) is the most susceptible to noise? Defend your answer. 6. Define constellation diagram and its role in analog transmission. 7. What are the two components of a signal when the signal is represented on a con- . stellation diagram? Which component is shown on the horizontal axis? Which is shown on the vertical axis? 8. Define analog-to-analog conversion? 9. Which characteristics of an analog signal are changed to represent the lowpass analog signal in each of the following analog-to-analog conversions? a. AM b. FM c. PM ] 0. Which of the three analog-to-analog conversion techniques (AM, FM, or PM) is the most susceptible to noise? Defend your answer. Exercises 11. Calculate the baud rate for the given bit rate and type of modulation. a. 2000 bps, FSK b. 4000 bps, ASK c. 6000 bps, QPSK d. 36,000 bps, 64-QAM SECTION 5.6 PRACTICE SET 159 12. Calculate the bit rate for the given baud rate and type of modulation. a. 1000 baud, FSK b. 1000 baud, ASK c. 1000 baud, BPSK d. 1000 baud, 16-QAM 13. What is the number of bits per baud for the following techniques? a. ASK with four different amplitudes b. FSK with 8 different frequencies c. PSK with four different phases d. QAM with a constellation of 128 points. 14. Draw the constellation diagram for the following: a. ASK, with peak amplitude values of 1 and 3 b. BPSK, with a peak amplitude value of 2 c. QPSK, with a peak amplitude value of 3 d. 8-QAM with two different peak amplitude values, I and 3, and four different phases. 15. Draw the constellation diagram for the following cases. Find the peak amplitude value for each case and define the type of modulation (ASK, FSK, PSK, or QAM). The numbers in parentheses define the values of I and Q respectively. a. Two points at (2, 0) and (3, 0). b. Two points at (3, 0) and (-3, 0). c. Four points at (2, 2), (-2, 2), (-2, -2), and (2, -2). d. Two points at (0 , 2) and (0, -2). 16. How many bits per baud can we send in each of the following cases if the signal constellation has one of the following number of points? a. 2 b. 4 c. 16 d. 1024 17. What is the required bandwidth for the following cases if we need to send 4000 bps? Let d = 1. a. ASK b. FSK with 2~f = 4 KHz c. QPSK d. 16-QAM 18. The telephone line has 4 KHz bandwidth. What is the maximum number of bits we can send using each of the following techniques? Let d = O. a. ASK b. QPSK c. 16-QAM d.64-QAM 160 CHAPTER 5 ANALOG TRANSMISSION 19. A corporation has a medium with a I-MHz bandwidth (lowpass). The corporation needs to create 10 separate independent channels each capable of sending at least 10 Mbps. The company has decided to use QAM technology. What is the minimum number of bits per baud for each channel? What is the number of points in the constellation diagram for each channel? Let d = O. 20. A cable company uses one of the cable TV channels (with a bandwidth of 6 MHz) to provide digital communication for each resident. What is the available data rate for each resident if the company uses a 64-QAM technique? 21. Find the bandwidth for the following situations if we need to modulate a 5-KHz voice. a. AM =5) PM (set ~ = 1) b. PM (set ~ c. 22. Find the total number of channels in the corresponding band allocated by FCC. a. AM b. FM CHAPTER 4 Digital Transmission Solutions to Review Questions and Exercises Review Questions 1. The three different techniques described in this chapter are line coding, block coding, and scrambling. 2. A data element is the smallest entity that can represent a piece of information (a bit). A signal element is the shortest unit of a digital signal. Data elements are what we need to send; signal elements are what we can send. Data elements are being carried; signal elements are the carriers. 3. The data rate defines the number of data elements (bits) sent in 1s. The unit is bits per second (bps). The signal rate is the number of signal elements sent in 1s. The unit is the baud. 4. In decoding a digital signal, the incoming signal power is evaluated against the baseline (a running average of the received signal power). A long string of 0s or 1s can cause baseline wandering (a drift in the baseline) and make it difficult for the receiver to decode correctly. 5. When the voltage level in a digital signal is constant for a while, the spectrum creates very low frequencies, called DC components, that present problems for a system that cannot pass low frequencies. 6. A self-synchronizing digital signal includes timing information in the data being transmitted. This can be achieved if there are transitions in the signal that alert the receiver to the beginning, middle, or end of the pulse. 7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransition coding. 8. Block coding provides redundancy to ensure synchronization and to provide inherent error detecting. In general, block coding changes a block of m bits into a block of n bits, where n is larger than m. 9. Scrambling, as discussed in this chapter, is a technique that substitutes long zerolevel pulses with a combination of other levels without increasing the number of bits. 1 2 10. Both PCM and DM use sampling to convert an analog signal to a digital signal. PCM finds the value of the signal amplitude for each sample; DM finds the change between two consecutive samples. 11. In parallel transmission we send data several bits at a time. In serial transmission we send data one bit at a time. 12. We mentioned synchronous, asynchronous, and isochronous. In both synchronous and asynchronous transmissions, a bit stream is divided into independent frames. In synchronous transmission, the bytes inside each frame are synchronized; in asynchronous transmission, the bytes inside each frame are also independent. In isochronous transmission, there is no independency at all. All bits in the whole stream must be synchronized. Exercises 13. We use the formula s = c × N × (1/r) for each case. We let c = 1/2. a. r = 1 → s = (1/2) × (1 Mbps) × 1/1 = 500 kbaud b. r = 1/2 → s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaud c. r = 2 → s = (1/2) × (1 Mbps) × 1/2 = 250 Kbaud d. r = 4/3 → s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud 14. The number of bits is calculated as (0.2 /100) × (1 Mbps) = 2000 bits 15. See Figure 4.1. Bandwidth is proportional to (3/8)N which is within the range in Table 4.1 (B = 0 to N) for the NRZ-L scheme. Figure 4.1 Solution to Exercise 15 Average Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8 B (3 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 Case b Case d 16. See Figure 4.2. Bandwidth is proportional to (4.25/8)N which is within the range in Table 4.1 (B = 0 to N) for the NRZ-I scheme. 17. See Figure 4.3. Bandwidth is proportional to (12.5 / 8) N which is within the range in Table 4.1 (B = N to B = 2N) for the Manchester scheme. 18. See Figure 4.4. B is proportional to (12/8) N which is within the range in Table 4.1 (B = N to 2N) for the differential Manchester scheme. 3 Figure 4.2 Solution to Exercise 16 Average Number of Changes = (0 + 9 + 4 + 4) / 4 = 4.25 for N = 8 B (4.25 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 Case b Figure 4.3 Case d Solution to Exercise 17 Average Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8 B (12.5 / 8) N Case a 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 Case b Figure 4.4 Case c 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 Case d Solution to Exercise 18 Average Number of Changes = (16 + 8 + 12 + 12) / 4 = 12 for N = 8 B (12 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 Case b Case d 4 19. See Figure 4.5. B is proportional to (5.25 / 16) N which is inside range in Table 4.1 (B = 0 to N/2) for 2B/1Q. Figure 4.5 Solution to Exercise 19 Average Number of Changes = (0 + 7 + 7 + 7) / 4 = 5.25 for N = 16 B (5.25 / 8) N Case a 00 +3 +1 −1 00 00 00 00 00 00 00 Case c 01 +3 +1 −1 10 01 10 01 10 01 10 11 00 11 00 11 00 11 −3 −3 11 11 11 11 11 11 11 11 +3 +1 −1 00 +3 +1 −1 −3 −3 Case b Case d 20. See Figure 4.6. B is proportional to (5.25/8) × N which is inside the range in Table 4.1 (B = 0 to N/2) for MLT-3. Figure 4.6 Solution to Exercise 20 Average Number of Changes = (0 + 7 + 4 + 3) / 4 = 4.5 for N = 8 B (4.5 / 8) N Case a Case c 0 0 0 0 0 0 0 0 +V +V −V 1 0 1 0 1 0 1 0 0 0 1 1 0 0 0 −V 1 1 1 1 1 1 1 1 +V +V −V −V Case b 0 Case d 21. The data stream can be found as a. NRZ-I: 10011001. b. Differential Manchester: 11000100. c. AMI: 01110001. 22. The data rate is 100 Kbps. For each case, we first need to calculate the value f / N. We then use Figure 4.6 in the text to find P (energy per Hz). All calculations are approximations. 5 a. f /N = 0/100 =0 → P = 1.0 b. f /N = 50/100 = 1/2 → P = 0.5 c. f /N = 100/100 = 1 → P = 0.0 d. f /N = 150/100 = 1.5 → P = 0.2 23. The data rate is 100 Kbps. For each case, we first need to calculate the value f/N. We then use Figure 4.8 in the text to find P (energy per Hz). All calculations are approximations. a. f /N = 0/100 =0 → P = 0.0 b. f /N = 50/100 = 1/2 → P = 0.3 c. f /N = 100/100 = 1 → P = 0.4 d. f /N = 150/100 = 1.5 → P = 0.0 24. a. The output stream is 01010 11110 11110 11110 11110 01001. b. The maximum length of consecutive 0s in the input stream is 21. c. The maximum length of consecutive 0s in the output stream is 2. 25. In 5B/6B, we have 25 = 32 data sequences and 26 = 64 code sequences. The number of unused code sequences is 64 − 32 = 32. In 3B/4B, we have 2 3 = 8 data sequences and 24 = 16 code sequences. The number of unused code sequences is 16 − 8 = 8. 26. See Figure 4.7. Since we specified that the last non-zero signal is positive, the first bit in our sequence is positive. Figure 4.7 Solution to Exercise 26 a. B8ZS 1 1 1 0 0 0 0 0 0 B 0 V 1 1 1 0 0 0 0 0 0 0 0 0 0 0 V B 0 B 0 0 0 V V b. HDB3 27. a. In a low-pass signal, the minimum frequency 0. Therefore, we have fmax = 0 + 200 = 200 KHz. → fs = 2 × 200,000 = 400,000 samples/s 6 b. In a bandpass signal, the maximum frequency is equal to the minimum frequency plus the bandwidth. Therefore, we have fmax = 100 + 200 = 300 KHz. → fs = 2 × 300,000 = 600,000 samples /s 28. a. In a lowpass signal, the minimum frequency is 0. Therefore, we can say fmax = 0 + 200 = 200 KHz → fs = 2 × 200,000 = 400,000 samples/s The number of bits per sample and the bit rate are N = 400 KHz × 10 = 4 Mbps nb = log21024 = 10 bits/sample b. The value of nb = 10. We can easily calculate the value of SNRdB SNRdB = 6.02 × nb + 1.76 = 61.96 c. The value of nb = 10. The minimum bandwidth can be calculated as BPCM = nb × Banalog = 10 × 200 KHz = 2 MHz 29. The maximum data rate can be calculated as Nmax = 2 × B × nb = 2 × 200 KHz × log24 = 800 kbps 30. We can first calculate the sampling rate (fs) and the number of bits per sample (nb) fmax = 0 + 4 = 4 KHz → fs = 2 × 4 = 8000 sample/s We then calculate the number of bits per sample. → nb = 30000 / 8000 = 3.75 We need to use the next integer nb = 4. The value of SNRdB is SNRdB = 6.02 × nb + 1.72 = 25.8 31. We can calculate the data rate for each scheme: a. NRZ b. Manchester c. MLT-3 d. 2B1Q → → → → N=2 N=1 N=3 N=4 ×B=2 ×B=1 ×B=3 ×B=4 × 1 MHz = 2 Mbps × 1 MHz = 1 Mbps × 1 MHz = 3 Mbps × 1 MHz = 4 Mbps 32. a. For synchronous transmission, we have 1000 × 8 = 8000 bits. b. For asynchronous transmission, we have 1000 × 10 = 10000 bits. Note that we assume only one stop bit and one start bit. Some systems send more start bits. c. For case a, the redundancy is 0%. For case b, we send 2000 extra for 8000 required bits. The redundancy is 25%. CHAPTER 5 Analog Transmission Solutions to Review Questions and Exercises Review Questions 1. Normally, analog transmission refers to the transmission of analog signals using a band-pass channel. Baseband digital or analog signals are converted to a complex analog signal with a range of frequencies suitable for the channel. 2. A carrier is a single-frequency signal that has one of its characteristics (amplitude, frequency, or phase) changed to represent the baseband signal. 3. The process of changing one of the characteristics of an analog signal based on the information in digital data is called digital-to-analog conversion. It is also called modulation of a digital signal. The baseband digital signal representing the digital data modulates the carrier to create a broadband analog signal. 4. a. ASK changes the amplitude of the carrier. b. FSK changes the frequency of the carrier. c. PSK changes the phase of the carrier. d. QAM changes both the amplitude and the phase of the carrier. 5. We can say that the most susceptible technique is ASK because the amplitude is more affected by noise than the phase or frequency. 6. A constellation diagram can help us define the amplitude and phase of a signal element, particularly when we are using two carriers. The diagram is useful when we are dealing with multilevel ASK, PSK, or QAM. In a constellation diagram, a signal element type is represented as a dot. The bit or combination of bits it can carry is often written next to it.The diagram has two axes. The horizontal X axis is related to the in-phase carrier; the vertical Y axis is related to the quadrature carrier. 7. The two components of a signal are called I and Q. The I component, called inphase, is shown on the horizontal axis; the Q component, called quadrature, is shown on the vertical axis. 8. The process of changing one of the characteristics of an analog signal to represent the instantaneous amplitude of a baseband signal is called analog-to-analog con- 1 2 version. It is also called the modulation of an analog signal; the baseband analog signal modulates the carrier to create a broadband analog signal. 9. a. AM changes the amplitude of the carrier b. FM changes the frequency of the carrier c. PM changes the phase of the carrier 10. We can say that the most susceptible technique is AM because the amplitude is more affected by noise than the phase or frequency. Exercises 11. We use the formula S = (1/r) × N, but first we need to calculate the value of r for each case. a. r = log22 b. r = log22 c. r = log24 d. r = log264 =1 =1 =2 =6 → → → → S = (1/1) × (2000 bps) S = (1/1) × (4000 bps) S = (1/2) × (6000 bps) S = (1/6) × (36,000 bps) = 2000 baud = 4000 baud = 3000 baud = 6000 baud 12. We use the formula N = r × S, but first we need to calculate the value of r for each case. a. r = log22 b. r = log22 c. r = log22 d. r = log216 =1 =1 =1 =4 → → → → N = (1) × (1000 bps) N = (1) × (1000 bps) N = (1) × (1000 bps) N = (4) × (1000 bps) = 1000 bps = 1000 bps = 1000 bps = 4000 bps 13. We use the formula r = log2L to calculate the value of r for each case. a. log24 b. log28 c. log24 d. log2128 =2 =3 =2 =7 14. See Figure 5.1. a. We have two signal elements with peak amplitudes 1 and 3. The phase of both signal elements are the same, which we assume to be 0 degrees. b. We have two signal elements with the same peak amplitude of 2. However, there must be 180 degrees difference between the two phases. We assume one phase to be 0 and the other 180 degrees. c. We have four signal elements with the same peak amplitude of 3. However, there must be 90 degrees difference between each phase. We assume the first phase to be at 45, the second at 135, the third at 225, and the fourth at 315 degrees. Note that this is one out of many configurations. The phases can be at 3 Figure 5.1 Solution to Exercise 14 a. ASK b. BPSK Q Q I 1 I –2 3 2 Q Q 3 3 3 3 1 1 1 1 I 3 I 3 3 c. QPSK 3 d. 8-QAM 0, 90, 180, and 270. As long as the differences are 90 degrees, the solution is satisfactory. d. We have four phases, which we select to be the same as the previous case. For each phase, however, we have two amplitudes, 1 and 3 as shown in the figure. Note that this is one out of many configurations. The phases can be at 0, 90, 180, and 270. As long as the differences are 90 degrees, the solution is satisfactory. 15. See Figure 5.2 Figure 5.2 Solution to Exercise 15 a. b. Q Q I 2 I –3 3 3 Q Q 2 2 –2 2 I I –2 –2 c. d. a. This is ASK. There are two peak amplitudes both with the same phase (0 degrees). The values of the peak amplitudes are A1 = 2 (the distance between 4 the first dot and the origin) and A2= 3 (the distance between the second dot and the origin). b. This is BPSK, There is only one peak amplitude (3). The distance between each dot and the origin is 3. However, we have two phases, 0 and 180 degrees. c. This can be either QPSK (one amplitude, four phases) or 4-QAM (one amplitude and four phases). The amplitude is the distance between a point and the origin, which is (22 + 22)1/2 = 2.83. d. This is also BPSK. The peak amplitude is 2, but this time the phases are 90 and 270 degrees. 16. The number of points define the number of levels, L. The number of bits per baud is the value of r. Therefore, we use the formula r = log2L for each case. a. log22 b. log24 c. log216 d. log21024 =1 =2 =4 = 10 17. We use the formula B = (1 + d) × (1/r) × N, but first we need to calculate the value of r for each case. a. r = 1 b. r = 1 c. r = 2 d. r = 4 → → → → B= (1 + 1) × (1/1) × (4000 bps) B = (1 + 1) × (1/1) × (4000 bps) + 4 KHz B = (1 + 1) × (1/2) × (4000 bps) B = (1 + 1) × (1/4) × (4000 bps) = 8000 Hz = 8000 Hz = 2000 Hz = 1000 Hz 18. We use the formula N = [1/(1 + d)] × r × B, but first we need to calculate the value of r for each case. a. r = log22 = 1 b. r = log24=2 c. r = log216= 4 d. r = log264= 6 → → → → N= [1/(1 + 0)] × 1 N = [1/(1 + 0)] × 2 N = [1/(1 + 0)] × 4 N = [1/(1 + 0)] × 6 × (4 KHz) × (4 KHz) × (4 KHz) × (4 KHz) = 4 kbps = 8 kbps = 16 kbps = 24 kbps 19. First, we calculate the bandwidth for each channel = (1 MHz) / 10 = 100 KHz. We then find the value of r for each channel: B = (1 + d) × (1/r) × (N) → r = N / B → r = (1 Mbps/100 KHz) = 10 We can then calculate the number of levels: L = 2r = 210 = 1024. This means that that we need a 1024-QAM technique to achieve this data rate. 20. We can use the formula: N = [1/(1 + d)] × r × B = 1 × 6 × 6 MHz = 36 Mbps 21. a. BAM = 2 × B = 2 × 5 = 10 KHz 5 b. BFM = 2 × (1 + β) × B = 2 × (1 + 5) × 5 c. BPM = 2 × (1 + β) × B = 2 × (1 + 1) × 5 = 60 KHz = 20 KHz 22. We calculate the number of channels, not the number of coexisting stations. a. n = (1700 - 530) KHz / 10 KHz b. n = (108 - 88) MHz / 200 KHz = 117 = 100 208 CHAPTER 7 TRANSMISSION MEDIA Applications The infrared band, almost 400 THz, has an excellent potential for data transmission. Such a wide bandwidth can be used to transmit digital data with a very high data rate. The Infrared Data Association (IrDA), an association for sponsoring the use of infrared waves, has established standards for using these signals for communication between devices such as keyboards, mice, PCs, and printers. For example, some manufacturers provide a special port called the IrDA port that allows a wireless keyboard to communicate with a PC. The standard originally defined a data rate of 75 kbps for a distance up to 8 m. The recent standard defines a data rate of 4 Mbps. Infrared signals defined by IrDA transmit through line of sight; the IrDA port on the keyboard needs to point to the PC for transmission to occur. Infrared signals can be used for short-range communication in a closed area using line-of-sight propagation. 7.3 RECOMMENDED READING For more details about subjects discussed in this chapter, we recommend the following books. The items in brackets [...] refer to the reference list at the end of the text. Books Transmission media is discussed in Section 3.8 of [GW04], Chapter 4 of [Sta04], Section 2.2 and 2.3 of [Tan03]. [SSS05] gives a full coverage of transmission media. 7.4 KEY TERMS angle of incidence Bayone-Neil-Concelman (BNC) connector cladding IrDA port line-of-sight propagation microwave MT-RJ coaxial cable multimode graded-index fiber core multimode step-index fiber omnidirectional antenna optical fiber parabolic dish antenna critical angle electromagnetic spectrum fiber-optic cable gauge Radio Government (RG) number ground propagation radio wave guided media horn antenna infrared wave reflection refraction RJ45 SECTION 7.6 shielded twisted-pair (STP) single-mode fiber sky propagation straight-tip (ST) connector subscriber channel (SC) connector transmission medium 7.5 PRACTICE SET 209 twisted-pair cable unguided medium unidirectional antenna unshielded twisted-pair (UTP) wireless communication SUMMARY o Transmission media lie below the physical layer. D A guided medium provides a physical conduit from one device to another. Twistedpair cable, coaxial cable, and optical fiber are the most popular types of guided media. D Twisted-pair cable consists of two insulated copper wires twisted together. Twistedpair cable is used for voice and data communications. D Coaxial cable consists of a central conductor and a shield. Coaxial cable can carry signals of higher frequency ranges than twisted-pair cable. Coaxial cable is used in cable TV networks and traditional Ethernet LANs. Fiber-optic cables are composed of a glass or plastic inner core surrounded by cladding, all encased in an outside jacket. Fiber-optic cables carry data signals in the form of light. The signal is propagated along the inner core by reflection. Fiberoptic transmission is becoming increasingly popular due to its noise resistance, low attenuation, and high-bandwidth capabilities. Fiber-optic cable is used in backbone networks, cable TV networks, and Fast Ethernet networks. D Unguided media (free space) transport electromagnetic waves without the use of a physical conductor. Wireless data are transmitted through ground propagation, sky propagation, and lineof-sight propagation.Wireless waves can be classified as radio waves, microwaves, or infrared waves. Radio waves are omnidirectional; microwaves are unidirectional. Microwaves are used for cellular phone, satellite, and wireless LAN communications. D Infrared waves are used for short-range communications such as those between a PC and a peripheral device. It can also be used for indoor LANs. o o 7.6 PRACTICE SET Review Questions 1. 2. 3. 4. 5. What is the position of the transmission media in the OSI or the Internet model? Name the two major categories of transmission media. How do guided media differ from unguided media? What are the three major classes of guided media? What is the significance of the twisting in twisted-pair cable? 210 CHAPTER 7 6. 7. 8. 9. 10. TRANSMISSION MEDIA What is refraction? What is reflection? What is the purpose of cladding in an optical fiber? Name the advantages of optical fiber over twisted-pair and coaxial cable. How does sky propagation differ from line-of-sight propagation? What is the difference between omnidirectional waves and unidirectional waves? Exercises 11. Using Figure 7.6, tabulate the attenuation (in dB) of a 18-gauge UTP for the indicated frequencies and distances. Table 7.5 Attenuation/or I8-gauge UTP Distance dB at 1 KHz dRat 10KHz dB at 100 KHz 1 Krn lOKm 15 Krn 20Km 12. Use the result of Exercise 11 to infer that the bandwidth of a UTP cable decreases with an increase in distance. 13. If the power at the beginning of a 1 KIn 18-gauge UTP is 200 mw, what is the power at the end for frequencies 1 KHz, 10 KHz, and 100 KHz? Use the result of Exercise 11. 14. Using Figure 7.9, tabulate the attenuation (in dB) of a 2.6/9.5 mm coaxial cable for the indicated frequencies and distances. Table 7.6 Attenuation/or 2.6/9.5 mm coaxial cable Distance dB at 1 KHz dB at 10 KHz dB at 100KHz 1 Km lOKrn 15Km 20Km 15. Use the result of Exercise 14 to infer that the bandwidth of a coaxial cable decreases with the increase in distance. 16. If the power at the beginning of a 1 KIn 2.6/9.5 mm coaxial cable is 200 mw, what is the power at the end for frequencies 1 KHz, 10KHz, and 100 KHz? Use the result of Exercise 14. 17. Calculate the bandwidth of the light for the following wavelength ranges (assume a propagation speed of 2 x 108 m): a. 1000 to 1200 nm b. 1000 to 1400 nm SECTION 7.6 PRACTICE SET 211 18. The horizontal axes in Figure 7.6 and 7.9 represent frequencies. The horizontal axis in Figure 7.16 represents wavelength. Can you explain the reason? lfthe propagation speed in an optical fiber is 2 x 108 ill, can you change the units in the horizontal axis to frequency? Should the vertical-axis units be changed too? Should the curve be changed too? 19. Using Figure 7.16, tabulate the attenuation (in dB) of an optical fiber for the indicated wavelength and distances. Table 7.7 Attenuation for optical fiber Distance dB at 800 nm dB at 1000 nm dB at 1200nm 1 Km lOKm 15Km 20Km 20. A light signal is travelling through a fiber. What is the delay in the signal if the length of the fiber-optic cable is 10 m, 100 m, and 1 Km (assume a propagation speed of 2 x 108 ill)? 21. A beam oflight moves from one medium to another medium with less density. The critical angle is 60°. Do we have refraction or reflection for each of the following incident angles? Show the bending of the light ray in each case. a. 40° b. 60° c. 80 0 236 CHAPTER 8 SWITCHING multistage switch output port packet-switched network routing processor setup phase space-division switching switch switching switching fabric 8.7 o o o o o o o o table lookup teardown phase time-division switching time-slot interchange (TSI) time-space-time (TST) switch trap virtual-circuit identifier (VCI) virtual-circuit network SUMMARY A switched network consists of a series of interlinked nodes, called switches. Traditionally' three methods of switching have been important: circuit switching, packet switching, and message switching. We can divide today's networks into three broad categories: circuit-switched networks, packet-switched networks, and message-switched. Packet-switched networks can also be divided into two subcategories: virtual-circuit networks and datagram networks A circuit-switched network is made of a set of switches connected by physical links, in which each link is divided into n channels. Circuit switching takes place at the physical layer. In circuit switching, the resources need to be reserved during the setup phase; the resources remain dedicated for the entire duration of data transfer phase until the teardown phase. In packet switching, there is no resource allocation for a packet. This means that there is no reserved bandwidth on the links, and there is no scheduled processing time for each packet. Resources are allocated on demand. In a datagram network, each packet is treated independently of all others. Packets in this approach are referred to as datagrams. There are no setup or teardown phases. A virtual-circuit network is a cross between a circuit-switched network and a datagram network. It has some characteristics of both. Circuit switching uses either of two technologies: the space-division switch or the time-division switch. A switch in a packet-switched network has a different structure from a switch used in a circuit-switched network.We can say that a packet switch has four types of components: input ports, output ports, a routing processor, and switching fabric. 8.8 PRACTICE SET Review Questions I. Describe the need for switching and define a switch. 2. List the three traditional switching methods. What are the most common today? SECTION 8.8 PRACTICE SET 237 3. What are the two approaches to packet-switching? 4. Compare and contrast a circuit-switched network and a packet-switched network. 5. What is the role of the address field in a packet traveling through a datagram network? 6. What is the role of the address field in a packet traveling through a virtual-circuit network? 7. Compare space-division and time-division switches. 8. What is TSI and its role in a time-division switching? 9. Define blocking in a switched network. 10. List four major components of a packet switch and their functions. Exercises 11. A path in a digital circuit-switched network has a data rate of I Mbps. The exchange of 1000 bits is required for the setup and teardown phases. The distance between two parties is 5000 km. Answer the following questions if the propagataion speed is 2 X 108 m: a. What is the total delay if 1000 bits of data are exchanged during the data transfer phase? b. What is the total delay if 100,000 bits of data are exchanged during the data transfer phase? c. What is the total delay if 1,000,000 bits of data are exchanged during the data transfer phase? d. Find the delay per 1000 bits of data for each of the above cases and compare them. What can you infer? 12. Five equal-size datagrams belonging to the same message leave for the destination one after another. However, they travel through different paths as shown in Table 8.1. Table 8.1 Exercise 12 Datagram Path Length Visited Switches 1 3200Km 1,3,5 2 11,700 Km 1,2,5 3 12,200 Km 1,2,3,5 4 10,200 Km 1,4,5 5 10,700 Km 1,4,3,5 We assume that the delay for each switch (including waiting and processing) is 3, 10, 20, 7, and 20 ms respectively. Assuming that the propagation speed is 2 x 108 m, find the order the datagrams arrive at the destination and the delay for each. Ignore any other delays in transmission. 238 CHAPTER 8 SWITCHING 13. Transmission of information in any network involves end-to-end addressing and sometimes local addressing (such as YCI). Table 8.2 shows the types of networks and the addressing mechanism used in each of them. Table 8.2 Exercise 13 Network Setup Circuit-switched End-ta-end Teardown End-ta-end End-ta-end Datagram Virtual-circuit Data Transfer End-to-end Local End-to-end Answer the following questions: a. Why does a circuit-switched network need end-to-end addressing during the setup and teardown phases? Why are no addresses needed during the data transfer phase for this type of network? b. Why does a datagram network need only end-to-end addressing during the data transfer phase, but no addressing during the setup and teardown phases? c. Why does a virtual-circuit network need addresses during all three phases? 14. We mentioned that two types of networks, datagram and virtual-circuit, need a routing or switching table to find the output port from which the information belonging to a destination should be sent out, but a circuit-switched network has no need for such a table. Give the reason for this difference. 15. An entry in the switching table of a virtual-circuit network is normally created during the setup phase and deleted during the teardown phase. In other words, the entries in this type of network reflect the current connections, the activity in the network. In contrast, the entries in a routing table of a datagram network do not depend on the current connections; they show the configuration of the network and how any packet should be routed to a final destination. The entries may remain the same even if there is no activity in the network. The routing tables, however, are updated if there are changes in the network. Can you explain the reason for these two different characteristics? Can we say that a virtual-circuit is a connectionoriented network and a datagram network is a connectionLess network because of the above characteristics? 16. The minimum number of columns in a datagram network is two; the minimum number of columns in a virtual-circuit network is four. Can you explain the reason? Is the difference related to the type of addresses carried in the packets of each network? 17. Figure 8.27 shows a switch (router) in a datagram network. Find the output port for packets with the following destination addresses: Packet 1: 7176 Packet 2: 1233 Packet 3: 8766 Packet 4: 9144 18. Figure 8.28 shows a switch in a virtual circuit network. SECTION 8.8 Figure 8.27 PRACTICE SET 239 Exercise 17 Destination address Output port 1233 3 2 1 4 2 1456 3255 4470 7176 8766 9144 4 2 3 3 2 Figure 8.28 Exercise 18 Incoming Outgoing Port VCI Port VCI 1 2 2 3 3 14 71 92 58 78 56 3 22 41 45 43 70 11 4 4 1 2 2 3 4 2 Find the output port and the output VCI for packets with the following input port and input VCI addresses: Packet 1: 3, 78 Packet 2: 2, 92 Packet 3: 4, 56 Packet 4: 2, 71 19. Answer the following questions: a. Can a routing table in a datagram network have two entries with the same destination address? Explain. b. Can a switching table in a virtual-circuit network have two entries with the same input port number? With the same output port number? With the same incoming VCls? With the same outgoing VCls? With the same incoming values (port, VCI)? With the same outgoing values (port, VCI)? 20. It is obvious that a router or a switch needs to do searching to find information in the corresponding table. The searching in a routing table for a datagram network is based on the destination address; the searching in a switching table in a virtualcircuit network is based on the combination of incoming port and incoming VCI. Explain the reason and define how these tables must be ordered (sorted) based on these values. 2]. Consider an n X k crossbar switch with n inputs and k outputs. a. Can we say that switch acts as a multiplexer if n > k? b. Can we say that switch acts as a demultiplexer if n < k? 240 CHAPTER 8 SWITCHING 22. We need a three-stage space-division switch with N = 100. We use 10 crossbars at the first and third stages and 4 crossbars at the middle stage. a. Draw the configuration diagram. b. Calculate the total number of crosspoints. c. Find the possible number of simultaneous connections. d. Find the possible number of simultaneous connections if we use one single crossbar (100 x 100). e. Find the blocking factor, the ratio of the number of connections in c and in d. 23. Repeat Exercise 22 if we use 6 crossbars at the middle stage. 24. Redesign the configuration of Exercise 22 using the Clos criteria. 25. We need to have a space-division switch with 1000 inputs and outputs. What is the total number of crosspoints in each of the following cases? a. Using one single crossbar. b. Using a multi-stage switch based on the Clos criteria 26. We need a three-stage time-space-time switch with N = 100. We use 10 TSIs at the first and third stages and 4 crossbars at the middle stage. a. Draw the configuration diagram. b. Calculate the total number of crosspoints. c. Calculate the total number of memory locations we need for the TSIs. CHAPTER 7 Transmission Media Solutions to Review Questions and Exercises Review Questions 1. The transmission media is located beneath the physical layer and controlled by the physical layer. 2. The two major categories are guided and unguided media. 3. Guided media have physical boundaries, while unguided media are unbounded. 4. The three major categories of guided media are twisted-pair, coaxial, and fiberoptic cables. 5. Twisting ensures that both wires are equally, but inversely, affected by external influences such as noise. 6. Refraction and reflection are two phenomena that occur when a beam of light travels into a less dense medium. When the angle of incidence is less than the critical angle, refraction occurs. The beam crosses the interface into the less dense medium. When the angle of incidence is greater than the critical angle, reflection occurs. The beam changes direction at the interface and goes back into the more dense medium. 7. The inner core of an optical fiber is surrounded by cladding. The core is denser than the cladding, so a light beam traveling through the core is reflected at the boundary between the core and the cladding if the incident angle is more than the critical angle. 8. We can mention three advantages of optical fiber cable over twisted-pair and coaxial cables: noise resistance, less signal attenuation, and higher bandwidth. 9. In sky propagation radio waves radiate upward into the ionosphere and are then reflected back to earth. In line-of-sight propagation signals are transmitted in a straight line from antenna to antenna. 10. Omnidirectional waves are propagated in all directions; unidirectional waves are propagated in one direction. 1 2 Exercises 11. See Table 7.1 (the values are approximate). Table 7.1 Solution to Exercise 11 Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz 1 Km −3 −5 −7 10 Km −30 −50 −70 15 Km −45 −75 −105 20 Km −60 −100 −140 12. As the Table 7.1 shows, for a specific maximum value of attenuation, the highest frequency decreases with distance. If we consider the bandwidth to start from zero, we can say that the bandwidth decreases with distance. For example, if we can tolerate a maximum attenuation of 50 dB (loss), then we can give the following listing of distance versus bandwidth. Distance 1 Km 10 Km 15 Km 20 Km Bandwidth 100 KHz 50 KHz 1 KHz 0 KHz 13. We can use Table 7.1 to find the power for different frequencies: 1 KHz 10 KHz 100 KHz P2 = P1 ×10−3/10 P2 = P1 ×10−5/10 P2 = P1 ×10−7/10 dB = −3 dB = −5 dB = −7 = 100.23 mw = 63.25 mw = 39.90 mw The table shows that the power is reduced 5 times, which may not be acceptable for some applications. 14. See Table 7.2 (the values are approximate). Table 7.2 Solution to Exercise 14 Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz 1 Km −3 −7 −20 10 Km −30 −70 −200 15 Km −45 −105 −300 20 Km −60 −140 −400 15. As Table 7.2 shows, for a specific maximum value of attenuation, the highest frequency decreases with distance. If we consider the bandwidth to start from zero, we can say that the bandwidth decreases with distance. For example, if we can tol- 3 erate a maximum attenuation of 50 dB (loss), then we can give the following listing of distance versus bandwidth. Distance 1 Km 10 Km 15 Km 20 Km Bandwidth 100 KHz 1 KHz 1 KHz 0 KHz 16. We can use Table 7.2 to find the power for different frequencies: 1 KHz 10 KHz 100 KHz dB = −3 dB = −7 dB = −20 P2 = P1 ×10−3/10 P2 = P1 ×10−7/10 P2 = P1 ×10−20/10 = 100.23 mw = 39.90 mw = 2.00 mw The table shows that power is decreased 100 times for 100 KHz, which is unacceptable for most applications. 17. We can use the formula f = c / λ to find the corresponding frequency for each wave length as shown below (c is the speed of propagation): a. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1200 × 10−9] = 33 THz b. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1400 × 10−9] = 57 THz 18. a. The wave length is the inverse of the frequency if the propagation speed is fixed (based on the formula λ = c / f). This means all three figures represent the same thing. b. We can change the wave length to frequency. For example, the value 1000 nm can be written as 200 THz. c. The vertical-axis units may not change because they represent dB/km. d. The curve must be flipped horizontally. 19. See Table 7.3 (The values are approximate). Table 7.3 Solution to Exercise 19 Distance dB at 800 nm dB at 1000 nm dB at 1200 nm 1 Km −3 −1.1 −0.5 10 Km −30 −11 −5 15 Km −45 −16.5 −7.5 20 Km −60 −22 −10 20. The delay = distance / (propagation speed). Therefore, we have: a. Delay = 10/(2 × 108) = 0.05 ms b. Delay = 100/(2 × 108) = 0.5 ms c. Delay = 1000/(2 × 108) = 5 ms 4 21. See Figure 7.1. Figure 7.1 Solution to Exercise 21 Refraction a. 40 degrees Critical angle = 60 Critical angle Refraction b. 60 degrees Critical angle = 60 Critical angle Reflection c. 80 degrees Critical angle = 60 Critical angle a. The incident angle (40 degrees) is smaller than the critical angle (60 degrees). We have refraction.The light ray enters into the less dense medium. b. The incident angle (60 degrees) is the same as the critical angle (60 degrees). We have refraction. The light ray travels along the interface. c. The incident angle (80 degrees) is greater than the critical angle (60 degrees). We have reflection. The light ray returns back to the more dense medium. CHAPTER 8 Switching Solutions to Review Questions and Exercises Review Questions 1. Switching provides a practical solution to the problem of connecting multiple devices in a network. It is more practical than using a bus topology; it is more efficient than using a star topology and a central hub. Switches are devices capable of creating temporary connections between two or more devices linked to the switch. 2. The three traditional switching methods are circuit switching, packet switching, and message switching. The most common today are circuit switching and packet switching. 3. There are two approaches to packet switching: datagram approach and virtualcircuit approach. 4. In a circuit-switched network, data are not packetized; data flow is somehow a continuation of bits that travel the same channel during the data transfer phase. In a packet-switched network data are packetized; each packet is somehow an independent entity with its local or global addressing information. 5. The address field defines the end-to-end (source to destination) addressing. 6. The address field defines the virtual circuit number (local) addressing. 7. In a space-division switch, the path from one device to another is spatially separate from other paths. The inputs and the outputs are connected using a grid of electronic microswitches. In a time-division switch, the inputs are divided in time using TDM. A control unit sends the input to the correct output device. 8. TSI (time-slot interchange) is the most popular technology in a time-division switch. It used random access memory (RAM) with several memory locations. The RAM fills up with incoming data from time slots in the order received. Slots are then sent out in an order based on the decisions of a control unit. 9. In multistage switching, blocking refers to times when one input cannot be connected to an output because there is no path available between them—all the possible intermediate switches are occupied. One solution to blocking is to increase the number of intermediate switches based on the Clos criteria. 1 2 10. A packet switch has four components: input ports, output ports, the routing processor, and the switching fabric. An input port performs the physical and data link functions of the packet switch. The output port performs the same functions as the input port, but in the reverse order. The routing processor performs the function of table lookup in the network layer. The switching fabric is responsible for moving the packet from the input queue to the output queue. Exercises 11. We assume that the setup phase is a two-way communication and the teardown phase is a one-way communication. These two phases are common for all three cases. The delay for these two phases can be calculated as three propagation delays and three transmission delays or 3 [(5000 km)/ (2 ×108 m/s)]+ 3 [(1000 bits/1 Mbps)] = 75 ms + 3 ms = 78 ms We assume that the data transfer is in one direction; the total delay is then delay for setup and teardown + propagation delay + transmission delay a. b. c. d. 78 + 25 + 1 = 104 ms 78 + 25 + 100 = 203 ms 78 + 25 + 1000 = 1103 ms In case a, we have 104 ms. In case b we have 203/100 = 2.03 ms. In case c, we have 1103/1000 = 1.101 ms. The ratio for case c is the smallest because we use one setup and teardown phase to send more data. 12. We assume that the transmission time is negligible in this case. This means that we suppose all datagrams start at time 0. The arrival timed are calculated as: First: Second: Third: Fourth: Fifth: (3200 Km) / (2 × 108 m/s) (11700 Km) / (2 × 108 m/s) (12200 Km) / (2 × 108 m/s) (10200 Km) / (2 × 108 m/s) (10700 Km) / (2 × 108 m/s) + (3 + 20 + 20) + (3 + 10 + 20) + (3 + 10+ 20 + 20) + (3 + 7 + 20) + (3 + 7 + 20 + 20) = = = = = 59.0 ms 91.5 ms 114.0 ms 81.0 ms 103.5 ms The order of arrival is: 3 → 5 → 2 → 4 → 1 13. a. In a circuit-switched network, end-to-end addressing is needed during the setup and teardown phase to create a connection for the whole data transfer phase. After the connection is made, the data flow travels through the already-reserved resources. The switches remain connected for the entire duration of the data transfer; there is no need for further addressing. b. In a datagram network, each packet is independent. The routing of a packet is done for each individual packet. Each packet, therefore, needs to carry an endto-end address. There is no setup and teardown phases in a datagram network (connectionless transmission). The entries in the routing table are somehow permanent and made by other processes such as routing protocols. 3 c. In a virtual-circuit network, there is a need for end-to-end addressing during the setup and teardown phases to make the corresponding entry in the switching table. The entry is made for each request for connection. During the data transfer phase, each packet needs to carry a virtual-circuit identifier to show which virtual-circuit that particular packet follows. 14. A datagram or virtual-circuit network handles packetized data. For each packet, the switch needs to consult its table to find the output port in the case of a datagram network, and to find the combination of the output port and the virtual circuit identifier in the case of a virtual-circuit network. In a circuit-switched network, data are not packetized; no routing information is carried with the data. The whole path is established during the setup phase. 15. In circuit-switched and virtual-circuit networks, we are dealing with connections. A connection needs to be made before the data transfer can take place. In the case of a circuit-switched network, a physical connection is established during the setup phase and the is broken during the teardown phase. In the case of a virtual-circuit network, a virtual connection is made during setup and is broken during the teardown phase; the connection is virtual, because it is an entry in the table. These two types of networks are considered connection-oriented. In the case of a datagram network no connection is made. Any time a switch in this type of network receives a packet, it consults its table for routing information. This type of network is considered a connectionless network. 16. The switching or routing in a datagram network is based on the final destination address, which is global. The minimum number of entries is two; one for the final destination and one for the output port. Here the input port, from which the packet has arrived is irrelevant. The switching or routing in a virtual-circuit network is based on the virtual circuit identifier, which has a local jurisdiction. This means that two different input or output ports may use the same virtual circuit number. Therefore, four pieces of information are required: input port, input virtual circuit number, output port, and output virtual circuit number. 17. Packet 1: 2 Packet 2: 3 Packet 3: 3 Packet 4: 2 18. Packet 1: 2, 70 Packet 2: 1, 45 Packet 3: 3, 11 Packet 4: 4, 41 19. a. In a datagram network, the destination addresses are unique. They cannot be duplicated in the routing table. b. In a virtual-circuit network, the VCIs are local. A VCI is unique only in relationship to a port. In other words, the (port, VCI) combination is unique. This means that we can have two entries with the same input or output ports. We can 4 have two entries with the same VCIs. However, we cannot have two entries with the same (port, VCI) pair. 20. When a packet arrives at a router in a datagram network, the only information in the packet that can help the router in its routing is the destination address of the packet. The table then is sorted to make the searching faster. Today’s routers use some sophisticated searching techniques. When a packet arrives at a switch in a virtual-circuit network, the pair (input port, input VCI) can uniquely determined how the packet is to be routed; the pair is the only two pieces of information in the packet that is used for routing. The table in the virtual-circuit switch is sorted based on the this pair. However, since the number of port numbers is normally much smaller than the number of virtual circuits assigned to each port, sorting is done in two steps: first according to the input port number and second according to the input VCI. 21. a. If n > k, an n × k crossbar is like a multiplexer that combines n inputs into k outputs. However, we need to know that a regular multiplexer discussed in Chapter 6 is n × 1. b. If n < k, an n × k crossbar is like a demultiplexer that divides n inputs into k outputs. However, we need to know that a regular demultiplexer discussed in Chapter 6 is 1 × n. 22. a. See Figure 8.1. Figure 8.1 Solution to Exercise 22 Part a … 4 × 10 Stage 2 n = 10 … … … … … … Stage 1 10 × 10 n = 10 N = 100 … 10 × 4 n = 10 … n = 10 4 × 10 4 × 10 … 10 × 4 10 × 10 10 Crossbars … N = 100 n = 10 10 × 4 4 Crossbars … n = 10 … 10 Crossbars Stage 3 b. The total number of crosspoints are Number of crosspoints = 10 (10 × 4) + 4 (10 × 10) + 10 (4 × 10) = 1200 c. Only four simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is 40. d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100 simultaneous connections. e. The blocking factor is 40/100 or 40 percent. 5 23. a. See Figure 8.2. Figure 8.2 Solution to Exercise 23 Part a … 6 × 10 Stage 1 Stage 2 n = 10 N = 100 n = 10 … … … … … … 10 × 6 n = 10 … 6 × 10 … n = 10 10 × 10 6 × 10 … 10 × 6 10 × 10 10 Crossbars … N = 100 n = 10 10 × 6 6 Crossbars … n = 10 … 10 Crossbars Stage 3 b. The total number of crosspoints are Number of crosspoints = 10 (10 × 6) + 6 (10 × 10) + 10 (6 × 10) = 1800 c. Only six simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is 60. d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100 simultaneous connections. e. The blocking factor is 60/100 or 60 percent. 24. According to Clos, n = (N/2)1/2 = 7.07. We can choose n = 8. The number of crossbars in the first stage can be 13 (to have similar crossbars). Some of the input lines can be left unused. We then have k = 2n − 1 = 15. Figure 8.3 shows the configuration. Figure 8.3 Solution to Exercise 24 Part a … 15 × 8 Stage 2 We can calculate the total number of crosspoints as n=8 … … … … … … Stage 1 13 × 13 n=8 … 8 × 15 n=8 … n=8 15 × 8 15 × 8 … 8 × 15 13 × 13 13 Crossbars … N = 104 n = 8 8 × 15 15 Crossbars … n=8 … 13 Crossbars Stage 3 N = 104 6 13 (8 × 15) + 15 (13 × 13) + 13 (15 × 8) = 5655 The number of crosspoints is still much less than the case with one crossbar (10,000). We can see that there is no blocking involved because each 8 input line has 15 intermediate crossbars. The total number of crosspoints here is a little greater than the minimum number of crosspoints according to Clos using the formula 4N[(2N)1/2 − 1], which is 5257. 25. a. Total crosspoints = N2 = 10002 = 1,000,000 b. Total crosspoints ≥ 4Ν[(2Ν)1/2 −1] ≥ 174,886. With less than 200,000 crosspoints we can design a three-stage switch. We can use n = (N/2)1/2 =23 and choose k = 45. The total number of crosspoints is 178,200. 26. We give two solutions. a. We first solve the problem using only crossbars and then we replace the crossbars at the first and the last stage with TSIs. Figure 8.1 shows the solution using only crossbars. We can replace the crossbar at the first and third stages with TSIs as shown in Figure 8.4. The total number of crosspoints is 400 and the total number of memory locations is 200. Each TSI at the first stage needs one TDM multiplexer and one TDM demultiplexer. The multiplexer is 10 × 1, but the demultiplexer is 1 × 4. In other words, the input frame has 10 slots and the output frame has only 4 slots. The data in the first slot of all input TSIs are directed to the first switch, the output in the second slot are directed to the second switch, and so on. Figure 8.4 First solution to Exercise 26 10 TSIs 10 TSIs 4 Crossbars 10 × 10 10 × 10 10 × 10 10 × 10 Stage 1 Stage 2 Stage 3 7 b. We can see the inefficiency in the first solution. Since the slots are separated in time, only one of the switches at the middle stage is active at each moment. This means that, instead of 4 crossbars, we could have used only one with the same result. Figure 8.5 shows the new design. In this case we still need 200 memory locations but only 100 crosspoints. Figure 8.5 Second solution to Exercise 26 10 TSIs 10 TSIs 1 Crossbars 10 × 10 Stage 1 Stage 2 Stage 3 SECTION 9. 7 4. 5. 6. 7. KEY TERMS 261 CMs and CMTSs for the minislots used for timesharing of the upstream channels. We will learn about this timesharing when we discuss contention protocols in Chapter 12. The CM sends a packet to the ISP, asking for the Internet address. The CM and CMTS then exchange some packets to establish security parameters, which are needed for a public network such as cable TV. The CM sends its unique identifier to the CMTS. Upstream communication can start in the allocated upstream channel; the CM can contend for the minislots to send data. Downstream Communication In the downstream direction, the communication is much simpler. There is no contention because there is only one sender. The CMTS sends the packet with the address of the receiving eM, using the allocated downstream channel. 9.6 RECOMMENDED READING For more details about subjects discussed in this chapter, we recommend the following books. The items in brackets [...] refer to the reference list at the end of the text. Books [CouOl] gives an interesting discussion about telephone systems, DSL technology, and CATV in Chapter 8. [Tan03] discusses telephone systems and DSL technology in Section 2.5 and CATV in Section 2.7. [GW04] discusses telephone systems in Section 1.1.1 and standard modems in Section 3.7.3. A complete coverage of residential broadband (DSL and CATV) can be found in [Max99]. 9.7 KEY TERMS 56Kmodem 800 service 900 service ADSL Lite ADSLmodem analog leased service analog switched service asymmetric DSL (ADSL) cable modem (CM) cable modem transmission system (CMTS) cable TV network common carrier community antenna TV (CATV) competitive local exchange carrier (CLEC) Data Over Cable System Interface Specification (DOCSIS) demodulator digital data service (DDS) digital service digital subscriber line (DSL) digital subscriber line access multiplexer (DSLAM) discrete multitone technique (DMT) distribution hub 262 CHAPTER 9 USING TELEPHONE AND CABLE NETWORKS FOR DATA TRANSMISSION downloading downstream data band end office fiber node head end high-bit-rate DSL (HDSL) hybrid fiber-coaxial (HFC) network in-band signaling incumbent local exchange carrier (ILEC) interexchange carrier (IXC) ISDN user port (ISUP) local access transport area (LATA) local exchange carrier (LEC) local loop long distance company message transport port (MTP) level modem modulator out-of-band signaling plain old telephone system (POTS) point of presence (POP) ranging regional cable head (RCH) regional office 9.8 o o o o server control point (SCP) signal point (SP) signal transport port (STP) signaling connection control point (SCep) Signaling System Seven (SS7) switched/56 service switching office symmetric DSL (SDSL) tandem office telephone user port (TUP) transaction capabilities application port (TCAP) trunk uploading upstream data band Y.32 Y.32bis Y.34bis Y.90 Y.92 very-high-bit-rate DSL (VDSL) video band V-series wide-area telephone service (WATS) SUMMARY The telephone, which is referred to as the plain old telephone system (POTS), was originally an analog system. During the last decade, the telephone network has undergone many technical changes. The network is now digital as well as analog. The telephone network is made of three major components: local loops, trunks, and switching offices. It has several levels of switching offices such as end offices, tandem offices, and regional offices. The United States is divided into many local access transport areas (LATAs). The services offered inside a LATA are called intra-LATA services. The carrier that handles these services is called a local exchange carrier (LEC). The services between LATAs are handled by interexchange carriers (lXCs). In in-band signaling, the same circuit is used for both signaling and data. In out-ofband signaling, a portion of the bandwidth is used for signaling and another portion SECTION 9.9 o o o o o o PRACTICE SET 263 for data. The protocol that is used for signaling in the telephone network is called Signaling System Seven (SS7). Telephone companies provide two types of services: analog and digital. We can categorize analog services as either analog switched services or analog leased services. The two most common digital services are switched/56 service and digital data service (DDS). Data transfer using the telephone local loop was traditionally done using a dial-up modem. The term modem is a composite word that refers to the two functional entities that make up the device: a signal modulator and a signal demodulator. Most popular modems available are based on the V-series standards. The V.32 modem has a data rate of 9600 bps. The V32bis modem supports 14,400-bps transmission. V90 modems, called 56K modems, with a downloading rate of 56 kbps and uploading rate of 33.6 kbps are very common. The standard above V90 is called V92. These modems can adjust their speed, and if the noise allows, they can upload data at the rate of 48 kbps. Telephone companies developed another technology, digital subscriber line (DSL), to provide higher-speed access to the Internet. DSL technology is a set of technologies, each differing in the first letter (ADSL, VDSL, HDSL, and SDSL. ADSL provides higher speed in the downstream direction than in the upstream direction. The high-bitrate digital subscriber line (HDSL) was designed as an alternative to the T-l line (1.544 Mbps). The symmetric digital subscriber line (SDSL) is a one twisted-pair version of HDSL. The very high-bit-rate digital subscriber line (VDSL) is an alternative approach that is similar to ADSL. Community antenna TV (CATV) was originally designed to provide video services for the community. The traditional cable TV system used coaxial cable end to end. The second generation of cable networks is called a hybrid fiber-coaxial (HFC) network. The network uses a combination of fiber-optic and coaxial cable. Cable companies are now competing with telephone companies for the residential customer who wants high-speed access to the Internet. To use a cable network for data transmission, we need two key devices: a cable modem (CM) and a cable modem transmission system (CMTS). 9.9 PRACTICE SET Review Questions 1. 2. 3. 4. What are the three major components of a telephone network? Give some hierarchical switching levels of a telephone network. What is LATA? What are intra-LATA and inter-LATA services? Describe the SS7 service and its relation to the telephone network. S. What are the two major services provided by telephone companies in the United States? 6. What is dial-up modem technology? List some of the common modem standards discussed in this chapter and give their data rates. 264 CHAPTER 9 USING TELEPHONE AND CABLE NETWORKS FOR DATA TRANSMISSION 7. What is DSL technology? What are the services provided by the telephone companies using DSL? Distinguish between a DSL modem and a DSLAM. 8. Compare and contrast a traditional cable network with a hybrid fiber-coaxial network. 9. How is data transfer achieved using CATV channels? 10. Distinguish between CM and CMTS. Exercises 11. Using the discussion of circuit-switching in Chapter 8, explain why this type of switching was chosen for telephone networks. 12. In Chapter 8, we discussed the three communication phases involved in a circuitswitched network. Match these phases with the phases in a telephone call between two parties. 13. In Chapter 8, we learned that a circuit-switched network needs end-to-end addressing during the setup and teardown phases. Define end-to-end addressing in a telephone network when two parties communicate. 14. When we have an overseas telephone conversation, we sometimes experience a delay. Can you explain the reason? 15. Draw a barchart to compare the different downloading data rates of common modems. 16. Draw a barchart to compare the different downloading data rates of common DSL technology implementations (use minimum data rates). 17. Calculate the minimum time required to download one million bytes of information using each of the following technologies: a. V32 modem b. V32bis modem c. V90 modem 18. Repeat Exercise 17 using different DSL implementations (consider the minimum rates). 19. Repeat Exercise 17 using a cable modem (consider the minimum rates). 20. What type of topology is used when customers in an area use DSL modems for data transfer purposes? Explain. 21. What type of topology is used when customers in an area use cable modems for data transfer purposes? Explain. CHAPTER 9 Using Telephone and Cable Networks Solutions to Review Questions and Exercises Review Questions 1. The telephone network is made of three major components: local loops, trunks, and switching offices. 2. The telephone network has several levels of switching offices such as end offices, tandem offices, and regional offices. 3. A LATA is a small or large metropolitan area that according to the divestiture of 1984 was under the control of a single telephone-service provider. The services offered by the common carriers inside a LATA are called intra-LATA services. The services between LATAs are handled by interexchange carriers (IXCs). These carriers, sometimes called long-distance companies, provide communication services between two customers in different LATAs. 4. Signaling System Seven (SS7) is the protocol used to provide signaling services in the telephone network. It is very similar to the five-layer Internet model. 5. Telephone companies provide two types of services: analog and digital. 6. Dial-up modems use part of the bandwidth of the local loop to transfer data. The latest dial-up modems use the V-series standards such as V.32 and V.32bis (9600 bps), V.34bis (28,800 or 33,600 bps), V.90 (56 kbps for downloading and 33.6 kbps for uploading), and V.92. (56 kbps for downloading and 48 kbps for uploading. 7. Telephone companies developed digital subscriber line (DSL) technology to provide higher-speed access to the Internet. DSL technology is a set of technologies, each differing in the first letter (ADSL, VDSL, HDSL, and SDSL). The set is often referred to as xDSL, where x can be replaced by A, V, H, or S. DSL uses a device called ADSL modem at the customer site. It uses a device called a digital subscriber line access multiplexer (DSLAM) at the telephone company site. 8. The traditional cable networks use only coaxial cables to distribute video information to the customers. The hybrid fiber-coaxial (HFC) networks use a combination of fiber-optic and coaxial cable to do so. 1 2 9. To provide Internet access, the cable company has divided the available bandwidth of the coaxial cable into three bands: video, downstream data, and upstream data. The downstream-only video band occupies frequencies from 54 to 550 MHz. The downstream data occupies the upper band, from 550 to 750 MHz. The upstream data occupies the lower band, from 5 to 42 MHz. 10. The cable modem (CM) is installed on the subscriber premises. The cable modem transmission system (CMTS) is installed inside the distribution hub by the cable company. It receives data from the Internet and passes them to the combiner, which sends them to the subscriber. The CMTS also receives data from the subscriber and passes them to the Internet. Exercises 11. Packet-switched networks are well suited for carrying data in packets. The end-toend addressing or local addressing (VCI) occupies a field in each packet. Telephone networks were designed to carry voice, which was not packetized. A circuit-switched network, which dedicates resources for the whole duration of the conversation, is more suitable for this type of communication. 12. The setup phase can be matched to the dialing process. After the callee responds, the data transfer phase (here voice transfer phase) starts. When any of the parties hangs up, the data transfer is terminated and the teardown phase starts. It takes a while before all resources are released. 13. In a telephone network, the telephone numbers of the caller and callee are serving as source and destination addresses. These are used only during the setup (dialing) and teardown (hanging up) phases. 14. The delay can be attributed to the fact that some telephone companies use satellite networks for overseas communication. In these case, the signals need to travel several thousands miles (earth station to satellite and satellite to earth station). 15. See Figure 9.1. Figure 9.1 Solution to Exercise 15 60 kbps 56 kbps 50 kbps 40 kbps 30 kbps 20 kbps 14.4 kbps 10 kbps 9600 bps V.32 V.32bis V.90 3 16. See Figure 9.2. Figure 9.2 Solution to Exercise 16 55 Mbps 54 Mbps 25 Mbps 24 Mbps 6 Mbps 5 Mbps 4 Mbps 3 Mbps 2 Mbps 1 Mbps ADSL Lower ADSL Lite HDSL SDSL VDSL Higher 17. → → → a. V.32 b. V.32bis c. V.90 Time = (1,000,000 × 8) /9600 Time = (1,000,000 × 8) / 14400 Time = (1,000,000 × 8) / 56000 ≈ 834 s ≈ 556 s ≈ 143 s 18. a. ADSL b. ADSL Lite c. HDSL d. SDSL e. VDSL → → → → → Time = (1,000,000 × 8) / 1,500,000 Time = (1,000,000 × 8) / 1,500,000 Time = (1,000,000 × 8) / 1,500,000 Time = (1,000,000 × 8) / 768,000 Time = (1,000,000 × 8) / 25,000,000 ≈ 5.3 s ≈ 5.3 s ≈ 5.3 s ≈ 10.42 s ≈ 0.32 s 19. We can calculate time based on the assumption of 10 Mbps data rate: Time = (1,000,000 × 8) / 10,000,000 ≈ 0.8 seconds 20. The DSL technology is based on star topology with the hub at the telephone office. The local loop connects each customer to the end office. This means that there is no sharing; the allocated bandwidth for each customer is not shared with neighbors. The data rate does not depend on how many people in the area are transferring data at the same time. 4 21. The cable modem technology is based on the bus (or rather tree) topology. The cable is distributed in the area and customers have to share the available bandwidth. This means if all neighbors try to transfer data, the effective data rate will be decreased.