Solutions to Review Questions and Exercises

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11 An internet is an interconnection of networks The Internet is the name of a spe. cific 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 net. 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, 14 Unicode uses 32 bits to represent a symbol or a character We can define 232 differ. ent symbols or characters, 15 With 16 bits we can represent up to 216 different colors. a Cable links n n 1 2 6 5 2 15,b Number of ports n 1 5 ports needed per device. a Mesh topology If one connection fails the other connections will still be work. 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 How. ever 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, 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 sensi. tive 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 dedi. cated line is established between them The connection is point to point. Figure 1 1 Solution to Exercise 21,Station Station Station Station.
Repeater Repeat er,Station Station Station Station. Figure 1 2 Solution to Exercise 22,Station Station. 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. Sol 02 fm Page 1 Saturday January 21 2006 10 27 AM. Network Models,Solutions to Review Questions and Exercises. Review Questions, 1 The Internet model as discussed in this chapter include physical data link net.
work 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. 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, Sol 02 fm Page 2 Saturday January 21 2006 10 27 AM.
d detection and correction of damaged and lost frames. 10 The network layer is concerned with delivery of a packet across multiple net. works therefore its responsibilities include,a providing host to host addressing. 11 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. 12 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, 13 The application layer services include file transfer remote access shared data. base management and mail services, 14 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 corre. spond to the Internet model layers, 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 com. munications is the Open Systems Interconnection OSI model. 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. 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.
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. d Responsibility for carrying frames between adjacent nodes data link layer. 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. Sender B 42 C 82 D 80,42 40 A D Data T2 80 82 A D Data T2. 21 See Figure 2 2,Figure 2 2 Solution to Exercise 21. Sender B 42 C 82 D 80,42 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 net. work the packet is lost If the corrupted destination address matches one of the sta. tions 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. 24 Most protocols issue a special error message that is sent back to the source in this. 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,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 fre. quency 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 modu. lation 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. 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 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 nonperi.
odic 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 dis. 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. 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 s. c T 1 f 1 140 KHz 0 00000714 s 7 14 10 6 s 7 14 s,a f 1 T 1 5 s 0 2 Hz. b f 1 T 1 12 s 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.
a 90 degrees 2 radian,b 0 degrees 0 radian,c 90 degrees 2 radian. 19 See Figure 3 1,Figure 3 1 Solution to Exercise 19. Frequency domain,0 20 50 100 200,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. Frequency domain,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. 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, c bit rate 1 bit duration 1 20 s 10 1 2 s 500 Kbps. a 10 1000 s 0 01 s,b 8 1000 s 0 008 s 8 ms,c 100 000 8 1000 s 800 s. 24 There are 8 bits in 16 ns Bit rate is 8 16 10 9 0 5 10 9 500 Mbps. 25 The signal makes 8 cycles in 4 ms The frequency is 8 4 ms 2 KHz. 26 The bandwidth is 5 5 25 Hz, 27 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. 28 The signal is nonperiodic so the frequency domain is made of a continuous spec. trum of frequencies as shown in Figure 3 4,Figure 3 4 Solution to Exercise 28.
10 volts 10 volts,KHz KHz 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. 33 100 000 bits 5 Kbps 20 s,34 480 s 300 000 km s 144 000 000 km. 35 1 m 1000 1000 m 1 mm,36 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 W 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,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,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 duratio. 1 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

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