数据通信与网络第四版答案.pdf

CHAPTER 1IntroductionSolutions to Review Questions and ExercisesReview Questions1.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 data-bases,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 connec-tion)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:bus7.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 secureb.Bus:easy installationc.Star:robustd.Ring:easy fault isolation9.The number of cables for each type of network is:a.Mesh(“一 1)/2b.Star.nc.Ring:n-1d.Bus:one backbone and n drop lines10.The general factors are size,distances(covered by the network),structure,and ownership.211.An internet is an interconnection of networks.The Internet is the name of a spe-cific worldwide network12.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-work.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.Exercises14.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.16.a.Cable links:n(n-1)/2=(6 x 5)/2=15b.Number of ports:(n-1)=5 ports needed per device 17.a.Mesh topology,.If one connection fails,the other connections will still be work-ing.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 1.122.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 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.3Figure 1.1 Solution to Exercise 21Figure 1.2 Solution to Exercise 2225.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.4CHAPTER 2Network ModelsSolutions to Review Questions and ExercisesReview Questions1.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 packets.5.Peer-to-peer processes are processes on two or more devices communicating at a same layer6.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 witha.physical characteristics of the mediab.representation of bitsc.type of encodingd.synchronization of bitse.transmission rate and modef.the way devices are connected with each other and to the links9.The data link layer is responsible fora.framing data bitsb.providing the physical addresses of the sender/receiverc.data rate control2d.detection and correction of damaged and lost frames10.The network layer is concerned with delivery of a packet across multiple net-works;therefore its responsibilities include a.providing host-to-host addressing b.routing11.The transport layer oversees the process-to-process delivery of the entire message.It is responsible fora.dividing the message into manageable segmentsb.reassembling it at the destinationc.flow and error control12.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.Exercises15.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.16.a.Route determination:network layerb.Flow control:data link and transport layersc.Interface to transmission media:physical layerd.Access for the end user:application layer 17.a.Reliable process-to-process delivery:transport layerb.Route selection:network layerc.Defining frames:data link layerd.Providing user services:application layere.Transmission of bits across the medium:physical layer 18.a.Communication with users application program:application layerb.Error correction and retransmission:data link and transport layers c.Mechanical,electrical,and functional interface:physical layer3d.Responsibility for carrying frames between adjacent nodes:data link layer 19.a.Format and code conversion services:presentation layerb.Establishing,managing,and terminating sessions:session layerc.Ensuring reliable transmission of data:data link and transport layersd.Log-in and log-out procedures:session layere.Providing independence from different data representation:presentation layer 20.See Figure 2.1.Figure 2.1 Solution to Exercise 2021.See Figure 2.2.Figure 2.2 Solution to Exercise 2122.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.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 3Data and SignalsSolutions to Review Questions and ExercisesReview Questions1.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 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(X=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-crete.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.2Exercises16.a.T=l/f=1/(24 Hz)=0.0417 s=41.7 x 10-3s=41.7 msb.T=l/f=l/(8 MHz)=0.000000125=0.125 x 10-6 s=0.125 四c.T=1/f=1/(140 KHz)=0.00000714 s=7.14 x 106s=7.14 gs 17.a.f=l/T=l/(5 s)=0.2 Hzb.f=l/T=1/(12|LIS)=83333 Hz=83.333 x 103Hz=83.333 KHzc.f=l/T=l/(220 ns)=4550000 Hz=4.55x 106Hz=4.55 MHz 18.a.90 degrees(n/2 radian)b.0 degrees(0 radian)c.90 degrees(兀/2 radian)19.See Figure 3.1Figure 3.1 Solution to Exercise 19Frequency domainBandwidth=200 一 0=20020.We know the lowest frequency,100.We know the bandwidth is 2000.The highest frequency must be 100+2000=2100 Hz.See Figure 3.2Figure 3.2 Solution to Exercise 2020 Frequency domainj100 2100Bandwidth=2100-100=200021.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 Kbpsb.bit rate=1/(bit duration)=1/(2 ms)=500 bps3c.bit rate 二 l/(bit duration)=1/(20 jus/10)=1/(2 jus)=500 Kbps 23.a.(10/1000)s=0.01sb.(8/1000)s=0.008 s=8 msc.(100,000 x 8)/1000)s=800 s24.There are 8 bits in 16 ns.Bit rate is 8/(16 x IO-9)=0.5 x 10-9=500 Mbps25.The signal makes 8 cycles in 4 ms.The frequency is 8/(4 ms)=2 KHz26.The bandwidth is 5 x 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 2728.The signal is nonperiodic,so the frequency domain is made of a continuous spec-trum of frequencies as shown in Figure 3.4.Using the first harmonic,data rate=2x6 MHz=12 MbpsUsing three harmonics,data rate=(2x6 MHz)/3=4 MbpsUsing five harmonics,data rate=(2x6 MHz)/5=2.4 Mbps30.dB=10 log10(90/100)=-0.46 dB31.-10=10 log10(P2/5)t log10(P2/5)=1 f(P2/5)=KT1fP2=0.5 W32.The total gain is 3 x 4=12 dB.The signal is amplified by a factor 10L2=15.85.433.100,000 bits/5 Kbps=20 s34.480 s x 300,000 km/s=144,000,000 km35.1|nm x 1000=1000 jum=1 mm36.We have4,000 log2(1+1,000)x 40 Kbps37.We have4,000 log2(1+10/0.005)=43,866 bps38.The file contains 2,000,000 x 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 x 1000 x 10=12,000,000 bits40.We haveSNR=(200 mW)/(10 x 2 x juW)=10,000We then haveSNRdB=10 log10 SNR=4041.We haveSNR=(signal power)/(noise power).However,power is proportional to the square of voltage.This means we haveSNR=(signal voltage)2/(noise voltage)2=(signal voltage)/(noise voltage)2=202=400We then haveSNRdB=10 log10 SNR x 26.0242.We can approximately calculate the capacity asa.C=B x(SNRdB/3)=20 KHz x(40/3)=267 Kbpsb.C=B x(SNRdB/3)=200 KHz x(4/3)=267 Kbpsc.C=B x(SNRdB/3)=1 MHz x(20/3)=6.67 Mbps43.a.The data rate is doubled(C2=2 x CJ.b.When the SNR is doubled,the data rate increases slightly.We can say that,approximately,(C2=Cx+1).44.We can use the approximate formulaC=Bx(SNRdB/3)or SNRdB=(3 x C)/BWe can say that the minimumSNRdB=3 x 100 Kbps/4 KHz=755This means that the minimumSNR=10 SNRdB/10=107 5 31,622,77645.We havetransmission time=(packet length)/(bandwidth)=(8,000,000 bits)/(200,000 bps)=40 s46.We have(bit length)=(propagation speed)x(bit duration)The bit duration is the inverse of the bandwidth.a.Bit length=(2 xlO8 m)x(1/(1 Mbps)=200 m.This means a bit occupies 200 meters on a transmission medium.b.Bit length=(2 xlO8 m)x(1/(10 Mbps)=20 m.This means a bit occupies 20 meters on a transmission medium.c.Bit length=(2 xlO8 m)x(1/(100 Mbps)=2 m.This means a bit occupies 2 meters on a transmission medium.47.a.Number of bits 二 bandwidth x delay 二 1 Mbps x 2 ms=2000 bitsb.Number of bits=bandwidth x delay=10 Mbps x 2 ms=20,000 bitsc.Number of bits 二 bandwidth x delay=100 Mbps x 2 ms=200,000 bits48.We haveLatency=processing time+queuing time+transmission time+propagation timeProcessing time=10 x 1 jus=10 jus=0.000010 sQueuing time=10 x 2 JLIS=20 jus=0.000020 sTransmission time=5,000,000/(5 Mbps)=1 sPropagation time=(2000 Km)/(2 x 108)=0.01 sLatency=0.000010+0.000020+1+0.01=1.01000030 sThe transmission time is dominant here because the packet size is huge.6CHAPTER 4Digital TransmissionSolutions to Review Questions and ExercisesReview Questions1.The three different techniques described in this chapter are line coding,block cod-ing,and scrambling.2.A data element is t