计算机网络第一次习题课(4——7).ppt

习题课（47）,2007.3.13,Ch4,Ch4(续),4.2Investigatenetworkconnectionsusedatyoursites.Whattypesofmediaareused?一般是双绞线光纤,Ch4(续),4.3Whatmediacanbeusedforanetworkconnectionthatpassesnearapowerfulelectricmotor?屏蔽双绞线：由一根屏蔽层围绕的双绞线组成。用于防止电磁辐射。同轴电缆：有屏蔽层。光纤：不会干扰也不会被干扰。（课本P47第二段末句）,Ch4(续),4.4Somecommercialnetworkservicesoffersatellitenetworkconnectionstoindividuals.Eachsubscriberisgivenasmalldishantennathatisusedtoreceivedata;thesubscriberisalsogivenadialuptelephonemodemthatisusedtosenddata.Findoutwhyasubscribercannotsenddatatothesatellite.天线的大小决定了通讯的距离。,Ch5,Ch5(续),5.2DrawthewaveformdiagramthatresultswhenthewordbitissentinASCIIacrossanRS232connection.idlestart0100011stopstart1001011stopstart0010111stopidle1:-150：+15idle:-15start:+15stop:-15,Ch5(续),5.3Supposeonesent100007-bitcharactersacrossanRS-232connectionthatoperatedat9600baud.Howlongwouldthetransmissionrequire?波特率：每秒钟硬件产生的电信号变化次数，对RS232来说，就等于每秒钟传输的位数。10000（72）/96009.375（s）,Ch5(续),5.6Theserialportsonsomecomputersuseconnectorswithfewerthan25pins.ReadabouttheDB-9connectorusedforserialportsonPCs.WhenRS-232isusedwithaDB-9connector.whichsignalsareomitted?相关的传输控制信号被忽略了。,Ch5(续),5.10MostRS-232hardwareallowsthecomputertowhichitattachestospecifythedataratetouseaswellasthenumberofstopbitstouse.Ifatransmitterisprogrammedtousetwostopbits,butareceiverisprogrammedtorequireonlyone,willdatabereceivedcorrectly?Ifso,whatisthedisadvantageofusinganextrastopbit?正确，但是会减缓传输速度。,Ch5(续),什么是RS-232?RS-232有什么特点？RS232是用于计算机与调制解调器、键盘或终端之类设备之间传输字符的串行接口的一种标准。特点：它要求发送器在传输字符各位之前先传输额外一位的0（起始位），在每个字符之后有一位1（终止位）。,Ch6,Ch6(续),6.1Considerusingamplitudemodulationwithasinewavethatoperatesatafrequencyof4000Hz.Howmanybitspersecondcanbeencoded?Why?4000调频与调幅技术每发送一个信号位都需要至少一个载波周期。,Ch6(续),6.2Explainwhyeachradiostationinanareamustbeassignedauniquecarrierfrequency.两个或多个使用不同载波频率的信号在相同介质中能够同时传输而互不干扰。详见6.6。,Ch6(续),6.7Theauthorusesamultiplexoranddemultiplexorthatallowsthreeseparatesourcesof8-bitdatatosendtothreeseparatedestinationsoverasinglechannel.Thechannelcanonlysend8-bitcharacters,andanysourcecantransferanypossible8-bitvalueatanytime.Deviseamultiplexingmechanismthatworkswith8-bithardwaretoacceptdatafromthreesourcesandsenditoverasharedchannel.,Ch6(续),图可以参考书上F6.6可以采用设置不同的频率告诉接收者数据源来自于哪里。注意三者之间的频率不要接近或成整数倍。具体见书上P74P75.,Ch6(续),什么是调制与解调？调制与解调有那些基本方法？调制：将数字信号与音频载波组合，产生适合于电话线上传输的音频信号（模拟信号）解调：从音频信号中恢复出数字信号方法：调幅(amplitudemodulation，AM)调频(frequencymodulation，FM)相位移动调制(phaseshiftmodulation),Ch6(续),载波复用技术有哪几种？各有什么特点？FDM（frequencydivisionmultiplexing）载波在不同的频率上传输，每个接收源只接受给定载波频率的信号WDM(wavedivisionmultiplexing)频分在光传输中的应用TDM(timedivisionmultiplexing)数据源“轮流”共享使用介质具体见书P7476。,Ch7,Ch7(续),7.1Assumethattwocomputersareusingtimedivisionmultiplexingtotaketurnssending1000-bytepacketsoverasharedchannelthatoperatesat64000bitspersecond.Ifthehardwaretakes100microsecondafteronecomputerskipssendingbeforetheothercanbegin,howlongwillittakeforeachofthecomputerstosendaonemegabytedatafile?,Ch7(续),所需传输的数据包（packet）数：10241024/1000=1048.576即1049个每个数据包的传输时间：10008/64000=0.125(s)总时间中用于传输两台机器的数据包所用的时间：(10492-1)0.125=262.125(s)总时间中的间隔时间：(1049-1)210010-6=0.2096(s)总时间：262.125+0.2096=262.3346(s),Ch7(续),7.2Inthepreviousexercise,calculatethetimerequiredfortransmissionifthecomputerstransmitserially.率先传输的机器的耗时：10490.125+1048510-6=131.13024(s)后传输的机器的耗时：131.130242+10010-6=262.26058(s),Ch7(续),7.7Supposemalfunctioninghardwareinacharacter-orientedtransmissionsystemsetsallbitstransferredtozero.Doesaparitybitcatchtheproblem?Whyorwhynot?不一定，这种方式不一定能检验出错误，如果有偶数位出错无法查出，如果有奇数位出错，可以检验出。见书7.7节。,Ch7(续),什么是校验和？什么是CRC？校验和见7