数据通信与网络chapter27
完整版数据通信与网络重点总结
《数据通信与计算机网络》复习要点第1章概述1.以数字信号来传送消息的通信方式称为数字通信,而传输数字信号的通信方式称为数字通信系统。
2.数据通信包括数据传输、数据交换和数据处理。
3.数据通信网络按照覆盖的物理范围可分为广域网、局域网和城域网。
4.计算机网络是通信技术与计算机技术密切结合的产物。
5.计算机网络已经历了由单一网络向互联网发展的过程。
6.计算机网络具有三个主要的组成部分(三大组成要素),即①能向用户提供服务的若干主机;②由一些专用的通信处理机(即通信子网中的结点交换机)和连接这些结点的通信链路所组成的一个或数个通信子网;③为主机与主机、主机与通信子网,或者通信子网中各个结点之间通信而建立的一系列协议。
7.计算机网络按通信方式分为广播网络和点到点网络。
8.计算机网络主要性能指标,包括速率、带宽、吞吐量、时延、时延带宽积和利用率。
9. OSI模型有七个层次,分别是:(7)应用层;(6)表示层;(5)会话层;(4)运输层;(3)网络层;(2)数据链路层;(1)物理层10.因特网使用的TCP/IP参考模型的四个层次是:应用层、传输层、互联网层、网络接入层。
11.协议是控制两个对等实体进行通信的规则的集合。
协议的语法定义了所交换信息的结构或格式;协议的语义定义了源端和目的端所要完成的操作。
12.服务指同一开放系统中某一层向它的上一层提供的操作,但不涉及这些操作的具体实现。
13.面向连接服务具有连接建立、数据传输和连接释放三个阶段。
14.标准是由标准化组织、论坛以及政府管理机构共同合作制订的。
标准可分为两大类:法定标准和事实标准。
一 .数据通信网络由哪些部分组成?答:从系统设备级的构成出发,可以认为数据通信系统由下面三个子系统组成:(1)终端设备子系统,由数据终端设备及有关的传输控制设备组成。
(2)数据传输子系统,由传输信道和两端的数据电路终接设备组成。
(3)数据处理子系统,指包括通信控制器在内的电子计算机。
数据通信与网络概论
数据通信与网络的发展正在改变人类的生活方式和日常活动。
今天的商业、工业、科学、教育、卫生、体育与娱乐,都要依赖于数据通信与网络互联。
一场巨大的技术革命正发生在数据通信与网络领域。
遍布全球的电话网、无线电广播和电视、计算机网络和通信卫星正在迅速地融合,信息收集、数据传输、存储和处理之间的差别正逐步消失。
数据通信的信道带宽越来越大。
网络的服务功能更加人性化、智能化。
人们的心愿总有一天会实现:在世界上任何地方,任何时刻都能登录Internet,快速而准确地下载和上传信息。
本章将讨论数据通信与网络的定义,计算机网络的组成、分类和拓扑结构,网络协议和标准,网络参考模型。
1.1 数据通信1.1.1 数据通信的组成数据通信(data communication)是指两台设备之间通过某种形式的传输介质进行的数据交换。
这些通信设备必然成为由软件与硬件组成的通信系统的一部分。
一个通信系统要达到的最基本目的是完成双方的数据交换。
通信双方可以借助于电话机、工作站、服务器或其他数据处理设备、交换传输设备等实现通信的目的。
一个简化的通信模型如图1-1所示。
1.通信模型的主要功能现将该模型的几个要素及其主要作用说明如下:•源站(source)源站设备产生传输的数据。
•发送器(transmitter)将源站的数据转化并编码适应于传输系统的电磁信号,如电话网上传输的模拟信号。
•传输系统(transmission system)可能是一根单独的传输线,也可能是复杂的网络通信系统。
数据通信与网络应用2•接收器(receiver)接收器接收来自传输系统的信号,并转换成目的设备能处理的信息。
如将模拟信号转换成数字比特流。
•目的站(destination)目的站设备从接收器获取信息并对其加以处理。
图1-1 简化的通信模型2.通信过程的主要任务上述简单的描述掩盖了通信技术的复杂性。
实际上,通信系统应在不同层次上完成多项复杂的任务。
•传输系统的复用技术,为防止服务过载,要求引入拥塞控制。
数据通信与网络
PPT文档演模板
2004年2月-2005年1月
数据通信与网络
课程主题
Computer Communications and Networking Technologies
– various networking schemes that have been devised
PPT文档演模板
2004年2月-2005年1月
数据通信与网络
教学内容
– 通信模型与术语 – 通信系统硬件
• 传输媒体,数据通信接口,网络设备 – 数据传输
• 信号分析,信道传输特性,数据编码,数据传输方式, 线路复用,链路控制
通信与计算机的关系
• 电信主要包括话音通信、图像通信、数据通信、 视频通信。数据及计算机通信是为计算机之间的 远程信息交换、分布处理、资源共享提供服务的。
• 数据及计算机通信是通信技术与计算机技术相结 合的产物,通信系统的性能提高及其技术的迅猛 发展得益于计算机硬件与软件技术的不断进步。
• 按照定义,计算机网络由通信子网与资源子网构 成。今天面对的通信子网以电信业运营的网络为 主干。
• 根据吉尔德定律与摩尔定律得知,通信网络的发展 大大快于微处理器的发展。 问题2:根据吉尔德定律,假如目前网络最大速率为
10G bps,5年后预计最大速率为多少?
– 网络带宽每年增4倍,5年后增至45=1024倍,达到10.24Tbps。
PPT文档演模板
2004年2月-2005年1月
数据通信与网络
数据通信与网络
数据通信与网络数据通信与网络数据通信指的是通过某种媒介(如电波、光纤等)传输数据的过程。
而网络则是由多台计算机和其他设备相互连接而成的数据通信系统,可以实现信息和资源的共享。
数据通信与网络的发展始于20世纪60年代,随着计算机技术的发展和各种通信技术的成熟,数据通信与网络得到了迅猛的发展。
如今,数据通信与网络在各个领域都有着广泛的应用,无论是个人生活、工作,还是大型企业的运营,都离不开数据通信与网络。
数据通信与网络的基本组成包括数据发送方、数据接收方和媒介。
数据发送方将要传输的数据进行编码和调制后,通过媒介发送给接收方,接收方则进行解码和调制,得到原始数据。
传输中所用的媒介可以是有线的,如电缆、光纤等,也可以是无线的,如无线电波、红外线等。
不同的媒介具有不同的传输速度、传输距离和传输可靠性。
网络是由多个互连的计算机和其他设备组成的,各个设备之间通过网络协议进行通信和数据交换。
网络可以分为局域网、城域网、广域网和因特网等不同规模和范围的网络。
局域网通常用于办公楼或校园内部,城域网用于城市范围内的通信,广域网则覆盖更大范围,如跨城市、跨国家等。
而因特网是全球最大的网络,连接了世界上各个地区的计算机和网络设备。
数据通信与网络的应用非常广泛。
在个人生活中,我们使用手机进行语音通信、发短信和上网;在办公场所,我们使用局域网与同事进行文件共享和协同办公;在医疗领域,医生可以通过远程医疗技术为患者进行诊断和治疗;在工业控制领域,通过物联网技术可以实现远程监控和控制;在金融领域,我们可以通过网上银行进行转账和支付等。
数据通信与网络的应用不仅提高了工作效率和生活质量,还促进了各行各业的发展和创新。
数据通信与网络也面临着一些挑战与问题。
首先是安全性问题,随着人们对网络的依赖程度越来越高,网络安全问题变得日益严峻。
网络攻击、数据泄露和个人隐私保护等问题亟待解决。
其次是网络拥堵问题,随着用户数量的增加和数据传输量的增大,网络的带宽变得紧张,容易出现拥堵现象,影响用户的体验和数据传输的效率。
数据通信与网络课件
?数据传输的速率(比特 /秒)
?比特持续时间或长度 1/R
?发送方发送一个比特所需的时间
?比率r :每个信号元素承载的数据元素的数量(见 P68图) ?调制速率 (modulation rate )
?信号电平改变的速率 ?以波特(baud) 为单位 = 每秒信号元素数
?“传号” (mark ) 和“空号” (space )
?分别是二进制数字 1和0
信号的解读 Interpreting Signals
?接收方必须知道:
?各个比特的定时方式 ——何时起始,何时结束 ?每个比特信号电平的状态 ——是高或低 ?这两项任务都是通过在每个比特间隔的中间位
置采样来进行的
?影响信号成功解读的因素:
?数据率提高会增加误码率 ?信噪比提高会降低误码率 ?带宽增加可提高数据率 ?亦可通过编码方案提高传输性能
单极性编码存在的问题
Problems for Unipolar Encoding
两个问题使得单极性编码在信号传输应用中使用不多:
?直流分量(DC Component )
?信号的平均振幅不是零。 ?不能由没有处理直流分量能力的媒体传输,如微波或变
压器。 ?主要用于光纤传输。
?同步(Synchronization )
?在一个码元时间内,不是有电压(或电流),就是无电压(或 电流) ,电脉冲之间没有间隔,不易区分识别。所以接 收方不能正确识别每一个比特何时开始、何时结束 (原始 数据中出现连续的1或0时)。
同步:问题在哪里 ?
到底有几个” 1”(用高电平表示) ?
当一台设备发送一个比特的数字信号时,它将在一定的
传输不同步:换一种方式理解
我发了几个大 箱几个小箱?
数据通信与网络cha课件
电路交换适用于实时、连续的数 据传输,如电话通信。
报文交换
报文交换是一种基于存储-转发原理的交换方式,通信双方 发送完整的报文给交换节点,交换节点根据报文的目的地 址选择输出链路,将报文发送给下一节点。
报文交换的特点是通信链路是动态的,即通信双方在通信 过程中需要多次建立链路。
报文交换适用于非实时的数据传输,如电子邮件、网页浏 览等。
多媒体通信协议
如H.323、SIP等,支持音视频通话、 数据协作等多媒体通信功能,满足不 同应用场景的需求。
04
数据交换技术
电路交换
电路交换是一种传统的通信交换 方式,通过建立一条专用的通信 电路,实现数据或语音的传输。
电路交换的特点是通信链路是静 态的,即通信双方在通信过程中 始终占用该链路,直到通信结束。
详细描述
随着互联网的普及和重要性的提 高,网络安全问题也日益突出。 TCP/IP协议族在设计时主要考虑 的是互联互通性,安全性不是其 首要考虑的问题。因此,在使用 TCP/IP协议族时,需要采取各种 安全措施和技术来保护数据的安 全和隐私。
HTTP协议
总结词
HTTP协议是一种应用层协议,用于在网络中传输超文本 (例如网页)。
入侵检测技术
入侵检测技术
入侵检测技术用于检测和识别网络中异常行为和攻击行为,及时发 出警报和处理。
误用检测
误用检测是指通过预先定义攻击模式和特征库的方式,对网络流量 和日志进行分析和匹配,以检测入侵行为。
异常检测
异常检测是指通过建立正常行为模型的方式,对网络流量和日志进行 分析和比较,以检测异常行为和入侵行为。
分组交换
1
分组交换是一种动态分配通信链路的方式,它将 数据划分为多个数据包(或分组),每个数据包 独立选择路径进行传输。
数据通信与网络课后习题答案上
CHAPTER 115. With 16 bits, we can represent up to 216 different colors.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.2d. Ring Topology: The failed connection may disable the whole network unless itis a dual ring or there is a by-pass mechanism.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 cancontinue its operation.21. See Figure 1.123.a. E-mail is not an interactive application. Even if it is delivered immediately, itmay 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 sensitiveto delay.c. Surfing the Internet is the an application very sensitive to delay. We except toget access to the site we are searching.25. The telephone network was originally designed for voice communication; the Internet was originally designed for data communication. The two networks aresimilar 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.Figure 1.1 Solution to Exercise 21StationStation StationRepeat erStationStation StationRepeat erStationStation StationRepeaterHub1CHAPTER 2Exercises15. The International Standards Organization, or the International Organization ofStandards, (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.217.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 layer19.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 layer21. See Figure 2.1.23. 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 protocolwill inform the source to resend the packet.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 bedetected by the data link layer.Figure 2.1 Solution to Exercise 21B/42 C/82A/40SenderSenderLAN1 LAN2R1D/8042 40 A D i j Data T2 80 82 A D i j Data T2CHAPTER 317.a. f = 1 / T = 1 / (5 s) = 0.2 Hzb. f = 1 / T = 1 / (12 μs) =83333 Hz = 83.333 ×103 Hz = 83.333 KHzc. f = 1 / T = 1 / (220 ns) = 4550000 Hz = 4.55×106 Hz = 4.55 MHz19. See Figure 3.121. Each signal is a simple signal in this case. The bandwidth of a simple signal iszero. So the bandwidth of both signals are the same.a. (10 / 1000) s = 0.01 sb. (8 / 1000) s = 0. 008 s = 8 ms2c. ((100,000 ×8) / 1000) s = 800 s25. The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz27. The signal is periodic, so the frequency domain is made of discrete frequencies. as shown in Figure 3.2.29.Using the first harmonic, data rate = 2 ×6 MHz = 12 MbpsUsing three harmonics, data rate = (2 ×6 MHz) /3 = 4 MbpsUsing five harmonics, data rate = (2 ×6 MHz) /5 = 2.4 Mbps31. –10 = 10 log10 (P2 / 5) →log10 (P2 / 5) = −1 →(P2 / 5) = 10−1 →P2 = 0.5 W33. 100,000 bits / 5 Kbps = 20 s35. 1 μm ×1000 = 1000 μm = 1 mm37. We have4,000 log2 (1 + 10 / 0.005) = 43,866 bps39. 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 bits41. We haveSNR= (signal power)/(noise power).However, power is proportional to the square of voltage. This means we have Figure 3.1 Solution to Exercise 19Figure 3.2 Solution to Exercise 270 20 50 100 200Frequency domainBandwidth = 200 −0 = 200Amplitude10 voltsFrequency30KHz10KHz...SNR = [(signal voltage)2] / [(noise voltage)2] =[(signal voltage) / (noise voltage)]2 = 202 = 400We then haveSNR dB = 10 log10 SNR ≈26.0243.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).45. We havetransmission time = (packet length)/(bandwidth) =(8,000,000 bits) / (200,000 bps) = 40 s47.a. Number of bits = bandwidth ×delay = 1 Mbps ×2 ms = 2000 bitsb. Number of bits = bandwidth ×delay = 10 Mbps ×2 ms = 20,000 bitsc. Number of bits = bandwidth ×delay = 100 Mbps ×2 ms = 200,000 bits1CHAPTER 413. 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 kbaudb. r = 1/2 →s = (1/2) ×(1 Mbps) ×1/(1/2) = 1 Mbaudc. r = 2 →s = (1/2) ×(1 Mbps) ×1/2 = 250 Kbaudd. r = 4/3 →s = (1/2) ×(1 Mbps) ×1/(4/3) = 375 Kbaud15. 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.17. See Figure 4.2. 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.219. See Figure 4.3. B is proportional to (5.25 / 16) N which is inside range in Table 4.1(B = 0 to N/2) for 2B/1Q.21. The data stream can be found asa. NRZ-I: 10011001.b. Differential Manchester: 11000100.c. AMI: 01110001.23. 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 = 0.0b. f /N = 50/100 = 1/2 →P = 0.3c. f /N = 100/100 = 1 →P = 0.4d. f /N = 150/100 = 1.5 →P = 0.0Figure 4.1 Solution to Exercise 15Figure 4.2 Solution to Exercise 170 0 0 0 0 0 0 01 1 1 1 1 1 1 1 0 0 1 1 0 0 1 10 1 0 1 0 1 0 1Case aCase bCase cCase dAverage Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8B (3 / 8) N0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 0 0 1 1 0 0 1 10 1 0 1 0 1 0 1Case aCase bCase cCase dAverage Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8B (12.5 / 8) N325. In 5B/6B, we have 25 = 32 data sequences and 26 = 64 code sequences. The numberof unused code sequences is 64 −32 = 32. In 3B/4B, we have 23 = 8 datasequences and 24 = 16 code sequences. The number of unused code sequences is16 −8 = 8.27a. In a low-pass signal, the minimum frequency 0. Therefore, we havef max = 0 + 200 = 200 KHz. →f s = 2 ×200,000 = 400,000 samples/sb. In a bandpass signal, the maximum frequency is equal to the minimum frequencyplus the bandwidth. Therefore, we havef max = 100 + 200 = 300 KHz. →f s = 2 ×300,000 = 600,000 samples /s29. The maximum data rate can be calculated asN max = 2 × B ×n b = 2 ×200 KHz ×log24 = 800 kbps31. We can calculate the data rate for each scheme:Figure 4.3 Solution to Exercise 19a. NRZ →N = 2 ×B = 2 ×1 MHz = 2 Mbpsb. Manchester →N = 1 ×B = 1 ×1 MHz = 1 Mbpsc. MLT-3 →N = 3 ×B = 3 ×1 MHz = 3 Mbpsd. 2B1Q →N = 4 ×B = 4 ×1 MHz = 4 Mbps11 11 11 11 11 11 11 1100 00 00 00 00 00 00 00 01 10 01 10 01 10 01 10+3+1−3−1+3+1−3−1+3+1−3−100 11 00 11 00 11 00 11+3+1−3−1Case aCase bCase cCase dAverage Number of Changes = (0 + 7 + 7 + 7) / 4 = 5.25 for N = 16B (5.25 / 8) N1CHAPTER 511. We use the formula S = (1/r) ×_______________N, but first we need to calculate thevalue of r foreach case.a. r = log22 = 1 →S = (1/1) ×(2000 bps) = 2000 baudb. r = log22 = 1 →S = (1/1) ×(4000 bps) = 4000 baudc. r = log24 = 2 →S = (1/2) ×(6000 bps) = 3000 baudd. r = log264 = 6 →S = (1/6) ×(36,000 bps) = 6000 baud213. We use the formula r = log2L to calculate the value of r for each case.15. See Figure 5.1a. 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 betweenthe first dot and the origin) and A2= 3 (the distance between the second dot andthe origin).b. This is BPSK, There is only one peak amplitude (3). The distance between eachdot 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 amplitudeand four phases). The amplitude is the distance between a point and theorigin, 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 and270 degrees.17. We use the formula B = (1 + d) ×(1/r) ×N, but first we need to calculate thevalue of r for each case.a. log24 = 2b. log28 = 3c. log24 = 2d. log2128 = 7Figure 5.1 Solution to Exercise 15a. r = 1 →B= (1 + 1) ×(1/1) ×(4000 bps) = 8000 Hzb. r = 1 →B = (1 + 1) ×(1/1) ×(4000 bps) + 4 KHz = 8000 Hzc. r = 2 →B = (1 + 1) ×(1/2) ×(4000 bps) = 2000 Hzd. r = 4 →B = (1 + 1) ×(1/4) ×(4000 bps) = 1000 Hz2 3 –3 3–222–22a. b.II IQQQc. d.IQ319.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) = 10We can then calculate the number of levels: L = 2r = 210 = 1024. This means thatthat we need a 1024-QAM technique to achieve this data rate.21.a. B AM = 2 ×B = 2 ×5 = 10 KHzb. B FM = 2 ×(1 + β) ×B = 2 ×(1 + 5) ×5 = 60 KHzc.B PM = 2 ×(1 + β) ×B = 2 ×(1 + 1) ×5 = 20 KHz1CHAPTER 613. To multiplex 10 voice channels, we need nine guard bands. The required bandwidthis then B = (4 KHz) ×10 + (500 Hz) ×9 = 44.5 KHz15.a. Group level: overhead = 48 KHz −(12 ×4 KHz) = 0 Hz.b. Supergroup level: overhead = 240 KHz −(5 ×48 KHz) = 0 Hz.2c. Master group: overhead = 2520 KHz −(10 ×240 KHz) = 120 KHz.d. Jumbo Group: overhead = 16.984 MHz −(6 ×2.52 MHz) = 1.864 MHz.a. Each output frame carries 2 bits from each source plus one extra bit for synchronization. Frame size = 20 ×2 + 1 = 41 bits.b. Each frame carries 2 bit from each source. Frame rate = 100,000/2 = 50,000frames/s.c. Frame duration = 1 /(frame rate) = 1 /50,000 = 20 μs.d. Data rate = (50,000 frames/s) ×(41 bits/frame) = 2.05 Mbps. The output datarate here is slightly less than the one in Exercise 16.e. In each frame 40 bits out of 41 are useful. Efficiency = 40/41= 97.5%. Efficiencyis better than the one in Exercise 16.19. We combine six 200-kbps sources into three 400-kbps. Now we have seven 400- kbps channel.a. Each output frame carries 1 bit from each of the seven 400-kbps line. Framesize = 7 ×1 = 7 bits.b. Each frame carries 1 bit from each 400-kbps source. Frame rate = 400,000frames/s.c. Frame duration = 1 /(frame rate) = 1 /400,000 = 2.5 μs.d. Output data rate = (400,000 frames/s) ×(7 bits/frame) = 2.8 Mbps. We can also calculate the output data rate as the sum of input data rate because there is no synchronizing bits. Output data rate = 6 ×200 + 4 ×400 = 2.8 Mbps.21. We need to add extra bits to the second source to make both rates = 190 kbps. Now we have two sources, each of 190 Kbps.a. The frame carries 1 bit from each source. Frame size = 1 + 1 = 2 bits.b. Each frame carries 1 bit from each 190-kbps source. Frame rate = 190,000frames/s.c. Frame duration = 1 /(frame rate) = 1 /190,000 = 5.3 μs.d. Output data rate = (190,000 frames/s) ×(2 bits/frame) = 380 kbps. Here theoutput bit rate is greater than the sum of the input rates (370 kbps) because ofextra bits added to the second source.23. See Figure 6.1.25. See Figure 6.2.Figure 6.1 Solution to Exercise 23O L E L Y I E B H HTDM327. The number of hops = 100 KHz/4 KHz = 25. So we need log225 = 4.64 ≈5 bits29. Random numbers are 11, 13, 10, 6, 12, 3, 8, 9 as calculated below:Figure 6.2 Solution to Exercise 25N1 = 11N2 =(5 +7 ×11) mod 17 −1 = 13N3 =(5 +7 ×13) mod 17 −1 = 10N4 =(5 +7 ×10) mod 17 −1 = 6N5 =(5 +7 ×6) mod 17 −1 = 12N6 =(5 +7 ×12) mod 17 −1 = 3N7 =(5 +7 ×3) mod 17 −1 = 8N8 =(5 +7 ×8) mod 17 −1 = 90000000110001010101001111010000010100111TDM41CHAPTER 711. See Table 7.1 (the values are approximate).13. We can use Table 7.1 to find the power for different frequencies:Table 7.1 Solution to Exercise 11Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz1 Km −3 −5 −710 Km −30 −50 −7015 Km −45 −75 −10520 Km −60 −100 −1401 KHz dB = −3 P2 = P1 ×10−3/10 = 100.23 mw10 KHz dB = −5 P2 = P1 ×10−5/10 = 63.25 mw2The table shows that the power for 100 KHz is reduced almost 5 times, which maynot be acceptable for some applications.15. We first make Table 7.2 from Figure 7.9 (in the textbook).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. 17. We can use the formula f = c / λto 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 THzb. B = [(2 ×108)/1000×10−9] −[(2 ×108)/ 1400 ×10−9] = 57 THz19. See Table 7.3 (The values are approximate).21. See Figure 7.1.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.100 KHz dB = −7 P2 = P1 ×10−7/10 = 39.90 mwTable 7.2 Solution to Exercise 15Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz1 Km −3 −7 −2010 Km −30 −70 −20015 Km −45 −105 −30020 Km −60 −140 −400Distance Bandwidth1 Km 100 KHz10 Km 1 KHz15 Km 1 KHz20 Km 0 KHzTable 7.3 Solution to Exercise 19Distance dB at 800 nm dB at 1000 nm dB at 1200 nm1 Km −3 −1.1 −0.510 Km −30 −11 −515 Km −45 −16.5 −7.520 Km −60 −22 −103c. 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.Figure 7.1 Solution to Exercise 21Critical angle = 60Critical angle = 60Critical angle = 60Refractionb. 60 degreesReflectionc. 80 degreesCritical angleCritical anglea. 40 degreesRefractionCritical angle1CHAPTER 811. 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 or3 [(5000 km)/ (2 ×108 m/s)]+ 3 [(1000 bits/1 Mbps)] = 75 ms + 3 ms = 78 msWe assume that the data transfer is in one direction; the total delay is thendelay for setup and teardown + propagation delay + transmission delaya. 78 + 25 + 1 = 104 msb. 78 + 25 + 100 = 203 ms2c. 78 + 25 + 1000 = 1103 msd. 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.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 isdone 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.c. In a virtual-circuit network, there is a need for end-to-end addressing duringthe 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 whichvirtual-circuit that particular packet follows.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 caseof 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 twotypes 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 receivesa packet, it consults its table for routing information. This type of network is considereda connectionless network.17.Packet 1: 2Packet 2: 3Packet 3: 3Packet 4: 219.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. Thismeans that we can have two entries with the same input or output ports. We canhave two entries with the same VCIs. However, we cannot have two entrieswith the same (port, VCI) pair.321.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 Chapter6 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 inChapter 6 is 1 ×n.a. See Figure 8.1.b. The total number of crosspoints areNumber of crosspoints = 10 (10 ×6) + 6 (10 ×10) + 10 (6 ×10) = 1800c. 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 thesame time, which means 100 simultaneous connections.e. The blocking factor is 60/100 or 60 percent.25.a. Total crosspoints = N2 = 10002 = 1,000,000b. Total crosspoints ≥4Ν[(2Ν)1/2 −1] ≥174,886. With less than 200,000 crosspointswe can design a three-stage switch. We can use n = (N/2)1/2 =23 andchoose k = 45. The total number of crosspoints is 178,200.Figure 8.1 Solution to Exercise 23 Part aStage 1 Stage 210 ×610 ×1010 ×1010Crossbars10Crossbars6Crossbars… …… ……… … …Stage 3…n = 10n = 10n = 10N = 100… … …n = 10n = 10n = 1010 ×6 N = 10010 ×66 ×106 ×10…CHAPTER 911. 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.213. 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.15. See Figure 9.1.17.19. We can calculate time based on the assumption of 10 Mbps data rate:Time = (1,000,000 ×8) / 10,000,000 ≈0.8 seconds21. 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.Figure 9.1 Solution to Exercise 15a. V.32 →Time = (1,000,000 ×8) /9600 ≈834 sb. V.32bis →Time = (1,000,000 ×8) / 14400 ≈556 sc. V.90 →Time = (1,000,000 ×8) / 56000 ≈143 sV.32 V.32bis V.9010 kbps 9600 bps14.4 kbps56 kbps20 kbps30 kbps40 kbps50 kbps60 kbps1CHAPTER 1011. We can say that (vulnerable bits) = (data rate) ×(burst duration)Comment: The last example shows how a noise of small duration can affect so many bits if the data rate is high.13. The codeword for dataword 10 is 101. This codeword will be changed to 010 if a 3-bit burst error occurs. This pattern is not one of the valid codewords, so the receiver detects the error and discards the received pattern.15.a. d (10000, 00000) = 1b. d (10101, 10000) = 2c. d (1111, 1111) = 0d. d (000, 000) = 0Comment: Part c and d show that the distance between a codeword and itself is 0.17.a. 01b. errorc. 00d. error19. We check five random cases. All are in the code.21. We show the dataword, codeword, the corrupted codeword, the syndrome, and the interpretation of each case:a. Dataword: 0100 →Codeword: 0100011 →Corrupted: 1100011 →s2s1s0 = 110 Change b3 (Table 10.5) →Corrected codeword: 0100011 →dataword: 0100The dataword is correctly found.b. Dataword: 0111 →Codeword: 0111001 →Corrupted: 0011001 →s2s1s0 = 011 Change b2 (Table 10.5) →Corrected codeword: 0111001→dataword: 0111The dataword is correctly found.c. Dataword: 1111 →Codeword: 1111111 →Corrupted: 0111110 →s2s1s0 = 111 Change b1 (Table 10.5) →Corrected codeword: 0101110→dataword: 0101The dataword is found, but it is incorrect. C(7,4) cannot correct two errors.a. vulnerable bits = (1,500) ×(2 ×10−3) = 3 bitsb. vulnerable bits = (12 ×103) ×(2 ×10−3) = 24 bitsc. vulnerable bits = (100 ×103) ×(2 ×10−3) = 200 bitsd. vulnerable bits = (100 ×106) ×(2 ×10−3) = 200,000 bitsI. (1st) ⊕(2nd) = (2nd)II. (2nd) ⊕(3th) = (4th)III. (3rd) ⊕(4th) = (2nd)IV. (4th) ⊕(5th) = (8th)V. (5th) ⊕(6th) = (2nd)3d. Dataword: 0000 →Codeword: 0000000 →Corrupted: 1100001 →s2s1s0 = 100 Change q2 (Table 10.5) →Corrected codeword: 1100101→dataword: 1100The dataword is found, but it is incorrect. C(7,4) cannot correct three errors.23. We need to find k = 2m −1 −m ≥11. We use trial and error to find the right answer:a. Let m = 1 k = 2m −1 −m = 21 −1 −1 = 0 (not acceptable)b. Let m = 2 k = 2m −1 −m = 22 −1 −2 = 1 (not acceptable)c. Let m = 3 k = 2m −1 −m = 23 −1 −3 = 4 (not acceptable)d. Let m = 4 k = 2m −1 −m = 24 −1 −4 = 11 (acceptable)Comment: The code is C(15, 11) with d min = 3.25.a. 101110 →x5 + x3 + x2 + xb. 101110 →101110000 (Three 0s are added to the right)c. x3 ×(x5 + x3 + x2 + x) = x8 + x6 + x5 + x4d. 101110 →10 (The four rightmost bits are deleted)e. x−4 ×(x5 + x3 + x2 + x) = x (Note that negative powers are deleted)27. CRC-8 generator is x8 + x2 + x + 1.a. It has more than one term and the coefficient of x0 is 1. It can detect a single-bit error.b. The polynomial is of degree 8, which means that the number of checkbits (remainder) r = 8. It will detect all burst errors of size 8 or less.c. Burst errors of size 9 are detected most of the time, but they slip by with probability (1/2)r−1 or (1/2)8−1≈0.008. This means 8 out of 1000 burst errors of size 9are left undetected.d. Burst errors of size 15 are detected most of the time, but they slip by with probability (1/2)r or (1/2)8 ≈0.004. This means 4 out of 1000 burst errors of size 15are left undetected.29. We need to add all bits modulo-2 (XORing). However, it is simpler to count the number of 1s and make them even by adding a 0 or a 1. We have shown the paritybit in the codeword in color and separate for emphasis.31. Figure 10.1 shows the generation of the codeword at the sender and the checkingof the received codeword at the receiver using polynomial division.Dataword Number of 1s Parity Codeworda. 1001011 →4 (even) →0 0 1001011b. 0001100 →2 (even) →0 0 0001100c. 1000000 →1 (odd) →1 1 1000000d. 1110111 →6 (even) →0 0 1110111433. Figure 10.2 shows the checksum to send (0x0000). This example shows that the checksum can be all 0s. It can be all 1s only if all data items are all 0, whichmeans no data at all.Figure 10.1 Solution to Exercise 31Figure 10.2 Solution to Exercise 33Codewordx7 + x5 + x2 + x + 1x7+x4+x3+ x + 1x4+x2+ x + 1 x11+x9 + x6+x5+x4x11+x9+x6+x5+x4 +x11+x9+x8+x7x8+x7+x6+x5+x4x8 +x6+x5+x4x7x7 +x5+x4+x3x5+x4+x3x5 +x3+x2+ xx4 + x2+ xx4 +x2+ x + 111DatawordSenderQuotientDivisorRemainderCodewordx7 + x5 + x2 + x + 1x4 x3 x 11x7+ + + +x4+x2+ x + 1 x11+x9 + x6+x5+ x4 +x11+x9+x6+x5+x4 +x11+x9+x8+x7x8+x7+x6+x5+x4x8 +x6+x5+x4x7x7 +x5+x4+x3x5+x4+x3x5 +x3+x2+ xx4 + x2+ xx4 x2 x 11+ + ++1DatawordQuotientDivisorRemainderReceiverChecksum (initial)Sum4 5 6 7B A 9 8F F F F0 0 0 0 Checksum (to send)0 0 0 01CHAPTER 1113. We give a very simple solution. Every time we encounter an ESC or flag character, we insert an extra ESC character in the data part of the frame (see Figure 11.1).15. We write two very simple algorithms. We assume that a frame is made of a onebyte beginning flag, variable-length data (possibly byte-stuffed), and a one-byteending flag; we ignore the header and trailer. We also assume that there is no error during the transmission.。
数据通信与网络
数据通信与网络一、引言数据通信与网络是现代社会中不可或缺的一部分。
它负责实现信息的传输和共享,极大地改变了人们的生活和工作方式。
本文将讨论数据通信与网络的基本概念、功能以及一些常见的应用。
二、数据通信的基本概念1. 数据通信的定义数据通信是指将信息从一个地方传送到另一个地方的过程。
这种传送需要借助设备和协议来确保信息的准确和完整传输。
2. 数据通信的基本要素数据通信包括发送方、接收方、传输介质和传输协议四个基本要素。
发送方将原始信息转换为信号,并通过传输介质将信号发送给接收方。
传输协议规定了信号传输的方式和规则。
3. 数据通信的模型数据通信可以使用OSI模型进行描述,包括物理层、数据链路层、网络层、传输层、会话层、表示层和应用层七个层次。
每个层次有不同的功能和任务,通过层与层之间的协议进行通信。
三、数据通信的功能1. 数据传输数据通信的主要功能是将原始数据转换为可传输的信号,并通过传输介质将信号传输给接收方。
这样接收方就可以恢复原始数据。
2. 数据交换数据交换指的是在网络中传输数据的过程。
数据通信通过数据交换的方式实现信息的共享。
3. 错误检测和纠正数据通信通过添加冗余信息和使用校验算法来检测和纠正传输过程中可能发生的错误,确保信息的准确性和完整性。
4. 数据压缩与加密数据通信可以利用压缩算法将数据压缩,减少传输的数据量。
同时,通过加密算法可以保护敏感数据的安全性,防止信息被非法获取。
四、数据通信的应用1. 互联网互联网是当前最常见和广泛使用的数据通信网络。
它连接了全球各地的计算机和设备,实现了信息的全球化传输和共享。
2. 移动通信移动通信是指通过移动网络进行数据通信的方式。
它包括手机通信、移动互联网、物联网等多种应用,为人们提供了便捷的通信手段。
3. 在线购物数据通信使得在线购物成为可能。
人们可以通过互联网将购物需求传输给电商平台,实现网上购物和订单交付。
4. 远程办公与视频会议数据通信为远程办公和视频会议提供了支持,使得人们可以通过网络进行远程工作和交流,提高了工作效率和便利性。
12春第4讲 基础3 数据通信与网络PPT课件
在数据通信系统中,信源和信宿是各种 类型的计算机终端,简称 DTE
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1.4.2 数据通信系统的组成
3 数据链路 数据终接设备 DCE 和传输信道
DCE (Data Circuit-terminal Equipment)
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1.4.3 数据通信的信息交换方式
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1.4.1 数据通信的特点
1 什么是数据通信
用二进制代码表示的多媒体信息统称 为数据,数据通信是计算机和通信相结合 的一种通信方式,以计算机系统为主体的 网络通信系统称为数据通信系统。
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一 请在这里输入您的主要叙述内容
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电路交换(以传统电话网为例)
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要将 5 部电话机两两相连, 则需要 10 对电线。
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当电话机的数量很大时, 就必须使用电话交换机进行连接。
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• 可能只要经过一个交换机(如 A 到 B)
• 可能要经过多个交换机(如 C 到 D)
网络何种控制信息,完成 何种动作,以及作出何种响应。
(2)语法:指数据与控制信息的结构或格式。
(3)时序:指事件的执行顺序。
三 请在这里输入您的主要叙述内容
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数据通信与网络答案17
CHAPTER 17SONET/SDHSolutions to Odd-Numbered Review Questions and ExercisesReview Questions1.The ANSI standard is called SONET and the ITU-T standard is called SDH. Thestandards are nearly identical.3.STS multiplexers/demultiplexers mark the beginning points and endpoints of aSONET link. An STS multiplexer multiplexes signals from multiple electricalsources and creates the corresponding optical signal. An STS demultiplexerdemultiplexes an optical signal into corresponding electric signals. Add/drop mul-tiplexers allow insertion and extraction of signals in an STS. An add/drop multi-plexer can add an electrical signals into a given path or can remove a desired signalfrom a path.5.Pointers are used to show the offset of the SPE in the frame or for justification.SONET uses two pointers show the position of an SPE with respect to an STS.SONET use the third pointer for rate adjustment between SPE and STS.7.A regenerator takes a received optical signal and regenerates it. The SONETregenerator also replaces some of the existing overhead information with newinformation.9.The path layer is responsible for the movement of a signal from its source to itsdestination. The line layer is responsible for the movement of a signal across aphysical line. The section layer is responsible for the movement of a signal acrossa physical section. The photonic layer corresponds to the physical layer of the OSImodel. It includes physical specifications for the optical fiber channel. SONETuses NRZ encoding with the presence of light representing 1 and the absence oflight representing 0.Exercises11.Each STS-n frame carries (9 ×n× 86) bytes of bytes. SONET sends 8000 framesin each second. We can then calculate the user data rate as follows:STS-3→ 8000 × (9 ×3× 86) × 8 = 148.608 Mbps12STS-9→ 8000 × (9 ×9× 86) ×8=445.824 MbpsSTS-12→8000 × (9 ×12× 86) × 8 = 594.432 Mbps13.The user data rate of STS-1 is (8000 × 9 × 86 × 8) = 49.536 Mbps. To carry a loadwith a data rate 49.540, we need another 4 kbps. This means that we need to insert4000 / 8 = 500 bytes into every 8000 frames. In other words, 500 out of every 8000frames need to allow the H3 byte to carry data. For example, we can havesequences of 16 frames in which the first frame is an overloaded frame and then 15frames are normal.15.In answering this question, we need to think about the three upper layers inSONET. The path layer is responsible for end-to-end communication. The linelayer is responsible between multiplexers. The section layer is responsible betweendevices.a.A1 and A2 are used as aligners (synchronizers). They perform the same job as apreamble or flag field in other networks. We can call them framing bytes. Thesebytes are set and renewed at each device to synchronize the two adjacentdevices. There is no need for these bytes at the line or path layer.b.C1 is used at the section layer to identify multiplexed STSs. This idea can becompared to statistical TDM in which each slot needs an address. In otherwords, C1 is the address of each STS-1 in an STS-n. C2 is like the port numbersin other protocols. When different processes need to communicate over thesame network, we need port addresses to distinguish between them. There is noneed for C byte at the line layer.c.D bytes are used for SONET administration. SONET requires two separatechannels at the section (device-to-device) and line (multiplexer-to-multiplexer)layers.No administration is provided at the line layer.d.E byte creates a voice communication channel between two devices at the endsof a section.e.F bytes also create a voice communication. F1 is used between two devices atthe end of a section; F2 is used between two ends. No bytes are assigned at theline layer.f.The only G bytes are used for status reporting. A device at the end of the pathreports its status to a device at the beginning of the path. No other layer needsthis byte.g.H bytes are the pointers. H1 and H2 are used to show the offsetting of the SPEwith respect to STS-1. H3 is used to compensate for a faster or slower user data.All three are used in the line layer because add/drop multiplexing is done at thislayer. H4 is used at the path layer to show a multiframe payload. Obviously wedo not need an H byte in the section layer because no multiplexing or demulti-plexing happens at this layer.h.The only J byte is at the path layer to show the continuous stream of data at thepath layer (end-to-end). The user uses a pattern that must be repeated to showthe stream is going at the right destination. There is no need for this byte at theother layers.3 i.As we discussed, K bytes are used for automatic protection switching, whichhappens at the line layer (multiplexing). Other layers do not need these bytes. j.Z bytes are unused bytes. All of the bytes in SOH are assigned, but in LOH and POH some bytes are still unused.4。
数据通信与网络技术精品PPT课件
22
PAP和CHAP原理
1. PPP提供了两种可选的身份认证方法:口令验证协议(Password Authentication Protocol,PAP)和质询握手协议(Challenge Handshake Authentication Protocol,CHAP)。
2. 信道服务单元(CSU)/数据服务单元(DSU)类似数据终端设 备到数据通信设备的复用器,可以提供以下几方面的功能:信 号再生,线路调节,误码纠正,信号管理,同步和电路测试等。
12
常用的广域网设备
1. ISDN终端适配器是用来连接ISDN基本速率接口(BRI)到其它 接口,如EIA/TIA-232的设备。从本质上说,ISDN终端适配器 就相当于一台ISDN调制解调器。
2. 下面我们将简单介绍几种常用的PPP、ISDN、ATM、帧中继、 SDH和PPPOE等广域网技术和协议。
15
16
17
18
PPP协议的特点
1. 能够控制数据链路的建立。 2. 能够对IP地址进行分配和使用。 3. 允许同时采用多种网络层协议。 4. 能够配置和测试数据链路。 5. 能够进行错误检测。 6. 有协商选项,能够对网络层的地址和数据压缩等进行协商。
23
PAP和CHAP原理
24
ISDN概述
1. ISDN(Integrated Service Digital Network)是一种由交换机 和数字信道构成,可提供语音、数据、图像等综合业务信息传 输的数字通信网络。
2. 它可以使用户通过一条通信线路获得各种电信服务。ISDN有三 个基本特征: 1) 端到端的数字连接。 2) 综合的业务。 3) 标准的入网接口(两种速率标准—基本速率和基群速率)。
《数据通信与计算机网络(第三版)》(1-3章)课件
2.2.2模拟数据的数字传输
模拟信号数字化过程主要包括三个步骤:抽样、量化、编码,如图212所示。
图2-12模拟信号的数字传输
2.2 数据编码与数据传输
【例2-3】一路电话信号的频带为300~3400Hz,则抽样频率应该 是什么样的范围?
解:由题意得fH=3400Hz,因为电话信号属于低通信号,故用低通 抽样定理满足要求,即fS≥2fH= 2×3400=6800Hz,所以按照6800Hz的 抽样频率对300~3400Hz的电话信号抽样,则可以不失真地还原成原来 的话音信号。
2.数据通信系统模型 数据通信系统是通过数据电路将分布在远地的数据终端设备与计 算机系统连接起来,实现数据传输、交换、存储和处理的系统。从计 算机网络的角度看,数据通信系统是把数据源计算机所产生的数据迅 速、可靠、准确地传输到数据宿(目的)计算机或专用外设。
图2-2数据通信系统模型
2.1.2数据通信方式
1.按照数字信号排列顺序分类 数据在信道上传输有两种基本的通信方式一种是串行通信, 另一种是并行通信,如图2-3所示。
图2-3串行、并行通信方式
2.1.2数据通信方式
2.按照消息传递的方向与时间的关系分类
发端
发端
收端 发端
(a)全双工通信 数据
收端 发端
收端 发端
(b)半双工通信 数据
(c)单工通信
收端 收端
图2-4 全双工、半双工、单工通信方式
2.1.3数据通信系统的主要技术指标
1.数据传输速率
(1)波特率RB也称调制速率,以波形每秒的振荡数来衡量。如
果数据不压缩,波特率等于每秒钟传输的数据位数,如果数据进行 了压缩,那么每秒钟传输的数据位数通常大于调制速率。在信息传 输通道中,携带数据信息的信号单元叫码元,单位时间内通过信道 传输的码元数目称为码元传输速率,简称波特率。单位是“波特 ”(Baud),记为B。波特率与码元长度有关:
数据通信与计算机网络(第二版)课件:物理层
3.1 物理层概述 3.2 传输介质 3.3 物理层协议举例 3.4 ADSL技术
3.1 物理层概述
在物理信道实体之间合理地通过中间系统, 为比特传输所需的物理连接的激活、保持和拆 除提供机械的、电气的功能特性和规程特性的 手段。
特别要指出的是,物理层并不是指连接计算 机的物理设备或具体的传输媒体,而是指在物 理媒体上的为上一层(数据链路层)提供一个 传输原始比特流的物理连接。
无线电 微波 红外线
X 射线
γ射线
可见光 紫外线
3.2.3 无线传输介质
1.无线电波通信 2.微波通信 3.红外通信
3.2 传输介质
3.2.1 传输介质的特性 3.2.2 有线传输介质 1.双绞线 2.同轴电缆 3.光缆 4.有线传输介质的比较 3.2.3 无线传输介质
本章首页
3.3 物理层协议举例
3.4 ADSL技术
3.4.2 ADSL基本原理
ADSL使用普通电话线作为传输介质,虽然 传统的MODEM也是使用电话线传输的,但 它只使用了0~4kHz的低频段,而电话线理 论上有接近2MHz的带宽,ADSL正是使用了 26kHz以后的高频段。经ADSL MODEM编 码后的信号通过电话线传到电话局后再通过 一个信号识别分离器,如果是语音信号就传 到交换机上,如果是数字信号就接入Internet。
3.1 物理层概述
物理层协议主要包括机械、电气、功能和规程4个 特性。
(1)机械特性。定义接口部件的形状、尺寸、规 格、引脚数量和排列顺序等。
(2)电气特性。定义接口部件的信号高低、脉冲 宽度、阻抗匹配、传输速率和传输距离等。
(3)功能特性。定义接口部件的引脚功能、数据 类型和控制方式等。
(4)规程特性。定义接口部件的信号线在建立、 维持、释放物理连接和传输比特流时的时序。
ICT全部课程(2024)
互联网与物联网的融合
分析互联网和物联网在技术和应用方面的融合趋势,如智能家居、 智能交通等。
7
02
计算机硬件与软件技术
Chapter
2024/1/29
8
计算机组成原理及硬件结构
01
02
03
计算机的基本组成
包括中央处理器(CPU) 、内存、输入/输出设备等 。
12
03
网络通信与互联网应用技术
Chapter
2024/1/29
13
局域网组建与维护技术
2024/1/29
局域网基本概念及拓扑结构
介绍局域网的定义、特点、分类以及常见拓扑结构,如星 型、环型、总线型和网状拓扑等。
局域网传输介质与设备
详细讲解局域网中使用的传输介质(如双绞线、同轴电缆 、光纤等)以及各种网络设备(如交换机、路由器、集线 器等)的功能和配置方法。
探讨有线通信(如电话线、光纤)和无线通信(如移动通信、卫 星通信)的原理和应用。
通信网络与协议
涉及计算机网络的组成、结构和协议,包括局域网、广域网和互 联网等。
6
互联网与物联网技术
2024/1/29
互联网基础与应用
介绍互联网的发展历程、基本原理和主要应用,如万维网、电子 邮件、社交媒体等。
物联网概念与技术
广域网优化与故障排除
介绍广域网性能优化方法以及常见故障排除技巧,帮助学生提高网络维护能力。
15
网络安全策略及防护措施
2024/1/29
网络安全基本概念及威胁
介绍网络安全的定义、特点以及常见威胁类型,如病毒、 蠕虫、木马、黑客攻击等。
数据通信与网络网络互联设备PPT学习教案
路由表产生的方式----动态路由
动态路由协议学习到的路由
在大型网络环境下,依靠路由协 议比如OSPF、RIP路由协议学习
第34页/共51页
路由器接口
▪ 配置接口
Console口 AUX口
第35页/共51页
路由器接口
局域网接口
AUI接口 RJ-45接口 SC接口
第36页/共51页
路由器接口
广域网接口
高速同步串口 异步串口
第37页/共51页
ISDN BRI端口
路由器的硬件连接
路由器与局域网的连接
100Mbps FastEthernet: 100BaseTX----使用5类UTP或1类
STP,传输100M,距离100米。 100BaseFX----使用光纤,传输
100M,距离550米。
F0/2: 0260.8c01.2222 F0/3: 0260.8c01.3333 F0/4: 0260.8c01.4444
F0/1
F0/3
B
0260.8c01.3333
C
F0/2
0260.8c01.2222
F0/4
D
0260.8c01.4444
未知单播帧,广播帧:
执行广播操作Flooding(泛洪)
二层交换机属数据链路层设备, 可以识别数据包中的MAC地址 信息,根据MAC地址进行转发, 并将这些MAC地址与对应的端 口记录在自第1己4页/共内51页部的一个地址 表中。
交换机MAC地址表学习(一)
MAC 地址表
A
0260.8c01.1111 F0/1
C
F0/2
0260.8c01.2222
F0/3 F0/4
0260.8c01.1111 F0/1
数据通信与通信网
电子商务
商家和客户可以通过数据通信技术进行在线 交易和支付,实现便捷的商业活动。
物联网
通过数据通信技术,实现各种物理设备之间 的互联互通,提高生产和生活效率。
02
CATALOGUE
数据通信技术
有线数据通信技术
数字信号传输
通过数字信号传输技术,将数据 转换为二进制形式进行传输,具 有抗干扰能力强、传输距离远等
码分复用(CDMA)
利用不同的码字在同一信道上同时传 输多个信号。
05
CATALOGUE
通信网应用
物联网通信
物联网通信是实现物联网应用的基础 ,通过各种传感器、控制器等设备采 集数据,并通过通信网络传输到数据 中心进行处理和分析。
物联网通信具有低功耗、低成本、高 可靠性等特点,广泛应用于智能家居 、智能交通、智能工业等领域。
04
CATALOGUE
通信网技术
宽带接入技术
光纤接入技术
利用光纤传输高速数据,提供高带宽、低延迟的接入服务。
无线宽带接入技术
利用无线通信技术,如Wi-Fi、4G/5G等,实现灵活、便捷的接入。
有线宽带接入技术
利用双绞线、同轴电缆等传输介质,提供稳定的宽带接入服务。
宽带卫星接入技术
利用卫星通信技术,实现全球范围内的宽带接入。
卫星通信
利用卫星作为中继站实现地球上不同地点之间的通信。
数据交换技术
01
电路交换
在通信过程中保持通信链路状态 ,适用于实时、连续的数据传输 。
报文交换
02
03
分组交换
将数据打包成报文,根据地址逐 个节点传输,适用于少量、非实 时的数据传输。
将数据分成若干个分组,每个分 组独立传输,适用于大量、实时 的数据传输。
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Figure 27.3 URL
27.5
27-2 WEB DOCUMENTS
The documents in the WWW can be grouped into three broad categories: static, dynamic, and active. The category is based on the time at which the contents of the document are determined.
Topics discussed in this section:
Static Documents Dynamic Documents Active Documents
27.6
Figure 27.4 Static document
27.7
Figure 27.5 Boldface tags
27.8
Figure 27.6 Effect of boldface tags
27.9
Figure 27.7 Beginning and ending tags
27.10
Figure 27.8 Dynamic document using CGI
27.11
Figure 27.9 Dynamic document using server-site script
Chapter 27 WWW and HTTP
27.1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
27-1 ARCHITECTURE
The WWW today is a distributed client/server service, in which a client using a browser can access a service using a server. However, the service provided is distributed over many locations called sites.
27.34
Example 27.3 (continued)
27.35
Note
HTTP version 1.1 specifies a persistent connection by default.
27.36
27.15
Note
Active documents are sometimes referred to as client-site dynamic documents.
27.16
27-3 HTTP
The Hypertext Transfer Protocol (HTTP) is a protocol used mainly to access data on the World Wide Web. HTTP functions as a combination of FTP and SMTP.
27.12
Note
Dynamic documents are sometimes referred to as server-site dynamic documents.
27.13
Figure 27.10 Active document using Java applet
27.14
Figure 27.11 Active document using client-site script
27.18
Figure 27.12 HTTP transaction
27.19
Figure 27.13 Request and response messages
27.20
Figure 27.14 Request and status lines
27.21
Table 27.1 Methods
Topics discussed in this section:
Client (Browser) Server Uniform Resource Locator Cookies
27.2
Figure 27.1 Architecture of WWW
27.3
Figure 27.2 Browser
27.4
27.32
Figure 27.17 Example 27.2
27.33
Example 27.3
HTTP uses ASCII characters. A client can directly connect to a server using TELNET, which logs into port 80 (see next slide). The next three lines show that the connection is successful. We then type three lines. The first shows the request line (GET method), the second is the header (defining the host), the third is a blank, terminating the request. The server response is seven lines starting with the status line. The blank line at the end terminates the server response. The file of 14,230 lines is received after the blank line (not shown here). The last line is the output by the client.
27.27
Table 27.5 Response headers
27.28
Table 27.6 Entity headers
27.29
Example 27.1
This example retrieves a document. We use the GET method to retrieve an image with the path /usr/bin/image1. The request line shows the method (GET), the URL, and the HTTP version (1.1). The header has two lines that show that the client can accept images in the GIF or JPEG format. The request does not have a body. The response message contains the status line and four lines of header. The header lines define the date, server, MIME version, and length of the document. The body of the document follows the header (see Figure 27.16).
27.22
Байду номын сангаас
Table 27.2 Status codes
27.23
Table 27.2 Status codes (continued)
27.24
Figure 27.15 Header format
27.25
Table 27.3 General headers
27.26
Table 27.4 Request headers
Topics discussed in this section:
HTTP Transaction Persistent Versus Nonpersistent Connection
27.17
Note
HTTP uses the services of TCP on wellknown port 80.
27.30
Figure 27.16 Example 27.1
27.31
Example 27.2
In this example, the client wants to send data to the server. We use the POST method. The request line shows the method (POST), URL, and HTTP version (1.1). There are four lines of headers. The request body contains the input information. The response message contains the status line and four lines of headers. The created document, which is a CGI document, is included as the body (see Figure 27.17).