同济大学通信系统原理(英文班)练习题附答案
通信电路原理习题答案xiti_2
2-10．请用巴特沃思逼近法涉及一个低通滤波器，要求在频率0~4kHz范围内衰减小于1dB，频率高于20kHz的范围衰减大于35dB，信源与负载阻抗为600Ω。
解：Ap=1dB As=35dB 估计一个带宽比：20/4=5由巴特沃思滤波器计算曲线可得：阶次n=3当通带内衰减为1dB时，其对应归一化频率是0.8由此可以得出截止频率为4/0.8=5 重新计算带宽比20/5=4查表得：阶次为3的衰减As结果为32dB 不满足要求重新估计带宽比：20/5=4由巴特沃思滤波器计算曲线可得：阶次n=4当通带内衰减为1dB时，其对应归一化频率是0.85由此可以得出截止频率为5/0.85=5.88, 重新计算带宽比20/5.88=3.4查表得：阶次为4的衰减As结果为36dB 满足要求可查表得归一化元件值并根据公司计算出实际元件值。
所得滤波器实际电路图(略)2-11．请用巴特沃思逼近法涉及一个低通滤波器，要求在频率0~3kHz范围内衰减小于2dB，频率高于30kHz的范围衰减大于35dB，信源与负载阻抗为600Ω。
解：Ap=2dB As=35dB 估计一个带宽比：30/3=10由巴特沃思滤波器计算曲线可得：阶次n=2当通带内衰减为2dB时，其对应归一化频率是0.88由此可以得出截止频率为3/0.88≈3.409 重新计算带宽比30/3.409≈8.80查表得：阶次为2的衰减As结果为36dB 满足要求查表得：归一化元件值为C1'=1.4142 L2'=1.4142 Rs'=1.0000实际元件值：L=RL⨯L'/ωc=600⨯1.4142/(2π⨯3.409⨯10^3)≈39.6mHC=C'/(RL⨯ωc)=1.4142/(600⨯2π⨯3.409⨯10^3)≈0.110μF所得滤波器实际电路图(略)。
校园通信试题及答案英语
校园通信试题及答案英语一、选择题（每题2分，共20分）1. What does "communication" mean in English?A. 交流B. 通讯C. 交通D. 教育2. The most common way of communication in a campus is:A. EmailB. Phone callC. Face-to-faceD. Social media3. Which of the following is NOT a benefit of effective communication?A. Improved understandingB. Enhanced relationshipsC. Increased isolationD. Better collaboration4. What is the best way to handle a disagreement in a group discussion?A. Ignoring itB. Arguing loudlyC. Listening to others' opinionsD. Interrupting others5. The term "non-verbal communication" refers to:A. Communication without wordsB. Communication through writingC. Communication via technologyD. Communication in a foreign language6. What is the purpose of a campus newsletter?A. To inform students about campus eventsB. To provide entertainmentC. To advertise productsD. To promote academic research7. In a group project, which of the following is the most important?A. Individual performanceB. Group collaborationC. DeadlinesD. Personal interests8. What is the role of a campus radio station?A. To broadcast lecturesB. To play musicC. To provide campus news and updatesD. To offer job opportunities9. How can you improve your listening skills in a classroom setting?A. By talking moreB. By taking notesC. By daydreamingD. By avoiding eye contact10. Which of the following is a good practice for effective communication?A. Interrupting othersB. Being open-mindedC. Being judgmentalD. Ignoring feedback二、填空题（每题1分，共10分）11. Effective communication requires both the sender and the ________ to be clear and concise.12. Non-verbal cues, such as body language and tone of voice, can ________ the message being conveyed.13. In a campus setting, ________ is often used to share information quickly among a large group of people.14. Active listening involves giving the speaker your full attention, ________, and responding appropriately.15. Conflict resolution in a group can be facilitated by________ and finding common ground.16. The campus intranet is a valuable resource for students to access ________ and other educational materials.17. Public speaking is an important skill that can be improved through ________ and practice.18. A well-structured email should include a clear subject line, a greeting, the ________, and a closing.19. The use of social media for campus communication has increased due to its ________ and accessibility.20. Feedback is essential for continuous improvement and should be given in a ________ and constructive manner.三、简答题（每题5分，共20分）21. What are the key elements of effective communication?22. Describe the process of a successful group project from initiation to completion.23. Explain the importance of feedback in the learning process.24. Discuss the role of technology in facilitating campus communication.四、论述题（每题15分，共30分）25. Analyze the impact of social media on campus communication and its implications for students and faculty.26. Discuss the challenges and opportunities of cross-cultural communication in a diverse campus environment.五、案例分析题（每题20分）27. A student group is working on a project and has a disagreement on the direction of the project. One member insists on their idea, while others want to explore different options. How would you suggest they resolve this conflict and ensure the project's success?答案：一、选择题1-5: A B C C A6-10: A B C B B二、填空题11. receiver12. enhance or distort13. social media14. nodding15. communication16. course materials17. feedback18. body of the message19. interactivity20. timely三、简答题21. Key elements of effective communication include clarity, conciseness, active listening, non-verbal cues, and appropriate feedback.22. A successful group project involves clear goals, division of labor, open communication, regular meetings, conflict resolution, and timely completion of tasks.23. Feedback is important in the learning process as it helps students understand their strengths and weaknesses, encourages improvement, and fosters a growth mindset.24. Technology plays a significant role in campus communication by providing platforms for instant messaging, social。
通信英语(第五版)练习题(大全)
第一部分通信工程师（新技术新业务）习题第一章电信网新技术与新业务第一部分通信工程师（新技术新业务）习题第一章电信网新技术与新业务一、填空题1 软交换网络独立于（传输网络），主要完成呼叫控制、资源分配、协议处理、路由认证、带宽管理和计费等功能。
2 软交换技术是由一个或者一族（控制处理机）来完成用户管理、业务逻辑、信令分析处理和路由选择等核心功能。
3 软交换技术和原先的电路交换技术不同之处是交换组织由原先的TDM时隙交换网络替换为（包／信元）交换网络。
4 软交换通过优化网络结构不但实现了网络的融合，更重要的是实现了（业务）的融合。
5.软交换与外部的接口必须采用（开放）的协议。
6.在三层MPLS VPN中，当一个站点同时属于多个VPN时，它必须具有一个在所有VPN 中（唯一）的地址空间。
7.MPLS VPN技术以（宽带IP网络）为基础，采用MPLS技术，在公共IP网络上构建企业IP专网。
8.三层MPLS VPN是一种基于MPLS技术的IP VPN，是在网络路由和交换设备上应用MPLS 技术，简化核心（路由器）的路由选择方式。
9.链路绑定是指将那些属性相同或相似的平行链路绑定为一个特定的链路束，而在（链路状态）数据库中用这个绑定的链路束来代表所有这些平行的链路。
10.GMPLS的无编号链路是指不用（IP地址）标识链路而采用其他的替代方法，具体地是在每个网络节点对链路进行本地编号，以链路经过设备的Id号或接口号作为链路的识别标志。
11.开放式智能网业务平台供应商提供的业务生成环境SCE本身也是一个（专用）系统，用一个智能网设备供应商提供的SCE开发出的业务逻辑只能在该供应商的业务平台上运行。
12.在传统智能网中，业务交换平台和业务控制平台分布在不同的结点上，在某种意义上可以说传统智能网也是（分布）式的。
13.目前研究人员比较一致的看法是分布式智能网中的业务平台应该是基于（CORBA）平台的大型分布式处理系统。
(完整)通信原理习题汇总,推荐文档
通信原理习题班级：14电信姓名：王斌学号：20141151046教师：董建娥第一章绪论习题1一、填空题1、数字通信系统的主要性能指标是和。
码元速率R B定义是，单位。
信息速率定义是，单位。
2、数字通信系统的有效性用衡量，可靠性用衡量。
3、模拟通信系统的有效性用衡量，可靠性用衡量。
4、在等概条件下，八元离散信源能达到最大熵是，若该信源每秒钟发送2000个符号，则该系统的信息速率为。
5、通信系统的有效性衡量指标对于模拟通信系统为，对于数字通信系统为。
6、通信系统的可靠性衡量指标对于模拟通信系统为对于数字通信系统为。
7、一个M进制基带信号，码元周期为T S秒，则传码率为，若码元等概出现，一个码元所含信息量为。
8、通信系统模型中有两个变换，它们分别是之间的变换和之间的变换。
9、模拟信号是指信号的参量可取值的信号，数字信号是指信号的参量可取值的信号。
10、根据信道中所传输信号特征的不同，通信系统可分为通信系统和通信系统。
二、画图1、画出模拟通信系统的一般模型。
2、画出通信系统的一般模型。
三、计算题1、对于二电平数字信号，每秒传输300个码元，问此传码率R B等于多少？若该数字信号0和1出现是独立等概率的，那么传信率R b等于多少？2、现有一个由8个等概符号组成的信源消息符号集，各符号间相互独立，每个符号的宽度为0.1ms。
计算：（1）平均信息量；（2）码元速率和平均信息速率；（3）该信源工作2小时后所获得的信息量；（4）若把各符号编成二进制比特后再进行传输，在工作2小时后发现了27个差错比特（若每符号至多出错1位），求传输的误比特率和误符号率。
3、某消息源的符号集由32个等概的符号组成，每符号宽度为2ms，编为5位。
设该消息源以编组方式发送消息，每组30个符号，再间歇15ms，然后再发送下一组，试：（1）、求信息传输速率；（2）、若传输1小时后发现有72个符号出错。
若每符号至多出错1位，且间歇期无差错，求误信率和误码率。
通信原理课后答案
第一章习题习题1。
1 在英文字母中E 出现的概率最大，等于0。
105，试求其信息量. 解：E 的信息量：()()b 25.3105.0log E log E 1log 222E =-=-==P P I习题1.2 某信息源由A,B ，C ，D 四个符号组成，设每个符号独立出现，其出现的概率分别为1/4，1/4，3/16,5/16.试求该信息源中每个符号的信息量。
解：b A P A P I A 241log )(log )(1log 222=-=-==b I B 415.2163log 2=-= b I C 415.2163log 2=-= b I D 678.1165log 2=-=习题1。
3 某信息源由A ，B ，C ，D 四个符号组成，这些符号分别用二进制码组00,01，10，11表示。
若每个二进制码元用宽度为5ms 的脉冲传输，试分别求出在下列条件下的平均信息速率.（1） 这四个符号等概率出现； （2）这四个符号出现概率如习题1.2所示。
解：（1）一个字母对应两个二进制脉冲，属于四进制符号，故一个字母的持续时间为2×5ms 。
传送字母的符号速率为Bd 100105213B =⨯⨯=-R等概时的平均信息速率为s b 2004log log 2B 2B b ===R M R R(2)平均信息量为比特977.1516log 165316log 1634log 414log 412222=+++=H则平均信息速率为 s b 7.197977.1100B b =⨯==H R R习题1。
4 试问上题中的码元速率是多少? 解：311200 Bd 5*10B B R T -===习题1.5 设一个信息源由64个不同的符号组成，其中16个符号的出现概率均为1/32，其余48个符号出现的概率为1/96，若此信息源每秒发出1000个独立的符号,试求该信息源的平均信息速率。
解:该信息源的熵为96log 961*4832log 321*16)(log )()(log )()(22264121+=-=-=∑∑==i i i i Mi i x P x P x P x P X H=5。
《通信原理》习题解答-final
Q (4 nf
0
m
) ( 4 nf m )
若用 q(t ) 对 m(t ) 进行抽样，则已抽样信号 ms (t ) 为 ms (t ) m(t )q(t ) 其频谱为
ms ( ) 1 M ( ) Q ( ) 2
1 M ( ) Q0 (4 nf m ) ( 4 nf m ) T n
5-1、已知低通信号 m(t ) 的频谱 M ( f ) 为 f 1 M ( f ) 200 0 f 200 f 为其他值
(1) 假设以 f s 300 Hz 的速率对 m(t ) 进行理想抽样，试画出已抽样信号 ms (t ) 的频谱草图。 (2) 若用 f s 400 Hz 的速率抽样，重做上题。 解： (1) 由题意知，已抽样信号为 ms (t ) m(t ) T (t ) 其频谱函数为 M s ( f ) 1 T
因此码元宽度： Ts (3 ~ 5) m (9 ~ 15)ms 3-7、设宽度为 T，传号、空号相同的数字信号通过衰落信道，已知多径时延
T / 4 ，接收信号为两条路径信号之和，试画出展宽后的接收信号波形，并讨
论此时对信号波形应提出什么要求？ 解：通过展宽以及叠加的信号波形如下图：
2 第二章 信息论初步
1
《通信原理》课后习题解答
作者：熊沛、王宇
3 第三章 信道与干扰
3-1、一恒参信道点饿幅频特性为 H ( ) (1 cos T0 )e jtd ，td 为常数，试确定 信号 S（t）通过该信道后的输出波形，并讨论之。 解：根据系统传输特性，接收到的信号频率响应 Y(w)为：
5-6、已知模拟信号抽样值的概率密度 f(x)，如图所示，若按四电平进行均匀 量化，试计算信号量化噪声功率比。
通信英语综合练习题.doc
《通信英语》综合练习题（即课后练习题）（第一章）•请将下述词组译成英文：抽样量化与编码话路幅值抽样频率抽样速率脉冲流重复率编码过程模拟信号传输质量数字通信数字传输含噪声的环境传输路由信噪比信号电平地面系统噪声功率二进制传输反向操作8位码序列接收端帧格式同步宁二•请将下述词组译成中文：1.the schemes for performing these three functions2.the series of amplitude values3.the speech channel of telephone quality4.the sequence of 8~binary digits5.the minimum theoretical sampling frequency6.the voice channel occupying the range 300 Hz to 3. 4 kHz7.the 8~digits per sample value8.the sparking of a car ignition system9.the stream of the pulses with a repetition rate of 64 kHz10.the relationship of the true signal to the noise signal11.the signal received from a satellite12.the complete information about a particular message13.the shape of the transmitted signal14.the a/ttenuation introduced by transmission path15.the unit that converts sampled amplitude value to a set of pulses16.the sequence relating to channel 1, 2 and so on17.the unique sequence of pulses called synchronization word18.the terrestrial system19.the presence or absence of the pulse20.the high-speed electronic switch21.the time division multiplexer22.the Time Division Multiplexing五.请将下述短文译成中文：1.If we consider binary transmission, the complete information about aparticular message will always be obtained by simply detecting the presence or absence of the pulse. By comparison, most other forms of transmission systems convey the message information using the shape, or level of the transmittedsignal; parameters that are most easily affected by the noise and attenuation introduced by the transmission path. Consequently there is an inherentadvantage for overcoming noisy environments by choosing digital transmission.2.The reader may ask, how does the demultiplexer know which group of 8~digitsrelates to channel 1, 2, and so on? Clearly this is important! The problem is easily overcome by specifying a frame format, where at the start of each framea unique sequence of pulses called the frame code, or synchronization word, isplaced so as to identify the start of the frame. A circuit of thedemultiplexer is arranged to detect the synchronization word, and thereby itknows that the next group of 8~digits corresponds to channel 1.3.Noise can be introduced into transmission path in many different ways; perhapsvia a nearby lightning strike, the sparking of a car ignition system, or thethermal low-level noise within the communication equipment it self. It is the rela tio nship of the t rue signal to the noise signal, known as the signal一to-noise ratio, which is of most interest to the communication engineer・4.Basically, if the signal is very large compared to the noise level, then aperfect message can take place; however, this is not always the case. Forexample, the signal received from a satellite, located in far outer space, is very week and is at a level only slightly above that of the noise・Alternative examples may be found within terrestrial systems where, althoughthe message signal is strong, so is the noise power・5.So far we have assumed that each voice channel has a separate coder, the unitthat converts sampled amplitude values to a set of pulses; and decoder, theunit that performs the reverse operation. This need not be so, and systems are in operation where a single codec is shared between 24, 30, or even 120separate channels.6.A high-speed electronic switch is used to present the analog informationsignal of each channel, taken in tern, to the codec. The codec is thenarranged to sequentially sample the amplitude value, and code this value into the 8 - dig it sequence. Thus the output to the codec may be seen as asequence of 8 pulses relating to channel 1, then channel 2, and so on. Thisunit is called a time division multiplexer.（第九章）一.请将下述词组译成英文：个人通信通信标准固定电话业务网络容量移动交换中心国际漫游宽带业务接口转换频谱分配模拟方式蜂窝通信原理拥塞蜂窝裂变基站移动交换中心寄存器收费功能接入方法突发脉冲传输方式开销信息切换算法短消息服务技术规范二.请将下述词组译成中文：1.the total access communication system2.the global mobile communication system3.the time division multipie access4.the facsimile and short message service5.the fixed communication networks6.the more personalized system7.the cost and quality of the link8.the market growth9.the fixed telephone service10.the coaxial cable11.the interface conversion12.the cellular communication principle13.the frequency reuse and cell splitting14.the cochannel interference15.the theoretical spectral capability16.the micro-cellular system17.the base station transceiver18.the subscriber register19.the burst transmission mode20.the overhead information21.the advanced handover algorithms22.the facsimile and short message service23.the GSM technical specifications五.请将下述短文译成中文：1.The success of mobile systems across the world is a sign that communication ismoving to wards a more personalized, convenie nt sys tem. People who have to use a mobile phone on business soon begin to realize that the ability to phone any time, any place in one' s personal life rapidly becomes a necessity, not a convenience.2.The fixed telephone service is global and the interconnection varies fromcoaxial cable to optical fiber and satellite. The national standards aredifferent, but with common interfaces and interface conversion,interconnection can take place. For mobi 1 e the problem is far more complex, with the need to roam creating a need for complex networks and systems. Thus in mobile the question of standards is far more crucial to success than fixed systems.3.The GSM system is based on a cellular communications principle which was firstproposed as a concept in the 1940s by Bell System engineers in the US. Theidea came out of the need to increase net work capac ity and got round thefact that broadcast mobile networks, operating in densely populated areas,could be jammed by a very small number of simuItaneous calls. The power of the cellular system was that it allowed frequency reuse.4.The cellular concept is defined by two features, frequency reuse and cellsplitting. Frequency reuse comes into play by using radio channels on the same frequency in coverage areas tha/t are far enough apart not to cause cochannel interference. This allows handling of simuItaneous calls that exceed thetheoretical spectral capacity. Cell splitting is necessary when the trafficdemand on a cell has reached the maximum and the cell is then derived into amicro-cellular system.5.The cell coverage area is cont rolled by a base station which is it self madeup of two elements. The first element is the transmission system whichcommunicates out to the mobile and also receives information from it to set upand maintain calls when actually in operation. The base station transceiver iscontrolled by the base station controller, which communicates with the mobileswitching center --------------------------------- the essential link to thelocal public swit ched t elephone net work, and to the subscriber data whichis stored in registers within the system.6.The GSM system operates in a burst transmission mode with 124 radio channels inthe 900 MHz band, and these bursts can carry different types of information.The first type of information is speech, which is coded at 6. 5 kbit/s or 13kbit/s. The second type is data, which can be sent a/t 3. 6 kbit/s, 6 kbit/s or12.6 kbit/s. These two forms of information are the useful part of thetransmission, but have to be supported by overhead information which is sent incontrol channels.7.The use of dig ital radio t ransmission and the advanced handover algor it hmsbetween radio cells in GSM networks allows for significantly better frequentlyusage than in analogue cellular systems, thus increasing the number ofsubscribers that can be served. Since GSM provides common standard, cellularsubscribers will also be able to use their telephones over the entire GSMservice area. Roaming is fully autoina/tic bet ween and wit hin all countriescovered by GSM system.8.In addition to international roaming, GSM provides new user services, such ashigh speed data communication, Facsimile and short message service. The GSMtechnical specifications are designed to work in concert with other standards,e. g. ISDN. Interworking between the standards is in this way assured. In thelong term perspective cellular systems, using a digital technology, will becomethe universal method of telecommunication.《通信英语》综合练习题答案说明：山于《通信英语》的综合练习题全部选自教材中第一、九单元的课后练习，而这些练习又都出自课文，所以可在课文中查找答案，这里就不再给出答案。
(完整版)通信原理教程习题答案第四版
R B1亍 100Bd10等概时的平均信息速率为R b（2）平均信息量为1H log 24 R B log 2 MR B log 2 4200 b s沁4討罟 誣詈1.977比特符号R , R B H 100 1.977197.7 b s习题1.4试冋上题中的码兀速率是多少?1 1解：R B3 200 BdT B 5*10 3习题1.5设一个信息源由64个不同的符号组成，其中16个符号的出现概率均 为1/32,其余48个符号出现的概率为1/96,若此信息源每秒发出1000个独立的符号, 试求该信息源的平均信息速率。
解：该信息源的熵为第一章习题习题1.1在英文字母中E 出现的概率最大，等于0.105,试求其信息量log 2 P E log 2 0.105 3.25 b习题1.3某信息源由A ，B ，C ，D 四个符号组成，这些符号分别用二进制码组 00, 01，10，11表示。
若每个二进制码元用宽度为 5ms 的脉冲传输，试分别求出在 下列条件下的平均信息速率。
（1）这四个符号等概率出现；（2）这四个符号出现概率如习题1.2所示。
解：（1）一个字母对应两个二进制脉冲，属于四进制符号，故一个字母的持续时 间为2>5mso 传送字母的符号速率为E 的信息量：I E 习题1.2某信息源由现的概率分别为1/4，1/4, 解：A ，B ，C ，D 四个符号组成，设每个符号独立出现，其出3/16, 5/16 试求该信息源中每个符号的信息量。
I AlOg 2-P(A) 3log 2 2.415b16lOg 2 P(A)log 21 2b35lOg21362.415bI Dlog21s1.678bM 6411H (X)P(X i )log 2 P(X i )Pg )log 2 P(X i )16* — log ? 32 48* — log 2 96i 1i 132 96=5.79比特/符号因此，该信息源的平均信息速率& mH 1000*5.795790 b/s 。
通信英语（第四版）课后习题名词解释答案加翻译句子
通信英语（第四版）课后习题名词解释答案加翻译句子1.PCM原理抽样量化与编码:sampling,quantizing and coding话路:speech channel幅值: amplitude value抽样频率: sampling frequency抽样速率: sampling rate脉冲流: stream of pulses重复率: repetition rate编码过程: coding process模拟信号: analog signal传输质量: transmission quality数字通信: digital communication数字传输: digital transmission含噪声的环境: noisy environment传输路由: transmission path信噪比 :signal-to-noise ratio信号电平 :signal levels噪声功率: noise power地面系统: terrestrial system二进制传输: binary transmission反向操作: reverse operation8-位码序列: 8-digit sequence接受端: receiving terminal帧格式 :frame format同步字 :synchronization word实现这三项功能的方案:the schemes for performing these three functions一串幅值: a series of amplitude values电话质量的话路 a speech channel of telephone quality一个8位二进制码的序列: a sequence of 8-binary digits理论上的最小抽样频率:a minimum theoretical sampling frequency占据着300Hz到3.4kHz频率范围的话路: a voice channel occupying the range 300Hz to 3.4kHz 每个样值8-位码: 8-digits per sample value汽车点火系统的打火: the sparking of a car ignition system重复率为64kHz的脉冲流: the stream of the pulses with a repetition rate of 64kHz真实信号与噪声信号的关系: relationship of the true signal to the noise signal由卫星上接受到的信号 :the signal received from a satellite一条特定消息中的全部信息 :the complete informatian about a particular message被传信号的波形 :the shape of the transmitted signal由传输路由引入的衰减: the attenuation introduced by transmission path将抽样的幅值转换成一串脉冲的单元:the unit that converts sampled amplitude value to a set of pulses涉及到第一路，第二路及其他各路的序列: a sequence relating to channel 1,2 and so on被称为同步字的独特码序列: a unique sequence of pulses called synchronization word地面系统 :terrestrial system脉冲的“有”或“无” : the presence or absence of the pulses 高速的电子开关: a high-speed electronic switch时分多路复用器 :the time division multiplexer时分多路复用 :Time Division Multiplexer2.异步串行数据传输串行接口 serial interface显示终端 CRT terminal发送器与接收器 transmitter and receiver数据传输 data transmission数据流 data stream闲置状态 the idle state传号电平 mark level空号电位 space level起始位 start bit停止位 stop bitT秒的持续时间 duration of T seconds奇偶校检位 parity bit错误标志 error flag传输错误 transmission error下降沿 fallinf edge符号间的空格 intersymbol space接收机的定时 receiver timing本地时钟 local clock磁带 magnetic tape控制比特 control bit逻辑1电平 logical 1 level二进制数据 binary data明显的缺点 obvious disadvantage异步串行数据传输 asynchronous serial data transmission最为流行的串行接口 the most popular serial interface所传送的数据 the transmitted data发送器与接收器的时钟the clocks at the transmitter and receiver电传机的时代 the era of teleprinter一个字符的点和划 the dots and dashs of a character符号间空格持续时间的三倍three times the duration ofintersymbol space被称为字符的比特组 the group of bits called characters由7或8个比特的信息组成的固定单元the invariable units comprising 7 or 8 bits of information 由接收机本地产生的时钟 a clock generated locally by the receiver在字符后所收到的奇偶校检位 the received parity bit following the character起始位的下降沿 the falling edge of the start bit数据链路面向字符的特性 the character-oriented nature of the data link3.数据通信地下电缆 underground cable通信卫星 communication satellite微波设备 microwave facilities调制器与解调器 modulator and demodulator缓冲器 buffer定时信号 timing signals同步脉冲 synchronization pulses时隙 time slot移位寄存器 shift register传输媒体 transmission medium线形衰弱 linear attenuation信息安全 information security键盘 keyboard数据终端 data terminals某种类型的数据转换设备some type of data conversion equipment视频显示终端 visual display terminal称为数据调制解调器的双向数据发送接收机two-way data transmistter-receiver called a data modem 全双工的数据传输系统full-duplex data trandmission system由数据处理器的运算速率所决定的速率 the rate determined by the operating speed of the data processor由接口部件来的定时信号timing signals from the interface assembly磁心存储器 magnetic core memories线性衰减和时延特性linear attenuation and delay characteristics传输损伤 transmission impairments语音中的冗余特性 the redundant nature of speech在数据发送器中的编码过程coding process in the data transmitter二进制的不归零信号 binary nonreturn-to-zero signal4.互联网网络资源：network resource信息服务：information services远程终端：remote terminals互联的系统：interconnected systems命令：command电子邮件：electronic mail主机：host无线信道：wireless channels搜索工具：searching tools用户界面：user interface存取：access文本信息：textual messages协议：protocol超文本协议：hypertext protocol分布在全世界的计算机的巨大网络：gaint network of computers located all over the world主干系统：backbone system全国范围的网络：nationwild network电子会议：electronic conferences实时对话：live conversation最大的信息库the largest repository of the computers on the net网络设备资源：network facilities resources在网上的绝大多数计算机：the vast majority of the computer on the netUNIX操作系统：the UNIX operating system在因特网和你的PC机之间传送数据的方法：a way to move data between the internet and your PC 方便的搜索工具：the convenient searching tools联网的超文本协议：the network hypertext protocol5.光纤通信介绍光纤通信:optical fiber communications光源：light source波长：wavelength激光器：laser色散：dispersion传输介质：transmission medium多模光纤：multi-mode fiber长途干线：long-houl trunks单模光纤：singer-mode fiber带宽：bandwidth带宽用户：wideband subscriber纤维光学：fiber-optics商用技术：commercial technologe门限电流：threshod current光检测器：photodetector波分复用：wavelength multiplexing纤维光网络：fiber-optic network视频带宽：video bandwidth长途传输：long distance transmission中继距离：repeater spacing已装光纤的总长度：the total length of installed fiber长途通信系统：long-haul telecommunication system低衰减的石英纤维：the low-loss silica fiber衰减接近瑞利极限的光纤：fibers with losses approaching the Rayleigh limit室温下的门限电流：room temperature threshold currents较长波长区：the longer wavelength region用户接入工程：subscriber access project部件性能和可靠性的改进：improvements in component performance and reliability已安装的光纤系统的数据速率：data rates for installed fibre optic system每秒吉比特：gigabit per second range波分复用：wavelength multiplexing带宽用户环路系统：widebend subscriber loop system多纤连接器：multifibre connectors设计寿命：projected lifetime光源：light source单模光纤：single-mode fibre分布反馈式激光器：distributed-feedback laser信息容量：information capacity交换体系：switching hierarchy带宽业务：broadband services9.蜂窝式移动电话系统蜂窝式移动电话：cellular mobile telephone服务性能：services performance频谱：frequency spectrum频带：frequency band微处理器：microprocessor移动手机：mobile unit广播业务：broadcast servise天线：antenna子系统：subsystems移动用户：mobile subscriber服务能力：service capability利用率：utilization带宽：bandwidth单边带：single-sideband扩频：spread spectrum大规模集成电路：large scale integrated circuits蜂窝点：cellular site蜂窝交换机：cellular switch无线机架：radio cabinet呼叫处理：call processing频谱利用率：frequency spectrum utilization有限的指定频带：the limited assigend ferquency band 服务区：servise area复杂的特性和功能：complicated features and functions大规模集成电路技术：large-scale integraesd circuit technology试验性的蜂窝系统：developmental cellular system中央协调单元：central coordinating element蜂窝管理：cellular administration传统移动电话的运行限制：operational limitiation of conventional mobile telephone system 有限的服务能力：limitied service capability无线通信行业：radio communcation industry可用的无线电频谱：available radio frequency spectrum所分配的频带：the allocated frequency band移动收发信机：mobile transceiver技术上的可行性：techological feasibility严格的频谱限制：severe spectrum limitations调频广播业务：FM broadcasting services传播路径衰耗：propagration path loss多径衰耗：multipath fading电话公司地方局:telephone company zone offices 10.全球移动通信系统个人通信 personal communcation通信标准 communcation standrads固定电话业务 fixed telephone services网络容量 network capability移动交换中心 mobile switching center国际漫游 international roaming宽带业务 broadband services接口转换 interface conversion频谱分配 frequency allocation模拟方式 analogue mode蜂窝通信原理 cellular communcation principe拥塞 jamming蜂窝裂变 cellular splitting基站 base station寄存器 register收费功能 billing function接入方法 access method突发脉冲传输方式 brusty transimission mode开销信息 overhead information切换算法 handover algorithms短消息服务 short message services技术规范 technical specificationtotal access communcation system 全接入的通信系统global mobile communcation system 全球移动通信系统time division multiple access 时分多址facsimile and short message services 传真和短消息服务fixed communcation networks 固定通信网络a more personalized system 更加个性化的系统the cost and quality of the link 链路的价格和质量market growth 市场的发展fixed telephone service 固定电话服务coxial cable 同轴电缆interface convision 接口转换cellular communcation priciple 蜂窝通信原则frequency reuse and cell splitting 频率复用和蜂窝裂变cochannel interference 共信道干扰theoretical spectual capability 理论上的频谱容量micro-cellular system 微蜂窝系统base station transceiver 基站收发信机subscriber register 用户寄存器burst transmission mode 突发脉冲传输模式overhead information 开销信息advanced handover algorithms 先进的切换算法facsimile and short message services 传真和短消息服务the GSM technique specications GSM技术规范说明一1 . 研究二进制的传输可见, 只要简单地去判别脉冲的“有”和“无”, 我们就获得了一条消息的全部信息。
同济大学通信系统原理(英文班)练习题附答案
Chapter 1Homework:Fill-in Questions:log M) bits 1. In a M-ary communication system, each symbol contains ( 2information content.2. The purpose of communication is to ( transfer information ). Efficiency of the digital communication system can be measured by the specifications such as () , () , ( ). And reliability can be measured by ( ) , ().3. The basic factors for measuring the merit of a communication system are (efficiency ), and ( reliability ).4. The main influence of constant parameter channel on signal transmission are usually described by their ( Amplitude-Frequency ) and ( Phase-Frequency ) characteristics.5. Common characteristics of random parameter channels are (transmission attenuation of the signal is varying with time ), (transmission delay of the signal varies with time ), and (signal arrives at the receiver over several paths ).Multiple-Choice Questions:1.For analog and digital communications, which is (are) true? ( A B C D ).A. digital communication typically uses more bandwidth.B. analog communication cares more about fidelityC. digital communication cares more about probability for correct decisionD. digital communication typically uses analog carrier to carry baseband signalsSingle-Choice Questions:1. The symbol rate of 16-ary digital signal is 1200Bd, then the corresponding information rate is ( D ); If with the same information rate, the symbol rate of 8-ary digital signal is ( D ).A. 1600b/s, 1200BdB. 1600b/s, 3200BdC. 4800b/s, 2400BD. 4800b/s, 1600B2. Assume the delay difference of two paths in a random parameter channel is τ, the frequencies of maximum transmission loss are ( B ), the frequencies =1msof minimum transmission loss are ( B ).A. kHz, kHzB. kHz, kHzC. kHz, kHz C. kHz, kHzTrue or FalseWhen there is no input signal, the additive interference doesn’t exist, but the multiplicative interference still exists. ( F )Homework: 2.1, 2.4, 2.5, 2.6, 2.8, 2.11Fill-in Questions:1. AWGN noise is short term for ( Additive White Gaussian Noise ).2. The power spectral density and ( the autocorrelation function ) of a stationary random process are a pair of Fourier transform.3. If bandwidth is 10 MHz, signal-to-noise ratio is 20dB, the Shannon capacity is C=( 66.6 Mbps ).Single-Choice Questions:1. Assume is a generalized stationary random process with zero mathematical expectation, then the average power of is ( D ).A. B. C. D.2. For a narrow band random process 00()()cos ()sin C S X t X t t X t t ωω=-, if ()X t is a Gaussian process, then ( A ).(A) ()C X t and ()S X t are also Gaussian processes; (B) Only ()C X t is also Gaussian process;(C) ()C X t and ()S X t are not Gaussian processes; (D) Only ()S X t is also Gaussian processes.parison of AM, DSB, SSB and FM these four communication system,which has the best efficiency? Which has the best reliability? Which have the same efficiency?SSB has best efficiency. FM has the best reliability. AM and DSB have the same efficiency.2.What is the requirement on the characteristics of the filter for producing theVSB signal?3.What kind of baseband signal is suitable for VSB modulation?The analog baseband signal with D.C. component and low frequency component is suitable for VSB modulation.4.Let the expression of a FM modulated signal beFind: (1) the maximum frequency deviation of the modulated signal.(2) the frequency of the carrier.(3) the bandwidth and the average power of the modulated signal.(4) if the FM circuit constant is kHz/V, then the expression of the baseband signal can be written as:(1) ∵(2) Hz(3) HzW(4) ∵ ∴V∴the baseband signal can be written as:5. Assume the amplitude of a FM modulated signal is 10V , the instantaneous frequency of the FM signal is:Find: (1) the expression of the FM modulated signal.(2) the maximum frequency deviation , the modulation index, and the bandwidth of the modulated signal.(1)(2) , , kHzFill-in Questions:1. Assume the expression of a FM signal is )102sin 5102cos(536t t ⨯+⨯ππ, the carrier frequency of this FM signal is ( )Hz, the maximum frequency deviation is (5000) Hz, the bandwidth of this FM signal is ( ) Hz ，the average power is (12.5 )W.Single-Choice Questions:1.Which analog modulation has the highest spectrum efficiency? ( C )(A) AM (B) DSB(C) SSB (D) VSBCalculation Questions:1. Let a baseband modulating signal be a sinusoidal wave with the frequency 10kHz, and theamplitude 1V . It modulates the phase of a carrier with frequency 10MHz, and the maximum phase deviation of modulation is 10rad. (1) Calculate the approximate bandwidth of the phase modulated signal. (2)If the frequency of the modulating signal is changed to 5kHz, calculate the bandwidth of the phase modulated signal.Answer: (1) We has known that: m 10 kHz, A 1 V m f == , and maximum phase deviation max 10 rad ϕ=.The instantaneous phase of the carrier can be represented as:()()p t k m t ϕ=, where 10p k =.Then, the instantaneous frequency deviation of the carrier can be represented as: d ()sin p m m d t k t dtϕωω=, and maximum frequency deviation is p m k ωω∆=. Hence, the frequency modulation index is:10p m f p m m k m k ωωωω∆====Thus, the approximate bandwidth of the phase modulated signal is2(1)2(110)10220 kHz f m B m f =+=+⨯=(2) If the frequency of the modulating signal is changed to 5kHz, then the approximate bandwidth of the phase modulated signal is:2(1)2(110)5110 kHz f m B m f =+=+⨯=.Chapter4Fill-in Questions:1. The basic process, from( Sampling the analog signal ), ( the quantization ), and ( converting to binary symbols ) , is usually called as pulse code modulation (PCM).Single-Choice Questions:1. The purpose of nonuniform quantization is ( B ).A. Protect big signalB. Protect small signalC. remove quantization noiseD. Increase quantization levels 2. In Delta modulation, the best way to avoid overload quantization noise is ( A ).A. Increase sampling frequencyB. Increase quantization stepC. Reduce sampling frequencyD. Reduce quantization step3. The sampling theorem points out that if the highest frequency of a low-pass analog signal is H f , then the signal can be represented by its samples when the sampling frequency no less than ( B ).(A) H f (B) 2H f (C) 3H f (D) 5H f4. If the samples obey uniform distribution over [,]a a -, then the quantization noise power is determinedby( C )(A) a (B) a and quantization interval ∆ (C) quantization interval ∆ (D) other parameters. Multiple-Choice Questions:1. Digitization of analog signals includes ( B 、D 、E )(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). SamplingShort-Answer Questions:1. In Delta modulation system, why we increase sampling frequency s f rather than quantization step ∆ to avoid overload quantization noise?Answer:Given the sampling frequency and the quantization step ∆, then the slope of a step is:s k f =∆⋅It is the maximum possible slope of a step-shaped wave, or it is called the maximum tracking slope of the decoder. When the slope of the input signal of the delta modulator exceeds this maximum, overload quantization noise will be generated. Therefore, in order to avoid overload quantization noise, it is necessary to make the product of ∆ and s f large enough, so that the slope of the signal can not exceed this product. One the other hand, the value of ∆ is directly related to the magnitude of the basic quantization noise. If the value of ∆ is large, then the basic quantization noise must also be large. Therefore, only the method of increasing s f to increase the product s k f =∆⋅ can ensure that the basic quantization noise and the overload quantization noise do not exceed the limit.Fill-in Questions:1.( 2 )Baud/Hz is the highest possible unit bandwidth rate, and is also called as the Nyquistrate.2.The time-domain equalizer is used to overcome (intersymbol interference (ISI)).3.The basic process, from ( Sampling the analog signal), ( the quantization ), and( converting to binary symbols ) , is usually called as pulse code modulation (PCM). Single-Choice Questions:1. If the samples obey uniform distribution over [,]a a-, then the quantization noise power is determined by ( C)(A) a(B) a and quantization interval ∆(C) quantization interval ∆(D) other parameters.Multiple-Choice Questions:1. Digitization of analog signals includes ( B, E )(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). Sampling2. Which are symbol code type for baseband digital signals? (A, B, C, D, E )(A). HDB3 (B). AMI (C). CMI (D). Biphasic code (E). 5B6B (F). BPSK(G).MSKShort-Answer Questions:1.Which are the design principles of code for digital baseband signal?As we know, the performance of a practical baseband transmission system can be found in its eye pattern (refer to the figure below). Please find out the characteristics in the figure 1, and point out their impacts on transmission performance.Answer:(1) The location of the central perpendicular line is the optimum sampling instant.(2) The middle horizontal line represents the optimum decision threshold level.(3) The perpendicular height of the shadow region represents the distortion range of the received signal.(4) The slope of the bevel edge of the “eye” represents sensitivity of the sampling instant to the timing error.(5) Under no noise situation, the degree of the opening of the “eye is the noise tolerance; if noise at the sampling instant exceeds this tolerance, then error decision may happen.Calculation Questions:1. Assume there are 4 overall transfer characteristics of baseband transmission systems shown in Figure2. If symbol rate is 2000Bd, please illustrate whether they can transmit information without ISI? Which system has better transmission characteristic?Solution: ∵symbol rate is 2000Bd,∴ISI exists in system (a) and (c), ISI does not exist in system (b) and (d).Since system (b) can be realized physically, and itssystem (d) can ’t be realized physically, and itsso system (b) has better transmission characteristic.1000(Hz)(Hz)(Hz) 750(Hz) 2000Chapter61. Assume there is a space-ground communication system, the symbols rate is 0.5MB, the bandwidth of receiver is 1MHz. Antenna gains of ground station and space station are respectively 40dB and 6dB ，the path loss is dB, where is the distance (km). Given the transmitting power is 10W, the double-side power spectral density of white noise is W/Hz. Requiring the symbol error probability of the system is , try to find the maximum communication distance under these conditions as below.(1) Adopting 2FSK modulation and coherent demodulation. (2) Adopting 2DPSK modulation and coherent demodulation. 解：（1）采用2FSK 方式传输，进行相干解调，其误码率为 510221-==rerfcP e 可计算出所需要的信噪比为r=18.3，噪声功率为 ()W B n n 661202104101104--⨯=⨯⨯⨯==σ 接收端信号幅度平方为()W r a n 462210464.11043.1822--⨯=⨯⨯⨯==σ 传输中信号功率衰减为()dB a A 35.5110464.1102lg 1022lg 10422=⨯⨯=-根据给定条件，51.35+40+6=60+10lgd ，可以算出传输距离为 ()km d 543210735.3==(2) 采用2DPSK 方式传输，选择差分相干方式进行解调，其误码率5e 1e 102r P --==，可计算出所需要的信噪比r =10.82，接收端信号幅度平方为2265n 2210.824108.65610 W a r σ--==⨯⨯⨯=⨯传输中信号功率衰减为 225221010lg 10lg 53.63 dB 28.65610A a -⨯==⨯根据给定条件，53.63+40+6=60+10lgd ，可以算出传输距离3.963109183 km d ==。
(完整版)通信原理课后答案1
习题解答（一）1-4 一个由字母A ，B ，C ，D 组成的字。
对于传输的每一个字母用二进制脉冲编码，00代替A ，01代替B ，10代替C ，11代替D ，每个脉冲宽度为5ms 。
（1）不同的字母是等可能出现时，试计算传输的平均信息速率；（2）若每个字母出现的可能性分别为P A =1/5，P B =1/4，P C =1/4，P D =3/10试计算传输的平均信息速率。
)/(5.1981010985.1)/(985.1103log 10341log 4141log 4151log 51)]()[log ()(2)/(20010102/10/52)/(24log log )(1log )1(32222123222s bit t I R bit x P x P x H s bit t I R ms ms t bit M x P I b ni i i b =⨯≈∆=≈----=-==⨯=∆==⨯=∆====-=-∑则：符号）（则：符号）（符号）（间为：传输每个符号占用的时符号解：1-5 国际莫尔斯电码用点和划的序列发送英文字母，划用持续3单位的电流脉冲表示，点用持续1单位的电流脉冲表示；且划出现的概率是点出现概率的1/3：（1）计算点和划的信息量；（2）计算点和划的平均信息量。
符号））（故。
划出现的概率为，，所以点出现的概率为出现概率的因为划出现的概率是点解：/(81.0241415.0432)(241log log )(415.043log log 4/14/33/1)1(22112222212121bit I P I P H bit P I bit P I P P =⨯+⨯=+==-=-=≈-=-===1-6设一信息源的输出由128个不同的字符组成。
其中16个出现的概率为1/32，其余112个出现的概率为1/224。
信息源每秒发出1000个符号，且每个符号彼此独立。
试计算该信息源的平均信息速率。
1-9 如果二进制独立等概信号，码元宽度为0.5ms ，求R B 和R b ；有四进制信号，码元宽度为0.5ms ，求传码率R B和独立等概时的传信率R b 。
通信原理课后练习答案
E [ c o s (c t1 )c o s (c t2 ) ]
1
2 E { c o s [c(t1 t2 ) 2] c o s [c(t2 t1 )]}
1 2E {co s[c(t1t2)2 ]}1 2co sc
1-10 已知某四进制数字传输系统的传信率为2400b/s, 接收端在 0.5h 内共收到 216 个错误码元,试计算该系统 的误码率 Pe，
解：系统的码元速率为
Rb2400bit/s
RBlogR2bM1200Baud
0.5h 内共收到的码元数为：
1 2 0 0 0 .5 3 6 0 0 2 1 6 1 0 4
sS S B (第t) 51 2 章m 课(t)c 后o s 作ct业1 2m 解ˆ(t答)sin ct
5-3 已知调制信号 m ( t) c o s ( 2 0 0 0 t) c o s ( 4 0 0 0 t) ，载波 为
c(t)cos104t ，进行单边带调制，试确定该单边带信号 的表示式，并画出频谱图。 解：单边带信号的表示式为
输入噪声的第功率3谱章密课度：后Pn作(f)业n 20 解, 答f
3-8 一个中心频率为fc、带宽为B的理想带通滤波器如 下图所示，假设是 均值为零、功率谱密度 为n0/2 的高斯白噪声,试求：
H( f )
B
B
fc
0
f fc
⑴ 滤波器输出噪声的自相关函数；
⑵ 滤波器输出噪声的平均功率；
⑶ 输出噪声的一维概率密度函数，
第1章课后作业解答
1-4 一个由字母 A、B、C、D 组成的字,对于传输的每 一个字母用 二进制脉冲 编码,00 代替A, 01 代替B,10 代 替 C,11 代替 D ,每个脉冲宽度为 5ms ，
通信工程专业英语习题答案
通信工程专业英语习题答案练习参考答案第1单元信号与系统1. 完形填空(1) band-limited, cutoff frequencies, to, between, the information rate is proportional to the bandwidth of the channel, on, to, by, by simply increasing the number of levels, noise will cause the receiver to mistake one level for another, no matter how elaborately the data is coded, between, theoretical maxima.(2) from, a loss of information , the twisted pair, addition of noise to the signal, depends on the signal frequency, the high-frequency components of the signal, over, between, through, of, with, including.(3) for, to, In some cases, over, (microwave relay, coaxial cable, or fiber), ADC, to, thereby achieving a reduction in required channel bandwidth, of, signal recovery, (Techniques such as filtering, auto-correlation and convolution).2. 英汉互译(1) “电信”（telecommunication）一词来源于希腊语tele（含义为“遥远的”）和拉丁语communicatio（涵义为“连接”）。
通信原理练习题含答案.pdf
通信原理练习题含答案.pdf通信原理练习题⼀、填空题1、数字通信在信道中传输的是（数字信号）,模拟通信在信道中传输的是（模拟信号）2、数字通信系统的可靠性指标为（误码率或误信率）,模拟通信系统的可靠性指标为（信噪⽐）3、数字通信系统的有效性指标为（传输速率）和（频带利⽤率）,模拟通信系统的有效性指标为（有效传输带宽）4、模拟信号的最⾼频率分量为1000Hz ，对其取样后若信号频谱不混叠，则抽样频率为（ 2000Hz ）5、窄带⾼斯平稳随机过程的包络服从（瑞利）分布，相位服从（均匀）分布。
6、信道复⽤是指（在同⼀信道上传输多路信号⽽互不⼲扰的⼀种技术）7、PCM30/32的⼀次群速率为（ 2.048M bit/s ）8、PCM30/32的每⼀话路的速率为（ 64k bit/s ）9、PCM30／32系统第2l 路信令码的传输位置为（第F6帧TS16时隙的后4位）10、已知信源熵()=X H 3bit/符号，信源每秒钟传送100个符号，则信源的信息传输速率（ ()1003300bit/s b B R R H x =?=×= ）11、某数字通信系统1秒钟内传输了100个码元，其中错误l 个码元，则误码率Pe 为（ 2110100= ） 12、基带传输系统的码元频带利⽤率最⾼为（ 2 Baud/Hz ）。
13、在相同符号速率下，数字调制⽅式中频带利⽤率最低的是（ 2FSK ）。
14、调制器的作⽤是（频率的变换、频谱的搬移）。
信道编码的作⽤是（提⾼传输的可靠性）。
信源编码的作⽤是（提⾼传输的有效性）。
PCM 的功能是（模数转换）。
15、可能存在定时信号提取困难的传输码型是（ AMI 码），解决⽅案是采⽤（ HDB3编码）。
16、均匀量化是指（将输⼊信号的取值域按等距离分割的量化），缺点是（⼩信号的量化信噪⽐太⼩）。
17、⾮均匀量化的特点是（⼩信号的量化间隔⼩，量化误差也⼩；⽽⼤信号的量化间隔⼤，量化误差也⼤）。
通信与信息工程专业英语教程参考答案(Communication English)概要
参考答案( Communication English )第一章电子通信导论1.将表1-1译成中文。
表1-1 通信大事纪年代事件公元前3000年埃及人发明象形文字公元800年借鉴印度, 阿拉伯人使用我们的现行数制1440年约翰.戈登贝尔发明可移动的金属记录带1752年贝莱明.富兰克林的风筝证明了雷闪是电1827年欧姆发表欧姆定律( I = E / R)1834年 C.F.高斯和E.H.韦伯发明电磁电报1838年W.F.库克和C.维特斯通发明电报1844年S.F.B.莫尔斯演示巴尔的摩和华盛顿的电报线路1850年G.R.基尔赫夫发表基尔赫夫电路定律1858年铺设第一条越洋电缆, 并于26天后举办博览会1864年J.C.麦克斯威尔预言电磁辐射1871年电报工程师协会在伦敦成立1876年 A.G.贝尔发明电话并获专利1883年 A.爱迪生发现真空管中的电子流1884年美国电气工程师协会(AIEE)成立1887年H.赫兹证明麦克斯威的理论1900年G.马可尼传送第一个越洋无线电信号1905年R.芬森登利用无线电传送语音和音乐1906年L.弗雷斯特真空三极管放大器1915年贝尔系统完成美国大陆电话线路1918年 E.H.阿尔莫斯通发明超外差接收机电路1920年第一个定时无线广播J.R.卡尔松将取样用于广播1926年美国演示电视1927年H.布兰克在贝尔实验室发明负反馈放大器1931年电传打字机投入运营1933年 E.H.阿尔莫斯通发明调频1935年R.A.沃特森-瓦特发明第一个实用雷达1936年英国广播公司(BBC)开办第一个电视广播1937年 A.雷弗斯提出脉冲编码调制(PCM)1941年J.V.阿当拉索夫在依俄华州立大学发明计算机1945年ENIAC电子数字计算机研发于宾夕伐尼亚大学1947年布雷登、巴登和肖克利在贝尔实验室研制晶体管S.O.莱斯在贝尔实验室研究噪声的统计表征1948年 C.E.香农发表他的信息理论1950年时分多路用于电话1953年美国提出NTSC彩色电视1957年苏联发射第一个地球卫星Sputnik I1958年 A.L.肖洛和C.H.托莱斯发表激光原理仙童公司的R.诺依斯生第一硅集成电路1961年美国开始立体声调频广播1962年第一个有源卫星, Telstar I , 实现美国与欧洲的电视中继1963年贝尔系统推出按键式电话电气与电子工程师协会(IEEE)成立1963~66年研究纠错码和自适应均衡1964年电子电话交换系统(No.1ESS)投入运营1965年笫一个商用通信卫星,Early Bird,发射1968年开发电缆电视系统1971年Intel公司研制第一个单片微处理器—40041972年摩托罗拉向美国联邦通信委员会(FCC)演示蜂窝式电话1976年推出个人计算机1979年64-kb随机存取存储器标志着进入VLSI时代1980年贝尔系统研发FT3光纤通信, Philips和Sony研发光碟(CD) 1984年苹果公司研发Macintosh计算机1985年传真机普及1989年摩托罗拉推出袖珍移动电话1990~现在用微处理器进行数字信号处理的时代, 数字示波器, 数字调谐接收机, 扩频系统, ISDN, HDTV, 数字卫星系统2.将表1-2译成中文。
通信原理试题及答案(完整版).doc
所以，信息量为解：若系统传送二进制码元的速率为1200Baud,则系统的信息速率为: ^, = 1200x10^2 = 1200 bM若系统传送十六进制码元的速率为2400Baud,则系统的信息速率为： 傀=2400x10^16 = 9600bit/s该恒参信道的传输函数为//((,)) = ,4/八叫"08心=A0Z •严“几=/(I + "sin )e " = 41 + — 27-4^■祁j "树jj ■心 _♦币)_， 2 24h 4b^(/) = ^(/-//) + —<5(/-/^ + ?；)——<?(/_/“ _石)冲激响应为2 2IT 1-3解：每个消息的平均信息量为…、1, 1 , r 1 1 f 1// (x) = - - log 2 - - 2 x - log 2 - - - log 2 -=1. 75bit/符号解：(1)每个字母的持续时间为2X10ms,所以字母传输速率为R R . = ------- ----- = 50Bciud842x10x10」不同字母等可能出现时，每个字母的平均信息量为H(x) = log2 4 = 2 川/符号平均信息速率为砖 7?4(兀)=100 bit/s(2)每个字母的平均信息量为H(X)=—— lo a — lo<> ]oa ---------------- -- |oa W5 82 5 4 82 4 4 “4 10 82 10 =1.985 bit/符号 所以平均信息速率为 R h = R B SH(x) = 99.25 (bit/s)1 1 3 3H(x) = 一一log. — 一-log. -« 0.811解：(1) 464464 bit/符号 m 个0和100-m 个1)出现的概率为⑴1-5 (2)某一特定序列(例如:in z o x 100-w /(八禺,…，俎logP (陷)=-log ⑴<4>24>=200-(100-m)log 2 3 (bit) (3)序列的燜H (屮)=1 OOH (X) = 8 IM / 序列 1-61-8解：4b Ah=+ 兀)一-H/-/“ -石)log 10 —= 30J51-10 解：⑴ 因为 S/N =30dB,即 10 N ,得：S/N=1000由香农公式得信道容量C = 51og ?(l+^)・ N= 3400xlog 2 (1 + 1000)-33.89X 103W Z /5(2)因为最大信息传输速率为4800b/s,即信道容量为4800b/so 由香农公式 C = B10g2(l + * 2 = 2万-1 = 2硕-1 = 2.66-1 = 1.66 得：N o 则所需最小信噪比为1.66。
通信专业英语考试习题
通信专业英语考试习题通信专业英语考试习题是许多学习通信的学生常常会遇到的挑战。
这些考试题旨在测试学生对通信学科的理解和运用能力。
下面将介绍一些常见的通信专业英语考试习题，希望能对大家备考有所帮助。
一、选择题1. 下面哪个选项是正确的数据通信的定义？A. 使用数字信号而非模拟信号进行信息传输。
B. 使用光纤而非电缆进行信息传输。
C. 使用卫星通信而非微波通信进行信息传输。
正确答案：A2. 通过无线电波进行通信的主要问题是什么？A. 信号的传输距离有限。
B. 频谱资源有限。
C. 安全性问题。
正确答案：B3. 下面哪个选项是正确的带宽定义？A. 数据的传输速率。
B. 可用频谱范围。
C. 数据的传输距离。
正确答案：B4. 频率调制（FM）是通过改变哪个参数来传输信息？A. 频率B. 幅度C. 相位正确答案：A5. 下面哪个选项是正确的对称数字用户线路（DSL）的定义？A. 一种提供高速互联网接入的传输技术。
B. 一种用于长距离电话通信的传输技术。
C. 一种用于数字音视频传输的传输技术。
正确答案：A二、填空题1. OSI模型中，物理层主要负责______________。
答案：将数据转换为适合在通信信道上传输的形式。
2. ___________________是一种广泛用于局域网连接的传输媒介。
答案：双绞线（Twisted Pair）。
三、简答题1. 请简要解释一下分组交换和电路交换的区别和优缺点。
答案：分组交换和电路交换是两种不同的数据传输方式。
电路交换在通信双方建立连接后，会一直占用通信资源直到通信结束；而分组交换将数据划分为小的数据包（分组）进行传输，传输过程中不需要提前建立连接。
电路交换的优点是传输稳定，可以保证实时性，适用于实时语音和视频通信；缺点是资源占用高且不灵活。
分组交换的优点是资源利用率高，适合不同时刻多个用户共享资源；缺点是传输延迟较大，对实时性要求较高的应用不太适用。
2. 请简要解释一下TCP/IP协议族中的IP协议和TCP协议的作用。
通信原理第十章习题解答
解： PSK 最佳接收机误码率为：()61001100.41021102121-⨯=≈=⎪⎪⎭⎫⎝⎛=eerfc n E erfc P b e π而对于实际接收机来说接受信噪比()3561001====n E Bn T E N s r b b i因此实际误码率为()22104.3352121-⨯=⎪⎪⎭⎫⎝⎛==erfc r erfcP e 两误码率之比为8500104104.36212=⨯⨯=--e e P P解：二进制双极型信号是确知信号，且信号“0”和“1”电波形的相关系数1-=ρ，故经过最佳基带传输系统后，其误码率为：()⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛-=00212121n E erfc n E erfc P b b e ρ其中()HZW n T dtt sE s T bs4002102,-⨯===⎰因此，系统最高传输速率为：()[]sbit P erfcn T R e ss 555102912114210=⨯⨯===--解：（1）匹配滤波器形式的最佳接收机结构如图10-22（a ）所示。
（2）取最大信噪比时刻T t =0，此时匹配滤波器的单位冲激响应()()()()t T s t h t T s t h -=-=2211其波形分别如图10-24（b ）、（c ）所示。
由于()t s i 可能为()t s 1或()t s 2，所以共有四种可能的输出()()()t h t s t y i 11*=()T t A -2023Tt T ≤≤()220T t A - T t T ≤≤2⎪⎭⎫ ⎝⎛-t T A 232023T t T << ()()=-*t T s t s 12()()=-*t T s t s 11()t TA-220Tt T 223<<=()()()t h t s t y i 22*=以上四种波形分别如图10-24（d ）、（e ）、（f ）、（g ）所示。
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Chapter 1Homework:Fill-in Questions:log M) bits 1. In a M-ary communication system, each symbol contains ( 2information content.2. The purpose of communication is to ( transfer information ). Efficiency of the digital communication system can be measured by the specifications such as (R B) , (R b) , ( η). And reliability can be measured by ( P e) , (P b).3. The basic factors for measuring the merit of a communication system are (efficiency ), and ( reliability ).4. The main influence of constant parameter channel on signal transmission are usually described by their ( Amplitude-Frequency ) and ( Phase-Frequency ) characteristics.5. Common characteristics of random parameter channels are (transmission attenuation of the signal is varying with time ), (transmission delay of the signal varies with time ), and (signal arrives at the receiver over several paths ).Multiple-Choice Questions:1.For analog and digital communications, which is (are) true? ( A B C D ).A. digital communication typically uses more bandwidth.B. analog communication cares more about fidelityC. digital communication cares more about probability for correct decisionD. digital communication typically uses analog carrier to carry baseband signalsSingle-Choice Questions:1. The symbol rate of 16-ary digital signal is 1200Bd, then the corresponding information rate is ( D ); If with the same information rate, the symbol rate of 8-ary digital signal is ( D ).A. 1600b/s, 1200BdB. 1600b/s, 3200BdC. 4800b/s, 2400BD. 4800b/s, 1600B2. Assume the delay difference of two paths in a random parameter channel is τ, the frequencies of maximum transmission loss are ( B ), the frequencies =1msof minimum transmission loss are ( B ).A. n kHz, (n+0.5)kHzB. (n+0.5)kHz, n kHzC. n kHz, (n+0.2)kHz C. (n+0.2)kHz, n kHzTrue or FalseWhen there is no input signal, the additive interference doesn’t exist, but the multiplicative interference still exists. ( F )Homework: 2.1, 2.4, 2.5, 2.6, 2.8, 2.11Fill-in Questions:1. AWGN noise is short term for ( Additive White Gaussian Noise ).2. The power spectral density and ( the autocorrelation function ) of a stationary random process are a pair of Fourier transform.3. If bandwidth is 10 MHz, signal-to-noise ratio is 20dB, the Shannon capacity is C=( 66.6 Mbps ).Single-Choice Questions:1. Assume X(t) is a generalized stationary random process with zero mathematical expectation, then the average power of X(t) is ( D ).A. E[X (t )]B. E 2[X (t )]C. R(∞)D. D[X (t )]2. For a narrow band random process 00()()cos ()sin C S X t X t t X t t ωω=-, if ()X t is a Gaussian process, then ( A ).(A) ()C X t and ()S X t are also Gaussian processes; (B) Only ()C X t is also Gaussian process;(C) ()C X t and ()S X t are not Gaussian processes; (D) Only ()S X t is also Gaussian processes.parison of AM, DSB, SSB and FM these four communication system,which has the best efficiency? Which has the best reliability? Which have the same efficiency?SSB has best efficiency. FM has the best reliability. AM and DSB have the same efficiency.2.What is the requirement on the characteristics of the filter for producing theVSB signal?[H(f+f0)+H(f−f0)]=C3.What kind of baseband signal is suitable for VSB modulation?The analog baseband signal with D.C. component and low frequency component is suitable for VSB modulation.4.Let the expression of a FM modulated signal be10cos(2π×106t+5sin103πt)Find: (1) the maximum frequency deviation of the modulated signal.(2) the frequency of the carrier.(3) the bandwidth and the average power of the modulated signal.(4) if the FM circuit constant is k f=5kHz/V, then the expression of the baseband signal can be written as:=5,∴∆f=m f∙f m=5×500=2500Hz(1) ∵m f=∆ff m(2) f0=106Hz(3) B=2(m f+1)f m=2×(5+1)×500=6000HzP =A 22=1022=50 W(4) ∵∆f =k f ∙A m ∴A m =∆fk f =25005000=0.5 V∴the baseband signal can be written as:m (t )=0.5cos 103πt5. Assume the amplitude of a FM modulated signal is 10V , the instantaneous frequency of the FM signal is:f (t )=106+104cos (2π×103t ) (Hz)Find: (1) the expression of the FM modulated signal.(2) the maximum frequency deviation , the modulation index, and the bandwidth of the modulated signal.(1) s (t )=10cos [2π×106t +10sin (2π×103t )](2) ∆f =104, m f =10, B =2(m f +1)f m =2×(10+1)×1000=22 kHzFill-in Questions:1. Assume the expression of a FM signal is )102sin 5102cos(536t t ⨯+⨯ππ, the carrier frequency of this FM signal is (106)Hz, the maximum frequency deviation is (5000) Hz, the bandwidth of this FM signal is (12×103 ) Hz ，the average power is (12.5 )W.Single-Choice Questions:1.Which analog modulation has the highest spectrum efficiency? ( C )(A) AM (B) DSB(C) SSB (D) VSBCalculation Questions:1. Let a baseband modulating signal be a sinusoidal wave with the frequency 10kHz, and theamplitude 1V . It modulates the phase of a carrier with frequency 10MHz, and the maximum phase deviation of modulation is 10rad. (1) Calculate the approximate bandwidth of the phase modulated signal. (2)If the frequency of the modulating signal is changed to 5kHz, calculate the bandwidth of the phase modulated signal.Answer: (1) We has known that: m 10 kHz, A 1 V m f == , and maximum phase deviation max 10 rad ϕ=.The instantaneous phase of the carrier can be represented as:()()p t k m t ϕ=, where 10p k =.Then, the instantaneous frequency deviation of the carrier can be represented as: d ()sin p m m d t k t dtϕωω=, and maximum frequency deviation is p m k ωω∆=. Hence, the frequency modulation index is:10p m f p m m k m k ωωωω∆====Thus, the approximate bandwidth of the phase modulated signal is2(1)2(110)10220 kHz f m B m f =+=+⨯=(2) If the frequency of the modulating signal is changed to 5kHz, then the approximate bandwidth of the phase modulated signal is:2(1)2(110)5110 kHz f m B m f =+=+⨯=.Chapter4Fill-in Questions:1. The basic process, from( Sampling the analog signal ), ( the quantization ), and ( converting to binary symbols ) , is usually called as pulse code modulation (PCM).Single-Choice Questions:1. The purpose of nonuniform quantization is ( B ).A. Protect big signalB. Protect small signalC. remove quantization noiseD. Increase quantization levels 2. In Delta modulation, the best way to avoid overload quantization noise is ( A ).A. Increase sampling frequencyB. Increase quantization stepC. Reduce sampling frequencyD. Reduce quantization step3. The sampling theorem points out that if the highest frequency of a low-pass analog signal is H f , then the signal can be represented by its samples when the sampling frequency no less than ( B ).(A) H f (B) 2H f (C) 3H f (D) 5H f4. If the samples obey uniform distribution over [,]a a -, then the quantization noise power is determinedby( C )(A) a (B) a and quantization interval ∆ (C) quantization interval ∆ (D) other parameters. Multiple-Choice Questions:1. Digitization of analog signals includes ( B 、D 、E )(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). SamplingShort-Answer Questions:1. In Delta modulation system, why we increase sampling frequency s f rather than quantization step ∆ to avoid overload quantization noise?Answer:Given the sampling frequency and the quantization step ∆, then the slope of a step is:s k f =∆⋅It is the maximum possible slope of a step-shaped wave, or it is called the maximum tracking slope of the decoder. When the slope of the input signal of the delta modulator exceeds this maximum, overload quantization noise will be generated. Therefore, in order to avoid overload quantization noise, it is necessary to make the product of ∆ and s f large enough, so that the slope of the signal can not exceed this product. One the other hand, the value of ∆ is directly related to the magnitude of the basic quantization noise. If the value of ∆ is large, then the basic quantization noise must also be large. Therefore, only the method of increasing s f to increase the product s k f =∆⋅ can ensure that the basic quantization noise and the overload quantization noise do not exceed the limit.Fill-in Questions:1.( 2 )Baud/Hz is the highest possible unit bandwidth rate, and is also called as the Nyquistrate.2.The time-domain equalizer is used to overcome (intersymbol interference (ISI)).3.The basic process, from ( Sampling the analog signal), ( the quantization ), and( converting to binary symbols ) , is usually called as pulse code modulation (PCM). Single-Choice Questions:1. If the samples obey uniform distribution over [,]a a-, then the quantization noise power is determined by ( C)(A) a(B) a and quantization interval ∆(C) quantization interval ∆(D) other parameters.Multiple-Choice Questions:1. Digitization of analog signals includes ( B, E )(A). Modulation (B). Quantization (C). Multiplexing (D). Source coding (E). Sampling2. Which are symbol code type for baseband digital signals? (A, B, C, D, E )(A). HDB3 (B). AMI (C). CMI (D). Biphasic code (E). 5B6B (F). BPSK(G).MSKShort-Answer Questions:1.Which are the design principles of code for digital baseband signal?As we know, the performance of a practical baseband transmission system can be found in its eye pattern (refer to the figure below). Please find out the characteristics in the figure 1, and point out their impacts on transmission performance.Answer:(1) The location of the central perpendicular line is the optimum sampling instant.(2) The middle horizontal line represents the optimum decision threshold level.(3) The perpendicular height of the shadow region represents the distortion range of the received signal.(4) The slope of the bevel edge of the “eye” represents sensitivity of the sampling instant to the timing error.(5) Under no noise situation, the degree of the opening of the “eye is the noise tolerance; if noise at the sampling instant exceeds this tolerance, then error decision may happen.Calculation Questions:1. Assume there are 4 overall transfer characteristics of baseband transmission systems shown in Figure2. If symbol rate is 2000Bd, please illustrate whether they can transmit information without ISI? Which system has better transmission characteristic?Solution: ∵symbol rate is 2000Bd,∴ISI exists in system (a) and (c), ISI does not exist in system (b) and (d).Since system (b) can be realized physically, and its η=1.33 Bd/Hzsystem (d) can ’t be realized physically, and its η=1 Bd/Hzso system (b) has better transmission characteristic.1000 -1000(Hz)(Hz) 1500 -1500 500 -500 (Hz) 750 -750(Hz) 2000 -2000Chapter61. Assume there is a space-ground communication system, the symbols rate is 0.5MB, the bandwidth of receiver is 1MHz. Antenna gains of ground station and space station are respectively 40dB and 6dB ，the path loss is (60+10lg d )dB, where d is the distance (km). Given the transmitting power is 10W, the double-side power spectral density of white noise is 2×10−12W/Hz. Requiring the symbol error probability of the system is P e =10−5, try to find the maximum communication distance under these conditions as below.(1) Adopting 2FSK modulation and coherent demodulation. (2) Adopting 2DPSK modulation and coherent demodulation. 解：（1）采用2FSK 方式传输，进行相干解调，其误码率为 510221-==rerfcP e 可计算出所需要的信噪比为r=18.3，噪声功率为 ()W B n n 661202104101104--⨯=⨯⨯⨯==σ 接收端信号幅度平方为()W r a n 462210464.11043.1822--⨯=⨯⨯⨯==σ 传输中信号功率衰减为()dB a A 35.5110464.1102lg 1022lg 10422=⨯⨯=-根据给定条件，51.35+40+6=60+10lgd ，可以算出传输距离为 ()km d 543210735.3==(2) 采用2DPSK 方式传输，选择差分相干方式进行解调，其误码率5e 1e 102r P --==，可计算出所需要的信噪比r =10.82，接收端信号幅度平方为2265n 2210.824108.65610 W a r σ--==⨯⨯⨯=⨯传输中信号功率衰减为 225221010lg 10lg 53.63 dB 28.65610A a -⨯==⨯根据给定条件，53.63+40+6=60+10lgd ，可以算出传输距离3.963109183 km d ==。