琼州学院 09通信工程专业《专业英语》试卷 2011 ——2012学年度第 1 学期期末考试（C ）卷答案 注意事项： 1、考前请将密封线内填写清楚 2、所有答案请直接答在试卷上（或答题纸上） 3、考试形式：（闭）卷 4、本试卷共 大题，满分100分。考试时间120分钟 5 一、科技常用词组翻译（本题共20 小题，每小题 1.5 分，共 30 分） 1、英翻中 (1)coaxial cable 同轴电缆 (2)router 路由器 (3)hub 集线器 (4)destination address 目的地址 (5)datagram 数据报 (6)amplify 放大 (7)medium 媒介 (8)information 信息 (9)system file 系统文件 (10)library file 库文件 2、中翻英 (1)微处理器 microprocessor (2)芯片 chip (3)电子商务 E-commerce (4)适配器 adapter (5)调制解调器 modem (6)中国电信 China Telecom (7)数字业务 digital service (8)虚拟现实 virtual reality (9)分组交换网 packet-switching network (10)设备控制器device controller 二、英、中词组正确搭配（每组 1 分，共 10 分） 班
级
姓名
学
号
密
封
装
订
线
1.Bearer channel 数据信道5.
2.Terminal Adapter 处理信号4.
3.analog system 用户设备6.
4.handle signaling 载波信道1.
5.data channel 模拟系统3.
6.customer equipment 终端适配器2.
7.service type 端到端数字连接10.
8.Integrated Services Digital Network 数字交换系统 9.
9.digital switching system 综合业务数字网8.
10.end-to-end digital connectivity 业务类型7.
三、把下面句子翻译成中文（本题共 5 小
题，每题 4 分，共 20 分）
1、Keep in mind that size of the software you plan to run
will dicate how much memory your computer should have.
你要运行的软件大小将决定计算机需要多少内存,这一点需要记住.
2.The system unit is a box-like case made of metal or
plastic.
系统单元是金属或塑料制成的象盒般的箱子.
https://www.doczj.com/doc/3418216135.html,puters are good for storing and manipulating both
structured and unstructured data.
计算机有助于存储和处理结构化数据和非结构化数据.
4．Data is a collection of un_organized facts.
数据是无组织的事实的集合.
5.A network protocol is a set of rule for communicating
between computers.
网络协议是计算机之间进行通信的一套规则.
四、选择填空（本题共 4 小题，每题 5 分，
共 20 分）
1.Which of the following is an “output device?”( )
A.CD ROM
B.Microphone
C.Printer
D.All the above
2.A computer calculate using the ( ) system.
A.fast
B.trianary
C.hexadecimal
D.binary
3.Internet is simply what? ( )
A.The WWW
B.A worldwide system of computer networks
C.A modem
D. An ISP or Internet Service Provider
4.The software used to control the computer and run
programs.( )
A.application
B.systew software
C.games
D.virus
5.A special sequence of number or letters that limits access
to eletronic mail boxes:( )
A.Authentication
B.code
C.Encryption
D.Password
五、阅读填空（本题共 5 空，每空 4 分，
共 20 分）
Upgrade a computer
You can Upgrade a computer to enhance it’s 1 ..
For many upgrades,you will need the assistance of an
experienced computer 2 erson.Upgrading usually
refer to replacing an old or obsolete computer with a newer
one to improve the efficiency of the computer.Upgrading can
also include adding a new 3 ,such as a tape drive
or DVD-ROM drive,to increse the capabilities of the computer.
You should always determine the cost of an upgrade before
performing it.If you are planning a major upgrade,such as 4 the system board or CPU ,it may be less expensive to purchase a new computer.
Increasing the 5 of memory in a computer is one of the most effetive upgrade you can perform.Doubling the existing memory in a computer can significantly increase its performance.
1.A.cost B.height C.performance D.experience
2.A.repair B.check C.administer D.keep
3.A.member B.tape C.drive https://www.doczj.com/doc/3418216135.html,ponent
4.A.replace B.replacing C.to replace D.replaced
5.A.amount B.cost C.speed D.quality
湖北科技职业学院10级汽车英语期末考试———————————————————————————————————— 一、单词互译(每空1 分，共20 分) Section A Directions: This section is to test how many words you know.Please translate english into chinese.(1*10) 1.Vehicle body 2.brake system 3.steering system 4.dash board 5.camshaft 6.Automatic Transmission 7.inlet valve 8.timing belt 9.cylinder 10.electronic controller Section B Directions: This section is to test how many words you know.Please translate chinese into english .(1*10) 1. 底盘2后悬架 3．转向盘 4 发动机 5 分电器6油底壳 7 活塞8 轴承 9手动变速器10冷却系统 二、阅读与理解（20题，每题2分，共40分） Part 1 The engine is the source of power that makes the wheels go around and the car move. The automobile engine is an internal combustion engine because the fuel (gasoline) is burned inside it. The burning of gasoline inside the engine produces high pressure in the engine combustion chamber .This high pressure forces piston to move, the movement is carried by connecting rods to the engine crankshaft. The crankshaft is thus made to rotate: the rotary motion is carried through the power train to the car wheels so that they rotate and the car moves. What do we see from the first paragraphs? ( ) The engine is source of power for a car. The engine can not make wheels go around. The engine is resource of oil. Which one Statement is true according the above first paragraph? ( ) The automobile engine is an international combustion engine. The automobile engine is an external combustion engine. The automobile engine is an inner combustion engine What do we see from the second paragraphs? ( ) High pressure is source of power inside a engine. Crankshaft is rotated without any force. Combustion chamber is not a part of the engine. Which one Statement is true according the above second paragraph? ( ) The movement of piston is carried by connecting rods to the engine crankshaft.
全国通信专业技术人员职业水平考试 2015年中级通信工程师通信专业实务（互联网技术）真题 试题一（14分） 回答下列关于网络操作系统的问题 1、（4分） Linux操作系统是目前最流行的操作系统之一，特点包括：易于移植，用户界面良好，请写出Linux系统的其他4个主要特点。 2、（2分） 在常用的台式计算机中，Linux内核开机是从何处加载到内存中，Linux内核功能是否能够扩展？ 3、（8分） 用户通过命令行与Linux系统进行交互，完成各种配置及资源操作，请按下面要求写出完整的Linux命令 答案： 1、源代码公开 完全免费 完全的多任务和多用户 适应多种硬件平台 稳定性好 用户界面良好 具有强大的网络功能 2、从磁盘加载到内存，可以扩展 3、ping https://www.doczj.com/doc/3418216135.html,; cd /home/zhang; rename test https://www.doczj.com/doc/3418216135.html,或mv test https://www.doczj.com/doc/3418216135.html,; grep ‘passwd’ test.c或cat test|grep 'passwd' 试题二（20分）
1、IP报头中的Time To Live（TTL）字段有什么作用？ 2、 IP报头中那些字段和分片有关？ 3、当计算机收到一个数据报时，如何判断这个IP数据报未经历过分片？ 4、当计算机收到一个IP数据报时，如何确定应提交给高层的UDP协议处理程序还是TCP 协议处理程序？ 5、 ICMP一般认为是IP层协议，ICMP协议有什么作用？ 6、 TCP报头中不含长度字段，如何确定TCP报文段中携带了多少字节的数据？ 7、一个携带1024字节应用层数据的TCP报文段Sequence number值16进制表示为3A470B7C，那么当收到对方发来的TCP报文中Acknowlegement number（Ack）是什么值时则表示这1024字节数据接收方已经正确接收到？ 8、 TCP协议提供了流量控制，发送方不能发出超过接收方数据缓冲区大小的数据，在TCP 报文段中，通过哪个字段向对方通知这个TCP连接在本地接收缓冲区的大小？ 9、 TCP端口号占16比特，216=65536，所以每台计算机允许同时工作的TCP连接数最大为65536.这种说法对吗？为什么？ 10、当计算机A与B建立TCP连接之后，A突然掉电后迅速重启动，B以为TCP连接仍存在，这就是“半开连接”问题。此时，B向A发出携带数据的TCP报文头，A使用TCP报头中的哪个字段通知B该链接已经不复存在？ 答案： 1、TTL表示此IP包最多能经过的跳数。p131 2、与分段有关的：Identification，Flags，Framgent Offset 3、DF=1；MF=0 4、根据Protocol字段值（6为TCP，17为UDP） 5、在发生特定错误时ICMP向源主机报告，并提供一组请求/应答消息，检测主机和网络的
2012—2013学年度第一学期 应用化学专业《专业英语》课程试卷（A ） 注意事项：1. 考生务必将自己姓名、学号、专业名称写在指定位置； 2. 密封线和装订线内不准答题。 一、词汇填空 （写出下列每个词汇对应的英 汉单词）（共20小题，每空1分，共20分） 1、化学性质 （chemical property ） 2、物理性质 （physical property ） 3、溶解度 （solubility ） 4、密度 （density ） 5、沸点 （boiling point ） 6、熔点 （melting point ） 7、反应 （reaction ） 8、无机的 （inorganic ） 9、有机的 （organic ） 10、化合物 （c ompound ） 11、烷烃 （alkane ） 12、乙醇 （ethanol ） 13、烯烃 （alkene ） 14、炔烃 （alkyne ） 15、ester （ 酯 ） 16、ether （ 醚 ） 17、acetone （ 丙酮 ） 18、formaldehyde （ 甲醛 ） 19、ammonia （ 氨 ）
20、benzene （ 苯 ） 二、给下列无机化合物的英语名称（共10小题, 每小题2分，共20分） 1、CaO calcium oxide 2、HClO 4 perchloric acid 3、CuSO 4 copper sulfate 4、NaBr sodium bromide 5、NaCl sodium chloride 6、HNO 3 nitric acid 7、HNO 2 nitrous acid 8、Al 2O 3 aluminum oxide 9、KNO 3 potassium nitrate 10、FeBr 3 ferric bromide 三、给下列有机化合物的英语名称（共5小题, 每小题4分，共20分） 1.辛烷 octane 2.CH 2=CHCH 2CH 3 1-butene 3.CH 3CH 2CH 2CH 2OH butanol 4.CH 3CH 2OCH 3
专业英语四级考试试题及答案 2005年英语专业四级考试试题 TEST FOR ENGLISH MAJORSTEST FOR ENGLISH MAJORS (2005) -GRADE FOUR- TIME LIMIT: 130 MIN PART IDICTATION [15 MIN ] Listen to the following passage. Altogether the passage will be read to you four times. During the first reading, which will be read at normal speed, listen and try to understand the meaning. For the second and third readings, the passagewill be read sentence by sentence, or phrase by phrase, with intervals of 15 seconds. The last reading will be read at normal speed again and during thistime you should check your work. You will then be given 2 minutes to check through your work once more. Please write the whole passage on ANSWER SHEET ONE. PART IILISTENING COMPREHENSION [15 MIN] In Sections A, B and C you will hear everything once only. Listen carefully and then answer the questions that follow. Mark the correct answer to each question on your answer sheet. SECTIONA CONVERSATIONS In this section you will hear several conversations. Listen to the conversations carefully and then answer the questions that follow.
(coaxial) cable (同轴)电缆 AC transmission system 交流输电系统active load （PLoad）有功负载active loss有功损耗 active power 有功功率 ammeter电流表 anode (cathode) 阳极（阴极） arc discharge 电弧放电 armature 电枢 control控制 autotransformer 自耦变压器Shutdown 断电、停电 breakdown （电）击穿 breaker (circuit breaker) 断路器brush 电刷 brushless DC motor无刷直流电机bus bar (bus) 母线 calibrate 校准 capacitor bank 电容器组 conductor 导线 copper loss 铜损 corona 电晕 CT 电流互感器 dielectric constant 介质常数dielectric loss 介质损耗 direct current （DC）直流电discharge/charge 放电/充电disconnector 隔离开关 double-circuit lines on the same tower双回同杆并架double-column transformer双绕组变压器 earth(ground) wire 接地线 earthing switch 接地开关 eddy current 涡流 electric field 电场 electrical circuit 电路 electrical machine control控制电机 electrical machinery电机 electromagnetic field 电磁场 excitation励磁 exciting winding 激磁绕组 excitor 励磁机 extra-high voltage (EHV) 超高压 fault clearing time 故障切除时间 feeder 馈线 flashover 闪络 (fossil-fired) power plant 火电厂 fuse 保险丝，熔丝 (GIS)气体绝缘变电站 generator terminal机端 generator 发电机 grounding 接地 high voltage高压 high-voltage engineering 高电压工程 impedance阻抗
2012年考研专业课自测试卷及答案：通信工程 一、填空题 1.卫星通信是利用人造地球卫星作为（）来转发无线电波而进行的两个或多个地球站之间的通信。 2.同步通信卫星是指在地球赤道上空约（）km的太空中围绕地球的圆形轨道上运行的通信卫星。 3.国际卫星的通信方式已由以模拟频分方式为主，转向以（）方式为主。 4.卫星通信网是由一个或数个通信卫星和指向卫星的若干个（）组成的通信网。 5.通信卫星的温控方式有两种，即（）和无源控制。 6.卫星通信地球站天线设备由主反射面、副反射面、馈源及跟踪伺服系统组成，一般采用修正型（）天线。 7.卫星通信地球站的频分多路复用是多个话音通道在基带频率上按频率顺序排列合并成一个（）的过程。 8.卫星通信地球站的时分多路复用是多个话音PCM信道在时间域内顺序排列合并成一个单一（）的过程。 9.通信卫星是作为无线电通信（）的人造地球卫星。 10.卫星通信地球站是指在地球的陆上、水上、空中设置的能通过通信卫星传输信息的（），简称地球站。 11.甚小天线地球站（）网根据其网结构可分为星状网、网状网以及（）。 二、单项选择题 1.模拟卫星通信单路单载波/调频/频分多址方式采用门限扩展技术可降低门限（C/N 值）（）dB。 A.1~2 B.2~3 C.3~4 D.4~5
2.国内卫星通信一般的TDMA方式的速率在（）Mbit/s以下。 A.60 B.120 C.180 D.200 3.卫星通信地球站通过赤道上空的通信卫星的转发进行通信，视地球站纬度高低，其1跳的单程空间距离为72000~（）km。 A.80000 B.82000 C.90000 D.92000 4.常用的固定卫星以C频段的居多，C频段为上行6GHz、下行（）GHz。 A.4 B.6 C.8 D.10 5.Ku频段卫星上行为（）GHz，下行为11GHz/12GHz。 A.10 B.14 C.18 D.22 6.卫星移动通信业务使用（）频段。 A.L B.S C.C D.Ku 7.卫星通信系统包括通信转发器和（）。 A.收信机 B.发信机 C.通信天线 D.基站 8.通信卫星中自旋稳定卫星的太阳能电池阵有效功率为三轴稳定卫星太阳能电池帆板有效功率的（）左右。 A.1/2 B.1/3 C.1/4 D.1/5 9.模拟卫星通信单路单载波/调频/频分多址方式采用语音激活技术，可提高转发器功率利用率（）倍。 A.1 B.15 C.2 D.25 三、多项选择题 1.卫星通信网按卫星的服务区域划分，有（）和国内卫星通信网等。 A.国际卫星通信网 B.区域卫星通信网
医学专业英语 I.Choose the best answer from the four choices marked A, B, C and D (40points) 1.Which of the following suffixes does not mean “pertaining to”? A. -ic B. -al C. -ar D. -our 2.Which of the following combining forms means cell? A. -cyte B. cyt/o C. cel/o D. both a and b 3.Which of the following forms means tissue? A. erythr/o B. immun/o C.hist/o D. vascul/o 4.A- means _____ . A. up B. down C. apart D. without 5.The suffix -itis means _____ . A. infammation B.inflammation C.inflammazed D. instrument 6.The combining form of radi/o means _________ . A. radar B. x-ray C. radio D. both B and C 7.The instrument for viewing is –scope whereas the instrument for cutting is ______ . A. –scopy B. –tome C. –meter D. –graph 8.Which of the following forms refers to “surgical incision o f ?” A. –tomy B. -stomy C. –tome D. –ectomy 9. A record of electric wave occurring in the brain is called _______ . A. electroencephogram B. electromyogram C. electrocardiogram D. electroencephalogram 10.Stethalgia means the pain of chest, which can also be expressed in ____ . A. stethodynia B. thoracalgia C. thoracodynia D. all of the above 11.Which of the following forms means fungus? A. my/o B. myel/o C. myc/o D. none of the above 12.The combining form meaning “the skin” is _______ . A. dem/o B. derm/o C. dermat/o D. both B and C 13.Which of the following combining forms means blood vessels ? A. vascul/o B. angi/o C. chromat/o D. Both A and B 14.The term meaning “pertaining to secretion” is _______ . A. crinogenic B. endocrine C. crinologic D. endocardial 15.The disease of having too many white blood cells is referred to as ________ . A. leukocyte B. leukemia C. leukogenesis D. erythrocyte 16.Inflammation of liver is known in English as ________ . A. bronchitis B. pneumonitis C. gastritis D. hepatitis 17.-plasm means ______ . A. treatment B. knowledge C. diagnosis D. growth 18.Mammo/o refers to ______ . A. mother B. thymus gland C. thyroid gland D. breast 19.Which of the following does not mean “within or in”? A. en- B. endo- C. intra- D. none of the above 20._____ means surgical repair of an organ. A. phag/o B. –plasm C. –plasty D. –pathy
Exercise 2 一、阅读理解（20points ）: Passage one The circuit element used to model the current-resisting behavior of a material is the ① . It is the simplest ② element. George Simon Ohm is credited with formulating the ③ relationship for a resistor based on experiments performed in 1826. This relationship is known as Ohm's Law. Ohm's Law states that the voltage across a resistor is ④ the current flowing through the resistor. The constant of proportionality is the resistance value of the resistor in ohms. The circuit symbol for the resistor is shown in Fig.1. For the current and voltage shown, Ohm's Law is u(t) = Ri(t) (1) where R≥0 is the resistance in ohms. Since R is constant, Eq.(1) is the equation of a straight line. For this reason, the resistor is called a linear resistor. A graph of u(t) versus i(t) for a linear resistor is a line passing through the origin with a slope of R. Resistors whose resistances do not remain constant for different terminal current are known as ⑤ resistors. For such a resistor, the resistance is a function of the current flowing in the device. In reality, all practical resistors are nonlinear because the electrical characteristics of all conductors are affected by environmental factors such as temperature. Many materials, however, closely approximate an ideal linear resistor over a desired operating region. Choose the best answer into the blanks. 1 A. capacitor B. inductor u (t ) i (t ) + _ Fig.1
苏州经贸职业技术学院 2010—2011学年第一学期 《通信技术概论》期终试卷（A）答案 班级：＿＿＿＿姓名：＿＿＿＿学号：＿＿＿＿成绩：＿＿＿＿ 一、填空题（15分） 1、微波通信分为模拟微波通信和数字微波通信两类，已成为一种重要传输传输手段，并与卫星通信和光纤通信一起，作为当今三大传输手段。 2、光纤是由中心的纤芯和外围的包层同轴组成的双层同心圆柱体。光缆是由被覆光纤、缓冲层、加强元件和外护套组成的。 3、信息是指消息中包含的有意义的内容，它是通过消息来表达的。 4、ADSL在一条线上既可以上网传输数据，又可以同时教学常规的语音服务，其原因是利用电话话音传输以外的频率传输数据。 5、卫星移动通信系统通常主要由空中段的通信卫星，端面段的关口站和用户段的众多远程终端设备三大部分组成。 二、名词解释（12分） 1．空分交换——在数字交换网络中，任一输入线与任一输出线（或任一复用线与任一复用线）之间的信息交换，称为空分交换。 2．移动通信——移动通信是指通信的双方，至少有一方是在移动中进行信息交换的。 3．分组交换——分组交换是用分组来传输和交换信息的。它是将用户传送的数据划分成一定长度，每个部分叫一个分组，在每个分组的前面 加一个分组头。
4．多址连接——多址连接是指在卫星的覆盖区内，各地球站通过同一个卫星，能同时实现多个方向多个地球站之间相互联系通信的一种方 式。 5．同步卫星——卫星的运转周期与地球自转周期相同，与地球上任一点都保持相对静止，这样的卫星称为同步卫星 三、判断题将下列描述正确的打“ ”，将下列描述错误的打“ ”（8分） 1、消息是信息的载体，一般将语言、文字、图像或数据等统称为消息。（ ） 2、大区制就是把整个服务区域划分成若干个小区，每个小区分别设置一个基站，负责本小区移动通信的联络和控制。（ ） 3、交织技术就是将一条相继比特以非相继方式被发送，将成串的差错变成单个的比特，再用信道编码纠错功能纠正差错。（ ） 4、微波中继通信是利用微波为载波携带信息。（ ） 5、发信站发送的信号只经过一次卫星转发后就被接收站接受的卫星通信线路叫做单跳单工通信线路。（ ） 6、CDMA是不同的移动台占用同一频率，每一移动台被分配一个独特的随机码序列，与其他台的码序列不同，以便区分不同的移动台。（ ） 7、移动用户没有开机，没有通信，MS不能应答寻呼消息，MS也不能通知网络其所处位置，此时MS被认为是“附着”。（ ） 8、分组拆装设备PAD是非分组终端接入分组网的接口设备。非分组终端必须经过分组拆装设备转换，才能进行分组交换。（ ） 四、计算题（10分） 1、普通电话线路带宽约3KHz，在保证不失真的情况下，其最大码元速率是多少？如果采用16进制调制方式，最大信息传输速率是多少？ 根据P18的公式2.1有： R B=2f N=2B N 则有：R B＝2×3＝6 KB 如果采用16进制调制方式，根据P11的公式1.3有： Rb=R B log 2 N 则有：R b＝R B log 2 N =6 ×log 2 16＝24Kb/s 答：最大码元速率是6 KHz。最大信息传输速率是24Kb/s。
2009 —2010学年第一学期 化学与材料学院（系）07级应用化学专业 《专业英语》期末试卷 1.Write the structural formula or Chinese name for each of the following (2% for each answer): （1）barium ion: （2）chlorate ion: （3）potassium ion: （4）carbonic acid: （5）ammonium ion: （6）pyrrole:（吡咯） （7）polystyrene: (聚苯乙烯) （8）p-hydroxybenzoic acid:(对羟基苯甲酸)（9）benzonitrile (苄腈) （10）critical pressure: （临界压力）（11）methanal: （甲醛）（12）buffer solution :（缓冲溶液）（13）alkali burette:（碱式滴定管）（14）extract ：(萃取)（15）tetrasulfur dinitride: （S4N2）（16）aldose:（醛醣）（17）sodium dihydrogenphosphate (磷酸二氢钠) （18）zinc oxide: （19）6-ethyl-4-methyldecane: （20）quantitative analysis: （定量分析） 2.Write the English name for each of the following(2% for each answer): （1）IBr: （2）天平（balance）（3）阴离子(anion) （4）H2SO3 （5）滴液漏斗: （dropping funnel）（6）CuNO3: （7）AgF: （8）滴定（n.）:(titrate)（9）Ca(MnO4)2: （10）辛醇: （11）十三烷： （12）(CH3CH2)2Hg: (diethylmercury) （13） CH3CHCH CH2 3: （14） CH3CH2CHCOOH CH3: （15）
Model Test 1 Part III Cloze [15 MIN.] Decide which of the choices given below would best complete the passage if inserted in the corresponding blanks. Mark the best choice for each blank on your answer sheet. Geography is the study of the relationship between people and the land. Geographers compare and contrast __1__ places on the earth. But they also __2__beyond the individual places and consider the earth as a __3__. The word geography __4__from two Greek words: ge，the Greek word for "earth" and graphic, __5__means "to write". The English word __6__ means "to describe the earth". Some geography books focus __7__a small area __8__ a town or city. Others deal with a state, a region, a nation, or an __9__continent. Many geography books deal __10__the whole earth. Another __11__to divide the study of geography is to __12__between physical geography and cultural geography. The former focuses on the natural world; the __13__starts with human beings and __14__how human beings and their environment act __15__each other. But when geography is considered as a single subject, __16__branch can neglect the other. A geographer might be described as one who observes, records, and explains the __17__between places. If all places __18__alike, there would be little need for geographers. We know, however, __19__no two places are exactly the same. Geography, __20__, is a point of view, a special way of looking at places. 1. [A] similar [B] various [C] distant [D] famous 2. [A] pass [B] reach [C] go [D] set 3. [A] total [B] unit [C] part [D] whole 4. [A] falls [B] removes [C] results [D] comes 5. [A] what [B] that [C] which [D] it 6. [A] ge [B]graphy [C]geography [D]graphic 7. [A] to [B] up [C]above [D]on 8. [A] outside [B] except [C] like [D] as 9. [A] extensive [B] enormous [C] overall [D] entire 10. [A] with [B] in [C] to [D] of 11. [A] way [B] means [C] habits [D] technique 12. [A] split [B] distinguish [C] integrate [D] tell 13. [A] second [ B] latter [C] next [D] later 14. [A] learns [ B ] realizes [ C ] studies [ D ] understands 15. [A] upon [B] for [C]as [D] to 16. [A] either [B] neither [C] one [D] each 17. [A] exceptions [B] differences [C] sameness [D] divisions 18. [A] being [B] are [C] were [D] be 19. [A] although [B] whether [C] since [D] that 20. [A] then [ B ] nevertheless [ C ] still [ D ] moreover Model Test 2 Part III Cloze [15 MIN.] Decide which of the choices given below would best complete the passage if inserted in the corresponding
专业英语期末考试试卷A卷标准答案及评分标准（2009-2010第二学期） 一、Translate the following speciality vocabulary into Chinese (20×1=20) 标准答案：1.电气工程 2. 正电荷 3 .信道 4.家庭自动化 5.桌面应用软件 6.下降沿 7.波特率 8.红外线传感器9 数字信号10.激光技术 11光纤12.微处理器13振荡器14.半导体15.远程控制 16触发器17.阴极射线管18 代码错误19.波导管20.通信工程 评分标准：得分=做对的个数×1 二、Translate the following speciality vocabulary into English(10×1=10) 1.circuit 2.voltage 3.diode 4.triode 5.microwave oven 6.watchdog 7.sensor或transducer 8.current 9.transistor 10. magnifier或amplifier 评分标准：得分=做对的个数×1 四、Multiple choices. (2×10=20） 1-5 BCDBD 6-10 BAAAC 评分标准：得分=做对的个数×2 五、Translate the following sentences into Chinese (5×2=10) 1、电源有两种：电流源与电压源。 2、电子元器件分为有源元件与无源元件两类。 3、激光技术是用来测量物体的存在、传感器到物体之间的距离以及物体尺寸的一种技术。 4、计算机的“大脑”被称为微处理器。 5、串行口有四种工作模式。 六、Choose the right statements according to the following passage .（true(T) or false (F) (5×2=10)） P1: 1-5 T F T F T 评分标准：得分=做对的个数×2 七、Reading Comprehension(5×2=10)） 1-5 B C A D C 评分标准：得分=做对的个数×2
专业：移动通信科目：移动通信技术 一、单项选择题 1．GSM网络结构中，Abis接口是（）的接口 A．MSC与HLR B．MSC与VLR C．MSC与BSC D．BSC与BTS 答案：D 2．对讲机属于那种通信方式（） A．半双工通信 B．全双工通信 C．单工通信 D．三工通信 答案：A 3．GSM系统对于话务量密集的局部地区，可以采用六列向小区。此时需要采用（）度定向天线 A．360 B．60 C．180 D．120 答案：B 4．实际工程一般要求天线间距大于（）倍信号波长 A．2 B．5 C．10
答案：C 5．GSM网络一般采用列向小区，即天线采用（）度定向天线，把基站分成3个扇形小区 A．360 B．120 C．180 D．60 答案：B 6．CDMA系统容量是模拟系统的（）倍 A．1~2 B．100~200 C．1000~2000 D．10~20 答案：D 7．GSM系统容量是模拟系统的（）倍左右 A．4 B．2 C．3 D．1 答案：B 8．GSM系统信号带宽为（）KHz。 A．200 B．2 C．20
答案：A 9．有线电视属于那种通信方式（） A．全双工通信 B．单工通信 C．半双工通信 D．三工通信 答案：B 10．GSM规范中规定：邻频道干扰保护比，C/I > 负（）dB A．6 B．9 C．12 D．3 答案：B 11．无线电广播采用（）方式 A．CDMA B．SDMA C．TDMA D．FDMA 答案：D 12．GSM是一个典型的（）多址系统 A．FDMA B．TDMA C．SDMA D．CDMA
答案：B 13．GSM网络结构，A接口是（）之间的接口A．BSC与BTS B．MSC与VLR C．MSC与HLR D．MSC与BSC 答案：D 14．无线广播属于那种通信方式（） A．三工通信 B．单工通信 C．全双工通信 D．半双工通信 答案：B 15．GSM规范中规定：同频道干扰保护比，C/I >（）dB A．6 B．3 C．12 D．9 答案：D 二、多项选择题 1．3G技术要求有哪些（） A．支持多媒体业务 B．上下行不对称 C．速度按需分配 D．OFDM
1.Why are cast metal sheet ingots hot-rolled first instead of being cold-rolled? Because of cold rolling is to use hot rolled steel coils as the raw material, after acid pickling to remove oxide skin for cold rolling, the finished product is hard roll, because of cold work hardening caused by deformation of continuous cold rolling hard roll strength, increase hardness, toughness and plastic index decreased, so the stamping performance will deteriorate, can only be used for simple deformation of the parts 2.What type of heat treatment is given to the rolled metal sheet after hot and “warm” rolling? What is its purpose? 轧钢的热处理的类型？轧钢热处理的目的？ Heat treatment of the main types are annealing, normalizing, quenching and tempering, solution treatment and aging treatment, cold treatment, chemical treatment, etc. Annealing： The steel is heated to a certain temperature and heat preservation for a period of time, and then make it slowly cooling, called annealing. Steel annealing is a heat the steel to the phase change or part of the phase change temperature, slow cooling after heat preservation heat treatment method. The purpose of annealing is to eliminate tissue defects, improve the organization make composition uniformity and fine grains, increase mechanical properties of the steel, reduce residual stress; Can decrease the hardness at the same time, improve the plasticity and toughness, improve machinability. So before annealing in order to eliminate and improve both the legacy of tissue defects and internal stress, and to prepare for the follow-up process, so the annealing is belong to the intermediate heat treatment, also called heat treatment in advance Normalizing： Normalizing is heated to above the critical temperature of steel, to all into homogeneous austenitic steel, heat treatment and natural cooling in air. It can eliminate hypereutectoid steel mesh cementite, for hyposteel normalizing can refine crystal lattice, improve comprehensive mechanical properties, low requirements for the parts use the normalized instead of the annealing process is more economic. Quenching: Quenching is the steel is heated to above the critical temperature, heat preservation for a period of time, then quickly into the quenching medium, the temperature plummeted, rapid cooling at greater than the critical cooling rate of speed, which is mainly composed of martensite and unbalanced heat treatment method of the organization. Can increase strength and hardness of the steel quenching, but to reduce its plasticity. That is commonly used in quenching hardening agent are: water, oil, caustic soda, and salt solution, etc