2011南理工研究生复试上机真题

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2011年南京理工大学331社会工作原理[专业硕士]考研真题【圣才出品】

2011年南京理工大学331社会工作原理[专业硕士]考研真题【圣才出品】

2011年南京理工大学331社会工作原理[专业硕士]考研真题
南京理工大学2011年硕士学位研究生入学考试试题
科目代码:331 科目名称:社会工作原理满分:150分
一、解释以下名词(每题5分,共50分)
1.社会工作者
2.社会工作价值观
3.助人活动
4.助人方法
5.人类需要
6.社会环境
7.社会工作研究
8.社会政策
9.社会服务方案策划
10.社区照顾
二、社会工作干预理论或实践模式主要包括哪些?其主要内容和特点是什么?(20分)
三、请从系统理论的观点,分析家庭和青少年偏差行为(可自行选一种偏差行为为例)的关系。

(30分)
四、简述社会服务机构志愿者管理的主要内容。

(20分)
五、简述社会工作专业伦理的主要内容。

(15分)
六、简述社会工作者的能力要求。

(15分)。

会计硕士专业学位联考(MPAcc)会计学南京理工大学真题2011年

会计硕士专业学位联考(MPAcc)会计学南京理工大学真题2011年

会计硕士专业学位联考(MPAcc)会计学南京理工大学真题2011年一、名词解释1. 资产答案:资产是指企业过去的交易或者事项形成的、由企业拥有或者控制的、预期会给企业带来经济利益的资源。

2. 趋势分析法答案:趋势分析法是将两期或连续数期财务报告中相同指标进行对比,确定其增减变动的方向、数额和幅度,以说明企业财务状况及经营成果变动趋势的一种方法。

3. 营业周期答案:营业周期是指从外购承担付款义务,到收回因销售商品或提供劳务而产生的应收账款的这段时间。

也就是说从取得存货开始到销售存货并收回现金为止的这段时间。

4. 货币时间价值答案:货币时间价值是指货币经过一段时间的投资和再投资后所增加的价值。

货币时间价值实质是没有风险和没有通货膨胀条件下的社会平均资金利润率。

5. 成本性态答案:成本性态是指成本总额与特定的业务量之间在数量方面的依存关系,又称成本习性。

6. 安全边际答案:安全边际是指实际(预计)的销售量与保本点销售量或实际(预计)的销售额与保本点销售额之间的差额,它有安全边际量和安全边际额两种形式。

二、简答题1. 简述权责发生制原则及其应用。

答案:权责发生制是收入按现金收入及未来现金收入的发生来确认;费用按现金支出及未来现金支出的发生进行确认,而不是以现金的收入与支付来确认收入费用。

按照权责发生制原则,凡是本期已经实现收入和已经发生或应当负担的费用,不论其款项是否已经收付,都应作为当期的收入和费用处理;凡是不属于当期的收入和费用,即使款项已经在当期收付,也不应作为当期的收入和费用。

权责发生制原则广泛应用于会计确认和计量的实践中,如包括购原材料采购时,原材料已收到并验收入库,款未付对方也未开发票,但是按权责发生制原则也应确认原材料;当销售货物时,如果购货方未付款也未开具销售发票,但货物已发出且购货方已确认收到,也需按权责发生制原则确认为销售收入;还有当销售货物时,如果购货方已付款且企业已开具销售发票,但货物未发出也应按权责发生制原则确认为销售收入。

南京理工大学马克思主义理论教学研究部871马克思主义基本原理历年考研真题及详解专业课考试试题

南京理工大学马克思主义理论教学研究部871马克思主义基本原理历年考研真题及详解专业课考试试题

目 录2011年南京理工大学871马克思主义基本原理考研真题2011年南京理工大学871马克思主义基本原理考研真题及详解2010年南京理工大学871马克思主义基本原理考研真题2010年南京理工大学871马克思主义基本原理考研真题及详解2011年南京理工大学871马克思主义基本原理考研真题南京理工大学2011年硕士学位研究生入学考试试题考试科目:871马克思主义基本原理一、辨析题(每小题10分,其中辨别正误3分,说明理由7分。

共30分)。

根据马克思主义哲学有关原理,辨别下列各题是否正确,并说明理由。

1.哲学的基本问题是唯物主义和唯心主义的关系问题。

2.实践标准的相对性是指对于不同事物有不同的检验标准3.社会心理和社会意识形式的主要差别是后者包含前者。

二、简答题(每小题14分,共70分)1.简述意识的能动性。

2.什么是价值规律?3.真理的相对性的含义是什么?4.简述地理环境和人口因素在社会存在中的地位作用。

5.唯心史观流行的阶段根源和认识根源是什么?三、论述题(两题任选一题.本题25分。

如果两题都答,以第一题为准,第二题无效。

)1.试用社会基本矛盾的原理分析在建设和谐社会中进行中国社会制度改革和创新的重要性。

2.如何理解辩证否定观?谈谈你对中国悠久历史文化传统的看法。

四、材料分析题(本题25分)。

认真阅读下列材料,结合材料并根据所学马克思主义哲学原理回答问题。

材料:1.体育的本质是什么?除了更高更快更强,除了竞技中所体现的拼搏精神,除了成绩,还有参与,弱”来提振精神,但真正的强者不需要用符号来“证明”本届亚运会金牌,日本和韩国获得的金牌不及我们的零头,但不能证明日本和韩国是个贫弱国家,或者是人们的身体素质不如我们。

2.目前,中国有1.6亿人患高血压,1.6亿人患高血脂,有2亿人超重或肥胖;城市里,每5个5孩子就有1个小胖墩儿,高中生里85%以上的学生是小四眼儿,和日本孩子比,我们的中学生在身体素质多项数据上差很多;我们参加运动的体育人口只有28%,人均体育设施在世界上排百名开外......在亚洲,我们的体育设施人均拥有量绝对排不上前10名,就在中国竞技体育取得辉煌成就的这20年中,中国人的体质正在明显滑坡!然而,我们的金牌总数是世界第一,亚洲绝对第一,这难道不是讽刺吗?难道这不令我们汗颜吗?1.谈谈你对材料1的认识和理解。

南京理工大学考研复试计算机数据库试题1

南京理工大学考研复试计算机数据库试题1

一、选择题60(选择一个最合适的答案,在答题纸上涂黑)1.一个事务中的一组更新操作是一个整体,要么全部执行,要么全部不执行。

这是事务的:A.原子性B.一致性C.隔离性D.持久性2.在数据库的三级模式结构中,描述一个数据库中全体数据的全局逻辑结构和特性的是:A.外模式 B.内模式 C.存储模式D.模式3.关于联系的多重性,下面哪种说法不正确?A.一个多对多的联系中允许多对一的情形。

B.一个多对多的联系中允许一对一的情形。

C.一个多对一的联系中允许一对一的情形。

D.一个多对一的联系中允许多对多的情形。

4.考虑学校里的"学生"和"课程"之间的联系,该联系的多重性应该是:A. 一对一B. 多对一C. 一对多D. 多对多5.下面哪种约束要求一组属性在同一实体集任意两个不同实体上的取值不同。

A. 键(key)约束。

B. 单值约束。

C. 参照完整性。

D. 域(domain)约束6.关系模型要求各元组的每个分量的值必须是原子性的。

对原子性,下面哪种解释不正确:A.每个属性都没有内部结构。

B.每个属性都不可再分解。

C.各属性值应属于某种基本数据类型。

D.属性值不允许为NULL。

7.对于一个关系的属性(列)集合和元组(行)集合,下面哪种说法不正确:A.改变属性的排列次序不影响该关系。

B.改变元组的排列次序不影响该关系。

C.改变元组的排列次序会改变该关系。

D.关系的模式包括其名称及其属性集合。

8.若R是实体集R1与R2间的一个多对多联系,将其转换为关系R',哪种说法不正确:A.R'属性应包括R1与R2的所有属性。

B.R'属性应包括R1与R2的键属性。

C.R1与R2的键属性共同构成R'的键。

D.R'的属性应包括R自身定义的属性。

9.关于函数依赖的判断,下面哪种说法不正确?A.若任意两元组在属性A上一致,在B上也一致,则有A → B成立。

B.若任意两元组在属性A上一致,在B上不一致,则A → B不成立。

南京理工大学工程硕士英语课程考试2011.8试题及参考答案

南京理工大学工程硕士英语课程考试2011.8试题及参考答案

南京理工大学工程硕士英语课程考试试题(2011年8月)Directions:Write all your answers on the Answer Sheet.Part I Reading Comprehension (40 points)Section A Reading in depthDirections:In this section there are 2 passages. Each passage is followed by 5 questions or unfinished sentences. For each of them there are 4 choices marked A, B, C, and D. Choose the best answer and mark your choice on your Answer Sheet.Passage OneQuestions 1 to 5 are based on the following passage.All forms of life interact with the environment in which they exist, and also with other life forms. In the 20th century this premise gave rise to the concept of ecosystems, which can be defined as any situation where there is interaction between organisms and their environment.Ecosystems are composed of a variety of abiotic and biotic components that function in an interrelated way. The structure and composition is determined by various environmental factors that are interrelated. Variations of these factors will initiate dynamic modifications to the ecosystem. Some of the more important components are: soil, atmosphere, radiation from the sun, water, and living organisms.Each living organism has a continual relationship with every other element that makes up its environment. Within the ecosystem, species are connected and dependent upon one another in the food chain, and exchange energy and matter between themselves as well as with their environment.Every species has limits of tolerance to factors that affect its survival, reproductive success and ability to continue to thrive and interact sustainably with the rest of its environment, which in turn may have effects on these factors for many other species or even on the whole of life. The concept of an ecosystem is thus an important subject of study, as such study provides information needed to make decisions about how human life may interact in a way that allows the various ecosystems to be sustained for future use rather than used up or otherwise rendered ineffective. For the purpose of such study, a unit of smaller size is called a micro-ecosystem. For example, an ecosystem can be a stone and all the life under it. A macro-ecosystem might involve a whole ecological region, with its drainage basin.The following ecosystems are examples of the kinds currently under intensive study: oceanic ecosystems, continental ecosystems, such as “forest ecosystems”, meadow ecosystems such as steppes or savannas, or agro-ecosystems, systems in inland waters, such as lentic ecosystems such as lakes or ponds; or lotic ecosystems such as rivers. Another classification can be made by reference to its communities, such as in the case of a human ecosystem. Regional groupings of distinctive plants and animals best adapted to the region’s physical natural environment, latitude, altitude, and terrain are known as biomes. The broadest classification, today under wide study and analysis, and also subject to widespread arguments about its nature and validity, is that of the entire sum of life seen as analogous to a self-sustaining organism.1. Which of the following statements about an ecosystem is NOT true?A) An ecosystem is usually composed of some nonliving and living organisms.B) A stone and all the life under it can be regarded as an ecosystem.C) An ecosystem means the interaction between some nonliving and living organisms andtheir environment.D) Sunlight, air, soil, water, and human beings are some of the important elements within theecosystem.2. How is the relationship among all the living organisms?A) All the living organisms are closely related to each other.B) Each living organism depends on others.C) In order to thrive, they have to give and gain energy from others.D) All of the above.3. Why is it important to learn the concept of an ecosystem according to the text?A) Many species have limited capacities to continue to grow vigorously.B) We can learn how to protect the natural resources for future use.C) Apart from a micro-ecosystem, it is also important to learn the concept of amacro-ecosystem which usually involves a whole ecological region.D) We can make decisions for human life to interact with sustainable development.4. Which of the following statements is TRUE according to the passage?A) Rivers, lakes or ponds belongs to oceanic ecosystem.B) The classification that each life can be regarded as a self-sustaining organism is still underwidespread arguments.C) The broadest classification of ecosystem might involve a whole ecological region, such asin the case of a human ecosystem.D) All of the above.5. The possible meaning of the word “lentic” in the last paragraph is “of or relating to________”.A) dirty waters B) still watersC) deep waters D) fresh watersPassage TwoQuestions 6 to 10 are based on the following passage.Exactly where we will stand in the long war against disease by the year 2050 is impossible to say. But if developments in research maintain their current pace, it seems likely that a combination of improved attention to dietary and environmental factors, along with advances in genetherapy and protein-targeted drugs, will have virtually eliminated most major classes of disease.From an economic standpoint, the best news may be that these accomplishments could be accompanied by a drop in health-care costs. Costs may even fall as diseases are brought under control using pinpointed, short-term therapies now being developed. By 2050 there will be fewer hospitals, and surgical procedures will be largely restricted to the treatment of accidents and other forms of trauma(外伤). Spending on nonacute care, both in nursing facilities and in homes, will also fall sharply as more elderly people lead healthy lives until close to death.One result of medicine’s success in controlling disease will be a dramatic increase in life expectancy. The extent of that increase is a highly speculative matter, but it is worth noting that medical science has already helped to make the very old (currently defined as those over 85 years of age) the fastest-growing segment of the population. Between 1960 and 1995, the U. S.population as a whole increased by about 45%, while the segment over 85 years of age grew by almost 300%. There has been a similar explosion in the population of centenarians, with the result that survival to the age of 100 is no longer the newsworthy fear that it was only a few decades ago. U. S. Census Bureau projections already forecast dramatic increases in the number of centenarians in the next 50 years: 4 million in 2050, compared with 37,000 in 1990.Although Census Bureau calculations project an increase in average life-span of only eight years by the year 2050, some experts believe that the human life-span should not begin to encounter any theoretical natural limits before 120 years. With continuing advances in molecular medicine and a growing understanding of the aging process, that limit could rise to 130 years or more.6. The main idea of this passage is that ___________.A) the health-care costs will have been reduced by 2050B) most people can live 120 years or even more after 2050C) the development of medical science may help increase human life expectancyD) the development of medical science help the very old become the fastest-growing part ofthe population7. According to paragraph 1 and 2, which of the following is true?A) The drop of health-care costs may result in the increase of most diseases.B) The drop of health-care costs may result in the decrease of most diseases.C) As a result of a drop of health-care costs, most diseases may be eliminated.D) The elimination of most diseases may result in a drop in health-care costs.8. The author thinks that, by 2050, surgical procedures will be greatly limited to the treatment oftrauma because ________.A) there will be fewer hospitalsB) the health-care expending dropsC) most classes of diseases will be brought under controlD) people don’t like to go through the surgical procedure9. In paragraph 3, the sentence “the extent of that increase is a highly speculative matter …”implies that ______.A) people are not quite sure about how much human life expectancy can be increaseB) increasing human life-span venturousC) whether people’s life expectancy can be increased is doubtfulD) people spend much time pondering on if human life-span can be longer10. From paragraph 3, we can infer that, only a few decades ago, _______.A) it was not rare for a person to live for 100 yearsB) it was not a newsworthy fear for a person to live for 100 yearsC) few people could live for 100 yearsD) many people could live for 100 yearsSection B (Skimming and Scanning)Directions: In this part, you will have to go over the passage quickly and answer the questions.For questions 11-17, markY (for YES) if the statement agrees with the information given in the passage;N (for NO) if the statement contradicts the information given in the passage;NG (for NOT GIVEN) if the information is not given in the passage.For questions 18-20, complete the sentences with the information given in the passage.What Do Dogs and Cats Bring to Us?Man’s best friend is helping reform young offenders and along with cats, research shows, giving owners better health.∙Unwanted dogs bring criminals to heelFor a project designed to fight recidivism(累犯)(among some of the worst young offenders in the United States), it has a somewhat unexpected title. Project Pooch is not the sort of name destined to make children sleep easier in their beds at night.The project gives juvenile murderers, rapists and armed robbers a badly behaved dog to look after and see if it improves their behaviour.The project at Oregon’s MacLaren youth correctional facility has had a 100% success rate in cutting reoffending. In the period studied, between 1993 and 1999, around 100 young offenders aged between 18 and 24 cared for wayward dogs. So far, not one is back behind bars.“I’m really pleased with the success, but I’m not altogether surprised,” said Joan Dalton, who runs Project Pooch. “These were kids who had never had any responsibility and we were taking them and saying, ‘hey, this is something you're in charge of, something you’re responsible for’.” Project Pooch takes in dogs from local dog shelters that have behavioral problems, such as aggressiveness or excessive barking. They are then paired off with young offenders, most of whom have been convicted of serious crimes such as murder and sexual assault.The young offenders have to look after the dogs seven days a week, feed them, walk them, and take them through obedience(顺从)training, until they can be given to families who are looking for a dog.“I think the project works because the kids see the effect the bad behaviour of the dog has on them and it allows them to see the effect their bad b ehaviour has on others,” said Mrs Dalton. “You could really see the change in the kids, and that has obviously carried through when they left the correctional facility.”At the 10th International Conference on Human-Animal Interactions in Glasgow, Mrs Dalton said she hoped the project would be extended to other correctional institutions across the US. “But funding is a big problem,” she added.It is not, however, merely young offenders who benefit from having a dog ---- everyone does. ∙Pets make owners healthierAccording to research unveiled at the conference, pet owners visit their doctors between 15% and 20% less than people who do not own animals. In a study carried out among 11,000 people in Germany, 2,000 people in Australia and 1,000 people in China, researchers discovered that even after things like gender and income were taken into account, pet owners visited their doctors 8% less.“This is the first time a study has been done where people have been repeatedly interviewed and it has really been shown conclusively that pet owners have better health than non-pet owners,” said Bruce Headey, from the University of Melbourne. “I don’t think you can give any singlereason why pet owners live longer, but I think companionship has a lot to do with it.”For people intending to prolong their life by buying a goldfish, however, there is a note of caution: the study only shows conclusively that owning a cat or a dog helps cut visits to the doctor. But anyone taking the trouble to look after a dog who does fall seriously ill may find another benefit to pet ownership.A study in the US on people who suffered heart attacks, revealed that pet owners are up to four times more likely to survive than non-pet owners.Erika Friedman, from the University of Maryland, said that in her study dog owners had a surprisingly high survival rate compared with non-pet owners.“There are a number of studies which show that in a moderately stressful situation pets can lower your stress responses,” she said.“We think this c arries over into everyday life and, for dog owners, their pet can help them stay alive after a heart attack.”But it is not just dogs that relieve stress. In a more anecdotal UK survey carried out on 500 cat owners aged over 55 by the charity Cats Protection, most cat owners said their pet helped them overcome stress and loneliness.Perhaps more worryingly, some 75% of cat owners questioned said they preferred cats to humans.Dennis Turner, president of the conference, said the research unveiled yesterday revealed the dependency between pet owners and their pets. “The human-animal bond has indeed come of age,” he said.Animal magic· Forget spending money on unused vitamin supplements and energy on unwanted gym visits, the secret to better health is pet ---- or specifically cat and dog ---- ownership. The average pet owner visits their doctor 20% less than non-pet owners.·In the event of a cardiac arrest, it probably pays to call the pet shop shortly after the emergency services. Dog owners are four times more likely to stay alive following a heart attack than those who live a life without man’s best friend.· In a survey destined to do nothing to reverse clichés about who owns cats, cat owners aged over 55 said their pets alleviate loneliness and make them feel less stressed. So enamoured by their feline friends are the cat owners that 75% prefer their company to humans.11.Project Pooch aims to help juvenile reoffenders and brings peace to the troubledneighborhoods.12.The dogs in the project Pooch are brought in by their owners and would be returned to theowners after the training.13.The Project Pooch has brought a sense of responsibility to the kids who paired with thewayward dogs.14.Projects like “Project Pooch” research on the interaction betw een Human beings and Animals.15.According to the research, because of the companionship brought out by pet-raising, petowners are less likely to go to hospital.16.Pet can save owner who suffer from heart attacks and lower the owner’s stress respon ses.17.The writer doesn’t think it is good for the pet owners to prefer only pets as their lifecompanions.18.is a big problem for projects like “Pooh” to be carried out in thecorrectional institutions in United States.19.There isn’t any speci fic reason why pets can or save the owners’ lives, butcompanionship has contributed a lot to it.20.When one is diagnosed of cardiac diseases, it is suggested to call the pet shop because dogowners are four times more likely to after a heart attack than those without dogs.11. NG 12. N 13. Y 14. Y15. Y 16. Y 17. Y18. Funding19. prolong20. stay alivePart II Short Answer Questions (2 points each, 10 points)Directions:In this section there is one passage followed by 5 questions. Read the passage and answer the questions briefly.Britain’s legal system, long known for Latin tongue-twisting phrases and slow delivery of justice, started a brave new world of plain English this month.In the biggest shakeup for the legal system in a century, civil cases will come to court as a last resort and lawyers will have to use approved contemporary phrases instead of ancient Latin expressions. Lawyers will also be punished with fines for arguing over minor points while judges will have to keep strict deadlines. Lawyers said the new reform package would benefit the profession by speeding up legal process and cutting costs.But the lawyers dismissed the removal of Latin expressions from their vocabulary as unnecessary. “I deliberately learned Latin at school to help me at the Bar,” said one lawyer who declined to be named. “At least they haven’t ruled out ‘carborundum illegitimi’,: he added, a Latin expression thought to originate in the 1940s and meaning “don’t let them grind you down.”“Some of the terminology (to replace the Latin) is patronizing(照顾,惠顾) in the extreme,”said one London lawyer. “To substitute nine words where originally two would have been used is patronizing to the public,” he added.To help simplify the language for “plaintiffs (people who are bringing a claim to court),”Latin expressions such as “in camera”, which means in private, will come under the axe. Other phrases to return to history include “ex parte”(without notice) and an “Anton Piller order”–named after a case involving a German company in the mid-1970s -- becomes simply a search order. “The reforms are to make justice simpler and more accessible,”a spokeswoman for Britain’s chief judicial officer, the Lord Chancellor said.To the relief of many traditionalists, however, one part of Britain’s legal system will remain unchanged for now at least – the white horsehair wig.21.What is the reform of Britain’s legal system?____________________________________________________________________.22. What are the benefits of the reform?____________________________________________________________________.23. What does phrase “come under the axe” (paragraph 4, sentence 1) mean?____________________________________________________________________.24. What will not be changed in this reform?____________________________________________________________________.25. What is the main topic of this passage?____________________________________________________________________.测试习题集-第192页::21. Some of the ancient Latin expressions will not be used in court.22. To speed up legal process and cut cost.23. To be removed.24. The lawyers will still wear the white horsehair wig.25. British Courts choose plain English.Part III Translation (50 points)A.From English to Chinese (30%)1 The lock is the new symbol of America. For some reason we are satisfied when we think we are well-protected; it does not occur to us to ask ourselves: Why has this happened? Why are we having to braaicade ourselves against our neighbors ans fellow citizns, and when, exactly, did this start to take over our lives?We trust no one. We have become so smart about self-protection that, in the end, we all outsmarted ourselves. We may have locked the evils out, but in doing so we have blocked ourselves in.That may be the legacy we remember best when we look back on this age: In dealing with the unseen horrors among us, we became prisoners of ourselves. All of us prisoners, in this time of our troubles.锁成了美国的新的象征。

2011年南京理工大学自动控制原理模拟考研试题.

2011年南京理工大学自动控制原理模拟考研试题.

南京理工大学2011年硕士学位研究生入学考试模拟试题试题编号:2011010033考试科目:控制理论基础(满分150分)注意:所有答案(包括填空题)按试题序号写在答题纸上,写在试卷上不给分一、(15分)(1)试求图1所示运算放大器电路的传递函数()/()o iE s E s。

(5分)(2)控制信号流图如图2所示,试求传递函数C(s)/R(s)。

(10分)图1图2二、(15分)反馈控制系统如下图3所示。

图3(1)确定参数K和K t的值,使系统的最大超调量为20%,调整时间为1秒(对于5%的误差范围)。

(2)取题(1)的K值,改变速度反馈系数Kt的值,试分析K t变化对系统瞬态性能的影响。

三、(15分)单位反馈系统的开环传递函数为11(1)(1)23()(1)K sG ss s++=+(1)绘制K从0->∞变化时系统的根轨迹;(2)求系统在欠阻尼状态时K的取值范围;(3)由根轨迹求出系统具有最小阻尼比时的闭环极点。

四、(10分)控制系统的结构图如图所示,其中G2(s)的频率特性如图5所示(T>0,τ>0),试画出该系统的幅相频率特性曲线,并用奈氏稳定判据分析其稳定性。

图4图5五、(10分)设单位反馈系统的开环传递函数为()(1)(5)KG ss s s=++要求系统校正后满足下列指标:(1)稳态速度误差为10V K =; (2)相角稳定裕度≥35° 六、设系统如图6所示图6(1)求系统的闭环z 传递函数; (2)K=10,T=1s 时,判断系统的稳定性;七、(15分)一具有理想继电特性的非线性系统方块图如图7所示,其非线性特性如图8所示。

试分别在c 、c 及e 、e 平面上画出系统的相平面图。

(()1(),(0)(0)0)r t t cc ===图7图8八、(15分)设A 为2×2的常阵,对于系统的状态方程X AX = ,当2(0)1X ⎡⎤=⎢⎥⎣⎦时,2t t e X e −−⎡⎤=⎢⎥⎣⎦;当1(0)1X ⎡⎤=⎢⎥⎣⎦时,2t t t t e te X e te −−−−⎡⎤+=⎢⎥+⎣⎦。

南京理工大学2011年硕士学位研究生入学考试试题

南京理工大学2011年硕士学位研究生入学考试试题

南京理工大学2011年硕士学位研究生入学考试试题科目代码:821科目名称:电磁场与电磁波满分:150注意:○1认真阅读纸上的注意事项;○2所有答案必须写在答题纸上,写在本试题纸或草稿纸上均无效;○3本试题纸必须随答题纸一起装入试题纸袋中交回一、 填空题与选择题(每空2分,共18分)1. 已知电位函数=cos ye x -Φ,则电场强度E =。

2. 在无界理想媒质中传播的均匀平面电磁波,电场与磁场的相位(A .相同;B .不同),幅度随传播距离的增加而(A .不变;B .衰减)。

而在导电媒质中传播的均匀平面波,电场与磁场的相位(A .相同;B .不同),幅度随传播距离的增加(A .不变B .衰减.)3. 在自由空间中,一个孤立的点电荷,其产生的等电位面是。

4. 在良导体中,电磁波的趋肤效应随着频率的增加而(A .减小B .增大);随着电导率和磁导率的增加而(A .减小B .增大)。

5. 已知α,β>0,那些波描述了沿+Z 方向传播的均匀平面波。

A.()j zx e e αβ-- B.()j z x y e je e β-- C.100sin()2j zy e x e βπ- D.8cos(310)x e t z β--⨯-二、 简答题(共17分)1. 写出微分形式麦克斯韦方程组。

(5分)2. 分别写出时变电磁场在理想介质和理想导体分界面上的边界条件。

(6分)3. 描述镜像法的基本思想;写出应用镜像法求解静态场为题时确定镜像电荷需要遵循的原则。

(6分)三、 (1)写出洛伦兹条件;(2)说明为什么引入洛伦兹条件?(3)利用洛伦兹条件及矢量恒等式2()A A A ∇⨯∇⨯=∇∇-∇ 推导达朗贝尔方程222=-A A J tεμμ∂∇-∂ 和222=-t ϕρϕεμε∂∇-∂(20分)四、 证明通过任意闭合曲面的传导电流和位移电流的总量为0。

(10分)五、 如图所示,半径为a 的球体内均匀充满着密度为0ρ的体电荷,球体中有一半径为b 的小球空腔,其中O 和'O 分别为球体和空腔中任意一点,r为O 点指向P 点的位置矢量,r 为'O 点指向p 点的位置矢量,求小球空腔中任意点P 的电场分布。

南理工研究生工程光学复试(DOC)

南理工研究生工程光学复试(DOC)

填空1.光电传感器的作用()2光电效应的光电探测器()()()()等3具有光电倍增的器件有()()等4半导体对光的吸收有()()()()等半导体对光的吸收主要是()5 热电探测器是将辐射能转化为()能,然后再把它转化为()能的器件6光外差法的物理机理是()7光电探测器中主要的固有噪声有()()()()等8 简单光学目标的形位检测方法:几何中心法和()二概念说明1、积分响应率2、外光电效应3、等效噪声功率4、黑体三简述、计算题1 一种光电材料的逸出功为0.67Ev,试计算该材料的红限波长。

(普朗克常数h=6.626*10-34(J.s),光速C=2.998*108(m/s),电子电量e=1.6*10-19库仑)2、光敏电阻R与R L=20kΩ的负载电阻串联后接于Vb=12V的直流电源上,无光照时负载上的输出电压为V1=20mV,有光照时负载上的输出电压为V2=2V,求①光敏电阻的暗电阻和亮电阻值②若光敏电阻的光电导灵敏度Sg=6×10-6s/lx,求光敏电阻所受的照度.3、简述发光二极管的发光原理,发光二极管的外量子效率与哪些因素有关。

4、简述光盘存储工作原理。

5、为什么要引入胖零电荷?胖零电荷属于暗电流吗?能通过对CCD器件制冷消除胖零电荷吗?6、利用光电信息技术测量速度有几种方法?为什么说激光多普勒测速技术的精度高,可信度也高?05年试题1.LED的优点()A寿命短B体积大C功耗大D发光效率高2.光电检测的核心()A处理电路B发光管C 透镜D 光电传感器3.可作为单色光源的有()A he-ne激光B发光二级管C日光灯D卤钨灯4.内光电效应的光电探测器()A热释电探器B光电管C光电倍增管D雪崩二极管5.具有光电倍增的器件有()A光电管B光电倍增管C光电二极管D光敏电阻6.响应频率较高的探测器()A雪崩二极管B光敏电阻C硒光电池D热释电器件7.对于微弱的光信号的探测,采用的方法是()A单电路直读法B双光路比较法C指零法D光外差法8.热释电探器的工作的物理机理是()A光伏特效应B光电导效应C热电效应D外光电效应9.光外差法的物理机制是()A光的反射B光的偏振C光的干涉D光的折射10.光电探测器的主要作用是()A光信号的放大B光信号转换成电信号C 电信号转换成光信号D光信号的调制概念说明(30)1.光电导效应2.光谱响应率D转移效率4.等效噪声功率5.相干探测原理说明(60)1.介绍光电检测的5种基本形式2.说明雪崩二极管的工作原理3.说明CCD器件的工作原理4.说明激光器的工作原理5.说明光电倍增管的工作原理6.为什么光外差探测法有利于探测微弱光信号设计题1.设计一套测量材料透过率的光电测试自动装置,要求消除光源的不稳定性的因素的影响,说明工作原理画出原理框图2.设计一台有合作目标的光电测距装置,说明工作原理画出原理框图06年填空1.光电检测的核心()A处理电路B光电管C透镜D光电传感器2.光的能量()A与光的频率成反比B与光的波长成反比C与光的频率无关D 与光的波长无关3.可用作干涉光源的有()Ahe-ne激光B发光二级管C钠灯D卤钨灯4.内光电效应的光电探测器()A光电池B光电管C光电倍增管DLED5.具有光电倍增的器件()A光电管B光电二极管C雪崩二极管D光敏电阻6.响应频率较高的探测器()A雪崩二极管B光敏电阻C硒光电池D热释电器件7.可消除光源的影响,采用的方法有()A单光路直读法B双光路差动法C指零法D双光路补偿法8.热释电探器的探测是()A光强度B温度的变化C电压D温度的大小9.光外差法的可检测()A光的相位B光的强度C光电流D光电压10.光电探测器的主要作用是()A光信号的放大B光信号转换成电信号C电信号转换为光信号D光信号的调制概念1.光伏效应2.光电导效应3.调制4.光学多普勒效应5.萨那克效应原理说明1.说明光电倍增管的工作原理2.为什么光外差探测法有利于探测微弱光信号3.说明热释电探测器的工作原理4.说明莫尔条纹检测的工作原理5.说明但光路指零法的工作原理6.举例说明为什么光学干涉仪可以看做是光载波的光学调制和解器的结合,从信号调制的角度看,光学信息可做怎样的分类(求画图说明)设计题要求(1)绘出工作原理图(2)绘出原理框图(3)说明工作原理1.设计一套测量材料透过率的光电测试自。

南京理工大学11年复试卷子

南京理工大学11年复试卷子

说明:以下我所说的参考教材分别为:《光电检测技术》雷玉堂中国计量出版社简称《雷》注意:是第一版,不是第二版一.概念说明。

(5题,每题4分,共20分)1.居里温度答案见《雷》P862.光谱响应答案见《雷》P383.噪声等效功率答案见《雷》P414.负电子亲和势把半导体材料表面作特殊处理,使表面区域能带弯曲,真空能级下降到导带之下,从而是有效的电子亲和势为负值的这种现象就叫负电子亲和势。

补充电子亲和势:导带上的电子向真空逸出时所需的最低能量,数值上等于真空能级(真空中静止电子的能量)与导带能级只差5.丹培效应答案见《王》P23二.原理说明。

(5题,每题6分,共30分)1.弱辐射探测情况下,光电导灵敏度有何特点,把光敏电阻制成蛇形状有何作用。

答案见《王》P302.为什么光外差探测法有利于探测微弱光信号(即光外差探测法有哪些优点)?(1)探测能力强(2)灵敏度高(3)信噪比高(4)有空间滤波能力(5)有光谱滤波能力(6)稳定性和可靠性高以上每一点我就不具体论述了3.说明热释电器件的工作原理答案见《王》P1084.说明CCD的驱动脉冲频率的上下限受哪些条件限制答案见《王》P2125.光电信息的几种调制方法振幅调制(AM),频率调制(FM),相位调制(PM),偏振调制(POM),光波谱调制(SM)三.分析题和设计题。

(5题,每题10分,共50分)1.通过激光束扫描不透明物体来测量物体轮廓尺寸的激光扫描法原理如图所示,分析工作原理。

(图形和答案见《王》P167)2.光电耦合器件电荷检测如图所示,分析工作原理。

(图形和答案见《雷》P123图5-21(a))3.路灯自动点熄控制电路如图所示,分析工作原理(图形没找到,但资料上面有)答案资料上面有4.叙述3种利用光电信息技术测量速度的方法答案见《王》P177,书上只有两种:光学多普勒差频检测和双频激光干涉仪,还有一种是光电开光测转速的。

5.为了获得最大的光能利用效率,设计太阳能电池板自动跟踪太阳的光电检测系统。

【复试资料】南理工2011年《控制理论基础》硕士研究生复试笔试试卷

【复试资料】南理工2011年《控制理论基础》硕士研究生复试笔试试卷

南京理工大学2011年硕士学位研究生入学考试复试试题考试科目:控制理论基础考试时间:4月10日19:30—21:30一.(15分)已知某状态方程x(t)=[300130001--] x(t)+[011] y(t)=[110]x(t)1).求系统的可控性,可观性。

(不可控,不可观)2).求系统的传递函数。

(31+s ) 3).{-2,-2,-2}和{-1,-2,-3}哪个可以通过状态反馈实现,无需计算状态反馈。

二.(10分)某闭环离散采样系统如图所示,其中ZOH 为零阶持器,Gp(S)为连续对象的传递函数,已知采样周期T=1,Gp(S)=11).试确定闭环传递函数。

(264.0+z ) 2).当r (t )=1(t )时,试确定系统的终值。

(0.5)(参考2010年第8题)三.(10分)非线性系统结构如图所示,N(A)=15++A A ,G(S)=)2)(1(++s s s k 1).求系统稳定,不稳定,临界稳定时K 的取值范围。

(0<K<1.2,稳定1.2<K<6,临界稳定6<K ,不稳定)2).求系统临界稳定时的振幅和角频率。

(A=kk --615,w =√2) 四.(15分)此题为实验设计题,题目较长。

大概是设计实验求出一阶惯性系统的参数 G(S)=12+S T TI 五.(20分)微机填空题,每空一分。

(参考教材第四,第六,第九章)六.(15分)微机存储芯片扩展从1800h 开始连续分配4KB ROM(低地址),4KB RAM(高地址))所用芯片为2K*8 EPROM2716,1K*4 RAM2114。

完成硬件设计,并写出芯片地址范围。

(参考教材100页)七.(15分)一个由0809和8259和8088组成的系统。

(图与教材235页图9-22类似)完成从0809INT1通道连续采集20个数据程序的设计。

写出8259初始化程序,完成主程序的设计(包括中断向量初始化,主程序和中断服务子程序)。

南理工历年上机试题

南理工历年上机试题

2010上机题1. 输入两个字符串s和t,打印出t在s中的出现的位置,并输出出现的次数如输入:s=qwqwertqweqqqq,t=qw输出:1 3 532. 观察一个数的序列规律,1,2,3,4,5,6,8,9,10,12,15,16,18……现在输入一个数8(即上述序列中的第8个数),输出:9试编程实现。

注:上述序列皆是1,2,3,5,的倍数3.求二进制逆序如输入10,对应二进制数为1010,其逆序为0101,对应十进制输出为5(5即为10的逆序数)。

4.求满足条件n*n-m*m-n*m=1,且使m*m+n*n值最大的m,n,其中m,n的值都要小于k,k由用户输入。

5.已知一个数组如a【10】={1,2,3,4,5,6,7,8,9,0},输入p,则右移p位,如输入3后,输出a【10】={8,9,0,1,2,3,4,5,6,7},要求时间复杂度和空间复杂度尽量小。

6.已知一个斜三角:22 32 14 77 4512 34 37 2344 23 1534 5488从最左上角元素开始往右或往右下走,请问顺着哪条路所经过的值的总和最大,如可以有路线:22,32,34,23,54 ;22,12,44,34,88等等请求出满足值总和最大的那条线路。

2010上机原题答案毕竟回忆的,个别地方会有一点点的出入第六题没来得及做,第一题有点问题,大家自己调一下。

第一题#include<iostream.h>#include<iomanip.h>#include<math.h>#include<string.h>void main(){char s[70],t[100];int i=0,j,count=0,len;cout<<"输入两字符串:"<<endl; cin>>s;cout<<"输入第2个字符串:"<<endl; cin>>t;len=strlen(t);while(s!='\0'){for(j=0;j<=len;j++){if(s[i+j]==t[j])j++;else break;}if(j==len+1){count++;cout<<i+1;}i++;}cout<<endl<<count<<endl; }第二题#include<iostream.h>#include<iomanip.h>#include<math.h>int fun(int n){while(n%2==0)n=n/2;while(n%3==0)n=n/3;while(n%5==0)n=n/5;if(n==1)return 1;else return 0;}void main(){int j=0,n,a[1500];long i;for(i=1;i<1000000;i++) { if(fun(i))a[j++]=i;}cout<<"输入n(n<=1500):"<<endl; cin>>n;cout<<a[n-1]<<endl;}第三题#include<iostream.h>#include<iomanip.h>#include<math.h>void main(){int i=0,m,j=0,n=0,a[100];cout<<"输入一个十进制数:"<<endl; cin>>m;while(m!=0){a=m%2;m=m/2;i++;}while(j<i){n=n*2+a[j];j++;}cout<<"其二进制逆序数:"<<endl<<n<<endl; }第四题#include<iostream.h>#include<iomanip.h>#include<math.h>void main(){int m,n,k,maxm,maxn,max=0;cout<<"输入k:"<<endl;cin>>k;for(m=0;m<=k;m++)for(n=0;n<=k;n++)if(n*n-m*n-m*m==1&&max<m*m+n*n){ max=m*m+n*n;maxm=m;maxn=n;}cout<<maxm<<""<<maxn<<endl;}第五题#include<iostream.h>#include<iomanip.h>#include<math.h>void main(){int p,i,j,r[10]={1,2,3,4,5,6,7,8,9,10},s[10]; cin>>p;for(i=0;i<=p-1;i++)s=r;for(j=p;j<=10;j++)r[j-p]=r[j];for(j=0;j<p;j++)r[10-p+j]=s[j];cout<<"左移后的结果:"<<endl;for(i=0;i<10;i++)cout<<r<<setw(5);cout<<endl;}2007年A.1 递归方程编程。

2011年南京理工大学854社会调查方法考研真题(圣才出品)

2011年南京理工大学854社会调查方法考研真题(圣才出品)

2011年南京理工大学854社会调查方法考研真题
南京理工大学
2011年硕士学位研究生入学考试试题
科目代码:854
科目名称:社会调查方法
满分150分
一、名词解释(每题5分,若40分)
1.抽样
2.解释性研究
3.纵向研究
4.实验组与控制组
5.实地研究
6.样本规模
7.无结构访谈
8.简化论
二、问答题(每题l0分,共50分)
1.社会研究的般过程有哪些阶段?
2.自填问卷法有哪些优点和缺点?
3.举例说明四种测量尺度的特点。

4.如何进行分层抽样’
5.谈谈你做社会调查的体会。

三、计算题(每题l0分,共20分)
1.调查了l7位青年人一年来阅读小说书的数目,袁中为调查结果,试求样本的平均值。

2.一项调查得到下列结果,某地人均月收入为920元,标准差为l70元,人均住房面积7.5平方米,标准差为1.8平方米。

试比较该市人均收入和人均住房情况哪一个差异程度比较大。

四、应用题(每题20分,共40分)
1.试对“生育意愿”制作个李克特量表。

2.试就“大学生的就业制度”设计四个调查题。

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这是2011年复试上机自己整理的答案,个别题目还没理解什么意思,贴出来大家交流。也供2012的学弟学妹们参考,有问题大家指教~
/*
第一题
一个数平方之后 逆序与原来相等,求这个数
*/
#include <iostream.h>
#include <iomanip.h>
#include <math.h>
此题为原题
*/
#include <iostream.h>
#include <iomanip.h>
#include <math.h>
#include <string.h>
void main()
{
for(int a=0;a<9;a++)
for(int b=0;b<9;b++)
cin>>a[i][j];
}
}
for( i=0;i<m;i++)
{
for(int j=0;j<m;j++)
cout<<setw(5)<<a[i][j];
cout<<endl;
}
int *b=new int[2*m+2];
int k=0;
如 4是非素数 有质数因子2 则4=2+2
22是非素数 有质数因子2 11 则2+2=2+1+1
*/
#include <iostream.h>
#include <iomanip.h>
#include <math.h>
#include <string.h>
bool isprim(int x)
cout<<"The string s1 is:"<<s1<<endl;
cout<<"The string s2 is:"<<s2<<endl;
cout<<"The result is:";
cout<<rs;
cout<<endl;
}
/*
第四题
三位数abc+bcc=552 求abc 都是多少
a[i]=n;
}
}
void fun3(int *a,int n)
{
for(int i=0;i<M+counter;i++)
{
if(a[i]==n)
{
for(int j=i;j<M+counter;j++)
a[j]=a[j+1];
counter--;
for(i=0;i<m;i++)
{
sum+=a[i][m-i-1];
}
b[k++]=sum;
int n=2*m+2;
sort(b,b+n,compare);
for(i=0;i<n;i++)
{
cout<<b[i]<<" ";
}
cout<<endl;
{
//这个地方 我马虎了 应该是降序,我排出来的结果是升序
if(a>b)
return true;
return false;
}
void main()
{
int m,i;
int sum=0;
cout<<"input m:";
cin>>m;
int **a= (int **)new int[m];
for(int c=0;c<9;c++)
{
int m=a*100+b*10+c;
int n=b*100+c*10+c;
if(m+n==532)
cout<<setw(5)<<"a:"<<a<<setw(5)<<"b:"<<b<<setw(5)<<"c:"<<c<<endl;
for(i=0;i<m;i++)
{
int sum=0;
for(int j=0;j<m;j++)
{
sum+=a[i][j];
}
b[k++]=sum;
}
for(i=0;i<m;i++)
{
int sum=0;
}
}
}
void print(int *a)
{
for(int i=0;i<M+counter;i++)
{
cout<<a[i]<<" ";
}
cout<<endl;
}
void main()
{
int a[M+10]={12,34,2,15,26,8,40};
*/
#include <iostream.h>
#include <iomanip.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define M 20
bool compare(int a,int b)
for(i=0;i<m;i++)
a[i]=new int [m];
cout<<"input the number:"<<endl;
for(i=0;i<m;i++)
{
for(int j=0;j<m;j++)
{
cout<<i<<","<<j<<":";
#include <string.h>
bool fun(int x)
{
int sum=0;
int temp=x;
while(x!=0)
{
sum=sum*10+x%10;
x=x/10;
}
if(temp==sum)
return true;
else
for(int j=0;j<m;j++)
{
sum+=a[j][i];
}
b[k++]=sum;
}
sum=0;
for(i=0;i<m;i++)
{
sum+=a[i][i];
}
b[k++]=sum;
sum=0;
int counter=0;
void fun1(int *a,int x)
{
a[M+counter]=x;
counter++;
}
void fun2(int *a,int m,int n)
{
for(int i=0;i<M+counter;i++)
{
if(a[i]==m)
}
}
/*
第五题
一个数组 能够 1、在末尾添加 2、能够替换 3、能够所有删除指定的元素
*/
#include <iostream.h>
#include <iomanip.h>
#include <math.h>
#include <string.h>
# define M 7
int weishu(int x)
{
int sum=0;
while(x!=0)
{
sum+=x%10;
x=x/10;
}
return sum;
}
bool fun(int x)
{
intபைடு நூலகம்sum=0;
int a[100];
int k=0;
char c;
int m,n;
print(a);
while(1)
{
cout<<"输入字符(a或者c或者d,其他字符退出):";
cin>>c;
switch(c)
{
case 'a':
cout<<"输入一个数字:";
cin>>m;
return false;
}
void main()
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