Xxx大学《考研化工专业英语》
（闭卷 120 分钟）
学号姓名专业年级重修标记□
I.阅读理解（Reading and Understanding )
(i) This might be achieved in a number of ways, for example, improving the profitability(利润) of the process, increasing the capacity (产量) by introduction a new catalyst, or lowering the energy requirement of the process.
1. Which one is not mentioned in above section.( )
(A) increasing the profitalbility
(B) improving the capacity
(C) using large amount of catalyst
(D) lowing the energy requirement
(ii) This often means that there will be close liaison(联系) between the chemical companies’technical sales representatives(代表) and the customer, and the level of technical support for the customer can be a major factor in winning sales.
2. We can conclude that ( ) from the above section.
(A) there are close liaison between the products and customer.
(B) a major factor for winning sales is the level of technical support for customer
(C) a major factor in winning sales is the close liaison between the chemical companies’
technical sales representatives and the customer.
(D) the major factors in winning sales are the the close liaison between the chemical companies’
technical sales representatives and the customer and the level of technical support for the customer.
(iii) A major advance in polymer chemistry was provided by the work of Karl Ziegler and Giulio Natta, which led in 1955 to the introduction of some revolutionary (革命性的) catalysts which bear their name. During the late 1970s production of linear (线性) low-density polyethylene (聚乙烯) was commercialized.
3. The introduction of some revolutionary catalysts is due to the in polymer
chemistry.
4. This paragraph implies that（）
(A) During the late 1970s, the linear low-density polythylene are widely produced.
(B) In 1955, the linear polyethylene was commercialized.
(C) Some revolutionary catalysts were introduced during the late 1970s..
(D) The linear low-density polyethylene bears the name of Karl Ziegler and Giulio Natta. (iv) The basic components (构成) of a typical chemical process are shown in Fig. 1, in which each
一、阅读与理解（每小题2分，共20分）
二、翻译一——英译汉（每小题6分，共30分）
三、翻译二——汉译英（每小题6分，共30分）
四、翻译三——摘要（20 分）
block (方框) represents (代表) a stage in the overall (全部) process for producing a product from the raw materials. Fig. 1 represents a generalized (一般的) process; not all stages will be needed for any particular process and the complexity (复杂程度) of each stage will depend on the nature of the process.
5. This paragraph implies that ( )
(A）Figure 1 represents particular process.
(B）each block in Fig.1 represents a stage of particular process.
(C）the complexity of each stage will depend on the cost.
(D）not all stages in Fig.1 are included in every chemical process.
6. Fig 1 represents a process for producing a product from the. (v) There are over three million known compounds, with no end(没有尽头) in sight as to the number that can and will be prepared(制备,准备) in the future. Each compound is unique(唯一的) and has characteristic(特有的) physical and chemical properties(性质).
Let us consider in some detail two compounds—water and mercuric(汞,水银) oxide.
Water is a colorless, odorless(无嗅), tasteless (无味) liquid that can be changed to a solid, ice at 0℃and to a gas, steam(蒸汽) at 100℃. It is composed of two atoms of hydrogen(氢) and one atom of oxygen per molecule (分子), which represents(表现) 11.2 percent hydrogen and
88.8 percent oxygen by mass (质量). Water reacts chemically with sodium (钠) to produce
hydrogen gas and sodium hydroxide(氢氧化物), with lime (石灰) to produce calcium (钙) hydroxide, and with sulfur(硫) trioxide(三氧化物) to produce sulfuric(硫的,硫磺的) acid(酸).
No other compound has all these exact(精确的) physical and chemical properties; they are characteristic of water alone.
Mercuric oxide is a dense(致密的), orange-red powder composed of a ratio of one atom of mercury to one atom of oxygen. Its composition(组成) by mass is 92.6 precent mercury and
7.4 percent oxygen. When it is heated to temperatures greater than 360℃, a colorless gas,
oxygen, and a silvery(银子似的；有银色光泽的) liquid metal, mercury, are produced. Here again are specific(明确的,特定的) physical and chemical properties belonging to mercuric oxide and to no other substance. Thus, a compound may be identified(区别,识别,鉴定) and distinguished(区分) from all other compounds by its characteristic properties.
7. Which one is not correct as to the above passage. ( )
(A）There are more than 3 million compounds which are known at present.
(B）The number of compounds which can be prepared in the future could be calculated.
(C）Every compound is special and unique.
(D）There are characteristic physical and chemical properties for each compound.
8. There are two compounds are detailed introduced in this passage, they are
and , respectively.
9. This passage suggested that ( )
(A) Water is liquid, but can be changed into solid and gas under cetain conditions.
(B) Water is composed of one atom of hydrogen and one atom of oxygen per molecule.
(C) Water could react with sodium or lime, but could not react with sulfur trioxide.
(D) No other compound has exact physical and chemical properties expect water and mercuric
oxide.
10. Mercuric oxide has the following properties expect ( )
(A）dense (B）orange-red color
(C）turn to liquid at more than 360o C
(D）the liquid mercury are much darker in color than mercuric oxide
II.翻译一（Translating English to Chinese，英译汉)
1. Next reaction (反应) was designed based on the obtained experimental (实验的) data.
2. The explosive growth in petrochemistry (石油化学) was largely due to the enormous increase
of synthetic polymers (合成聚合物).
3. Because of the diversity (多样性) of operations (操作) and close links (联系) in many areas, there is no simple definition (定义) of the chemical industry.
4. The purpose of this meeting is to provide information and knowledge in order to reduce uncertainty (不确定性), solve problems and provide better data.
5. The basic components (构成) of a typical chemical process are shown in Fig. 1, in which each block (方框) represents (代表) a stage in the overall (全部) process for producing a product from the raw materials.
III.翻译二（Translating Chinese to English，汉译英)
1. 值得注意的是这些化合物(compounds)的大多数都是无机物(inorganic materials)。
2. 虽然催化剂(catalyst)A和B之间的区别(distinction)不明显，但我们仍尝试区分(separate)
它们。
3. 截止到战争结束，合成聚合物(synthetic polymer)的便利(facility)已经在三个国家广泛
(widely)推广(expend)。
4. 纳米工程技术（Nano-Engineering Technology）能制造出具有独特性质（property）的物质，
这将促进可再生（renewable）能源的开发。5. 工厂的开工率（the plant attainment），即工厂运行全年中的实际运行时数（available hours）
的百分数通常达到90％。
IV.翻译三(请写出下列英文摘要的相应中文翻译)
Ni-MH novel secondary battery（镍氢新型二次电池）has been the spotlight（焦点）in the research of battery. In this paper,nano-nickel hydroxide（纳米级氢氧化镍）was prepared through sedimentation transform（沉淀转化）method, and the reactive conditions were obtained by controlling the content of surfactant（表面活性剂）and experiment temperature. The structure of the sample was studied by XRD, TEM, meanwhile, the sample was characterized（测试）by electrochemical testing. As a result, it is showed that the prepared sample was spherical（球形的）, the particle was nanosized, the diameter was 30~50nm, and the crystalline was β-type. Compared with the typical nickel hydroxide, the electrochemical capacity（容量）could be improved by more than 20%.
一、单项选择题（在每小题的四个被选答案中，选出一个正确的答案，并将其 答案按顺序写在答题纸上，每小题2分，共40分） 1． 闭环控制系统的特点是 A 不必利用输出的反馈信息 B 利用输入与输出之间的偏差对系统进行控制 C 不一定有反馈回路 D 任何时刻输入与输出之间偏差总是零，因此不是用偏差来控制的 2．线性系统与非线性系统的根本区别在于 A 线性系统有外加输入，非线性系统无外加输入 B 线性系统无外加输入，非线性系统有外加输入 C 线性系统满足迭加原理，非线性系统不满足迭加原理 D 线性系统不满足迭加原理，非线性系统满足迭加原理 3． 2 22 )]([b s b s t f L ++=，则)(t f A bt b bt cos sin + B bt bt b cos sin + C bt bt cos sin + D bt b bt b cos sin + 4．已知 ) (1 )(a s s s F += ，且0>a ，则 )(∞f A 0 B a 21 C a 1 D 1 5．已知函数)(t f 如右图所示，则 )(s F A s s e s e s --+2211 B s s e s s 213 212+-- C )22121(1332s s s s se e e se s ------+ D )221(1s s s e e s e s ----+ 6．某系统的传递函数为 ) 3)(10() 10()(+++= s s s s G ，其零、极点是 A 零点 10-=s ，3-=s ；极点 10-=s B 零点 10=s ，3=s ；极点 10=s
C 零点 10-=s ；极点 10-=s ，3-=s D 没有零点；极点 3 =s
清华大学本科生考试试题专用纸 考试课程 控制工程基础 （卷） 年 月 日 . 设有一个系统如图所示，, , ()，当系统受到输入信号t t x i sin 5)(= 的作用时，试求系统的稳态输出)(t x o 。(分 ) i x o x K K D 图 . 设一单位反馈系统的开环传递函数为 ) 11.0(100 )(+= s s s G 现有三种串联校正装置，均为最小相位的，它们的对数幅频特性渐近线如图所示。 若要使系统的稳态误差不变，而减小超调量，加快系统的动态响应速度，应选取哪种校正装置？系统的相角裕量约增加多少？(分) ) L (w ) (d B ) () ) L (w ) (d B ) ) L (w ) (d B ) () ()
图 . 对任意二阶环节进行校正，如图，如果使用控制器， , 均为实数，是否可以实现闭环极点的任意配置？试证明之。(分) 图 . 一个未知传递函数的被控系统，先未经校正，构成单位反馈闭环。经过测试，得知闭环系统的单位阶跃响应如图所示。 问：() 系统的开环低频增益是多少？(分) () 如果用主导极点的概念用低阶系统近似该系统，试写出其近似闭环传递函数；(分) （）如果采用形式的串联校正()I c 1 K G s s =+，在什么范围内时，对原开环系统 相位裕量的改变约在 5.7~0-??之间？(分) 17/8 图 ．已知计算机控制系统如图所示，采用数字比例控制()D z K =，其中>。设采样周期 (i X s ) z 图 （）试求系统的闭环脉冲传递函数() ()() o c i X z G z X z =； (分) （）试判断系统稳定的值范围； (分) （）当系统干扰()1()n t t =时，试求系统由干扰引起的稳态误差。 (分)
1990 年英译汉试题 People have wondered for a long time how their personalities，and behaviors are formed. It is not easy to explain why one person is intelligent and another is not，or why one is cooperative and another is competitive. Social scientists are，of course，extremely interested in these types of questions. (61)They want to explain why we possess certain characteristics and exhibit certain behaviors. There are no clear answers yet，but two distinct schools of thought on the matter have developed. As one might expect，the two approaches are very different from each other. The controversy is often conveniently referred to as‖nature vs. nurture‖. (62)Those who support the ―nature‖side of the conflict believe that our personalities and behavior patterns are largely determined by biological factors. (63)That our environment has little， if anything，to do with our abilities，characteristics and behavior is central to this theory. Taken to an extreme，this theory maintains that our behavior is predetermined to such a great degree that we are almost completely governed by our instincts. Those who support the ―nurture‖ theory，that is，they advocate education，are often called behaviorists. They claim that our environment is more important than our biologically based instincts in determining how we will act. A behaviorist，B.F. Skinner，sees humans as beings whose behavior is almost completely shaped by their surroundings. The behaviorists maintain that，like machines，humans respond to environmental stimuli as the basis of their behavior. Let us examine the different explanations about one human characteristic，intelligence， offered by the two theories. Supporters of the ―nature‖theory insist that we are born with a certain capacity for learning that is biologically determined. Needless to say，they don‘t believe that factors in the environment have much influence on what is basically a predetermined characteristic. On the other hand，behaviorists argue that our intelligence levels are the product of our experiences. (64)Behaviorists suggest that the child who is raised in an environment where there are many stimuli which develop his or her capacity for appropriate responses will experience greater intellectual development. The social and political implications of these two theories are profound. In the United States， blacks often score below whites on standardized intelligence tests. This leads some ―nature‖ proponents to conclude that blacks are biologically inferior to whites. (65)Behaviorists，in contrast， say that differences in scores are due to the fact that blacks are often deprived of many of the educational and other environmental advantages that whites enjoy. Most people think neither of these theories can yet fully explain human behavior. 1991 年英译汉试题 The fact is that the energy crisis，which has suddenly been officially announced，has been with us for a long time now，and will be with us for an even longer time. Whether Arab oil flows freely or not，it is clear to everyone that world industry cannot be allowed to depend on so fragile a base. (71)The supply of oil can be shut off unexpectedly at any time，and in any case，the oil wells will all run dry in thirty years or so at the present rate of use. (72)New sources of energy must be found，and this will take time，but it is not likely to result in any situation that will ever restore that sense of cheap and plentiful energy we have had in the times past. For an indefinite period from here on，mankind is going to advance cautiously，and consider itself lucky that it can advance at all. To make the situation worse，there is as yet no sign that any slowing of the world‘s population is in sight. Although the birthrate has dropped in some nations，including the United States，the population of the world seems sure to pass six billion and perhaps even seven billion as the twenty-first century opens. (73)The food supply will not increase nearly enough to match this，which means that we are heading into a crisis in the matter of producing and marketing food. Taking all this into account，what might we reasonably estimate supermarkets to be like in the year2001? To begin with，the world food supply is going to become steadily tighter over the next thirty years—even here in the United States.By2001，the population of the United States will be at least two hundred fifty million and possibly two hundred seventy million，and the nation will find it difficult to expand food production to fill the additional mouths. (74)This will be particularly true since energy pinch will make it difficult to continue agriculture in the high energy American fashion that makes it possible to combine few farmers with high yields. It seems almost certain that by2001the United States will no longer be a great food exporting nation and that，if necessity forces exports，it will be at the price of belt tightening at home. In fact，as food items will end to decline in quality and decrease in variety，there is very likely to be increasing use of flavouring additives. (75)Until such time as mankind has the sense to lower its population
2015年武汉理工大学《控制工程基础》模拟题1 1、 选择填空（30分，每小题2分） （下列各题均给出数个答案，但只有一个是正确的，请将正确答案的序号写在空白 处） 1.1在下列典型环节中，属于振荡环节的是 。 (A) 101.010)(2++= s s s G (B) 1 01.01)(2 ++=s s s G (C) 101 )(+=s s G 1.2系统的传递函数定义为在零初始条件下输出量的Laplace 变换与输入量的Laplace 变换之比，其表达式 。 （A ）与输入量和输出量二者有关 （B ）不仅与输入量和输出量二者有关，还与系统的结构和参数有关 （C ）只与系统的结构和参数有关，与输入量和输出量二者无关 1.3系统峰值时间p t 满足 。 （A ） 0)(=p p o dt t dx （B ）)()(∞=o p o x t x （C ）)()()(∞??≤∞-o o p o x x t x 其中，)(t x o 为系统的单位阶跃响应。 1.4开环传递函数为G (s )的单位反馈系统的静态速度误差系数的计算式为 。 (A) )(lim 0 s G K s v →= (B) )(lim 2 s G s K s v →= (C) )(lim 0 s sG K s v →= 1.5最大百分比超调量(%)p M 的定义式为 。 （A ）)()(max (%)∞-=o o p x t x M (B) %100) () ()(max (%)∞∞-= o o o p x x t x M （C ）) () (max (%)t x t x M i o p = 其中，)(t x i 为系统的输入量，)(t x o 为系统的单位阶跃响应，)(max t x o 为)(t x o 的最大值。 1.6给同一系统分别输入)sin()(11t R t x i ω=和)sin()(2t R t x r i ω=这两种信号（其中， r ω是系统的谐振频率，1ω是系统正常工作频率范围内的任一频率），设它们对应的稳态输出分别为)sin()(1111?ω+=t C t x o 和)sin()(222?ω+=t C t x r o ，则 成立。 （A ）21C C > （B ）12C C > （C ）21C C = 1.7 若一单位反馈系统的开环传递函数为) ()(1220 a s a s a s G += , 则由系统稳定的必 要条件可知， 。 （A ）系统稳定的充分必要条件是常数210,,a a a 均大于0
桂 林 电 子 科 技 大 学 试 卷 2013－2014 学年第二学期 课程名称《控制工程基础》（A 卷.闭卷）适用年级或专业） 一、填空题（每题1分，共15分） 1、对自动控制系统的基本要求可以概括为三个方面，即：、快速性和 。 2、自动控制系统有两种基本控制方式，当控制装置与受控对象之间只有顺向作用而无反向联系时，称为；当控制装置与受控对象之间不但有顺向作用而且还有反向联系时，称为。含有测速发电机的电动机速度控制系统，属于。 3、控制系统的 称为传递函数。一阶系统传函标准形式是，二阶系统传函标准形式是。 4、两个传递函数分别为G 1(s)与G 2(s)的环节，以并联方式连接，其等效传递函数为()G s ，则G(s)为（用G 1(s)与G 2(s)表示）。 5、奈奎斯特稳定判据中，Z = P - R ，其中P 是指 ，Z 是指 ，R 指 。 6、若某系统的单位脉冲响应为 0.20.5()105t t g t e e --=+， 则该系统的传递函数G(s)为。 7、设系统的开环传递函数为 2 (1) (1) K s s Ts τ++，则其开环幅频特性为 ，相频特性为 。 二、选择题（每题2分，共20分） 1、关于传递函数，错误的说法是 ( ) s 的真分D.闭环传递函数的极点决定了系统的稳定性。 2、采用负反馈形式连接后，则 ( ) A 、一定能使闭环系统稳定；B 、系统动态性能一定会提高； C 、一定能使干扰引起的误差逐渐减小，最后完全消除； D 、需要调整系统的结构参数，才能改善系统性能。 3、已知系统的开环传递函数为50 (21)(5)s s ++，则该系统的开环增益为 ( )。 A 、50 B 、25 C 、10 D 、5 4、下列哪种措施对提高系统的稳定性没有效果 ( )。 A 、增加开环极点； B 、在积分环节外加单位负反馈； C 、增加开环零点； D 、引入串联超前校正装置。
一. 填空题(每小题2.5分，共25分) 1． 对控制系统的基本要求一般可以归纳为稳定性、 快速性 和 准确性 。 2． 按系统有无反馈，通常可将控制系统分为 开环系统 和 闭环系统 。 3． 在控制工程基础课程中描述系统的数学模型有 微分方程 、 传递函数 等。 4． 误差响应 反映出稳态响应偏离系统希望值的程度，它用来衡量系统 控制精度的程度。 5． 一阶系统 1 1 Ts 的单位阶跃响应的表达是 。 6． 有系统的性能指标按照其类型分为时域性能指标和 频域性能指标 。 7． 频率响应是线性定常系统对 谐波 输入的稳态响应。 8． 稳态误差不仅取决于系统自身的结构参数，而且与 的类型有关。 9． 脉冲信号可以用来反映系统的 。 10. 阶跃信号的拉氏变换是 。 二． 图1为利用加热器控制炉温的反馈系统（10分） 电压放大 功率放大 可逆电机 + -自偶调压器~220V U f +给定毫 伏信号 + -电炉热电偶加热器 U e U g 炉温控制系统 减速器 - 图1 炉温控制结构图 试求系统的输出量、输入量、被控对象和系统各部分的组成，且画出原理方框图，说明其工作原理。 三、如图2为电路。求输入电压i u 与输出电压0u 之间的微分方程， 并求该电路的传递函数（10分） 图2 R u 0 u i L C u 0 u i C u 0 u i R (a) (b) (c)
四、求拉氏变换与反变换(10分) 1．求[0.5]t te -（5分） 2．求1 3 [] (1)(2) s s s - ++ （5分） 五、化简图3所示的框图，并求出闭环传递函数（10分）
历年考研英语翻译真题及答案解析38 英译汉：考查考生理解所给英语语言材料并将其译成汉语的能力。要求译文准确、完整、通顺。 2014年考研英语二翻译真题及答案解析 Directions: Translate the following text from English into Chinese. Write your translation on ANSWER SHEET 2. (15 points) Most people would define optimism as endlessly happ y, with a glass that’s perpetually half fall. But that’s exactly the kind of false deerfulness that positive psychologists wouldn’t recommend. “Healthy optimists means being in touch with reality.” says Tal Ben-Shahar, a Harvard professor, According to Ben- Shalar,realistic optimists are these who make the best of things that happen, but not those who believe everything happens for the best. Ben-Shalar uses three optimistic exercisers. When he feels down-sag, after giving a bad lecture-he grants himself permission to be human. He reminds himself that mot every lecture can be a Nobel winner; some will be less effective than others. Next is reconstruction, He analyzes the weak lecture, leaning lessons, for the future about what works and wh at doesn’t. Final ly, there is perspective, which involves acknowledging that in the ground scheme of life, one lecture really doesn’t matter. 翻译 大多数人愿意把乐观定义为无尽的欢乐，就像一只总是装着半杯水的杯子。但那是一种绝不会为积极心理学家所推荐的虚假快乐。哈佛大学的Tal Ben-Shahar教授说，“健康的乐观主义，意味着要处于现实之中。”在Ben-Shahar看来，现实的乐观主义者，会尽最大努力做好一件事，而不是相信每件事都会有最好的结果。 Ben-Shahar 会进行三种乐观方面的练习。比如说，当他进行了一次糟糕的演讲，感到心情郁闷的时候，他会告诉自己这是人之常情。他会提醒自己：并不是每一
本资料由理硕教育整理，理硕教育是全国唯一专注于北理工考研辅导的学校，相对于其它机构理硕教育有得天独厚的优势。丰富的理工内部资料资源与人力资源确保每个学员都受益匪浅，确保理硕教育的学员初试通过率89%以上，复试通过率接近100%，理硕教育现开设初试专业课VIP一对一，假期集训营，复试VIP一对一辅导，满足学员不同的需求。因为专一所以专业，理硕教育助您圆北理之梦。详情请查阅理硕教育官网 一．基础概念 1.闭环系统：有反馈的系统（对应开环系统，一般的系统都是负反馈系统）。 2.控制系统的分类： （1）按照控制目标分类：恒值（目标值不变），随动（目标值变化且未知），过程（目标值变化且已知） （2）按照输入输出关系分类：线性（满足叠加关系的系统），非线性（不满足叠加关系的系统）（3）按照控制信号的形式分类：连续，离散，概率 （4）按照输入输出个数：经典（单输入单输出），现代（多输入多输出） 3.控制系统的组成： 4.传递函数：零初始条件下，输出与输入的拉氏变换之比。要注意传函一般是真分式，即分子阶数小于分母阶数。传函只与系统结构有关，与输入输出无关。 5.一阶系统：注意增益以及时间常数 6.二阶系统：注意阻尼比与固有频率（表达式，阻尼比小于1的系统叫欠阻尼系统），掌握相关量的计算（上升时间，超调量，调节时间，峰值时间），掌握其对阶跃、斜坡、加速度信号的响应。
7.稳态误差：系统达到稳态时与目标值的差。 8.幅相特性曲线：掌握画法。设传函为G(s)，令s=jw，则G(jw)的模随w的变化规律为幅频特性，G(jw)的相角随w的变化规律为相频特性。 9.奈氏判据：Z=P-2N(Z为传函右极点个数，P是开环传函右极点个数，N为奈氏图包含(-1,0j)的圈数。 10.伯德图：掌握画法，初始斜率-20v（v为积分器个数），找出交接频率，遇一阶系统斜率变化20，遇二阶系统斜率变化40，在分子上为增大，在分母上为减小。 11.稳定裕度与幅值裕度：计算方法。两者均正时系统稳定。 12.PID控制器：结构简单，稳定性好，工作可靠，调整方便。 （1）P：比例控制，控制量与误差量成比例。K增大时稳定性降低，但系统快速性与稳态精度变好。 （2）I：积分控制：控制量与误差量对时间的积分成比例。用来消除稳态误差。 （3）D：微分控制：控制量与误差量对时间的微分成比例。起预估作用，可以避免振荡，但是会使系统抗高频干扰的能力下降。 二．相关计算 1.拉氏变换：掌握阶跃，正余弦，指数，斜坡等常用函数的拉氏变换。 2.拉氏变换相关性质：初值定理，终值定理，延时定理。 3.梅森增益公式：求传函，看懂其中每一项的意义。
常用翻译技巧总结 翻译题里考察三方面内容： 1、专有名词(如operational research expert)、习惯用法(如depend on)及多义词的翻译(如school、set的多义) 2、一般性翻译技巧:包括词义选择，词序调整，词性转换和增词法等等 3、具体句型(定从、状从、主从、宾从、表从、同位从、强调结构、并列、比较、倒装、插入、被动、否定等) 其中2、3是大考点，具体内容可在论坛下XDF的翻译笔记来看，在此不赘述。 可看出，应对翻译题的主要武器是翻译技巧，下面正式进入正题(常用方法、被动语态译法、形容词译法、举例详解) 一、常用方法 二、 英汉两种语言在句法、词汇、修辞等方面均存在着很大的差异，因此在进行英汉互译时必然会遇到很多困难，需要有一定的翻译技巧作指导。常用的翻译技巧有增译法、省译法、转换法、拆句法、合并法、正译法、反译法、倒置法、包孕法、插入法、重组法和综合法等。这些技巧不但可以运用于笔译之中，也可以运用于口译过程中，而且应该用得更加熟练。 1增译法：指根据英汉两种语言不同的思维方式、语言习惯和表达方式，在翻译时增添一些词、短句或句子，以便更准确地表达出原文所包含的意义。这种方式多半用在汉译英里。汉语无主句较多，而英语句子一般都要有主语，所以在翻译汉语无主句的时候，除了少数可用英语无主句、被动语态或“There be…”结构来翻译以外，一般都要根据语境补出主语，使句子完整。英汉两种语言在名词、代词、连词、介词和冠词的使用方法上也存在很大差别。英语中代词使用频率较高，凡说到人的器官和归某人所有的或与某人有关的事物时，必须在前面加上物主代词。因此，在汉译英时需要增补物主代词，而在英译汉时又需要根据情况适当地删减。英语词与词、词组与词组以及句子与句子的逻辑关系一般用连词来表示，而汉语则往往通过上下文和语序来表示这种关系。因此，在汉译英时常常需要增补连词。英语句子离不开介词和冠词。另外，在翻译时还要注意增补一些原文中暗含而没有明言的词语和一些概括性、注释性的词语，以确保译文意思的完整。总之，通过增译，一是保证译文语法结构的完整，二是保证译文意思的明确。如： (1) What about calling him right away? 马上给他打个电话，你觉得如何？(增译主语和谓语) (2) If only I could see the realization of the four modernizations. 要是我能看到四个现代化实现该有多好啊！(增译主句) (3) Indeed, the reverse is true 实际情况恰好相反。(增译名词) (4)就是法西斯国家本国的人民也被剥夺了人权。 Even the people in the fascist countries were stripped of their human rights.(增译物主代词) (5)只许州官放火，不许百姓点灯。 While the magistrates were free to burn down house, the common people were forbidden to light
桂林电子科技大学试卷 2013－2014 学年第二学期 课程名称《控制工程基础》（A卷.闭卷）适用年级或专业） 考试时间 120 分钟班级学号姓名 一、填空题（每题1分，共15分） 1、对自动控制系统的基本要求可以概括为三个方面，即：、快速性和。 2、自动控制系统有两种基本控制方式，当控制装置与受控对象之间只有顺向作用而无反向联系时，称为；当控制装置与受控对象之间不但有顺向作用而且还有反向联系时，称为。含有测速发电机的电动机速度控制系统，属于。 3、控制系统的称为传递函数。一阶系统传函标准形式是，二阶系统传函标准形式是。 4、两个传递函数分别为G1(s)与G2(s)的环节，以并联方式连接，其等效传递函数为() G s，则G(s)为（用G1(s)与G2(s)表示）。 5、奈奎斯特稳定判据中，Z = P - R ，其中P是 指，Z是指，R指。 6、若某系统的单位脉冲响应为0.20.5 ()105 t t g t e e -- =+， 则该系统的传递函数G(s)为。 7、设系统的开环传递函数为 2 (1) (1) K s s Ts τ+ + ，则其开环幅频特性为，相频特性为。 二、选择题（每题2分，共20分） 1、关于传递函数，错误的说法是 ( ) A.传递函数只适用于线性定常系统； B.传递函数不仅取决于系统的结构参数，给定输入和扰动对 传递函数也有影响； C.传递函数一般是为复变量s的真分式； D.闭环传递函数的极点决定了系统的稳定性。 2、采用负反馈形式连接后，则 ( ) A、一定能使闭环系统稳定； B、系统动态性能一定会提高；
C 、一定能使干扰引起的误差逐渐减小，最后完全消除； D 、需要调整系统的结构参数，才能改善系统性能。 3、已知系统的开环传递函数为 50 (21)(5) s s ++，则该系统的开环 增益为 ( )。 A 、 50 B 、25 C 、10 D 、5 4、下列哪种措施对提高系统的稳定性没有效果 ( )。 A 、增加开环极点； B 、在积分环节外加单位负反馈； C 、增加开环零点； D 、引入串联超前校正装置。 5、系统特征方程为0632)(23=+++=s s s s D ，则系统 ( ) A 、稳定； B 、单位阶跃响应曲线为单调指数上升； C 、临界稳定； D 、右半平面闭环极点数2=Z 。 6、下列串联校正装置的传递函数中，能在1c ω=处提供最大相位 超前角的是 ( )。 A 、 1011s s ++ B 、1010.11s s ++ C 、210.51s s ++ D 、0.11101 s s ++ 7、已知开环幅频特性如图1所示， 则图中不稳定的系统是 ( )。 系统① 系统② 系统③ 图1 A 、系统① B 、系统② C 、系统③ D 、都不稳定 8、非单位负反馈系统，其前向通道传递函数为G(S)，反馈通道传递函数为H(S)，当输入信号为R(S)，则从输入端定义的误差E(S)为 ( ) A 、 ()()()E S R S G S =? B 、()()()()E S R S G S H S =?? C 、()()()()E S R S G S H S =?- D 、()()()() E S R S G S H S =- 9、开环频域性能指标中的相角裕度γ对应时域性能指标 ( ) 。 A 、超调%σ B 、稳态误差ss e C 、调整时间s t D 、峰值时间p t 10、已知下列负反馈系统的开环传递函数，应画零度根轨迹的是 ( )。 A 、*(2)(1)K s s s -+ B 、*(1)(5K s s s -+）
2012年考研英语翻译真题及答案 Part C Directions: Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written clearly on ANSWER SHEET 2. (10 points) Since the days of Aristotle, a search for universal principles has characterized the scientific enterprise. In some ways, this quest for commonalities defines science. Newton’s laws of motion and Darwinian evolution each bind a host of different phenomena into a single explicatory frame work. (1)In physics, one approach takes this impulse for unification to its extreme, and seeks a theory of everything—a single generative equation for all we see. It is beco mi ng less clear, however, that such a theory would be a simplification, given the dimensions and universes that it might entail, nonetheless, unification of sorts remains a major goal. This tendency in the natural sciences has long been evident in the social sciences too. (2)Here, Darwinism seems to offer justification for it all humans share common origins it seems reasonable to suppose that cultural diversity could also be traced to more constrained beginnings. Just as the bewildering
一、填空题（每题1分，共15分） 1、对于自动控制系统的性能要 求可以概括为三个方面， 即： 、 和 ，其中最基本的要求是 。 2、若某单位负反馈控制系统的前向传递函数为()G s ，则该系统的 开环传递函数 为 。 3、能表达控制系统各变量之间关系的数学表达式或表示方法，叫系统的数学模型，在古典控制理论中系统数学模型 有 、 等。 4、判断一个闭环线性控制系统是否稳定，可采 用 、 、 等方法。 5、自动控制系统有两种基本控制方式，当控制装置与受控对象之间只有顺向作用而无反向联系时，称 为 ；当控制装置与受控对象之间不但有顺向作用而且还有反向联系时，称为 。 6、设系统的开环传递函数为 12(1)(1) K s T s T s ++，则其开环幅 频特性为 ，相频特性 为 。 7、最小相位系统是 指 。 二、选择题（每题2分，共20分） 1、关于奈氏判据及其辅助函数 F(s)= 1 + G(s)H(s)，错误的说法是 ( ) A 、 F(s)的零点就是开环传递函数的极点 B 、 F(s)的极点就是开环传递函数的极点 C 、 F(s)的零点数与极点数相同 D 、 F(s)的零点就是闭环传递函数的极点 2、已知负反馈系统的开环传递函数 为221 ()6100 s G s s s +=++，则该系统的 闭环特征方程为 ( )。 A 、2 61000s s ++= B 、 2(6100)(21)0s s s ++++= C 、2 610010s s +++= D 、 与是否为单位反馈系统有关 3、一阶系统的闭环极点越靠近S 平面原点，则 ( ) 。 A 、准确度越高 B 、准确度越低 C 、响应速度越快 D 、响应速度越慢 4、已知系统的开环传递函数为 100 (0.11)(5) s s ++，则该系统的开环增 益为 ( )。 A 、 100 B 、1000 C 、20 D 、不能确定 5、若两个系统的根轨迹相同，则有相同的： A 、闭环零点和极点 B 、开环零点 C 、闭环极点 D 、阶跃响应 6、下列串联校正装置的传递函数中，能在1c ω=处提供最大相位超前角的是 ( )。 A 、 1011s s ++ B 、1010.11s s ++ C 、 210.51s s ++ D 、0.11 101 s s ++
控制工程基础期末试题 一、单项选择题(本大题共10小题，每小题2分，，共20分) 1.如果系统中加入一个微分负反馈，将使系统的超调量σp( ) A.增加 B.减小 C.不变 D.不定 2.运算放大器具有_____的优点。( ) A.输入阻抗高，输出阻抗低 B.输入阻抗低，输出阻抗高 C.输入阻抗高，输出阻抗高 D.输入、输出阻抗都低 3.在伯德图中反映系统抗高频干扰能力的是( ) A.低频段 B.中频段 C.高频段 D.无法反映 4.设开环系统频率特性G(jω)= ,当ω=1rad/s时，其频率特性幅值M(1)=( ) A. B.4 C. D.2 5.设开环传递函数G(s)H(s)= ,α>0,K>0，随着K增大，闭环系统 ( ) A.相对稳定性变差，快速性不变 B.相对稳定性变好，快速性不变 C.相对稳定性不变，快速性变好 D.相对稳定性变差，快速性变差 6.对于一阶、二阶系统来说，系统特征方程的系数都是正数是系统稳定的( ) A.充分条件 B.必要条件 C.充分必要条件 D.以上都不是 7.开环传递函数为G(s)H(s)=, 则实轴上的根轨迹为( ) A.〔－4,∞) B.〔－4,0〕 C.(－∞,－4) D. 〔0,∞〕 8.进行串联滞后校正后，校正前的穿越频率ωc与校正后的穿越频率的关系，通常是( ) A.ωc= B.ωc> C.ωc< D.ωc与无关 9.PID控制规律是____控制规律的英文缩写。( ) A.比例与微分 B.比例与积分 C.积分与微分 D.比例、积分与微分 10.比例环节的频率特性相位移θ(ω)=( ) A.90° B.－90° C.0° D.－180° 二、填空题(本大题共10小题，每小空1分，共15分) 1.根轨迹全部在根平面的__________部分时，系统总是稳定的。 2.设系统的频率特性G(jω)=R(ω)+JI(ω)，则相频特性∠G(jω)=__________。 3.随动系统中常用的典型输入信号是__________和__________。 4.超前校正装置的最大超前角处对应的频率ωm=__________。 5.根据系统给定值信号特点，控制系统可分为__________控制系统、__________控制系统和程序控制系统。
考研英语翻译模拟试题及答案解析 (1)Any discussion of the American educational system would be less than complete if it did not mention the emphasis that many colleges and universities place upon the nonacademic, social,“extracurricular”aspect of education, often defined as personal growth. Perhaps a useful way of viewing the notion of personal growth would be to picture the very large and general term“education” as being all-embracing, including as subsets within it academic and nonacademic components. This may be one of the most difficult concepts to convey to someone who is not intimately familiar with American higher education. Few educational systems in other countries place the same emphasis on this blend of academic and personal education. The majority of colleges and universities in the United States make some attempt to integrate personal and intellectual growth in the undergraduate years. (2) If the ultimate goal of undergraduate education in America were simply to convey a set body of knowledge, the term of studies could undoubtedly be reduced. Yet the terms of studies are extended in order to give students a chance to grow and develop in other ways. Numerous opportunities are made available to students to become involved in sports, student government, musical and dramatic organizations, and countless other organized and individual activities designed to enhance one’s personal growth and provide some recreation