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1、Manipulator is now used as a industrial robots in use, the control objectives often appear often in industrial automation. Industrial automation technology has gradually matured, as mature a technology line has been rapid development in industrial automation as a separate subject. Manipulator applic
2、ation began to filter into welding, logistics, mechanical processing, and other industries. Especially at high or very low temperatures, full of poisonous gases, high radiation case, robot in similar circumstances showed great use also brings great convenience to the staff. Precisely because of this
3、 robot to get peoples attention began to be a high degree of development. Labor rates, working conditions, labor intensive aspects of promoting development. Both at home and abroad to develop the PLC (programmable logic controller) is in various special circumstances and under special conditions set
4、 for mechanical devices. Now turned on the development of the microelectronics automatic control technology and the rapid development of the trains, the success of PLC hardware software and simulation control win big and successful development, now continues to develop as a factory automation standa
5、rds. Because robots are good development of the technology makes a good optimization of productive capital, and robot shows this unique advantages, such as: has good compatibility, wide availability, hardware is complete, and programming that can be mastered in a short time, so in the context of ind
6、ustrial PLC applications became ubiquitous. Manipulator in many developed country agriculture and industry has been applied, such as the use of mechanical harvesting large areas of farmland, repeated operations on the high-speed line that uses a robotic arm, and so on. Today, the high level of autom
7、ation combined with restrictions on the manipulator development level is slightly lower than the international. The design is mainly arm welding machine by PLC Automation control. This of design let designers on in school by learn of has a must of consolidation, understand has some usually didnt opp
8、ortunities awareness in world range within some leading level of knowledge has has must awareness, hope designers can in yihou of design in the can success of using in this design in the proceeds of experience 1.2 manipulator in both at home and abroad of research profile automation mechanical arm r
9、esearch began Yu 20th century medium-term, after years with with computer and automation technology of development, Makes mechanical arm on the Grand stage of industrial automation and shine, gradually became an industrial evaluation standards, and its importance can be seen. Now original robotic ar
10、m spent most of mass production and use on the production line, which is programmed robotic arm. As the first generation of manipulator position control systems main features, although not back several generations that can detect the external environment, but can still successfully complete like wel
11、ding, painting, delivery as well as for materials simple movements. Second generation mechanical arms are equipped with sensors and manipulators have the environment there is a certain amount of sense, when the mechanical arm is to use the program as a basis. Difference is that the robot begand学而思教育
12、学习改变命运 思考成就未来！ 高考网2008年普通高等学校招生全国统一考试（山东卷）文科数学第卷（共60分）参考公式：锥体的体积公式：，其中是锥体的底面积，是锥体的高球的表面积公式：，其中是球的半径如果事件互斥，那么一、选择题：本大题共12小题，每小题5分，共60分在每小题给出的四个选项中，只有一项是符合题目要求的1满足，且的集合的个数是（ ）A1B2C3D42设的共轭复数是，若，则等于（ ）ABCD3函数的图象是（ ）yxOyxOyxOyxOABCD4给出命题：若函数是幂函数，则函数的图象不过第四象限在它的逆命题、否命题、逆否命题三个命题中，真命题的个数是（ ）A3B2C1D0俯视图正(主)
13、视图侧(左)视图23225设函数则的值为（ ）ABCD6右图是一个几何体的三视图，根据图中数据，可得该几何体的表面积是（ ）ABCD7不等式的解集是（ ）ABCD8已知为的三个内角的对边，向量若，且，则角的大小分别为（ ）ABCD9从某项综合能力测试中抽取100人的成绩，统计如表，则这100人成绩的标准差为（ ）分数54321人数2010303010ABC3D10已知，则的值是（ ）ABCD11若圆的半径为1，圆心在第一象限，且与直线和轴相切，则该圆的标准方程是（ ）ABCDOyx12已知函数的图象如图所示，则满足的关系是（ ）ABCD第卷（共90分）二、填空题：本大题共4小题，每小题4分，共
14、16分13已知圆以圆与坐标轴的交点分别作为双曲线的一个焦点和顶点，则适合上述条件的双曲线的标准方程为 开始?是输入p结束输出否14执行右边的程序框图，若，则输出的 15已知，则的值等于 16设满足约束条件则的最大值为 三、解答题：本大题共6小题，共74分17（本小题满分12分）已知函数（，）为偶函数，且函数图象的两相邻对称轴间的距离为（）求的值；（）将函数的图象向右平移个单位后，得到函数的图象，求的单调递减区间18（本小题满分12分）现有8名奥运会志愿者，其中志愿者通晓日语，通晓俄语，通晓韩语从中选出通晓日语、俄语和韩语的志愿者各1名，组成一个小组（）求被选中的概率；（）求和不全被选中的概率1
15、9（本小题满分12分）ABCMPD如图，在四棱锥中，平面平面，是等边三角形，已知，（）设是上的一点，证明：平面平面；（）求四棱锥的体积20（本小题满分12分）将数列中的所有项按每一行比上一行多一项的规则排成如下数表： 记表中的第一列数构成的数列为，为数列的前项和，且满足（）证明数列成等差数列，并求数列的通项公式；（）上表中，若从第三行起，第一行中的数按从左到右的顺序均构成等比数列，且公比为同一个正数当时，求上表中第行所有项的和21（本小题满分12分）设函数，已知和为的极值点（）求和的值；（）讨论的单调性；（）设，试比较与的大小22（本小题满分14分）已知曲线所围成的封闭图形的面积为，曲线的内切
16、圆半径为记为以曲线与坐标轴的交点为顶点的椭圆（）求椭圆的标准方程；（）设是过椭圆中心的任意弦，是线段的垂直平分线是上异于椭圆中心的点（1）若（为坐标原点），当点在椭圆上运动时，求点的轨迹方程；（2）若是与椭圆的交点，求的面积的最小值2008年普通高等学校招生全国统一考试（山东卷）文科数学（答案）一、选择题1B解析：本小题主要考查集合子集的概念及交集运算。集合中必含有,则或.选B.2D解析：本小题主要考查共轭复数的概念、复数的运算。可设，由得选D.3A解析：本小题主要考查复合函数的图像识别。是偶函数，可排除B、D，由的值域可以确定.选A.4C解析：本小题主要考查四种命题的真假。易知原命题是真命题
17、,则其逆否命题也是真命题,而逆命题、否命题是假命题.故它的逆命题、否命题、逆否命题三个命题中, 真命题有一个。选C.5A解析：本小题主要考查分段函数问题。正确利用分段函数来进行分段求值。选A.6D 解析：本小题主要考查三视图与几何体的表面积。从三视图可以看出该几何体是由一个球和一个圆柱组合而成的，其表面及为选D。7D解析：本小题主要考查分式不等式的解法。易知排除B;由符合可排除C;由排除A, 故选D。也可用分式不等式的解法,将2移到左边直接求解。8C解析：本小题主要考查解三角形问题。，.选C. 本题在求角B时,也可用验证法.9B解析：本小题主要考查平均数、方差、标准差的概念及其运算。 选B.1
18、0C解析：本小题主要考查三角函数变换与求值。， 选C.11B解析：本小题主要考查圆与直线相切问题。设圆心为由已知得选B.12A 解析：本小题主要考查正确利用对数函数的图象来比较大小。由图易得取特殊点 .选A.二、填空题13解析：本小题主要考查圆、双曲线的性质。圆得圆与坐标轴的交点分别为则所以双曲线的标准方程为14解析：本小题主要考查程序框图。，因此输出152008解析：本小题主要考查对数函数问题。 1611 解析：本小题主要考查线性规划问题。作图(略)易知可行域为一个四角形,其四个顶点分别为验证知在点时取得最大值11.三、解答题17解：（）因为为偶函数，所以对，恒成立，因此即，整理得因为，且，
19、所以又因为，故所以由题意得，所以故因此（）将的图象向右平移个单位后，得到的图象，所以当（），即（）时，单调递减，因此的单调递减区间为（）18解：（）从8人中选出日语、俄语和韩语志愿者各1名，其一切可能的结果组成的基本事件空间，由18个基本事件组成由于每一个基本事件被抽取的机会均等，因此这些基本事件的发生是等可能的用表示“恰被选中”这一事件，则，事件由6个基本事件组成，因而（）用表示“不全被选中”这一事件，则其对立事件表示“全被选中”这一事件，由于，事件有3个基本事件组成，所以，由对立事件的概率公式得19（）证明：在中，由于，ABCMPDO所以故又平面平面，平面平面，平面，所以平面，又平面，故平
20、面平面（）解：过作交于，由于平面平面，所以平面因此为四棱锥的高，又是边长为4的等边三角形因此在底面四边形中，所以四边形是梯形，在中，斜边边上的高为，此即为梯形的高，所以四边形的面积为故20（）证明：由已知，当时，又，所以，即，所以，又所以数列是首项为1，公差为的等差数列由上可知，即所以当时，因此（）解：设上表中从第三行起，每行的公比都为，且因为，所以表中第1行至第12行共含有数列的前78项，故在表中第13行第三列，因此又，所以记表中第行所有项的和为，则21解：（）因为，又和为的极值点，所以，因此解方程组得，（）因为，所以，令，解得，因为当时，；当时，所以在和上是单调递增的；在和上是单调递减的（
21、）由（）可知，故，令，则令，得，因为时，所以在上单调递减故时，；因为时，所以在上单调递增故时，所以对任意，恒有，又，因此，故对任意，恒有22解：（）由题意得又，解得，因此所求椭圆的标准方程为（）（1）假设所在的直线斜率存在且不为零，设所在直线方程为，解方程组得，所以设，由题意知，所以，即，因为是的垂直平分线，所以直线的方程为，即，因此，又，所以，故又当或不存在时，上式仍然成立综上所述，的轨迹方程为（2）当存在且时，由（1）得，由解得，所以，解法一：由于，当且仅当时等号成立，即时等号成立，此时面积的最小值是当，当不存在时，综上所述，的面积的最小值为解法二：因为，又，当且仅当时等号成立，即时等号成
22、立，此时面积的最小值是当，当不存在时，综上所述，的面积的最小值为manipulator control mode and programmable controllers introduction 2.1 Select discussion with manipulator control 2.1.1 classification of control relays and discrete electronic circuit can control old industrial equipment, but also more common. Mainly these two relati
23、vely cheap and you can meet the old-fashioned, simple (or simple) industrial equipment. So he can see them now, however these two control modes (relay and discrete electronic circuits) are these fatal flaws: (1) cannot adapt to the complex logic control, (2) only for the current project, the lack of
24、 compatibility and (3) not reforming the system with equipment improvements. Spring for the development of Chinas modern industrial automation technology the substantial increase in the level of industrial automation, completed the perfect relay of the computer too much. In terms of controlling the
25、computer showed his two great advantages: (1) each of the hardware can be installed on one or more microprocessors; (2) the official designer of the software writing content control is all about. Now in several ways in the context of industrial automation can often be seen in three ways: (1) Program
26、mable Logical Controller (referred to as IPC); (2) Distributed Control System (DCS for short), and (3) the Programmable Logical Controller (PLC for short). 2.1.2 PLC and the IPC and DCS contrast contrast 1, each of the three technologies of origins and development requirements for fast data processi
27、ng makes it invented the computer. The men brought in terms of hardware there, using a high level of standardization, can use more compatibility tools, is a rich software resources, especially the need for immediacy in operational systems. So the computer can effectively control is used to control a
28、nd meet its speed, on the virtual model, real-time and in computational requirements. Distributed system started with a control system for industrial automatic instrument used to control, whereas now it is successfully developed into industrial control computer used as a central collection and distr
29、ibution system and transition of distributed control system in analogue handling, loop control, has begun to reflect the use of a huge advantage. Though distributed system has great advantages in loop regulation, but only as a means of continuous process control. Optimization of PLC is the correspon
30、ding relay needs was born, its main use in the work order control, early primary is replaced relay this hulking system, focused on the switch controlling the running order of functions. Marked by the microprocessor in the early 1970 of the 20th century emerged, micro-electronics technology has developed rapidly, people soon microelectronics processing technology will be used in the Programmable Logical Controller (that is15