1、2012/2013学年上期专业英语翻译实训计划书姓名: 班级: 学号: 时 间班 级地 点指导教师一、实习(训)目的及要求1通过实训,使学生能借助词典等工具,初步具备阅读与本专业相关的英文文章的能力;熟悉科技英语翻译的一般技巧和基本技能;掌握阅读和翻译专业科技资料的能力和方法.2翻译材料由老师指定.要求词汇量不得少于3000单词;翻译内容原则上须选用计算机控制技术、DCS系统、仪器仪表和自动控制原理及系统、火电厂热能动力设备及系统等与本专业相关的英语教材、教参、科技报告、专业期刊、产品说明书等。3翻译要求尊重原文,用词准确,语句通顺,逻辑清楚。4为便于规范和管理,实训报告建议打印(同时交电子稿
2、)。格式要求:标题格式为标题2;正文字号为小四,字体为仿宋。英文字体为Times New Roman,字号为小四。二、实习(训)内容翻译本专业相关的英语资料.三、实习(训)进程表1。热控10班:时 间实训任务实训地点12.24实训动员,并安排实训任务 教室12.2512.27翻译英语资料、编写和打印实训报告教室,图书馆电子阅览室12。28提交实习报告;答辩教室,图书馆电子阅览室四、纪律要求实习期间须严格遵守校规校纪,按要求作息。不得无故迟到、早退、旷课。凡缺课时间超过实训总学时三分之一以上,取消实训资格。五、成绩评定方法1考勤占20%;2实习报告和答辩占80。Section 2.8 Desig
3、n ExamplesTherefore, the output measurement isA plot of yt) is shown in Figure 2。46 for P=3。 We can see that y(t) is proportional to the force after 5 seconds. Thus in steady state, after 5 seconds, the response y i) is proportional to the acceleration, as desired。 If this period is excessively long
4、, we must increase the spring constant, k, and the friction, b, while reducing the mass, M。 If we are able to select the components so that b/M=12 and k/M =32, the accelerometer will attain the proportional response in 1 second。 (It is left to the reader to show this。) EXAMPLE 2。16 Design of a labor
5、atory robot In this example, we endeavor to show the physical design of a laboratory device and demonstrate its complex design. We will also exhibit the many components commonly used in a control system. A robot for laboratory use is shown in Figure 2。47。 A laboratory robots work volume must allow t
6、he robot to reach the entire bench area and access existing analytical instruments. There must also be sufficient area for a stockroom of supplies for unattended operation.The laboratory robot can be involved in three types of tasks during an analytical experiment。 The first is sample introduction,
7、wherein the robot is trained to accept a number of different sample trays, racks, and containers and to introduce them into the system。 The second set of tasks involves the robot transporting the samples between individual dedicated automated stations for chemical preparation and instrumental analys
8、is. Samples must be scheduled and moved between these stations as necessary to complete the analysis. In the third set of tasks for the robot, flexible automation provides new capability to the analytical laboratory。 The robot must be programmed to emulate the human operator or work with various dev
9、ices。 All of these types of operations are required for an effective laboratory robot. The ORCA laboratory robot is an anthropomorphic arm, mounted on a rail, designed as the optimum configuration for the analytical laboratory 14. The rail can be located at the front or back of a workbench, or place
10、d in the middle of a table when access to both sides of the rail is required。 Simple software commands permit moving the arm from one side of the rail to the other while maintaining the wrist position(to transfer open containers) or locking the wrist angle (to transfer objects in virtually any orien
11、tation)。 The rectilinear geometry, in contrast to the cylindrical geometry used by many robots, permits more accessories to be placed within the robot workspace and provides an excellent match to the laboratory bench。 Movement of all joints is coordinated through software, which simplifies the use o
12、f the robot by representing the robot positions and movements in the more familiar Cartesian coordinate space.Chapter 2 Mathematical Models of SystemsFIGURE 2。47 Laboratory robot used for sample preparation. The robot manipulates small objects, such as test tubes, and probes in and out of tight plac
13、es at relatively high speeds 15。 ( Photo courtesy of Beckman Coulter, Inc. ) Table 2.9 ORCA Robot Arm Hardware SpecificationsArticulated, Joy Stick withArm RailMounted Teach Pendant Emergency Stop Degrees of freedom 6 Cycle lime 4 s (move 1 inch up, 12 inch across, 1 inch down, and back) Reach 54 cm
14、 Maximum speed 75 cm/sHeight 78 cm Dwell time 50 ms typical (for moves within a motion)Rail 1 and 2 m Payload 0。5 kg continuous, 2.5 kg transient (with restrictions)Weight 8。0 kg Vertical deflection 1.5 mm at continuous payload 1Precision Crosssectional 0。25 mm work envelopeFinger travel 40 mm(gripp
15、er)Gripper rotation 77 revolutions The physical and performance specifications of the ORCA system are shown in Table 2.9。 The design for the ORCA laboratory robot progressed to the selection of the component parts required to obtain the total system. The exploded view of the component parts required
16、 to obtain the total system. The exploded view of the robot is shown in Figure 2.48. This device uses six DC motors, gears, belt drives, and a rail and carriage。 The specifications are challenging and require the designer tomodel the system components and their interconnections accurately.Section 2。
17、8 Design ExamplesFIGURE 2.48 Exploded view of the ORCA robot showing the components 15. ( Courtesy of Beckman Coulter, Inc。)I1 = (Vi Vi) G,I1= ( V2 V3) G,V2= (I1- I2) R,V3 = I2Z,Chapter 2 Mathematical Models of SystemsFIGURE 2.49 (a) Ladder network, (b) its signalflow graph, and (c) its block diagra
18、m。where G = 1/R, Z(s) = l/Cs, and I1 (s) = I1 (we omit the (s))。 The signalflow graph constructed for the four equations is shown in Figure 2.49(b), and the corresponding block diagram is shown in Figure 2.49(c). The three loops are L1= GR = -1, L2 = GR =1,and L3 = GZ。 All loops touch the forward pa
19、th。 Loops Li and L3 are nontouching. Therefore, the transfer function isIf one prefers to utilize block diagram reduction techniques, one can start at the output withV3(s) = ZI2(s)。But the block diagram shows thatI2(s) = G (V2(s) V3(s)。Therefore,V2(s) = ZGV2(s) - ZGV3(s)Section 2。9 The Simulation of
20、 Systems Using Control Design SoftwareSoWe will use this relationship between V3(s) and Vz(s) in the subsequent development。 Continuing with the block diagram reduction, we havebut from the block diagram, we see thatTherefore, Substituting for V2(s) yieldsBut we know that GR = 1; hence, we obtainNot
21、e that the DC gain is 1/2, as expected。 The pole is desired at p = 2(106.1) = 666。7 = 2000/3。 Therefore, we require RC = 0。001。 Select R = 1k and C = 1F。 Hence, we achieve the filter2。9 THE SIMULATION OF SYSTEMS USING CONTROL DESIGN SOFTWAREApplication of the many classical and modern control system
22、 design and analysis ools is based on mathematical models. Most popular control design software packages an be used with systems given in the form of transfer function descriptions。 In his book, we will focus on mfile scripts containing commands and functions to analyze and sign control systems。 Var
23、ious commercial control system packages re available for student use。 The mfiles described here are compatible with the ATLAB ontrol System Toolbox and the LabVIEW MathScript RT Module。See Appendix A for an introduction to MATLAB。See Appendix B for an introduction to LabVIEW MathScipt RT Module。Chap
24、ter 2 Mathematical Models of SystemsWe begin this section by analyzing a typical springmassdamper mathematical model of a mechanical system. Using an mfile script, we will develop an interactive analysis capability to analyze the effects of natural frequency and damping on the unforced response of t
25、he mass displacement。 This analysis will use the fact that we haVe an analytic solution that describes the unforced time response of the mass displacement。Later, we will discuss transfer functions and block diagrams。 In particular, we are interested in manipulating polynomials, computing poles and z
26、eros of transfer functions, computing closed-loop transfer functions, computing block diagram reductions, and computing the response of a system to a unit step input。 The section concludes with the electric traction motor control design of Example 2。14。The functions covered in this section are roots
27、, poly, conv, polyval, tf, pzmap, pole, zero, series, parallel, feedback, minreal, and step. Spring-MassDamper System. A springmassdamper mechanical system is shown in Figure 2。2. The motion of the mass, denoted by y(t), is described by the differential equationThe unforced dynamic response yt) of t
28、he springmassdamper mechanical system is where and The initial displacement is y(0)。 The transient system response is underdamped when 1, overdamped when 1, and critically damped when= 1. We can visualize the unforced time response of the mass displacement following an initial displacement of y(0).
29、Consider the underdamped case: The commands to generate the plot of the unforced response are shown in Figure 2。50。 In the setup, the variables y(0),t ,and are input at the command level。 Then the script unforced。m is executed to generate the desired plots。 This creates an interactive analysis capab
30、ility to analyze the effects of natural frequency and damping on the unforced response of the mass displacement。 One can investigate the effects of the natural frequency and the damping on the time response by simply entering new values ofand at the command prompt and running the script unforced。m a
31、gain. The time-response plot is shown in Figure 2.51. Notice that the script automatically labels the plot with the values of the damping coefficient and natural frequency. This avoids confusion when making many interactive simulations. Using scripts is an important aspect of developing an effective
32、 interactive design and analysis capability For the springmassdamper problem, the unforced solution to the differential equation was readily available。 In general, when simulating closedloop feedbackSection 2。9 The Simulation of Systems Using Control Design Softwarey0=0。15;nwn=sqrt(2); zeta=1/(2sqrt
33、(2));t=0:0.1:10;unforcedunforced。mFIGURE 2。50 Script to analyze the spring-massdamperFIGURE 2。51 Springmassdamper Unforced response。control systems subject to a variety of inputs and initial conditions, it is difficult to obtain the solution analytically。 In these cases, we can compute the solutions
34、 numerically and to display the solution graphically。Most systems considered in this book can be described by transfer functions. Since the transfer function is a ratio of polynomials, we begin by investigating how to manipulate polynomials, remembering that working with transfer functions means tha
35、t both a numerator polynomial and a denominator polynomial must be specified Chapter 2 Mathematical Models of Systems FIGURE 2.52 Entering the Polynomial p(p) = s3 + 3s2 + 4 and calculating its roots。Polynomials are represented by row vectors containing the polynomial coefficients in order of descen
36、ding degree. For example, the polynomialis entered as shown in Figure 2.52. Notice that even though the coefficient of the sterm is zero, it is included in the input definition of p(s)If p is a row vector containing the coefficients of p(s) in descending degree, then roots(p) is a column vector cont
37、aining the roots of the polynomial. Conversely, if r is a column vector containing the roots of the polynomial, then poly(r) is a row vector with the polynomial coefficients in descending degree. We can compute the roots of the polynomial p(s) = s3, + 3s2 + 4 with the roots function as shown in Figu
38、re 2。52. In this figure, we show how to reassemble the polynomial with the poly function.Multiplication of polynomials is accomplished with the conv function. Suppose we want to expand the polynomialThe associated commands using the conv function are shown in Figure 2.53. Thus, the expanded polynomi
39、al isFIGURE 2。53 Using conv and polyval to multiply and evaluate the polynomials (3s + 2s + 1) (s + 4)。Section 2.9 The Simulation of Systems Using Control Design SoftwareFIGURE 2.54 (a) The tf function。 (b) Using the tf function to create transfer function objects and adding them using t h e ” + ope
40、rator。The function polyval is used to evaluate the value of a polynomial at the given value of the variable。 The polynomial n(s) has the value n(5) = -66, as shown in Figure 2。53。Linear, timeinvariant system models can be treated as objects, allowing one to manipulate the system models as single ent
41、ities。 In the case of transfer functions, one creates the system models using the tf function; for state variable models one employs The ss function (see Chapter 3). The use of tf is illustrated in Figure 2。54(a). For example, if one has the two system modelsone can add them using the ”+ operator to
42、 obtainThe corresponding commands are shown in Figure 2.54(b) where sysl represents Gi(s) and sys2 represents G2Cs)。 Computing the poles and zeros associated with a transfer function is accomplished by operating on the system model object with the pole and zero functions, respectively, as illustrate
43、d in Figure 2。55.In the next example, we will obtain a plot of the polezero locations in the complex plane。 This will be accomplished using the pzmap function, shown in Figure 2.56。 On the pole-zero map, zeros are denoted by an ”o and poles are denoted by an ”X” 。If the pzmap function is invoked wit
44、hout left-hand arguments, the plot is generated automatically.Chapter 2 Mathematical Models of SystemsFIGURE 2。55 (a) The pole and zero functions。 (b) Using the pole and zero functions to compute the pole and zero locations of a linear system。FIGURE 2.56 The pzmap function。EXAMPLE 2。18 Transfer func
45、tionsConsider the transfer functionsUsing an mfile script, we can compute the poles and zeros of G(s), the characteristic equation of H(s), and divide G(s) by H(s)。 We can also obtain a plot of the polezero map of G(s)IH(s) in the complex plane。The polezero map of the transfer function G(s)IH(s) is
46、shown in Figure 2。57, and the associated commands are shown in Figure 2。58。 The pole-zero map shows clearly thefive zero locations, but it appears that there are only two poles. This Section 2。9 The Simulation of Systems Using Control Design SoftwareFIGURE 2。57 Polezero map for G(s)/H(s).FIGURE 2.58
47、 Transfer function example for Gs) and H(s)。cannot be the case, since we know that for physical systems the number of poles must be greater than or equal to the number of zeros。 Using the roots function, we can ascertain that there are in fact four poles at s = 1. Hence, multiple poles or multiple zeros at the same location cannot be discerned on the poleze