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专业英语翻译实训计划书.doc

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

2、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)。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 12.25~12.27 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

3、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 Design Examples Therefore, the output measurement is A plot of y{t) is shown in Figure 2。46 for P=3。 We can see that y(t) is proportional to the force after 5

4、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, 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

5、 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 laboratory robot In this example, we endeavor to show the physical design of a laboratory device and demonstrate its complex design. We will also exh

6、ibit the many components commonly used in a control system. A robot for laboratory use is shown in Figure 2。47。 A laboratory robot’s work volume must allow the robot to reach the entire bench area and access existing analytical instruments. There must also be sufficient area for a stockroom of sup

7、plies for unattended operation. The laboratory robot can be involved in three types of tasks during an analytical experiment。 The first is sample introduction, wherein the robot is trained to accept a number of different sample trays, racks, and containers and to introduce them into the syste

8、m。 The second set of tasks involves the robot transporting the samples between individual dedicated automated stations for chemical preparation and instrumental analysis. Samples must be scheduled and moved between these s

9、tations 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 devices。 All of these types of operations are required for

10、 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 placed in the middle of a table when access to both

11、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 orientation)。 The rectilinear geometry, in contrast

12、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 of the robot by representing the robot positions

13、 and movements in the more familiar Cartesian coordinate space. Chapter 2 Mathematical Models of Systems FIGURE 2。47 Laboratory robot used for sample preparation. The robot manipulates small objects, such as test tubes, and probes in and out of tight places at relatively high speeds [15]。 ( Photo

14、courtesy of Beckman Coulter, Inc. ) Table 2.9 ORCA Robot Arm Hardware Specifications Articulated, Joy Stick with Arm Rail—Mounted Teach Pendant Emergency Stop Degrees of freedom 6 Cycle lime 4 s (move 1 inch up, 12 in

15、ch across, 1 inch down, and back) Reach ±54 cm Maximum speed 75 cm/s Height 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

16、 8。0 kg Vertical deflection <1.5 mm at continuous payload 1㎡ Precision Cross—sectional ±0。25 mm work envelope Finger travel 40 mm (gripper) Gripper rotation ±77 revolutions The physical and perfo

17、rmance 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 to obtain the total system. The exploded view of the rob

18、ot 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。8 Design Examples FIGURE 2.48 Exploded view of the ORC

19、A 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 Systems FIGURE 2.49 (a) Ladder network, (b) its signal—flow graph, and (c) its block diagram。 where G = 1/R, Z(s) = l/Cs, an

20、d I1 (s) = I1 (we omit the (s))。 The signal—flow 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 path。 Loops Li and L3 are non

21、touching. Therefore, the transfer function is If one prefers to utilize block diagram reduction techniques, one can start at the output with V3(s) = ZI2(s)。 But the block diagram shows that I2(s) = G (V2(s) — V3(s))。 Therefore, V2(s) = ZGV2(s) - ZGV3(s) Section 2。9 The Simulation of Systems U

22、sing Control Design Software So We will use this relationship between V3(s) and Vz(s) in the subsequent development。 Continuing with the block diagram reduction, we have but from the block diagram, we see that Therefore, Substituting for V2(s) yields But we know that GR = 1; hence, we obtain

23、 Note that the DC gain is 1/2, as expected。 The pole is desired at p = 2π(106.1) = 666。7 = 2000/3。 Therefore, we require RC = 0。001。 Select R = 1kΩ and C = 1μF。 Hence, we achieve the filter 2。9 THE SIMULATION OF SYSTEMS USING CONTROL DESIGN SOFTWARE Application of the many classical and modern con

24、trol system 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 m—file scripts containing commands and functions to analyze and sign control

25、systems。 Various commercial control system packages re available for student use。 The m—files 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 MathSci

26、pt RT Module。 Chapter 2 Mathematical Models of Systems We begin this section by analyzing a typical spring—mass—damper mathematical model of a mechanical system. Using an m—file script, we will develop an interactive analysis capability to analyze the effects of natural frequency and damping on th

27、e unforced response of the mass displacement。 This analysis will use the fact that we ha Ve 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 polyno

28、mials, computing poles and zeros 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 functio

29、ns covered in this section are roots, poly, conv, polyval, tf, pzmap, pole, zero, series, parallel, feedback, minreal, and step. Spring-Mass—Damper System. A spring—mass—damper mechanical system is shown in Figure 2。2. The motion of the mass, denoted by y(t), is described by the differential equat

30、ion The unforced dynamic response y{t) of the spring—mass—damper mechanical system is where and The initial displacement is y(0)。 The transient system response is underdamped whenζ〈 1, overdamped whenζ〉 1, and critically damped whenζ= 1. We can visualize the unforced time response of the mass d

31、isplacement following an initial displacement of y(0). 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)ω,t ,and ζ are input at the command level。 Then the script unforced。m is executed to generate t

32、he desired plots。 This creates an interactive analysis capability 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 ofω

33、and ζ at the command prompt and running the script unforced。m again. 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.

34、Using scripts is an important aspect of developing an effective interactive design and analysis capability For the spring—mass—damper problem, the unforced solution to the differential equation was readily available。 In general, when simulating closed—loop feedback Section 2。9 The Simulation

35、of Systems Using Control Design Software »y0=0。15; ωn »wn=sqrt(2); ζ »zeta=1/(2*sqrt(2)); »t=[0:0.1:10]; »unforced unforced。m FIGURE 2。50 Script to analyze the spring-massdamper FIGURE 2。51 Spring—massdamper Unforced response。 control systems subject to a variety of inputs and initi

36、al conditions, it is difficult to obtain the solution analytically。 In these cases, we can compute the solutions 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, w

37、e begin by investigating how to manipulate polynomials, remembering that working with transfer functions means that 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 calc

38、ulating its roots。 Polynomials are represented by row vectors containing the polynomial coefficients in order of descending degree. For example, the polynomial is entered as shown in Figure 2.52. Notice that even though the coefficient of the s term is zero, it is included in the input definition

39、 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 containing 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

40、in descending degree. We can compute the roots of the polynomial p(s) = s3, + 3s2 + 4 with the roots function as shown in Figure 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 w

41、ant to expand the polynomial The associated commands using the conv function are shown in Figure 2.53. Thus, the expanded polynomial is FIGURE 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 Sof

42、tware FIGURE 2.54 (a) The tf function。 (b) Using the tf function to create transfer function objects and adding them using t h e ” + " operator。 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

43、 shown in Figure 2。53。 Linear, time—invariant system models can be treated as objects, allowing one to manipulate the system models as single entities。 In the case of transfer functions, one creates the system models using the tf function; for state variable models one employs The ss function (se

44、e Chapter 3). The use of tf is illustrated in Figure 2。54(a). For example, if one has the two system models one can add them using the ”+" operator to obtain The corresponding commands are shown in Figure 2.54(b) where sysl represents Gi(s) and sys2 represents G2Cs)。 Computing the poles and zero

45、s associated with a transfer function is accomplished by operating on the system model object with the pole and zero functions, respectively, as illustrated in Figure 2。55. In the next example, we will obtain a plot of the pole—zero locations in the complex plane。 This will be accomplished using th

46、e 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 without left-hand arguments, the plot is generated automatically. Chapter 2 Mathematical Models of Systems FIGURE 2。55 (a) The pole and zero f

47、unctions。 (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 functions Consider the transfer functions Using an m—file script, we can compute the poles and zeros of G(s), the characteristic equati

48、on of H(s), and divide G(s) by H(s)。 We can also obtain a plot of the pole—zero map of G(s)IH(s) in the complex plane。 The pole—zero map of the transfer function G(s)IH(s) is shown in Figure 2。57, and the associated commands are shown in Figure 2。58。 The pole-zero map shows clearly the five zero l

49、ocations, but it appears that there are only two poles. This Section 2。9 The Simulation of Systems Using Control Design Software FIGURE 2。57 Pole—zero map for G(s)/H(s). FIGURE 2.58 Transfer function example for G{s) and H(s)。 cannot be the cas

50、e, 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 pole—ze

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