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1、2012/20132012/2013 学年上期学年上期专业英语翻译实训专业英语翻译实训计计划划书书姓名姓名:班级：班级：学号：学号：时时间间班班级级地地点点指指导导教教师师一、实习（训一、实习（训)目的及要求目的及要求1 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
2、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、实习二、实习(训训)内容内容翻译本专业相关的英语资料。三、实习三、实习(训训)进程表进程表1.1.热控热控 1010 班：班：时 间12。2412.2512.27实训任务实训动员，并安排实训任务翻译英语资料、编写和打印实训报告实训地点教室教室,图书馆电子1阅览室教室，图书馆电子12。28四、纪律要求四、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
3、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、成绩评定方法五、成绩评定方法1考勤占 20;2实习报告和答辩占 80。提交实习报告；答辩阅览室2Section 2.8 Design ExamplesSection 2.8 Design ExamplesTherefore,the output measurement isA plot of yt)is shown in Figure 2。46 for P=3.We can see that y（t)isproportional to the force after 5 seconds.Thus in steady state,aft
4、er 5 seconds,theresponse y i)is proportional to the acceleration,as desired。If this period isexcessively long,we must increase the spring constant，k,and the friction,b，whilereducing the mass,M。If we are able to select the components so that b/M=12 andk/M=32，the accelerometer will attain the proporti
5、onal response in 1 second。(It isleft to the reader to show this。)EXAMPLE 2EXAMPLE 2。16 Design of a laboratory robot16 Design of a laboratory robotIn this example,we endeavor to show the physical design of a laboratory device anddemonstrate its complex design.We will also exhibit the many componentsc
6、ommonly used in a control system。A robot for laboratory use is shown in Figure 2。47.A laboratory robots workvolume must allow the robot to reach the entire bench area and access existinganalytical instruments。There must also be sufficient area for a stockroom of suppliesfor unattended operation。The
7、laboratory robot can be involved in three types of tasks during an analyticalexperiment。The first is sample introduction，wherein the robot is trained to accepta3number of different sample trays，racks，and containers and to introduce them intothe system。The second set of tasks involves the robot trans
8、porting the samplesbetweenindividual dedicated automated stations for chemical preparation and instrumentalanalysis.Samples must be scheduled and moved between these stations as necessaryto complete the analysis。In the third set of tasks for the robot,flexible automationprovides new capability to th
9、e analytical laboratory.The robot must be programmedto emulate the human operator or work with various devices.All of these types ofoperations are required for an effective laboratory robot.The ORCA laboratory robot is an anthropomorphic arm，mounted on a rail，designed as the optimum configuration fo
10、r the analytical laboratory 14。The railcan be located at the front or back of a workbench,or placed in the middle of a tablewhen access to both sides of the rail is required。Simple software commands permitmoving the arm from one side of the rail to the other while maintaining the wristposition（to tr
11、ansfer open containers)or locking the wrist angle(to transfer objects invirtually any orientation).The rectilinear geometry，in contrast to the cylindricalgeometry used by many robots,permits more accessories to be placed within the robotworkspace and provides an excellent match to the laboratory ben
12、ch.Movement of alljoints is coordinated through software,which simplifies the use of the robot byrepresenting the robot positions and movements in the more familiar Cartesiancoordinate space.Chapter 2 Mathematical Models of Systems4FIGURE 2。47 Laboratory robot used for sample preparation.The robot m
13、anipulatessmall objects,such as test tubes,and probes in and out of tight places at relativelyhigh speeds 15.（Photo courtesy of Beckman Coulter，Inc.)Table 2Table 2。9 ORCA Robot Arm Hardware Specifications9 ORCA Robot Arm Hardware SpecificationsArticulated，Joy Stick withArmRailMountedTeach PendantEme
14、rgency StopDegrees of freedom6Cycle lime4 s(move 1 inch up,12 inchacross，1 inch down，andback）Reach54 cmMaximum speed75 cm/sHeight78 cmDwell time50 ms typical（for moveswithin a motion）Rail1 and 2 mPayload0。5 kg continuous,2.5 kgtransient(with restrictions）Weight8.0 kgVertical deflection 1,and critica
15、lly damped when=1.We can visualize the unforced time response of the mass displacement following aninitial displacement of y（0).Consider the underdamped case:10The commands to generate the plot of the unforced response are shown in Figure2。50.In the setup，the variables y(0），t，and are input at the co
16、mmand level。Then the script unforced.m is executed to generate the desired plots.This creates aninteractive analysis capability to analyze the effects of natural frequency and dampingon the unforced response of the mass displacement。One can investigate the effectsof the natural frequency and the dam
17、ping on the time response by simply enteringnew values ofand at the command prompt and running the script unforced.m again.The timeresponse plot is shown in Figure 2.51。Notice that the script automaticallylabels the plot with the values of the damping coefficient and natural frequency。Thisavoids con
18、fusion when making many interactive simulations.Using scripts is animportant aspect of developing an effective interactive design and analysis capabilityFor the spring-massdamper problem，the unforced solution to thedifferential equation was readily available.In general，when simulating closedloopfeed
19、backSection 2Section 2。9 The Simulation of Systems Using Control Design Software9 The Simulation of Systems Using Control Design Softwarey0=0。15;wn=sqrt（2）;zeta=1/(2sqrt(2）;t=0：0.1：10;unforcednunforced。mFIGURE 2FIGURE 2。5050 Script to analyze the spring-massdamper11FIGURE 2FIGURE 2。5151 Spring-massd
20、amper Unforced response。control systems subject to a variety of inputs and initial conditions，it is difficult toobtain the solution analytically.In these cases,we can compute the solutionsnumerically and to display the solution graphically。Most systems considered in this book can be described by tra
21、nsfer functions。Since the transfer function is a ratio of polynomials,we begin by investigating how tomanipulate polynomials，remembering that working with transfer functions meansthat both a numerator polynomial and a denominator polynomial must be specifiedChapter 2 Mathematical Models of SystemsCh
22、apter 2 Mathematical Models of SystemsFIGURE 2FIGURE 2。5252 Entering the Polynomial p(p）=s3+3s2+4 and calculating itsroots.Polynomials are represented by row vectors containing the polynomialcoefficients in order of descending degree。For example，the polynomialis entered as shown in Figure 2.52.Notic
23、e 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,thenroots(p)is a column vector containing the roots of the polynomial.Conversely,if r isa column vector containing the ro
24、ots of the polynomial,then poly(r）is a row vectorwith the polynomial coefficients in descending degree。We can compute the roots ofthe 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.12Multiplicat
25、ion of polynomials is accomplished with the conv function.Supposewe want to expand the polynomialThe associated commands using the conv function are shown in Figure 2.53.Thus,theexpanded polynomial isFIGURE 2.53FIGURE 2.53 Using conv and polyval to multiply and evaluate the polynomials(3s+2s+1）（s+4)
26、。Section 2.9 The Simulation of Systems Using Control Design SoftwareSection 2.9 The Simulation of Systems Using Control Design SoftwareFIGUREFIGURE 2.542.54（a)The tf function。(b)Using the tf function to create transferfunction objects and adding them using t h e”+”operator.The function polyval is us
27、ed to evaluate the value of a polynomial at the given valueof the variable。The polynomial n(s)has the value n（5)=-66，as shown in Figure2。53.Linear,timeinvariant system models can be treated as objects,allowing one tomanipulate the system models as single entities。In the case of transfer functions,on
28、ecreates the system models using the tf function;for state variable models one employsThe ss function（see Chapter 3)。The use of tf is illustrated in Figure 2。54（a）.13For example，if one has the two system modelsone can add them using the”+”operator to obtainThe corresponding commands are shown in Fig
29、ure 2.54（b)where sysl representsGi（s)and sys2 represents G2Cs)。Computing the poles and zeros associated with atransfer function is accomplished by operating on the system model object with thepole and zero functions,respectively，as illustrated in Figure 2.55.In the next example，we will obtain a plot
30、 of the pole zero locations in thecomplex plane。This will be accomplished using the pzmap function,shown inFigure 2.56。On the pole-zero map，zeros are denoted by an”o”and poles aredenoted by an X”。If the pzmap function is invoked without lefthand arguments，the plot is generated automatically.Chapter
31、2 Mathematical Models of SystemsChapter 2 Mathematical Models of SystemsFIGURE 2.55（a)The pole and zero functions。（b)Using the pole and zerofunctions to compute the pole and zero locations of a linear system.FIGURE 2FIGURE 2。5656 The pzmap function.EXAMPLE 2EXAMPLE 2。1818 Transfer functionsTransfer
32、functionsConsider the transfer functionsUsing an mfile script，we can compute the poles and zeros of G（s）,thecharacteristic equation of H(s),and divide G(s）by H(s）。We can also obtain a plotof the polezero map of G(s）IH(s）in the complex plane.14The polezero map of the transfer function G(s）IH(s)is sho
33、wn in Figure 2.57,andthe associated commands are shown in Figure 2.58。The pole-zero map showsclearly thefive zero locations,but it appears that there are only two poles.This15Section 2.9 The Simulation of Systems Using Control Design SoftwareSection 2.9 The Simulation of Systems Using Control Design
34、 SoftwareFIGURE 2FIGURE 2。5757 Pole-zero map for G(s）/H（s）。FIGURE 2.58FIGURE 2.58 Transfer function example for Gs)and H（s）。cannot be the case，since we know that for physical systems the number ofpoles must be greater than or equal to the number of zeros.Using the roots function，we can ascertain tha
35、t there are in fact four poles at s=-1。Hence,multiple poles ormultiple zeros at the same location cannot be discerned on the pole-zero map.Chapter 2 Mathematical Models of SystemsChapter 2 Mathematical Models of SystemsFIGURE 2FIGURE 2。5959 Open-loop control system（without feedback）。Block Diagram Mo
36、delsBlock Diagram Models。Suppose we have developed mathematical models inthe form of transfer functions for a process，represented by G（s）,and a controller,represented by Gc(s），and possibly many other system components such as sensorsand actuators。Our objective is to interconnect these components to
37、form a controlsystem.A simple open-loop control system can be obtained by interconnecting a processand a controller in series as illustrated in Figure 2。59.We can compute the transferfunction from R（s)to Y(s），as follows。EXAMPLE 2EXAMPLE 2。1919 Series connectionLet the process represented by the tran
38、sfer function G（s)be16and let the controller represented by the transfer function Gc（s)beWe can use the series function to cascade two transfer functions G（s)and G2（s),as shown in Figure 2。60。The transfer function Gc（s）G(s)is computed using the series function as shownin Figure 2。61.The resulting tr
39、ansfer function iswhere sys is the transfer function name in the mfile script.FIGURE 2.60FIGURE 2.60（a）Block diagram.(b)The series function.Section 2Section 2。9 The Simulation of Systems Using Control Design Software9 The Simulation of Systems Using Control Design SoftwareFIGURE 2.61FIGURE 2.61 Appl
40、ication of theseries function。FIGURE 2.62FIGURE 2.62（a)Block diagram。（b)The parallel function。FIGURE 2.63FIGURE 2.63 A basic control system with unity feedback.Block diagrams quite often have transfer functions in parallel。In such cases,thefunction parallel can be quite useful.The parallel function
41、is described in Figure 2.62.We can introduce a feedback signal into the control system by closing the loopwith unity feedback，as shown in Figure 2.63。The signal Ea（s）is an error signal;the signal R(s）is a reference input.In this control system，the controller is in theforward path，and the closedloop
42、transfer function isChapter 2 Mathematical Models of SystemsChapter 2 Mathematical Models of Systems17FIGUREFIGURE 2.642.64(a)Block diagram.（b)The feedback function with unityfeedback。FIGURE 2FIGURE 2。6565(a)Block diagram。(b)The feedback function。We can utilize the feedback function to aid in the bl
43、ock diagram reductionprocess to compute closed loop transfer functions for single and multiple loopcontrol systems.It is often the case that the closed-loop control system has unity feedback,asillustrated in Figure 2.63。We can use the feedback function to compute theclosedloop transfer function by s
44、etting H(s）=1.The use of the feedback function forunity feedback is depicted in Figure 2。64.The feedback function is shown in Figure 2。65 with the associated systemconfiguration,which includes H(s)in the feedback path.If the input”sign”is omitted,then negative feedback is assumed。Section 2.9 The Sim
45、ulation of Systems Using Control Design SoftwareFIGURE 2FIGURE 2。6666（a)Block diagram.（b）Application of the feedback function.FIGURE 2.67FIGURE 2.67 A basic control system with the controller in the feedback loop。EXAMPLE 2.20EXAMPLE 2.20The feedback function with unity feedbackThe feedback function
46、with unity feedbackLet the process,G（s）,and the controller,Gc(s），be as in Figure 2.66(a）。To apply the18feedback function,we first use the series function to compute Gc（s）G（s)，followedby the feedback function to close the loop。The command sequence is shown inFigure 2.66(b).The closed-loop transfer fu
47、nction,as shown in Figure 2。66(b）,isAnother basic feedback control configuration is shown in Figure 2。67.In thiscase,the controller is located in the feedback path.The closedloop transfer functionisEXAMPLE 2.21EXAMPLE 2.21 The feedback functionThe feedback functionLet the process,G(s），and the contro
48、ller，H（s),be as in Figure 2。68（a)。To compute the closedloop transfer function with the controller in the feedbackloop，we useChapter 2 Mathematical Models of SystemsChapter 2 Mathematical Models of SystemsFIGURE 2.68FIGURE 2.68 Application of the feedback function:（a)block diagram，(b）mfile script。the
49、 feedback function.The command sequence is shown in Figure 2。68（b).Theclosed-loop transfer function isThe functions series，parallel,and feedback can be used as aids in blockdiagram manipulations for multiple-loop block diagrams.EXAMPLE 2.22EXAMPLE 2.22 Multiloop reductionMultiloop reductionA multilo
50、op feedback system is shown in Figure 2。26.Our objective is to computethe closed-loop transfer functionWhen19AndSection 2.9 The Simulation of Systems Using Control Design SoftwareSection 2.9 The Simulation of Systems Using Control Design SoftwareFIGURE 2FIGURE 2。6969 Multiple-loop block reduction。20