专业英语翻译实训计划书

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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。 1

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 12。24 实训任务实训动员，并安排实训任务实训地点教室教室，图书馆电子12.25~12。27 翻译英语资料、编写和打印实训报告阅览室教室,图书馆电子12。28 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％.

提交实习报告;答辩阅览室 Section 2。8 Design Examples

Y(s)ppp22ss12(s2)

Therefore， the output measurement is

1y(t)[p2pet(p2)e2t],t0.

2A 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 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 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 exhibit 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 supplies 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 system. 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 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 devices。 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 placed 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 orientation). 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 of the robot by representing the robot positions 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 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 inch 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） 5 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 1㎡ Precision Cross-sectional ±0。25 mm work envelope Finger travel 40 mm （gripper) Gripper rotation ±77 revolutions 6

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 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。8 Design Examples

FIGURE 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 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， and 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 nontouching。 Therefore， the transfer function is

T(s)V3P1GZV11(L1L2L3)L1L332GZ

11/(3RC).3RCs232/(3RC) 9

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 Using Control Design Software So

V2(s)1ZGV3(s) ZG

We will use this relationship between V3(s) and Vz（s) in the subsequent development。 Continuing with the block diagram reduction， we have

V3(s)ZGV3(s)ZGR(I1(s)I2(s)).

but from the block diagram, we see that

I1G(V1(s)V2(s)),I2V3(s). ZTherefore，

V3(s)ZGV3(s)ZG2R(V1(s)V2(s))GRV3(s).

Substituting for V2（s） yields

(GR)(GZ)V1(s).

12GRGZ(GR)(GZ)V3(s)But we know that GR = 1; hence， we obtain

V3(s)

GZV1(s).

32GZNote 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

333.3.

(s666.7)T(s) 11

2.9 THE SIMULATION OF SYSTEMS USING CONTROL DESIGN SOFTWARE

Application of the many classical and modern control 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 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 MathScipt 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 the 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 polynomials, 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 functions 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 equation

My(t)by(t)ky(t)r(t).

The unforced dynamic response y｛t) of the spring—mass—damper mechanical system is

y(t)y(0)12entsin(n12t),

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 displacement following an initial displacement of y(0）。 Consider the underdamped case：

y(0)0.15m,n2rad1kb,(2,1). secM22MThe 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 the 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 ofωand 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。 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 of Systems Using Control Design Software

»y0=0。15; »wn=sqrt(2)； »zeta=1/(2＊sqrt(2））； »t=［0:0。1：10］； »unforced ωn ζ 14

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 initial 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, we 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 calculating its

roots.

Polynomials are represented by row vectors containing the polynomial coefficients in order of descending degree。 For example， the polynomial

p(s)s33s24.

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 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 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 want to expand the polynomial

n(s)(3s22s1)(s4).

The associated commands using the conv function are shown in Figure 2.53. Thus, the expanded polynomial is

n(s)3s314s29s4.

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 Software

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 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 (see Chapter 3). The use of tf is illustrated in Figure 2。54(a)。

For example， if one has the two system models

G1(s)101andG(s). 22S2s5s1one can add them using the ”+” operator to obtain

s212s15G(s)G1(s)G2(s)3

s3s27s5.

The 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 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 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 without left-hand arguments, the plot is generated automatically.

Chapter 2 Mathematical Models of Systems

FIGURE 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 functions Consider the transfer functions

6s21(s1)(s2) G(s)3andH(s)2s3s3s1(s2i)(s2i)(s3).Using an m—file 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 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 locations， 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 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 pole—zero map。 Chapter 2 Mathematical Models of Systems

FIGURE 2.59 Open-loop control system （without feedback).

Block Diagram Models. Suppose we have developed mathematical models in the 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 sensors and actuators. Our objective is to interconnect these components to form a control system。

A simple open-loop control system can be obtained by interconnecting a process and a controller in series as illustrated in Figure 2。59。 We can compute the transfer function from R（s) to Y(s）， as follows。 EXAMPLE 2。19 Series connection

Let the process represented by the transfer function G(s） be

G(s)1 2500sand let the controller represented by the transfer function Gc（s） be

Gc(s)s1. s2We 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 shown in Figure 2。61 .The resulting transfer function is

Gc(s)G(s)s1sys

500s31000s2 25

where sys is the transfer function name in the m-file script.

FIGURE 2。60 (a) Block diagram。 (b) The series function。

Section 2。9 The Simulation of Systems Using Control Design Software

FIGURE 2.61 Application of the series function.

FIGURE 2.62 (a） Block diagram. （b） The parallel function。

FIGURE 2。63 A basic control system with unity feedback.

Block diagrams quite often have transfer functions in parallel. In such cases, the function parallel can be quite useful. The parallel function is described in Figure 2。62。

We can introduce a feedback signal into the control system by closing the loop with 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 the forward path, and the closed—loop transfer function is

Chapter 2 Mathematical Models of Systems

FIGURE 2. (a) Block diagram. （b） The feedback function with unity feedback.

FIGURE 2。65 （a) Block diagram。 (b) The feedback function.

We can utilize the feedback function to aid in the block diagram reduction process to compute closed—loop transfer functions for single- and multiple-loop control systems.

It is often the case that the closed-loop control system has unity feedback， as illustrated in Figure 2。63。 We can use the feedback function to compute the closedloop transfer function by setting H（s) = 1. The use of the feedback function for unity feedback is depicted in Figure 2.。

The feedback function is shown in Figure 2。65 with the associated system configuration, which includes H(s） in the feedback path。 If the input \"sign” is omitted， then negative feedback is assumed.

Section 2。9 The Simulation of Systems Using Control Design Software

FIGURE 2。66 (a) Block diagram。 (b) Application of the feedback function。

FIGURE 2。67 A basic control system with the controller in the feedback loop。 EXAMPLE 2.20 The feedback function with unity feedback

Let the process， G（s）, and the controller, Gc(s）， be as in Figure 2。66(a）. To apply the

feedback function, we first use the series function to compute Gc（s）G（s）, followed by the feedback function to close the loop。 The command sequence is shown in Figure 2。66(b). The closed—loop transfer function, as shown in Figure 2.66（b), is

T(s)s1sys.

500s31000s2s1Another basic feedback control configuration is shown in Figure 2.67. In this case, the controller is located in the feedback path。 The closed—loop transfer function is

G(s).

1G(s)H(s)T(s)EXAMPLE 2.21 The feedback function

Let the process, G(s）， and the controller, H（s), be as in Figure 2.68（a）. To compute the closed-loop transfer function with the controller in the feedback loop, we use

Chapter 2 Mathematical Models of Systems

FIGURE 2.68 Application of the feedback function：（a） block diagram， (b）

m—file script.

the feedback function。 The command sequence is shown in Figure 2。68（b）。 The closed-loop transfer function is

T(s)s1sys.

500s31000s2s1The functions series, parallel， and feedback can be used as aids in block diagram manipulations for multiple—loop block diagrams. EXAMPLE 2.22 Multiloop reduction

A multiloop feedback system is shown in Figure 2.26. Our objective is to compute the closed—loop transfer function

Y(s) R(s)T(s)

When

11,G2(s),s10s1 2s1s1G3(s)2,G4(s).s4s4s6G1(s)And

H1(s)s1,H2(s)2,andH3(s)1. s2Section 2.9 The Simulation of Systems Using Control Design Software

FIGURE 2.69 Multiple-loop block reduction.

For this example， a five-step procedure is followed: Step 1. Input the system transfer functions。 Step 2。 Move H2 behind G4. Step 3。 Eliminate the G3G^Hi loop. Step 4. Eliminate the loop containing H2。

Step 5。 Eliminate the remaining loop and calculate T(s）。

The five steps are utilized in Figure 2。69， and the corresponding block diagram reduction is shown in Figure 2。27. The result of executing the commands is

s54s46s35s2sys 6543212s205s1066s2517s3218s2196s712We must be careful in calling this the closed—loop transfer function。 The transfer function is defined as the input—output relationship after pole-zero cancellations. If we compute the poles and zeros of T（s）， we find that the numerator and denominator polynomials have (s + 1） as a common factor。 This must be canceled before we can claim we have the closed-loop transfer function. To assist us in the pole-zero cancellation, we will use the minreal function。 The minreal function， shown in Figure 2。70， removes common pole—zero factors of a transfer function. The final step in the block reduction process is to cancel out the common factors， as shown in Figure 2.71. After the application of the minreal function， we find that the order of the denominator polynomial has been reduced from six to five, implying one pole-zero cancellation.

Chapter 2 Mathematical Models of Systems

FIGURE 2。70 The mineral function。

FIGURE 2。71 Application of the minreal function.

EXAMPLE 2.23 Electric traction motor control

Finally, let us reconsider the electric traction motor system from Example 2.14。 The block diagram is shown in Figure 2。44（c)。 The objective is to compute the closed—loop transfer function and investigate the response of Highly oscillatory. We can investigate the response ω(t） to a reference input ， ωd(t)， by utilizing the step function。 The step function, shown in Figure 2。73,

calculates the unit step response of a linear system. The step function is very important， since control system performance specifications are often given in terms of the unit step response

FIGURE 2。72 Electric traction motor block reduction.

Section 2.9 The Simulation of Systems Using Control Design Software

FIGURE 2。73 The step function。

FIGURE 2。74 （a) Traction motor wheel velocity step response，（b） m-file script.

If the only objective is to plot the output,y（t), we can use the step function without left—hand arguments and obtain the plot automatically with axis labels。 If we need y｛f) for any purpose other than plotting， we must use the step function with left-hand arguments， followed by the plot function to plot y（t）。 We define t as a row vector containing the times at which we wish the value of the output variable y(t）. We can also select t - /gnai, which results in a step response from t — 0 to t — ffmai and the number of intermediate points are selected automatically。

The step response of the electric traction motor is shown in Figure 2。74。 As expected, the wheel velocity response, given by y(t)， is highly oscillatory. Note that the output is y(t)=ω(t）

Chapter 2 Mathematical Models of Systems

2。10 SEQUENTIAL DESIGN EXAMPLE: DISK DRIVE READ SYSTEM

In Section 1.10, we developed an initial goal for the disk drive system： to position the —4 reader head accurately at the desired track and to move from one track to another within 10 ms, if possible。 We need to identify the plant, the sensor, and the controller。 We will obtain a model of the plant G（s) and the sensor. The disk drive reader uses a permanent magnet DC motor to rotate the reader arm （see Figure 1。29)。 The DC motor is called a voice coil motor in the disk drive industry. The read head is mounted on a slider device， which is connected to the arm as shown in Figure 2。75. A flexure （spring metal) is used to enable the head to float above the disk at a gap of less than 100 nm.The thin—film head reads the magnetic flux and provides a signal to an amplifier. The error signal of Figure 2。76（a) is provided by reading the error from a prerecorded index track。 Assuming an accurate read head， the sensor has a transfer function H(s) = 1, as shown in Figure 2.76(b). The model of the permanent magnet DC motor and a linear amplifier is shown in Figure 2.76（b). As a good approximation， we use the model of the armature—controlled DC motor as shown earlier in

FIGURE 2。75 Head mount for reader， showing flexure.

FIGURE 2。76 Block diagram model of disk drive read system.

章节2.8 设计实例

ppp22ss12(s2)

Y(s)则，输出测量值为

1y(t)[p2pet(p2)e2t],t0.

2当P=3时，输出y（t)如图2.46，我们可以看到，5秒钟之后输出y（t）正比于力的大小。因此在稳定状态,5秒钟之后输出响应y（t)正如预期的那样正比于加速度。如果这段时间太长，我们必须加大弹簧常数k和摩擦力b，但是要减少质量M.如果我们能够选择组成部分，那么b/M=12且k/M=32。在一秒钟之内,加速计能够达到成比例的响应。（留给读者去表示这个）例2.16 设计的一个实验室机器人

在这个例题中，我们尽量展示一个实验室装置的物理设计和展示组成原理，我们也会展示许多组成部件在控制系统中一般的使用原理。

一个机器人实验室使用如图2。47，一个实验室的机器人的工作体积必须让机器人达到整个板凳区和访问有的分析仪器。这必须为为人操作提供足够的区域

在分析实验中，这种机器人能够涉及到三种任务。第一个是示例介绍其中机器人是被训练来接受许多不同的样品盘、机框和容器并且为了将其引入系统，第二组任务涉及机器人运输样品个人专用自动站之间的化学制剂和仪器分析.样品必须预定和之间转移这些电台作为必要完成分析。

在第三组任务的机器人,灵活的自动化提供了新的能力分析实验室。机器人必须

被编程来模拟人类操作员或使用各种设备。所有这些类型的操作是一个有效的实验室所需机器人。

ORCA实验室的机器人是一个拟人化的手臂，安装在一个扶手，为分析实验室作为最佳设计外形，扶手坐落在工作试验台的前面或者后面,当必须要接近扶手两边的时候，可以放在桌子中间,在保持手腕位置和手腕角度(为了传输对象在任何方向)一定的情况下，一个简单的指令允许手臂从扶手的一边移到另一边（为了转让打开容器）。直线几何和圆柱几何运用在许多机器人中，在机器人工作空间内部允许有许多配件并且给工作试验台提供一个极佳的配件。所有关节的运动通过软件协调，通过陈述机器人的位置可以简化机器人的使用并且在更熟悉的笛卡尔坐标空间运动. 第二章系统的数学模型

图2。47实验室的机器人用于样品制备。机器人操纵小对象,如试管，探测器在紧的地方在相对较高的速度[15］.(照片由Beckman Coulter公司。) 表2.9 ORCA机器人手臂的硬件规格

铰接操纵杆

手臂载于铁轨上的示教器紧急停机自由角度 6 循环时间 4s 范围 —54～+54cm 最大速度 75cm/s 高度 78cm 延迟时间 50ms 扶手 1到2米有效的负荷 0.5kg连2。5kg短暂的重量 8。0kg 竖直偏差在连续负荷下小于1.5mm 精度 -0.25～+0。25mm 代表性的工作包络手指经过（夹子） 40mm 夹子经过 —77~+77转速 ORCA物理和性能说明都展示在表2.9中,设计了ORCA实验室机器人发展到选择的部件需要获得整个系统.机器人部件分解图如图2.48所示,这个装置使用6个直流电动机、齿轮、皮带传动和一个扶手和托板,这个规范是具有挑战性的，需要设计是模型系统组件和他们之间准确的联系.

章节2。8 设计实例

图2。48,ORCA机器人部件分解图显示在【15】例2。17 设计低通滤波器

我们的目的是设计一阶低通滤波器,在频率为106.1HZ一下通过信号，在频率为106。1HZ以上衰减信号，另外直流增益应该是1/2。

一个带有能源存储元素的梯形网络如图2.49（a）所示,将作为一个一阶低通网络，注意直流增益将等于1/2（开路的电容器）.这个直流电压方程是

I1 = (Vi - Vi） G， I1= （ V2 — V3) G,

V2= (I1- I2） R, V3 = I2Z,

第二章系统的数学模型

图2.49 (a)梯形网络（b)信号流程图 (c)方框图

在这G = 1/R, Z（s） = l/Cs, and I1 （s） = I1（我们省略了s）四个方程的信号流动曲线如图2.49（b）所示,方框图如图2.49（c）所示，这三个循环是L1= -GR = —1， L2 = -GR =—1,and L3 = -GZ。所有循环向前接近，循环L1和L3没有接触，因此，传递函数是

V3P1GZV11(L1L2L3)L1L332GZ

11/(3RC).3RCs232/(3RC)T(s)如果一个人愿意用框图还原技术，可以开始在输出

V3(s） = ZI2（s)。

但是方框图显示

I2（s) = G （V2(s) — V3(s))。

因此

V2(s） = ZGV2(s） - ZGV3（s)

章节2.9 仿真系统的使用控制设计软件这样

V2(s)1ZGV3(s) ZG在后面的发展中，我们使用V2（s)和V3（s)之间的关系，继续变换减少框图,我们可以得出

V3(s)ZGV3(s)ZGR(I1(s)I2(s)).

但是从这个方框图中我们可以看到

I1G(V1(s)V2(s)),I2V3(s). Z因此

V3(s)ZGV3(s)ZG2R(V1(s)V2(s))GRV3(s).

取代V2(s)换算得

(GR)(GZ)V1(s).

12GRGZ(GR)(GZ)V3(s)但是我们知道GR=1，因此得出

V3(s)GZV1(s).

32GZ注意，积分器增益就像预期的那样是1/2，当p=2π（106。1）=666.7=2000/3，因此，我们要求RC=0。001,选择R=1kΩ并且C=1μF.一次我们达到了过滤的目的。

333.3.

(s666.7)T(s)2.9 仿真系统的使用控制设计软件

应用程序的许多古典和现代控制系统设计和分析工具是基于数学模型.最受欢迎的控制设计软件包可以使用系统形式给出传递函数的描述。在这本书中,我们将把重点放在m脚本包含命令和函数来分析和设计控制系统。各种商业控制系统

软件包供学生使用。这里描述的m文件兼容MATLAB控制系统工具箱和虚拟仪器RT单元。

查看附录A以介绍MATLAB 参见附录B以介绍虚拟仪器RB 章节2 系统的数学模型

我们通过分析一个典型的弹簧质量阻尼器的数学机械系统的模型。开始这个部分，使用一个m文件的脚本,我们将开发一个互动分析能力,分析了固有频率和阻尼的影响，在自然的反应的质量位移的情况下分析了固有频率和阻尼的影响.这些分析将会利用我们有一个解析解这个事实,描述了自然的时间响应的大规模的位移.

稍后，我们将讨论转移函数和方框图. 特别是,我们有兴趣操纵多项式,，计算波兰人和0的转移函数，计算闭环传递函数, 计算框图减少，和计算响应系统一个单位阶跃输入。这个部分提出了电力牵引电动机控制设计的例子2.14。

本节讨论的功能是求多项式的根、矩阵的特征多项式、多项式乘、计算多项式的值、零极点图、极点、零点、连续、平行、反馈、状态方程和步骤。弹簧质量阻尼器系统。一个弹簧质量阻尼器机械系统如图2.2所示。这个运动的质量,用y（t），所描述的微分方程为

My(t)by(t)ky(t)r(t).

自然动态响应y { t)的弹簧质量阻尼器机械系统是

y(t)y(0)12entsin(n12t),

这里and 初始位移是y（0)。系统的瞬态响应

是〈 1时欠阻尼的ζ,当ζ> 1时，过阻尼，和当ζ= 1时,极度阻尼。我们可以想象在初始位移的y(0)的跟踪下这样的时间响应的大规模位移。考虑在欠阻尼的案例实例：

rad1kb,(2,1). secM22My(0)0.15m,n2这个命令生成的情节非受迫性响应如图2.50所示。在设置中,变量y(0）、ω、f、ζ是输入在命令水平。然后脚本是非受迫性的。m是执行生成所需的情节。这将创建一个交互式分析能力，分析了在固有频率和阻尼的影响下自然反应的大规模位移。对人为的影响进行调查，显示自然频率和阻尼的时间响应通过简单地进入新一个ω值和£值在命令提示符并运行脚本的。这个时间响应图如图2。51所示。注意，该脚本自动标记的情节是带有阻尼系数和自然频率的。这将避免了混淆在很多交互式模拟中使用脚本是发展一个有效的互动设计和分析能力的重要方面的。

对于弹簧质量阻尼器的问题,自然解决微分方程是现成的。一般来说,当模拟闭环反馈。

章节2。9仿真系统的使用控制设计软件

图2。50 脚本分析弹簧图2.51 弹簧非受迫性反应

图2。50

图2.51

控制系统受到各种输入和初始条件的影响，很难获得解决方案来进行分析.在这些情况下，我们可以计算数值的解决方案并显示图形化的解决方案.

在这本书中，大多数系统可以通过传递函数来描述，因为传递函数的比例是多项式，我们开始研究如何操纵多项式，记住使用传递函数方法，这是一个必须被指定的分子多项式和分母多项式章节2系统的数学模型

图2。52进入多项式p（p)= s3 + 3 s2 + 4和计算其根

为了递减的顺序，多项式通过包含多项式系数的行向量来描述。例如，

多项式

p(s)s33s24.

输入如图2.52所示，请注意，尽管系数的术语是零但它包含在输入定义p(s).

如果p是一个包含系数p(s）在下降顺序的行向量,那么根（p）是一个列向量含有的根多项式。相反，如果r 一个含有的根多项式列向量,那么矩阵（r）是一个与多项式系数下降程度相对应的行向量。我们可以计算该多项式p(s）= s3,+ 3 s2 + 4的根且根功能如图2。52所示。在这个图中，我们展示了如何重新组装多项式与矩阵函数

乘法的多项式是由带有乘法函数来完成的假设我们想扩大多项式

n(s)(3s22s1)(s4).

相关的命令使用多项式乘法功能如图2.53所示。因此扩大多项式是

n(s)3s314s29s4.

图2。53 使用多项式乘法和计算多项式值来估计多项式（3s* + 2s + 1) （s + 4). 章节2。9仿真系统的使用控制设计软件

图2。54 (a）tf函数、（b)使用tf函数来创建传递函数对象和添加使用t h e“+”操作符的他们

这个求多项式值的函数是用来估计给出变量值的多项式，这个多项式有n(—5)=-66，如图2.53所示。

线性非时变系统模型可以看作对象，允许一个操作系统模型作为单一实体。对于传递函数,一个创建系统模型使用tf函数；为状态变量模型提供的函数(见第三章)。在图2。54中使用tf，例如,如果有两个系统模型。

G1(s)101andG(s). 22S2s5s1一个可以添加他们使用“+”操作符来获得

s212s15G(s)G1(s)G2(s)3 2s3s7s5.

相应的命令是如图2.54（b）, 这里sysl代表Gi(s)和sys2代表G 2（s）,计算两极和0关联到一个传递函数是由操作系统上完成模型对象极点和零点功能分别如图2.55

在下一个示例中,我们将在复杂的平面上获得极、零点位置的分布。这是通过使用零极点图功能,如图2.56所示，在零极点图上，零点用0表示,极点用×表示.如果零极点图函数被调用时没有左论点，,那么这个情节将会自动生成。章节2 系统的数学模型

图2.55 （a）极点和零点作用（b）运用极点和零点的作用计算极点和零点的系统位置

图2。56 零极点函数

实例2。18 传递函数考虑下传递函数

6s21(s1)(s2) G(s)3andH(s)2s3s3s1(s2i)(s2i)(s3).使用一个m文件脚本,我们可以计算函数G（s)的极点和零点，特征方程为H（s），通过H（s）划分G(s），在复杂的平面上，我们也可以获得G(s)/ H(s)的零极点图。

传递函数G（s）/ H（s）的零极点图如图2.57，相关的命令如图2。58,零极点图展示了5个零点位置，但似乎仅仅只有两个极点. 章节2。9 控制系统设计

图2。57 传递函数G（s）/ H（s）的零极点图

传递函数G（s）/ H（s）相关的命令图

不能如此，因为我们知道,对于物理系统极点的数量必须大于或等于零点的数量。使用根函数,我们可以确定,实际上，在s = 1时现在有四个极点。因此，在极零地图上多个极点或多个零在同一位置不能分辨。章节2 系统的数学模型

图2.59 开环控制系统（没有反馈）

框图模型。假设我们已经开发出的数学模型形式的传递函数为一个过程,由G(s)来表示，和一个控制器,通过Gc(s)来表示,和许多其它系统组件可能如传感器和致动器.我们的目标是连接这些组件来构建一个控制系统.

一个简单的开环控制系统可以通过互相连接的一个过程和控制器系列来实现,如图2.59所示。我们可以计算传递函数从R（s）到Y轴（s),如下所示。示例2。19系列连接

通过传递函数G（s）代表这个过程。

G(s)1 500s2通过传递函数Gc(s）代表控制器

Gc(s)s1. s2 58

我们可以使用系列函数来级联两个传输函数G \\（s）和G2（s),如图2.60所示。

传递函数的Gc(s)G（s）计算使用系列函数,如图2.61所示。由此产生的传递函数为

Gc(s)G(s)s1500s31000s2sys

在m脚本中，系统文件名是传递函数名

图2。60 (a)方框图（b)系列函数

章节2。9 仿真系统的使用控制设计软件

图2。61 应用系列函数

图2.62 （a）方框图（b)平行图

图2.63 带有反馈的基本控制系统

方框图通常有平行的传递函数，在这种情况下，函数平行可能非常有用。并行函数描述图2.62

我们可以介绍一个反馈信号控制系统通过关闭循环与单位反馈,如图2.63所示。信号Ea（s）是一个错误的信号，这个信号R（s）是一个标准输入。在此控制系统中，控制器是在前面径并且闭环传递函数是

G(s).

1G(s)H(s)T(s)章节2 系统的数学模型

图2. （a）方框图（b）带有反馈环节的反馈函数

图2。65（a）方框图（b）反馈函数

我们可以利用反馈函数来帮助减少框图过程来计算闭环传递函数，并且为单多个循环控制系统。

通常情况下，闭环控制系统具有统一的反馈，如图2。63。我们可以使用反馈闭环的函数来计算传递函数,通过设置H（s)= 1，使用反馈函数单位反馈如图2..

反馈功能如图2.65和相关的系统配置,,其中包括H(s）反馈路径。如果输入“符号”省略了，那么负面反馈是假定的。章节2.9 仿真系统的使用控制设计软件

图2.66 (a）方框图（b）应用反馈函数

图2。67 带有反馈环节的基本控制系统

示例2。20反馈函数与单位反馈

在这个过程中，G（s）和控制器,Gc（s),如图2。66所示.应用反馈功能,我们首先使用系列函数来计算Gc（s）G(s）,其次函数关闭的反馈循环。这个命令序列显示在图2.66（b)中。闭环传递函数,如图2.66(b)所示

T(s)s1sys.

500s31000s2s1 63

另一个基本的反馈控制配置如图2。67。在这种情况下，控制器是位于反馈路径.闭环传递函数为

G(s).

1G(s)H(s)T(s)示例2.21反馈函数

在这个过程中，G(s）和控制器H(s）,如图2。68所示。计算闭环传递函数与控制器反馈循环。章节2 系统的数学模型

图2.68 应用反馈函数 (a）方框图(b)m文件脚本

这个反馈函数.这个命令序列如图2。68（b）。这个闭环传递函数是

T(s)s1sys.

500s31000s2s1这个函数串联系统、并联系统和反馈可以用于协助框图对多个环方框图操纵。

示例2.22多回路的减少

一个多回路反馈系统如图2.26。我们的目标是计算闭环传递函数T(s)Y(s)R(s) 当

G11(s)s10,G12(s)s1,G(s)s21s1 3s24s4,G4(s)s6.并且

H11(s)ss2,H2(s)2,andH3(s)1.

章节2。9仿真系统使用控制设计软件

图2.69 多个循环快还原

对于这个示例，一个五步过程之后是：步骤1。输入系统传递函数. •步骤2。移动H2 G4后面。步骤3。消除G3G ^嗨循环. 步骤4。消除循环包含H2。

步骤5。消除剩余的循环和计算T（s)。

利用的五个步骤在图2.69,和相应的框图还原如图2.27.执行该命令的结果是

s54s46s35s2sys

12s6205s51066s42517s33218s22196s712我们必须注意在调用此闭环传递函数。传递函数定义为输入—输出关系在极零点取消。如果我们计算极点和零点 T（s），我们发现结果的分子和分母多项式有(s + 1）作为一个常见因素。这必须在我们可以要求我们有闭环传递函数之前取消。为了帮助我们取消零极点，我们必须使用状态方程。状态方程如图2。70所示，删除普通极点零点因素的传递函数，最后一步是取消块还原过程的共同因素，如图2.71.应用程序的状态方程之后,我们发现订单的分母多项式减少6到5,暗示一个极零点取消。章节2系统的数学模型

图2.70 状态方程函数

图2。71 状态方程应用函数

示例2.23电力牵引电动机控制

最后，让我们重新考虑电力牵引电动机系统从示例2。14。这个方块图如图2.44(c)。目标是计算闭环传递函数和调查的回应ω（s)到一个命令ωd（s）。第一步，如图2。72,是计算闭环传递函数ω（s）/ ωd(s= T(s)。闭环特征方程是二阶ωn= 52和ζ= 0.012。自阻尼低,我们预期响应高度振荡.我们可以调查响应ω（t）到一个参考输入ωd(t),利用这个步骤函数. 这个步骤的功能，如图2。73，计算单位阶跃响应的一个线性系统.这个步骤函数是非常重要的,因为控制系统性能规格通常给定的角度单位阶跃响应。

图2。72电力牵引电动机块还原.

章节2。9 仿真系统使用控制设计软件

图2.73 这个步骤函数

图2。74（a)牵引电动机轮速度阶跃响应,（b)m文件的脚本

如果唯一的目的是绘制输出，y（t），我们可以使用阶梯函数没有左侧的参数和获取情节自动与轴标签。如果我们需要y ｛ f)出于任何目的策划，我们必须使用步骤函数左论点,其次是情节函数绘制y（t）。我们定义t作为一个行包含情况向量我们希望输出变量的值y（t)。我们也可以选择t - 0 to t - ffmai结果在一阶跃响应从t - 0到和数量的中间点自动选择。

这一步反应的电力牵引电动机如图2.74.作为预期，车轮速度响应，给出了y(t)，是高度振荡。注意这个输出y(t)= ω(t）。章节2 系统的数学模型

2.10顺序设计的例子：磁盘驱动器读取系统

在1。10节，我们开发了一个初始目标磁盘驱动系统。位置4读头准确在所需的跟踪和在10毫秒，如果可能的话将从一个追踪到另一个。我们需要确定植物，传感器和控制器。我们将获得一个模型的植物G（s)和传感器。磁盘驱动器读者使用一个永磁直流电机旋转读者臂(见图1.29）。直流汽车被称为一个音圈电动机在磁盘驱动行业.读头安装在一个滑块设备，它是连接到手臂如图2。75。一曲(弹簧金属)是用来使头浮离磁盘在差距比100海里更少。这个薄膜磁头读取磁通和提供了一个信号放大器.错误的信号，图2.76（a）是由阅读错误从预先录制的指数跟踪。假设一个准确读头,该传感器具有一个转移，函数H(s)= 1,如图2.76（b).该模型的永久磁铁直流电机和线性放大器如图2.76（b）。作为一个好的近似值，我们使用模型的直流电机电枢可控正如之前所述。

图2.75 为读者展示安装头

图2。76 磁盘驱动器模型方框图

期望顶部位置偏差控制装置制动器和读取臂实际位置 - 放大器直流电机和手臂传感器读头（a）

R(s) E(s）放大器电动机和手臂 Y(s） Ka G(s)Km s(Js1)(LsR) 传感器 H(s）=1 （b）

图2.76 磁盘驱动器模型方框图

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