单倒置摆控制系统的状态空间设计例子.pdf

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1、现代控制理论基础期中作业题目：单倒置摆控制系统的状态空间设计姓名：谢海波学号：P091812925 专业班级：电气工程及其自动化3 班指导老师：刁晨西北民族大学电气工程学院单倒置摆控制系统的状态空间设计摘要：随着现代科学技术的迅速发展，对自动控制的精度、速度、范围及适应能力的要求越来越高，从而推动了自动控制理论和技术的发展。本次设计通过对单倒置摆系统的稳定性分析，建立出相应的状态空间状态模型；选取小车的位移及速度、摆的角位置和角速度作为状态变量、控制力作为输入变量、小车瞬时位置作为输出变量，建立状态空间描述，通过对状态空间描述的能空性和稳定性分析，可以确定当倒置摆向左或向右倾倒时，能否通过控制

2、作用使它回复到原来直立位置。通过对系统的稳定性分析发现，没有反馈的控制系统是不稳定的，所以要设置反馈；本次设计采用了全反馈，为了实现单倒置摆控制系统的全状态反馈，必须获取系统的全部状态，因而可以设计全维状态观测器来实现。然而控制系统的小车位移可以用传感器测量，实际中无需估计，所以又可以设计降维状态观测器来实现状态反馈。本次设计涉及到系统状态空间方程的建立、系统的运动分析、能控性和能观测判据的的使用，状态反馈配置系统极点的综合、状态观测器实现状态反馈等问题，体现了状态空间分析法在现代控制理论中的重要地位以及在工程中的具体运用。关键词：反馈、状态空间、观测器、matlab 仿真一、设计要求：如图

3、1所示，为单倒置摆系统的原理图。设摆的长度为L、质量为 m，用铰链安装在质量为 M的小车上。小车有一台直流电动机拖动，在水平方向对小车施加控制力u,相对参考系产生位移z。若不给小车施加控制力，则倒置摆会向左或向右倾倒，因此，它是一个不稳定系统。通过设计状态空间模型和状态反馈以及观测器，对控制系统进行控制；控制的目的是，当倒置摆无论出现向左或向右倾倒时，通过控制直流电动机，使小车在水平方向运动，将倒置摆保持在垂直位置上。图 1 二、设计需求分析：对所要设计的系统分析知道，系统中单倒置摆能否处于垂直位置是由控制力（即输入量）决定的，反过来输入量的大小，输入量作用的时间又是由单倒置摆的位置（输出量Z

4、 和角度）决定的，只有两者都取合适的变量值时，才能使系统真正处于稳定状态。三、设计步骤：（一）对设计系统进行力学分析（抽象出研究对象）：为简化问题，工程上可以忽略一些次要因素。在本例中，我们为了简化问题，方便研究系统空间的设计问题，忽略了摆杆质量、执行电动机惯性以及摆轴、轮轴、轮与接触面之间文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE

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6、码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9

7、 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1

8、ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文

9、档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1

10、E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I

11、1 ZE7B10G4M5V3的摩擦及风力。设小车的瞬时位置为z，倒置摆出现的偏角为，则摆心瞬时位置为)sin(lz。在控制力u 的作用下，小车及摆均产生加速运动，根据牛顿第二定律，在水平直线运动方向的惯性力应与控制力u 平衡，则有ulzdtdmdtzdM)sin(2222即umlmlzmMsincos)(2(1)由于绕摆旋转运动的惯性力矩应与重力矩平衡，因而有sincos)sin(22mgllzdtdm即sincossincoscos2gllz（2）式（1）、式（2）两个方程都是非线性方程，需作线性化处理。由于控制的目的是保持倒置摆直立，因此，在施加合适u 的条件下，可以认为、均接近于零，此时

12、sin，1cos,且可以忽略2项，于是有umlzmM)(（3）glz（4）连联立求解式（3）、式（4），可得uMMmgz1（5）uMlgMlmM1)(（6）消去中间变量，可得输入量为u、输出量为z 的微分方程为uMlguMzMlgmMz1)(4（7）综合上述的分析，可抽象出系统的研究对象为：位移 z、小车的速度z、摆的角速度 及其角速度的。系统的研究对象抽象成这四四个变量建立空间状态方程，并分析被控对象的特性。（二）建立倒置摆的状态空间模型：在上一步中，我们已经选取了四个研究对象作为状态变量，它们分别为：位移z、小车的速度z、摆的角速度 及其角速度的。Z 为输出变量，在考虑zzdtd，dtd及

13、式（5）、（6）、（7），可列出倒置摆的状态空间模型表达式为：文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U

14、2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G

15、4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3

16、I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N

17、6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B1

18、0G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:C

19、R3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3uMlMMlgmMMmg10100)(0010000000010 xx（8a）x0001y(8b)式中Tzzx为方便研究，假定系统的参数M=1kg,m=0.1kg,l=1m,2/81.9smg,则系统状

20、态方程中参数矩阵为：01100100001000010A,1010b，0001c（9）此时倒置摆的状态空间模型表达式为：u101001100100001000010 xx（10）x0001y画出状态结构图如图2 所示（三）对模型进行分析（即对被控对象进行分析）在建立完模型后我们需要对模型进行分析。作为被控制的倒置摆，当它向左或向右倾倒时，能否通过控制作用使它回复到原直立位置，这取决于其能控性。因此我们首先分析它的能控性。1.能控性分析根据能控性的秩判据，并将式（9）的有关数据带入该判据，可得4bAbAAbbM32rankrank（11）因此，单倒置摆的运动状态是可控的。换句话说，这意味着总存在

21、一控制作用u,将非零状态x转移到零。2.稳定性分析由单倒置摆系统的状态方程，可求的其特征方程为：0)11(22AI（12）文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B1

22、0G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:C

23、R3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ

24、3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7

25、B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码

26、:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9

27、HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3解得特征值为0,0，11，-11。四个特征值中存在一个正根，两个零根，这说明单倒置摆系统，即被控系统不稳定的。3.仿真:采用 matlab 对

28、被控对象进行仿真，按照图2 进行模块连接，仿真结果为图3 所示，由图 3 的曲线可知可知，系统不稳定，不能到达控制目的。图 2、单倒置摆开环系统结构图（仿真模型）仿真结果：文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I

29、5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6

30、U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10

31、G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR

32、3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3

33、N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B

34、10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3图 3 单倒置摆开环系统的个变量的阶跃响应曲线由上面两个方面对系统模型进行分析，可知被控系统

35、是具有能控性的，但是被控系统是不稳定的，需对被控系统进行反馈综合，使四个特征值全部位于根平面S左半平面的适当位置，以满足系统的稳定工作已达到良好、静态性能的要求。（四）确定方案为了满足系统的稳定工作已达到良好、静态性能的要求。因此我们需要设计两种控制器方案来使系统到达控制的目的，分别为：全维状态观测器的设计和降维观测器的设计采用全状态反馈。取状态变量z、z、为反馈信号，状态控制规律为kxvu（13）设3210kkkkk式中，30 kk分别为 z、z、反馈至参考输入v 的增益。则闭环控制系统的状态方程为vbxbkAx)(设置期望闭环极点为-1，-2，-1+i,-1-i 由状态方程的特征多项式和期

36、望特征多项式的系数相等可求得：0k=-0.4，1k=-1，2k=-21.4，3k=-6 画出状态反馈系统结构图如图4 所示，按照图4 仿真模型进行matlab 仿真，仿真结果曲线如图 5 所示。文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5

37、V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B

38、5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2

39、O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4

40、M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I

41、5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6

42、U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3图 4 单倒置摆全反馈系统结构图（仿真模型）仿真结果如下文档编码:CR

43、3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3

44、N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B

45、10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:

46、CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 H

47、J3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE

48、7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编

49、码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3文档编码:CR3I5B5U1E9 HJ3N6U2O4I1 ZE7B10G4M5V3图 5 单倒置摆全状态反馈的阶跃响应曲线由仿真结果可知，单倒置摆的全状态反馈为稳定的闭环系统。观察仿真曲线：单位阶跃的作用下，输出变量z 逐渐趋于某一常数(-2.5)，状态变量 则是逐渐趋于0。当参考输入v单位阶跃时，状态向量在单位阶跃的作用下相应逐渐趋于稳定，这时摆杆回到原始

50、位置（即=0），小车也保持稳定（即z=某一常数）。如果不将4 个状态变量全用作反馈，该系统则不能稳定。方案一：全维观测器的设计为实现单倒置摆控制系统的全状态反馈，必须获取系统的全部状态，即z、z、的信息。因此，需要设置z、z、的四个传感器。在实际的工程系统中往往并不是所有的状态信息都是能检测到的，或者，虽有些可以检测，但也可能由于检测装置昂贵或安装上的困难造成难于获取信息，从而使状态反馈在实际中难于实现，甚至不能实现。在这种情况下设计全维状态观测器，解决全维状态反馈的实现问题。（1）判定系统状态的能观测性将式（9）中的数值代入能观测性秩判据，得：4)()(TTTTTTrankrankcAcAc