倒立摆的中英文翻译.doc

《倒立摆的中英文翻译.doc》由会员分享，可在线阅读，更多相关《倒立摆的中英文翻译.doc（5页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、【精品文档】如有侵权，请联系网站删除，仅供学习与交流倒立摆的中英文翻译.精品文档.The Inverted Pendulum SystemThe inverted pendulum system is a popular demonstration of using feedback control to stabilize an open-loop unstable system. The first solution to this problem was described by Roberge 1 in his aptly named thesis, The Mechanical Se

2、al. Subsequently, it has been used in many books and papers as an example of an unstable system.Siebert 2, pages 177182 does a complete analysis of this system using the Routh Criterion, by multiplying out the characteristic equation as a polynomial of s and studying the coefficients. Although corre

3、ct, this approach is unnecessarily abstruse. This system is the ideal root-locus analysis example.Figure 1: Geometry of the inverted pendulum systemConsider the inverted pendulum system in Figure 1. At a pendulum angle of from vertical, gravity produces an angular acceleration equal to, and a cart a

4、cceleration of produces an angular acceleration of Writing these accelerations as an equation of motion, linearizing it, and taking its Laplace Transform, we produce the plant transfer function G(s), as follows:where the time constant is defined as This transfer function has a pole in the right half

5、-plane, which is consistent with our expectation of an unstable system.We start the feedback design by driving the cart with a motor with transfer function M(s) and driving the motor with a voltage proportional to the angle . Including the familiar motor transfer functionFigure 2: Root-locus plot of

6、 pendulum and motor, L(s) = M(s)G(s)with the plant G(s), we get a root locus with one pole that stays in the right half-plane. Using normalized numbers, we get the root-locus plot as is seen in Figure 2.In order to stabilize the system, we need to get rid of the remaining zero at the origin so that

7、the locus from the plant pole on the positive real axis moves into the left half-plane. Thus our compensator must include a pole at the origin. However, we should balance the added compensator pole with an added zero, so that the number of poles less the number of zeros remains equal to two, leaving

8、 the root-locus asymptotes at (otherwise, the asymptotes would be and which eventually lead the poles into the right half-plane). Thus we use a compensatorand we assume that M K L. The block diagram of the system is shown in Figure 3, and the root-locus plot becomes as in Figure 4 (note that since t

9、here is an inversion in G(s), we draw the block diagram with a positive summing junction).Figure 3: Block diagram of the compensated systemFigure 4: Root-locus plot of pendulum with integrating compensator, L(s) = K(s)M(s)G(s)Siebert explains that a physical interpretation for the need for this inte

10、grator arises from the fact that we are using a voltage-controlled motor. Without the integrator a constant angular error only achieves a constant cart velocity, which is not enough to make the pendulum upright. In order to get “underneath” the pendulum, the cart must be accelerated; therefore, we n

11、eed the integrator.This system is now demonstrably stable, however, the root locus is awfully close to the jw-axis. The resulting closed-loop system has a very low margin of stability and would have very oscillatory responses to disturbances. An easy fix to this problem is to decrease the motor time

12、 constant with velocity feedback, which moves the centroid of the asymptotes to the left. The root-locus plot of this system is seen in Figure 5.Unfortunately, there is still a problem with this system, albeit subtle. Consider the closed-loop transfer function from to in Figure 3.The poles at the or

13、igin makes the system subject to drift. With these integrators, Murphys Law guarantees that the time response of x(t) will grow without bound, and the cart will quickly run out of track.The solution is positive feedback around the motor and compensator. This feedback loop has the effect of moving th

14、e poles off the origin, thus preventing the pole/zero cancellations that are the source of this uncontrollable mode. The root-locus plot of the corrected system appears in Figure 6.Figure 5: Root-locus plot of pendulum with improved motor time constantFigure 6: Root-locus plot of pendulum with posit

15、ion compensationSiebert notes that this positive feedback causes the motor to initially make deviations in x(t) worse, but that this behavior is the desired effect. When balancing a ruler in your hand, to move the ruler to the right, you must first move your hand sharply to the left, pointing the ru

16、ler to the right, so that when you catch the ruler, you have moved both your hand and ruler to the right.Physically, the pendulum is stabilized at a small angle from vertical, such that it always points toward the center of the track. Thus, the pendulum is always falling toward the center of the tra

17、ck, and the only possible equilibrium is a vertical pendulum in the middle of the track. If the cart is to the left of the track center, the control will stabilize the pendulum pointing to the right, such that it then falls a little more to the right. To catch the falling pendulum, the cart must mov

18、e to the right (back toward the center). That motion is the desired behavior!References1 James K. Roberge. The mechanical seal. Bachelors thesis, Massachusetts Institute of Technology, May 1960.2 William McC. Siebert. Circuits, Signals, and Systems. MIT Press, Cambridge, Massachusetts, 1986.倒立摆系统倒立摆

19、系统的是一个使用反馈控制使开环不稳定系统稳定的典型实证。Roberge 1在他的名字叫“The Mechanical Seal”的论文中第一次解决了这个问题。随后，它被作为一个不稳定系统的例证用在了很多书和报纸上。Siebert 2, pages 177-182 用劳斯判据对这个系统做了一个完整的分析，增加了特征方程式作为多项式的s以及系数的研究。虽然正确，但是却没有必要用这个难解的方法。该系统是理想的根轨迹分析的例子。图1：倒立摆的几何图形考虑到图1中的倒立摆系统。在摆从垂直到时，重力产生一个重力加速度等于，与小车的速度产生一个角加速度。写出这些加速度作为一个运动方程，然后线性化，在进行拉普

20、拉斯变换，我们就得到了传递函数G(s)，如下：定义时间常数为。这个传递幻数有一个极点在右半平面，与我们的系统不稳定的期望是一致的。我们开始用传递函数M(s)和均衡的电压控制电机使小车移动设计反馈直到角度为。其中常见的电机传递函数为图2：摆和电机的根轨迹图，L(s) = M(s)G(s)用G(s)，我们到到一极停留在右半平面的根轨迹。采用标准化法，我们得到的根轨迹如图2所示。为了稳定系统，我们必须先除去其余的零点从而使轨迹从plant极的正实轴移动到左半平面上。但是，我们应该在增加的零点上平衡的添加补偿极点，使得极点的数量比零点的数量少2个，将根轨迹的渐近线留在（否则，最终导致右半平面的极点渐近

21、线在和）。从而，我们得到一个补偿器我们假设。该系统的框图如图3所示，根轨迹成为图4（注意到有一个G(s)的反向，我们绘制了真实的连接图）。图3：补偿系统框图图4：摆与整合补偿的根轨迹图，L(s) = K(s)M(s)G(s)西伯特解释说，事实上我们正在使用一个电压控制的电机物理解释我们需要额这种积分的出现。没有积分常数的转角误差只能实现恒定的车速度，而不足以使摆直立。为了使下面的摆直立，车必须加快速度，因此，我们需要积分。此系统现在证明是稳定的，然而，根轨迹是非常接近jw轴的。由此产生的闭环系统有一个非常低的稳定余裕，将在很大程度上受到振荡的干扰。有一个简单的办法来解决这个问题，就是降低电机的

22、时间常数与速度反馈，向左移动重心的渐近线。该系统的根轨迹图，在图5可见。不幸的是，虽然很模糊，但此系统依然存在一个问题。考虑到从到的闭环传递函数在原点的极点使系统漂移。墨菲定理和这些积分保证的响应时间将没有约束的增长，车将很快的开出跑道。它的解决办法是电机和补偿器的积极反馈。这个反馈回路用移动极点的办法关闭原点，从而在这个无法控制的模式下防止极点/原点的取消。校正系统的根轨迹图显示在图6。图5：摆改善时间常数的根轨迹图6：摆位置补偿的根轨迹图西伯特指出，这种积极的反馈，使电机最初的偏差更大，但这也是预期的效果。此时把标尺平放在你的手上，当你的手移动到标尺右边前，你必须向将你的手猛然的移向左边，将标尺指向右边，当你抓住标尺时，你必须将你的手和标尺一起移向右边。物理上，摆稳定在一个垂直的小角度，这样它总是指向接近中心的轨道。因此，摆总是指向接近中心的轨道，唯一可能的平衡就是一个摆垂直的立在轨道的中心。如果车在轨道中心的左侧，那么摆就会在它的控制下指向右侧，以致于它会向右偏离一些。为了赶上落下的摆，车必须向右移动（回向中心）。这种运动正是所期待的！参考文献1 James K. Roberge. 机械密封。学士论文，麻省理工学院，1960年5月2 William McC. Siebert. 电路、信号和系统。麻省理工学院出版社，剑桥，马萨诸塞州，1960年。