機械設計外文翻譯-倒立擺系統(tǒng)?【中文1360字】【PDF+中文WORD】,中文1360字,PDF+中文WORD,機械設計,外文,翻譯,倒立,系統(tǒng),中文,1360,PDF,WORD 【中文1360字】 倒立擺系統(tǒng) 倒立擺系統(tǒng)的是一個使用反饋控制使開環(huán)不穩(wěn)定系統(tǒng)穩(wěn)定的典型實證。Roberge [1]在他的名字叫“The Mechanical Seal”的論文中第一次解決了這個問題。隨后，它被作為一個不穩(wěn)定系統(tǒng)的例證用在了很多書和報紙上。 Siebert [2, pages 177-182] 用勞斯判據(jù)對這個系統(tǒng)做了一個完整的分析，增加了特征方程式作為多項式的s以及系數(shù)的研究。雖然正確，但是卻沒有必要用這個難解的方法。該系統(tǒng)是理想的根軌跡分析的例子。 圖1：倒立擺的幾何圖形 考慮到圖1中的倒立擺系統(tǒng)。在擺從垂直到時，重力產(chǎn)生一個重力加速度等于，與小車的速度產(chǎn)生一個角加速度。寫出這些加速度作為一個運動方程，然后線性化，在進行拉普拉斯變換，我們就得到了傳遞函數(shù)G(s)，如下： 定義時間常數(shù)為。這個傳遞幻數(shù)有一個極點在右半平面，與我們的系統(tǒng)不穩(wěn)定的期望是一致的。 我們開始用傳遞函數(shù)M(s)和均衡的電壓控制電機使小車移動設計反饋直到角度為。其中常見的電機傳遞函數(shù)為 圖2：擺和電機的根軌跡圖，L(s) = M(s)G(s) 用G(s)，我們到到一極停留在右半平面的根軌跡。采用標準化法，我們得到的根軌跡如圖2所示。 為了穩(wěn)定系統(tǒng)，我們必須先除去其余的零點從而使軌跡從plant極的正實軸移動到左半平面上。但是，我們應該在增加的零點上平衡的添加補償極點，使得極點的數(shù)量比零點的數(shù)量少2個，將根軌跡的漸近線留在（否則，最終導致右半平面的極點漸近線在和）。從而，我們得到一個補償器 我們假設。該系統(tǒng)的框圖如圖3所示，根軌跡成為圖4（注意到有一個G(s)的反向，我們繪制了真實的連接圖）。 圖3：補償系統(tǒng)框圖 圖4：擺與整合補償?shù)母壽E圖，L(s) = K(s)M(s)G(s) 西伯特解釋說，事實上我們正在使用一個電壓控制的電機物理解釋我們需要額這種積分的出現(xiàn)。沒有積分常數(shù)的轉(zhuǎn)角誤差只能實現(xiàn)恒定的車速度，而不足以使擺直立。為了使下面的擺直立，車必須加快速度，因此，我們需要積分。 此系統(tǒng)現(xiàn)在證明是穩(wěn)定的，然而，根軌跡是非常接近jw軸的。由此產(chǎn)生的閉環(huán)系統(tǒng)有一個非常低的穩(wěn)定余裕，將在很大程度上受到振蕩的干擾。有一個簡單的辦法來解決這個問題，就是降低電機的時間常數(shù)與速度反饋，向左移動重心的漸近線。該系統(tǒng)的根軌跡圖，在圖5可見。 不幸的是，雖然很模糊，但此系統(tǒng)依然存在一個問題?？紤]到從到的閉環(huán)傳遞函數(shù) 在原點的極點使系統(tǒng)漂移。墨菲定理和這些積分保證的響應時間將沒有約束的增長，車將很快的開出跑道。 它的解決辦法是電機和補償器的積極反饋。這個反饋回路用移動極點的辦法關閉原點，從而在這個無法控制的模式下防止極點/原點的取消。校正系統(tǒng)的根軌跡圖顯示在圖6。 圖5：擺改善時間常數(shù)的根軌跡 圖6：擺位置補償?shù)母壽E圖 西伯特指出，這種積極的反饋，使電機最初的偏差更大，但這也是預期的效果。此時把標尺平放在你的手上，當你的手移動到標尺右邊前，你必須向?qū)⒛愕氖置腿坏囊葡蜃筮?，將標尺指向右邊，當你抓住標尺時，你必須將你的手和標尺一起移向右邊。 物理上，擺穩(wěn)定在一個垂直的小角度，這樣它總是指向接近中心的軌道。因此，擺總是指向接近中心的軌道，唯一可能的平衡就是一個擺垂直的立在軌道的中心。如果車在軌道中心的左側(cè)，那么擺就會在它的控制下指向右側(cè)，以致于它會向右偏離一些。為了趕上落下的擺，車必須向右移動（回向中心）。這種運動正是所期待的！ 參考文獻 [1] James K. Roberge. 機械密封。學士論文，麻省理工學院，1960年5月 [2] William McC. Siebert. 電路、信號和系統(tǒng)。麻省理工學院出版社，劍橋，馬薩諸塞州，1960年。 The Inverted Pendulum System The 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 Seal.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 correct,this approach is unnecessarily abstruse.This system is the ideal root-locus analysis example.Figure 1:Geometry of the inverted pendulum system Consider the inverted pendulum system in Figure 1.At a pendulum angle of from vertical,gravity produces an angular acceleration equal to,and a cart acceleration 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-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 function Figure 2:Root-locus plot of 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 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 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 compensator and 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 there is an inversion in G(s),we draw the block diagram with a positive summing junction).Figure 3:Block diagram of the compensated system Figure 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 integrator 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 need 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 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 origin 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 the 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 constant Figure 6:Root-locus plot of pendulum with position compensation Siebert 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 ruler 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 track,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 move to the right(back toward the center).That motion is the desired behavior!References 1 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.