A Light Weight Rotary Double Pendulum: Maximizing the Domain of Attraction R. W. Brockett* and Hongyi Li* Engineering and Applied Sciences Harvard University Cambridge, MA 02138, USA brockett, hongyi @hrl.harvard.edu f g Abstract— The usual application of feedback stabilization is the structure. based on linearization and the assumption of linear controlla- 2) The problem of avoiding high feedback gains which bility. Least squares optimal control is often used to determine a can lead to torque saturation, noise amplification, and set of feedback gains that will stabilize the system. In addition to the commonly acknowledged challenges arising when this undesirable vibration. methodology is applied to open-loop unstable systems with sig- 3) A reduction in the size of the domain of asymptotic nificant nonlinearities, there are also significant problems which stability. are not given prominence in the literature. These include the The general problem of stabilization via feedback has role of high gain feedback in exciting vibrations in unmodeled modes and the possibility of lightly damped oscillatory modes received considerable attention in both a linear and nonlinear accentuating the effect of the nonlinearities. In this paper we context. Indeed, one of the central questions in classical report some experimental results relating to these points and control theory is the problem of avoiding instability in suggest some techniques for dealing with them. feedback systems. In this paper we are concerned with the I. INTRODUCTION problem of designing a control law that provides an adequate domain of attraction for a nonlinear problem of the form This paper describes an experimental effort concerned with the stabilization of a rotary double link pendulum. This x_ = f(x) + g(x)u problem, and several related ones, has a long history going under the hypothesis that f(0) = 0 ; g(0) = 0. We assume back several decades. In the literature the name of Furuta that f and g are smooth functions and that6 the linearized stands out because of his early interest in these problems system is controllable. Such questions are widely studied and the steady stream of experimental results reported by in the literature, being motivated by a variety of concrete his group [2], [5], [1]. Our approach differes from those problems including experimental work on stabilizing various previously reported not in terms of the basic form of the types of open loop unstable mechanical systems. mechanics, but rather in terms of important aspects of the Because we assume that the linearization of the system at experimental apparatus. Most notably, we are working with x = 0 is controllable, in theory there always exists a linear a system with the following attributes: control law that stabilizes the null solution. Moreover, there 1) All measurement signals are transmitted over a wireless exists a change of coordinates valid in a neighborhood of the link delivering 50 samples per second. This removes origin that makes g equal to a constant vector so we can carry the need for wires connecting the moving parts to a out the first step in the feedback linearization procedure. fixed platform but introduces some latency in the data A precise question along these lines can be formulated as path and limits the sampling frequency. follows. 2) The apparatus is light weight and has low inertia. This Linear Feedback Stabilization Problem: Given a system results in a structure with low stiffness and a tendency of the form x_ = f(x) + G(x)u with f(0) = 0 and g(0) = 0, to vibrate. find the linear time invariant control law u = Kx such6 that 3) The motor is a low torque, completely integrated the null solution of x_ = f(x) + G(x)Kx is asymptotically amplifier/motor/controller unit rather than a high per- stable and the domain of attraction of 0 is as large as possible formance direct drive motor. This makes it necessary in the sense that it contains a ball of the form x a with to design around the possibility of torque saturation. a maximal. jj jj ≤ In attempting to further develop the rotary link double pendulum stabilization technology, several challenges present II. A HIERARCHY OF MECHANICAL SYSTEMS themselves. The principal ones are. Our ultimate goal is to shed further light on the problem 1) The emergence of vibrational modes associated with of enlarging the domain of attraction for nonlinear control unmodeled dynamics associated with the elasticity of systems. The platform that we use for the experiments to be described later consists of a horizontal link, driven by a *This work was supported in part by the National Science Foundation servo motor, and two vertical links that moves freely in the under Yale prime CCR 9980058 and the US Army Research Office under Boston University prime GC169369 NGD, the National Science Foundation plane perpendicular to the horizontal link, as shown in Fig. 4. grant number EIA 0218411 Both this type of system and several other simpler mechanical x 5 systems have been extensively studied in the literature. In order to maximize the usefulness of this paper and to put our results in a broader context, we compare this system with a general version of the well known ball and beam system x 3 and the double pendulum on a cart system. We will see that these form a progression with the ball and beam being the least complex, the double pendulum being of intermediate . complexity and the rotary pendulum being the most complex. x 1 For more about classical mechanics, see Whittaker [4]. A. The Ball, Beam, Cart Fig. 1. Ball, Beam, Cart This is the least complicated system in our hierarchy. Consider the system ball and beam on a cart as shown in Linearization of the Ball-Beam on Cart Fig. 1. The kinetic energy is 1 2 2 F x¨ + V x = ub K = m5[(x _ 5 +x _ 1 cos x3) + (x5x_ 3 x_ 1 sin x3) ] 2 − 1 x_ 5 2 1 2 with + I5( x_ 3) + m3x_ 2 r5 − 2 1 1 0 0 1 2 2 3 + I3x_ 3 m3l3x_ 1x_ 3 cos x3 2 − 0 I + I I5 F = 6 5 3 r5 7 Where x5 is the position of the center of the ball relative 6 − 7 6 0 I5 m + I5 7 to the joint of the beam. r5 is the radius of the ball. x3 is 6 r5 5 2 7 4 − r5 5 the angular position of the beam. x1 is the position of the cart. All xi’s are on the same vertical plane. m3, m5, I3, 0 0 0 1 and I5 are the mass and momentum of inertia of the ball and 2 3 2 3 beam respectively. We assume to have direct control over the V = 0 m3l3g m5g ; b = m3l3 6 − 7 6 7 acceleration of the cart, i.e., x¨1 = u. The potential energy is 6 7 6 7 4 0 m5g 0 5 4 m5 5 − V = m5gx5 sin x3 + m3gl3 cos x3: Transfer function G(s) from u to x can be most easily The Euler-Lagrange equations take the form obtained from the linearization in second order form, F (x)¨x + G(x; x_) + V (x) = ub(x) F s2X(s) + VX(s) = U(s)b with x = [x ; x ; x ]T 1 3 5 This yields the transfer function 1 0 0 2 3 1 2 0 m x2 + I + I I5 2 s 3 F (x) = 6 5 5 5 3 r5 7 6 − 7 2 6 I5 I5 7 6 p22s + p20 7 6 0 m5 + 2 7 G(s) = 6 4 2 7 4 −r5 r 5 6 q4s + q2s + q0 7 5 6 7 6 2 7 6 p32s + p30 7 0 4 2 2 3 4 q4s + q2s + q0 5 G(x; x_) = 2m5x5x_ 3x_ 5 6 7 with 6 2 7 m5x5x_ 3 4 5 2 − p22 = m5r5I5 + m3l3(m5r + I5) − 5 0 2 2 2 3 p20 = m r g − 5 5 V (x) = m3l3g sin x3 + m5gx5 cos x3 2 6 − 7 p32 = m5r (I3 + I5) + m3l3I5 6 7 − 5 4 m5g sin x3 5 p30 = 2m3m5l3r5g and b(x) is given by 2 q4 = m5r5(I3 + I5) + I3I5 1 2 q2 = 2m5r5I5g m3l3g(m5r + I5) 2 3 − 5 b(x) = m3x5 sin x3 + m3l3 cos x3 2 2 6 7 q0 = m5r5g 6 7 − 4 m5 cos x3 5 − B. Cart with the double pendulum C. Rotary Double Pendulum The coupled equations of motion that describe the dynam- The coupled equations of motion that describe the dynam- ics of the pendulum, as shown in Fig. 2, are described by ics of the pendulum, as shown in Fig. 3, are described by the Euler-Lagrangian equation the Euler-Lagrangian equation F (θ)θ¨ + G(θ; θ_) + V (θ) = ub(θ) F (θ)θ¨ + G(θ; θ_) + V (θ) = ub(θ) with θ = [θ ; θ ; θ ]T T 1 2 3 with θ = [θ1; θ2; θ3] 1 0 0 2 2 3 F (θ) = 0 I2 + m3L m3l3L2 cos(θ2 θ3) F (θ) = 2 − 6 7 1 0 0 4 0 m3l3L2 cos(θ2 θ3) I3 5 2 2 3 − 0 I2 + m3L m3l3L2 cos(θ2 θ3) 2 − 6 7 0 4 0 m3l3L2 cos(θ2 θ3) I3 5 2 _ 3 − G(θ; θ_) = g2(θ; θ) 6 _ 7 0 4 g3(θ; θ) 5 2 _2 3 G(θ; θ_) = m3l3L2θ3 sin(θ2 θ3) 0 − 2 3 6 _2 7 (m2gl2 + m3gL2) sin θ2 4 m3l3L2θ2 sin(θ2 θ3) 5 V (θ) = − − − 6 7 4 m3gl3 sin θ3 5 0 − 2 3 and b(θ) is given by (m2gl2 + m3gL2) sin θ2 V (θ) = − 6 7 1 4 m3gl3 sin θ3 5 2 3 − (m2l2 + m3L2)L1 cos θ2 b(θ) = − and b(θ) is given by 6 7 4 m3l3L1 cos θ3 5 − 1 Where 2 3 b(θ) = (m2l2 + m3L2) cos θ2 _ 1 2 − g2(θ; θ) = [2(I2 + m3L2) sin 2θ2 + m3l3L2 cos θ2 6 m l cos θ 7 − 4 3 3 3 5 _2 _2 − sin θ3]θ + m3l3L2θ sin(θ2 θ3) 1 3 − Where I2 and I3 are the moments of inertia of link 2 and link _ 1 _2 g3(θ; θ) = ( I3 sin 2θ3 + m3l3L2 sin θ2 cos θ3)θ1 3 with respect to their joints p2 and p3 respectively.