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Unformatted text preview: ECE 360 System Modeling and Control Design Project Due: Monday, December 10, 2007 I. System Description The IP02 inverted pendulum from Quanser Consulting, Inc., consists of a motordriven cart which is equipped with two encoders. One encoder measures the position of the cart via a pinion which meshes with the track. The other encoder measures the angle of the pendulum which is free to swing in front of the cart. The objective of this project is to design a controller that starts with the pendulum in the “down” position then swings it up and maintains it upright. Thus, the desired controller has two parts: a “swingup” controller which will oscillate the cart until it has built up enough energy in the pendulum that it is almost upright. (This controller will be provided for your project.) When the pendulum is almost upright, a “balance” controller is turned on and is used to maintain the pendulum vertical. You need to design this stabilizing controller subject to several transient specifications. II. Nonlinear Model Figure 1: Simplified Picture of CartPendulum System The equations of motion for the cartpendulum system shown in Figure 1 can be derived from the EulerLagrange equations 1 F x L x L dt d = ∂ ∂ ∂ ∂ . (1) . = ∂ ∂ ∂ ∂ θ θ L L dt d (2) where x : Cart position (m) θ : Pendulum angle (rad) F : Input force to the cart (N) p M : Mass of pendulum rod (kg) c M : Mass of cart (kg) L : Length of rod (m) g : Gravitational constant (g = 9.81 m/s 2 ) and L is the Lagrangian of the system defined as V T L = (3) where T is the total kinetic energy of the system and V is the total potential energy of the system. According to the referential system in Figure 1, the total kinetic energy T of the system is . 2 . 2 . 2 2 1 2 1 2 1 θ J r M x M T p p c + + = (4) and the total potential energy of the system is cos 2 p L V M g θ = (5) and where p r : Position of the center of gravity of the pendulum (m) in the chosen referential p M : Mass of pendulum rod (kg) c M : Mass of cart (kg) J : Mass moment of inertia of pendulum rod around its center of gravity (kgm 2 ) g : Gravitational constant (g = 9.81 m/s 2 ) The mass moment of inertia J of a slender rod of length L and mass M p with respect to an axis perpendicular to the rod and passing through the center of gravity is 2 12 1 L M J p = (6) Assignment I : Using the following kinematic equations, θ sin 2 L x x p + = (7) 2 θ cos 2 L y p = (8) θ θ cos 2 . . . L x x p = (9) θ θ sin 2 . . L y p = (10) 2 . 2 . . p p p y x r + = (11) form the Lagrangian of the system and then derive the following nonlinear equations of motion of the cartpendulum system 2 . .. .....
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This document was uploaded on 11/01/2011 for the course ECE 360 at Boise State.
 Spring '08
 STAFF

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