U of M EE 4237 - State Space Control Laboratory (14 pages)

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State Space Control Laboratory



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State Space Control Laboratory

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Pages:
14
School:
University of Minnesota- Twin Cities
Course:
Ee 4237 - State Space Control Laboratory

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DEPARTMENT OF ELECTRICAL ENGINEERING UNIVERSITY OF MINNESOTA EE 4237 State Space Control Laboratory Experiment 10 Balancing robot using LEGO Mindstorms NXT Objective 1 To balance a small LEGO robot using a gyroscope sensor 2 To study the dynamics of the inverted pendulum Apparatus 1 2 3 LEGO robot PC RobotC software Prelab Report 1 What is an inverted pendulum 2 What is the application area of a self balancing robot Postlab Report Answer the questions asked within or at the end of the procedure 2 1 Modeling of Balancing Robot This section provides a modeling of two wheeled balancing robot First the equations of motion for the wheels are obtained The free body diagram of the robot is shown below The lists of used symbols are shown in Appendix A P H C M Hf x Fig 1 The free body diagram of wheeled balancing robot According to Newton s law of motion the sum of forces on the horizontal F x direction gives Ma x M x H f H Since there is no acceleration in the M 0 1 y direction the sum of Moments around the center of the wheel gives Ia I C H f r From the DC motor dynamics the output torque to the wheels can be obtained C I k d k m k e m Va dt Rm Rm 3 Thus plugging equation 3 into 2 2 3 Hf km ke k I m Va w Rm r Rm r r 4 Substituting equation 4 into 1 to get the equation for the wheels M x k m ke k I m Va w H Rm r Rm r r 5 The angular rotation can be transformed into linear motion by simple transformation r x r x It yields k k k I M w2 x m 2e x m Va H Rm r Rm r r 6 And next the robot chassis can be modeled as an inverted pendulum using Newton s law of motion The free body diagram of the chassis is shown below F xp x l 2 Fig 2 The free body diagram of the chassis The sum of forces in the horizontal direction F x 2 M p x H M p l cos M p l sin 7 4 The sum of forces perpendicular to the pendulum F xp M p x cos H cos P sin M p g sin M p l 8 The sum of moment around the center of mass of pendulum H l cos P l sin I p 9 Combining equation 8 and 9 the final equation is shown as below



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