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Unformatted text preview: 1) (60 Points) The second order equations of motion of the cartpole system are: 2 ) (cos cos 2 cos sin + + = m M l g m F x ) ) (cos ( cos sin 2 cos sin 2 + + + = m M l l g m m M m F where M denotes the mass of the cart, m the mass of the pole, g the gravity constant, and l the distance of the pole center of mass from the pole hinge. a) Reformulate the equations of motion as a set of first order differential equations, using T x x x = Let 1 x = x , 2 x = x , 3 x = , 4 x = + + + + + = = = x x x x x x x x x x x x x x x x x x m M l l g m m M m F m M l g m F x x x x x 3 2 4 4 4 2 3 4 4 1 2 4 4 2 3 4 4 4 3 2 1 ) ) (cos ( cos sin cos sin ) (cos cos cos sin 4 3 2 1 , b) Derive the equilibrium points of the system without the input, i.e., Ax x = The conditions of the equilibrium points: F=0, 4 3 2 1 = = T x x x x x 1 = x , 3 = x =&gt; ) (cos cos cos sin 2 4 4 2 3 4 4 = + + x x x x x m M l g m F (1) ) ) (cos ( cos sin cos sin 2 4 4 4 2 3 4 4 = + + + x x x x x x m M l l g m m M m F (2) Substitute 1 = x , 3 = x , F=0 in (1) and (2). We can get x x g 4 4 cos sin = 0 x g m m M 4 cos + = 0 Simplified equation: x 4 cos = 0 x 4 = 2 or 2  x 1 = 0, x 2 = arbitrary position, x 3 = 0, x 4 = 2 or 2  c) Give the liberalized equations of the system in form of the generic linear system equations: Bu Ax x + = + + + + + = = = x x x x x x x x x x x x x x x x x x m M l l g m m M m F m M l g m F x x x x x 3 2 4 4 4 2 3 4 4 1 2 4 4 2 3 4 4 4 3 2 1 ) ) (cos ( cos sin cos sin ) (cos cos cos sin 4 3 2 1 , , [ ] Bu Ax F m M l m m M m x x x x A x x x x x x x x + = + + + = = ) ) (cos ( sin 1 ) (cos 1 4 3 2 1 2 4 4 2 4 4 3 2 1 In order to get the linearized equations of matrix A to express x , we need to use Taylor expansion: ) )( ( ' ) ( ) ( a x a F a F x F + = The position of equilibrium point is x , T x x x x 4 3 2 1 is and [ ] T x x x x 4 3 2 1 is =&gt; x x = =&gt; x x + = =&gt; assume a = x , x = x + =&gt; ) )( ( ' ) ( ) ( x x x x x F F F + + = + = =&gt; x ) is equilibrium point,...
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 Spring '08
 Schaal

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