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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.004 Dynamics and Control II Fall 2007 Problem Set #10 Solution Posted: Friday, Dec. 7, ’07 1. Inverted Pundulum a) Answer: From the following free body diagram, the equation of motion can be found by torque balance between the torque due to gravity, mg × l ; inertial torque ml 2 × θ ¨ ; and the externally applied torque T ( t ). We find ml 2 θ ¨ ( t ) = mgl sin θ ( t ) + T ( t ) . ¨ g T ( t ) ⇒ θ ( t ) − sin θ ( t ) = . l ml 2 Approximating sin θ ( t ) ≈ θ ( t ), we find the linearized equation of motion as ¨ g T ( t ) ⇒ θ ( t ) − θ ( t ) = . l ml 2 b) Answer: Laplace transforming, we obtain s 2 Θ( s ) − g Θ( s ) = T ( s ) ⇒ Θ( s ) = 1 /ml 2 . l ml 2 T ( s ) s 2 − g/l Using the given numerical values and g = 9 . 81 ( m/s 2 ), we find the transfer function as Θ( s ) 1 1 = = . T ( s ) s 2 − 9 ( s + 3)( s − 3) 1 c) Answer: The transfer function has a pole in the right–hand plane, thus the system is unstable. Just like the example during lectures, stabilizing a stick so that it stands vertically without falling is a good example of an unstable system. The point is that if the stick were to stand perfectly vertical, it would not fall; but the slightest angular perturbation away from the vertical position (e.g. due to wind or slight motion of your hand, etc.) will grow rapidly causing the stick to fall from its vertical position. Because of its extreme sensitivity to perturbation the inverted pendulum is the archetypal unstable system. d) Answer: The block diagram with P controller. Imaginary Axis e) Answer: The root locus of the system is as follows: Root Locus −4 −3 −2 −1 0 1 2 3 4 −4 −3 −2 −1 0 1 2 3 4 Real Axis As the gain K increases, the closed–loop poles start moving from s = ± 3 along the real axis. After they meet at s = and move along the imaginary axis. When the poles are on the real axis, the system is unstable. When the poles are on the imaginary axis, the system is marginally stable. Thus this system cannot be stabilized with a proportional (P) controller. f) Answer: The block diagram with PD controller. 2 0 To move the closed–loop poles into the left–hand plane, you want to add a zero in left–hand plane. There are two possibilities: Root Locus Editor for Open Loop 1 (OL1) Root Locus Editor for Open Loop 1 (OL1) 5 1 4 0.8 3 0.6 2 0.4 1 0.2 −1 −0.2 −2 −0.4 −3 −0.6 −4 −0.8 −5 −1 −10 −8 −6 −4 −2 0 2 4 −10 −8 −6 −4 −2 0 2 4 Real Axis Real Axis The zero can be farther than s = − 3 or in between s = and s = − 3. Either one can stabilize the system with a big gain. The second configuration (the zero in between the two poles) will generally need a much bigger gain to stabilize. So the first one (zero to the left of the stable pole) is more practical choice....
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.004 taught by Professor Derekrowell during the Fall '08 term at MIT.
 Fall '08
 DerekRowell
 Mechanical Engineering

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