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Unformatted text preview: Lecture 12 – Wednesday, Oct. 3 2.004 Fall ’07 Velocity control system R 2 R 1 ≡ K ; R = 1; K m = K v = 1; J = 0 . 1; r = 0 . 1; D = 0 . 5; M = 9; f v = 5 . Lecture 12 – Wednesday, Oct. 3 2.004 Fall ’07 Block diagram V ( s ) . 1 K = V ref ( s ) s + 2 + 0 . 1 K Plant and controller + E ( s ) Input Actuating Output signal (error) Feedback R ( s ) G ( s ) C ( s ) Input 1 + G ( s ) H ( s ) Output Figure by MIT OpenCourseWare. V ref 0.1 K s+ 2 v(t) 1 V tach MIT Mechanical Engineering Department – 2.004 (a) Supplement to Lecture 12 Plant and controller E ( s ) + Input Output Actuating signal (error) Feedback G ( s ) 1 + G ( s ) H ( s ) C ( s ) R ( s ) Input Output Figure by MIT OpenCourseWare. V ref V tach 0.3162K s+ 2 1 v(t) MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.004 Dynamics and Control II Fall 2007 Supplement to Lecture 12 Dynamics of a DC Motor with Pinion Rack Load and Velocity Feedback The system given in Lecture 12 (Figure a on cover page) was a DC motor connected to a pinion rack with a mass–damper load. The motor was connected in a velocity– feedback configuration with a differential op–amp providing the error signal. Here we summarize the analysis of this system. Plant transfer function We begin by analyzing the plant, i.e. , the motor connected directly to a voltage source input v s ( t ) and velocity output v ( t ). The result that we are about to derive is referred to as the “plant model.” After we know the plant model, we can proceed to model the entire system which includes the velocity feedback loop. To analyze the plant, we begin by writing the by–now familiar electrical equation of motion that is derived from KVL on the electrical circuit including the motor: v s ( t ) i ( t ) R v e ( t ) = 0 ⇒ v s ( t ) = i ( t ) R + K v ω ( t ) , (1) where we used the relationship v e = K v ω for the motor’s back–emf. In the Laplace domain, this equation becomes V s ( s ) = I ( s ) R + K v Ω( s ) . (2) Next we must write the mechanical equation of motion as torque balance on the motor’s shaft. The motor’s torque is provided by the electrical current through the windings, and is K m i ( t ). There are three torques counteracting the motor: (i) the pinion inertia J , (ii) the rotational viscous damper D , and (iii) the torque exercised by the translational elements (mass M and translational viscous damper f v ) via the pinion gear. Since at the moment we do not know how to express torque (iii), let us denote it by T p . The time–domain equation of motion of the motor shaft then becomes K m i ( t ) = Jω ˙ ( t ) + Dω ( t ) + T p ( t ) . (3) In the Laplace domain, the equation of motion is re–written as K m I ( s ) = Js Ω( s ) + D Ω( s ) + T p ( s ) . (4) We still have to find T p ( s ). This is done by force–balance on the translational part of the load, i.e. the pinion’s rack that carries the mass M . If F p ( t ) denotes the force by the pinion on the rack, then...
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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

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