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Manufacturing Controls. FALL 2001 Lecture 3 . Syllabus. DATE TOPIC NOTES 1. Sep. 20 Mechatronics Design Process Ch. 1 2. Sep. 25 System Modeling and Simulation Ch. 2 3. Sep. 27 Laplace Transforms and Transfer Functions Ch. 2 4. Oct. 2 Electrical Examples Ch.2, Notes
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Manufacturing Controls FALL 2001 Lecture 3 (C) 2001, Ernest L. Hall, University of Cincinnati
Syllabus • DATE TOPIC NOTES • 1. Sep. 20 Mechatronics Design Process Ch. 1 • 2. Sep. 25 System Modeling and Simulation Ch. 2 • 3. Sep. 27 Laplace Transforms and Transfer Functions Ch. 2 • 4. Oct. 2 Electrical Examples Ch.2, Notes • 5. Oct. 4 Mechanical Examples Ch.2, Notes • 6. Oct. 9 Thermal and Fluid Examples, QUIZ 1 (Take Home) • 7. Oct. 11 Sensors and Transducers Ch. 3 • 8. Oct. 16 Advanced MATLAB • 9. Oct. 18 Analog and Digital Sensing Ch. 3, Notes • 10. Oct. 23 Actuating Devices Ch. 4 • 11. Oct. 25 DC Motor Model Ch. 4, Notes • 12. Oct. 30 Boolean Logic Ch. 5 • 13. Nov. 1 Programmable Logic Controllers Ch. 5, Notes • 14. Nov. 6 Stability and Compensators, P, PI and PD Ch. 6 • 15. Nov. 8 PID Controllers Ch. 7 • 16. Nov. 13 QUIZ 2 (In Class - Open Book) • 17. Nov. 15 Practical and Optimal Compensator Design Ch. 8 • 18. Nov. 20 Frequency Response Methods Ch. 9, Notes • 19. Nov. 22 THANKSGIVING HOLIDAY Ch. 9, Notes • 20. Nov. 27 Optimal Design of a Motion Control System Ch. 9, Notes • 21. Nov. 29 QUIZ 3 (In Class - Closed Book) • 22. Dec. FINAL EXAM (In Class - Closed Book) Comprehensive (C) 2001, Ernest L. Hall, University of Cincinnati
Today’s objective • To continue the introduction to systems theory and the use of Laplace transforms with Matlab by studying a simple example, the rigid link pendulum. • By the end of this class you will be able to describe the linear model of the rigid link pendulum, know the parameters of its model, and simulate the operation of this system. (C) 2001, Ernest L. Hall, University of Cincinnati
The Laplace transform F(s) of f(t) is: (C) 2001, Ernest L. Hall, University of Cincinnati
The Laplace convolution theorem states that (C) 2001, Ernest L. Hall, University of Cincinnati
Convolution reduced to product (C) 2001, Ernest L. Hall, University of Cincinnati
Block diagrams • This transform relationship may be used to develop block diagram representations and algebra for linear systems, which is very useful to simplify the study of complicated systems. (C) 2001, Ernest L. Hall, University of Cincinnati
Basic blocks (C) 2001, Ernest L. Hall, University of Cincinnati
Blocks in cascade (C) 2001, Ernest L. Hall, University of Cincinnati
Blocks in parallel (C) 2001, Ernest L. Hall, University of Cincinnati
Feedback block (C) 2001, Ernest L. Hall, University of Cincinnati
Jd2/dt2 Dd/dt (MgL/2)sin T (c) Figure 6. Rigid link pendulum structure diagram. (a) Physical diagram; (b) Components of weight vector; (c)Free body torque diagram; L/2 T L/2 Mg Mgcos (a) Mg Mgsin (b) Rigid link pendulum (C) 2001, Ernest L. Hall, University of Cincinnati
Dynamic response for the mechanical system model of the human leg • The model assumes an input torque, T(t), viscous damping, D at the hip joint, and inertia, J, around the hip joint. • Also, a component of the weight of the leg, W = Mg, where M is the mass of the leg and g is the acceleration of gravity, creates a non-linear torque. • Assume that the leg is of uniform density so that the weight can be applied at the centroid at L/2 where L is the length of the leg. (C) 2001, Ernest L. Hall, University of Cincinnati
For definiteness • Let D = 0.01 lb. s • J= 4.27 ft lb. s2 • W = Mg = 40 pounds • L=3 feet • We will use an input torque amplitude of • T(t)= 75 ft lbs. of torque. (C) 2001, Ernest L. Hall, University of Cincinnati
Robot • The pendulum gives us a good model for a robot arm with a single degree of freedom. • With a rigid link, it is natural to drive the rotation by a torque applied to the pinned end and to represent the mass at the center of mass of the link. • Other physical variations lead to different robot designs. For example, if we mount the rigid link horizontally and then articulate it, we reduce the effect of gravity on the motion. (C) 2001, Ernest L. Hall, University of Cincinnati
Apply the rotational form of Newton's second law about the pinned end to balance the torque • The angle of motion is shown with positive direction counter clockwise • The motion is resisted by three torques • the component of weight is (MgL/2)sin q • the damping torque is D(dq/dt) • and the inertial torque is Jd2q/dt2 • For a bob mass at the end of a link, the inertia is J = ML2. However, for a distributed link the inertia is only ML2/12. (C) 2001, Ernest L. Hall, University of Cincinnati
Jd2/dt2 Dd/dt (MgL/2)sin T (c) Figure 6. Rigid link pendulum structure diagram. (a) Physical diagram; (b) Components of weight vector; (c)Free body torque diagram; L/2 T L/2 Mg Mgcos (a) Mg Mgsin (b) (C) 2001, Ernest L. Hall, University of Cincinnati
Using the small angle approximation, sin q = q gives (C) 2001, Ernest L. Hall, University of Cincinnati
The previous equation is a linear form. Since it is linear, we can take the Laplace transform to obtain the transfer function between the output and input. (C) 2001, Ernest L. Hall, University of Cincinnati
Simplifications • It is also interesting to show how the equations simplify for a pendulum mounted in a horizontal plane rather than a vertical plane. • For a horizontally articulated pendulum or robot, the weight is perpendicular to the motion and does no work so the equation simplifies. (C) 2001, Ernest L. Hall, University of Cincinnati
Horizontal pendulum (C) 2001, Ernest L. Hall, University of Cincinnati
SCARA* Application - assembly & insertion Configuration – RRP Percentage - 15 Advantages - horizontal compliance; high speed; no gravity effect Disadvantage - complex kinematics; variable resolution; limited vertical motion *Selective Compliance Articulated Robot for Assembly (Source for the percent of use: V.D. Hunt, Robotics Sourcebook, New York: Elsevier, 1988.) (C) 2001, Ernest L. Hall, University of Cincinnati
Configurations fit applications • Cartesian • Application – assembly and machine loading • Configuration – PPP • Percentage – 18 • Advantage – equal resolution, simple kinematics • Disadvantage – Poor space utilization (C) 2001, Ernest L. Hall, University of Cincinnati
We can also develop a Matlab m-file solution to this linear differential equation. • J = 4.27; • D = 0.1; • M = 40/32.2; • g = 32.2; • L = 3; • num = [0, 180/3.14159]; % 180/3.14159 is to translate radians into degrees • den = [J, D, M*g*L/2 ]; • t= 0:0.1:10; • impulse(num,den,t); %find impulse response • grid on; • xlabel=(‘Degrees’); • ylabel=(‘Time(seconds)’); • title(‘Unit impulse response of the rigid link pendulum'); (C) 2001, Ernest L. Hall, University of Cincinnati
Impulse response (C) 2001, Ernest L. Hall, University of Cincinnati
Try it! • Copy m-file into Matlab. • Run the m-file • Compare the result • What is the period? • What is the frequency? • Change the time to 100 seconds. • Does the damping effect become more clear? (C) 2001, Ernest L. Hall, University of Cincinnati
Any questions? (C) 2001, Ernest L. Hall, University of Cincinnati