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# Meeting w4 chapter 2 part 2

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### Meeting w4 chapter 2 part 2

1. 1. Chapter 2 – Analog Control System (cont.) <ul><li>Electrical Elements Modelling </li></ul><ul><li>Mechanical Elements Modelling </li></ul>
2. 2. 4. Electrical Elements Modelling
3. 3. Example – RLC Network <ul><li>Determine the transfer function of the circuit. </li></ul><ul><li>Solution: </li></ul><ul><li>All initial conditions are zero. Assume the output is v c (t) . </li></ul><ul><li>The network equations are </li></ul>
4. 4. cont. <ul><li>Laplace transform the equation: </li></ul>Therefore,
5. 5. Potentiometer <ul><li>A potentiometer is used to measure a linear or rotational displacement. </li></ul>Linear Rotational
6. 6. Rotational Potentiometer <ul><li>The output voltage, </li></ul><ul><li>Where Kp is the constant in V/rad. </li></ul><ul><li>Where  max is the maximum value for  (t) . </li></ul><ul><li>The Laplace transform of the equation is </li></ul>
7. 7. Tachometer <ul><li>The tachometer produces a direct current voltage which is proportional to the speed of the rotating axis </li></ul>
8. 8. Operational Amplifier (Op-Amp)
9. 9. DC Motor <ul><li>Applications e.g. tape drive, disk drive, printer, CNC machines, and robots. </li></ul><ul><li>The equivalent circuit for a dc motor is </li></ul>
10. 10. DC Motor (cont.) <ul><li>Reduced block diagram </li></ul>The transfer function (consider TL(t) equals to zero)
11. 11. Example 1 <ul><li>Problem : Find the transfer function, G(s) = VL(s)/V(s) . Solve the problem two ways – mesh analysis and nodal analysis. Show that the two methods yield the same result. </li></ul>
12. 12. Example 1 (cont.)
13. 13. Now, writing the mesh equations, Nodal Analysis
14. 14. 5. Mechanical Elements Modelling <ul><li>The motion of mechanical elements can be described in various dimensions, which are: </li></ul><ul><li>1. Translational. </li></ul><ul><li>2. Rotational. </li></ul><ul><li>3. Combination of both. </li></ul>
15. 15. Translation <ul><li>The motion of translation is defined as a motion that takes place along or curved path. </li></ul><ul><li>The variables that are used to describe translational motion are acceleration, velocity , and displacement . </li></ul>
16. 16. Translational Mechanical System
17. 17. Example 1 <ul><li>Find the transfer function for the spring-mass-damper system shown below. </li></ul><ul><li>Solution: </li></ul><ul><li>1. Draw the free-body diagram of a system and assume the mass is traveling toward the right. </li></ul>Figure 2.4 a. Free-body diagram of mass, spring, and damper system; b. transformed free-body diagram
18. 18. cont. <ul><li>From free-body diagram, write differential equation of motion using Newton’s Law. Thus we get; </li></ul><ul><li>Laplace transform the equation: </li></ul><ul><li>Find the transfer function: </li></ul>
19. 19. Example 2 <ul><li>Find the transfer function, x o (s)/x i (s) for the spring-mass system. </li></ul><ul><li>Solution: </li></ul><ul><li>The ‘object’ of the above system is to force the mass (position x o (t)) to follow a command position x i (t). </li></ul><ul><li>When the spring is compressed an amount ‘x’m, it produces a force ‘kx’ N ( Hooke’s Law ). </li></ul>
20. 20. cont. <ul><li>When one end of the spring is forced to move an amount x i (t), the other end will move and the net compression in the spring will be </li></ul><ul><li>x(t) = x i (t) – x o (t) </li></ul><ul><li>So the force F acting on the mass are, </li></ul><ul><li>From Newton’s second law of motion, F = ma </li></ul><ul><li>Therefore, </li></ul><ul><li>Transforming the equation: </li></ul>
21. 21. Example 3 <ul><li>Find the transfer function for the spring-mass with viscous frictional damping. </li></ul><ul><li>Solution: </li></ul><ul><li>The friction force produced by the dash pot is proportional with velocity, which is; ƒ = viscous frictional constant N/ms-1 </li></ul>
22. 22. cont. <ul><li>The net force F tending to accelerate the mass is F= Fs – FD, </li></ul><ul><li>F = k ( Xi(t) – Xo(t) ) – ƒ </li></ul><ul><li>Free Body Diagram, </li></ul><ul><li>From N II, </li></ul><ul><li>F = ma </li></ul><ul><li>Laplace transform, </li></ul><ul><li>Ms 2 Xo(s) = k[Xi(s) – Xo(s)] – BsXo(s) </li></ul>F=ma K(Xi-Xo) m ƒ
23. 23. Rotational Mechanical System <ul><li>The rotational motion can be defined as motion about a fixed axis. </li></ul><ul><li>The extension of Newton’s Law of motion for rotational motion states that the algebraic sum of moments or torque about a fixed axis is equal to the product of the inertia and the angular acceleration about the axis where, </li></ul><ul><li>J = Inertia </li></ul><ul><li>T = Torque </li></ul><ul><li>θ = Angular Displacement </li></ul><ul><li>ω = Angular Velocity </li></ul><ul><li>where Newton’s second law for rotational system are, </li></ul>
24. 24. Modelling of Rotational Mechanical System
25. 25. Example 1 <ul><li>Rotary Mechanical System </li></ul>
26. 26. cont. <ul><li>The shaft has a stiffness k, which means, if the shaft is twisted through an angle θ, it will produce a torque kθ, where K – (Nm/rad). </li></ul><ul><li>For system above the torque produce by flexible shaft are, </li></ul><ul><li>Ts = K ( θ i (t)- θ o(t)) Nm </li></ul><ul><li>The viscous frictional torque due to paddle </li></ul><ul><li>Therefore the torque required to accelerating torque acting on the mass is </li></ul><ul><li>Tr = Ts - TD </li></ul>
27. 27. cont. <ul><li>From Newton’s second law for rotational system, </li></ul><ul><li>Therefore, </li></ul><ul><li>Transforming equation above, we get: </li></ul><ul><li>Transfer function of system: </li></ul>
28. 28. Example 2 <ul><li>Closed Loop Position Control System </li></ul>K s Load v a (t) Motor Amplifier Gears Load Handwheel Potentiometer K p Error Detector  i  o e(t) R L  m (t)
29. 29. cont. <ul><li>The objective of this system is to control the position of the mechanical load in according with the reference position. </li></ul><ul><li>The operation of this system is as follows:- </li></ul><ul><ul><li>A pair of potentiometers acts as an error-measuring device. </li></ul></ul><ul><ul><li>For input potentiometer, vi(t) = kpθi(t) </li></ul></ul><ul><ul><li>For the output potentiometer, vo(t) = kpθo(t) </li></ul></ul><ul><ul><li>The error signal, Ve(t) = Vi(t) – Vo(t) = kpθi(t) - kpθo(t) (1) </li></ul></ul><ul><ul><li>This error signal are amplified by the amplifier with gain constant, Ks. Va(t) = K s Ve(t) (2) </li></ul></ul>
30. 30. cont. <ul><li>Transforming equations (1) and (2):- </li></ul><ul><li>Ve(s) = Kpθi(s) - Kpθo(s) (3) </li></ul><ul><li>By using the mathematical models developed previously for motor and gear the block diagram of the position control system is shown below:- </li></ul>+ - B K t R+Ls  i (s) + - K s n s  o (s) K p V a (s) 1 J 1eq s+B 1eq T L (s) + -