Introduction to Control Systems

1. Introduction to Control Systems • Historical perspective • Introduction to Feedback Control Systems • Closed loop system examples

2. Historical Perspective • 13.7B BC Big Bang • 13.4B Stars and galaxies form • 5B Birth of our sun • 3.8B Early life begins • 700M First animals • 200M Mammals evolve • 65M Dinosaurs extinct • 600K First Trace of humans

3. Feedback Control Systems emerge rather recently • 1600 Drebbel Temperature regulator • 1781 Pressure regulator for steam boilers • 1765 Polzunov water level float regulator

4. Closed loop example: Polzunov’s Water level float regulator

5. Feedback Control Systems emerge rather recently • 1600 Drebbel Temperature regulator • 1681 Pressure regulator for steam boilers • 1765 Polzunov water level float regulator • 1769 James Watt’s Steam Engine and Governor

6. Closed loop example: James Watt’s flyball governor

7. Open loop and closed loop control systems Open Loop System

8. Open loop and closed loop control system models Open Loop System Closed Loop System

9. Example: Feedback in everyday life

10. Multivariable Control System Model

11. Multivariable Control System

12. Robotics A robot is a programmable computer integrated with a machine

13. Example: Disk Drive

14. Example: Automatic Parking Control

15. Feedback Control: Benefits and cost Benefits: Cost:

16. Feedback Control: Benefits and cost Benefits: Cost: • Reduction of sensitivity to process parameters • Disturbance rejection • More precise control of process at lower cost • Performance and robustness not otherwise achievable

17. Feedback Control: Benefits and cost Benefits: Cost: • Reduction of sensitivity to process parameters • Disturbance rejection • More precise control of process at lower cost • Performance and robustness not otherwise achievable • More mathematical sophistication • Large loop gain to provide substantial closed loop gain • Stabilizing closed loop system • Achieving proper transient and steady-state response

18. Ideal Control System Elements

19. Example: Mechanical system Determine y(t)

20. Example: Mechanical system Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0 r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2 r(t) = M d 2y(t)/dt 2+b dy(t)/dt + k y(t) L2CCDE

21. Example: Mechanical system Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0 r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2 r(t) = M d 2y(t)/dt 2+b dy(t)/dt + k y(t) Homogenous solution:yH (t) r(t) = 0 Particular solution: yp(t) = f [r(t)] Total solution: y(t) = yH(t) + yp(t)

22. Example: Mechanical system Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0 r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2 r(t) = M d 2y(t)/dt 2+b dy(t)/dt + k y(t)

23. Example: Mechanical system Assume the system is initially at rest: y(0-) = v(0-) = 0 r(t) – k ∫0-tv(τ)d τ – b v(t) = M dv(t) /dt r(t) = M dv(t) / dt +b v(t) + k ∫0-tv(τ)d τ

24. Example: Electrical system Determine v(t) Assume circuit initially at rest. What does this mean?

25. Example: Electrical system Write node equation r(t) = (1/L) + C dv(t)/dt + G v(t) Homogenous solution vH (t) : r(t) = 0 Particular solution: vp(t) = f [r(t)] Total solution: v(t) = vH(t) + vP (t) t ò v(τ)d τ 0-

26. How to describe a plant/system mathematically? if a dynamic system Differential Equation KVL

27. Consider a plant that can be described by an n-order differential equation are constant coefficients, and independent variable dependent variable There are 2 cases to be considered (1) If q(t)=0  Homogeneous Case (2) If q(t) ≠0  In-Homogeneous Case

28. Homogeneous Case Let a 2nd order differential equation be: Define a differential operator D

29. Homogeneous Case (cont.)

30. Homogeneous Case (cont.) Re-write In general If one can solve a first order D.E, then it can be used to solve nth order D.E

31. Homogeneous Case (cont.) Solve a 1st order differential equation General eqn. Notice that: Multiply both sides by eat

32. Homogeneous Case (cont) Integrate both sides of equation Zero Input Response Zero State Response due to initial condition due to input

33. Homogeneous Case (cont.) Consider Initial condition

34. Homogeneous Case (cont.)

35. Homogeneous Case (cont.) What are C1 & C2??? Based on the initial condition:

36. Homogenous (cont.) Thus

37. Homogeneous Case (cont.) Easiest method: Find the roots of the characteristic equations For second order system:

38. Homogeneous Case (cont.) Therefore, So how to find these values? Based on Initial Condition

39. Homogeneous Case (cont.) Example: Consider again We’ll have Thus, So, By considering the initial conditions: and We’ll get: Therefore:

40. 4 cases to be considered Case 1: Distinct real roots Case 2: Equal roots & real GENERAL SOLUTION Case 4: Complex conjugate roots Case 3: Imaginary roots

41. In-Homogeneous Case The solution: Particular solution Homogeneous solution Determine the particular solution is the most challenging!!! Homogeneous solution can be found by setting q(t)=0, and then find the solution as in homogeneous case

42. Finding the particular solution, xP(t) Undetermined Coefficient Method Assumption: Various derivative of q(t) have a finite of functional form • Functional forms: (1)exponential (eat) (2)sinosoidal (sin ωt, cos ωt) (3) polynomial (tm) (4)combinational of these function and their products (teat, eatsinwt) Then guess:

43. In-Homogeneous Case (cont.) Example: Initial condition

44. Finding Xp(t)

45. In-Homogenous case (cont.) So, we’ll have Now, we need to find c1,c2, A & B First, how to find A & B? Let us consider xp(t) again and substitute it into the original differential equation

46. Re-arrange: A = ½ B = -3/2 A-3B = 5 B+3A = 0 So,

47. In-Homogenous case (cont.) So, What about c1 & c2? Again, you must find them by considering the given INITIAL CONDITION

48. In Class exercise • In a group of two students, please solve the following problem within 10 minutes. Please submit them after right after 10 minutes ends. Intellectual discussion is highly encouraged. Find the solution the given differential equation provided that

49. Using Laplace Approach Consider the previous example again: Apply the Laplace transform to given diff. eqn Simplify it: X1(s) X2(s)

50. Using Laplace Approach (cont.) Partial fraction expansion: and Determine the values for A,B,C,D,E & F Then,