1. Control Systems Lect.3 Steady State Error Basil Hamed

2. Definition and Test Inputs Basil Hamed2

3. Application to Stable Systems Basil Hamed3

4. SS Error for Unit Step Input Basil Hamed4

5. 7.2 Steady-State Error for Unity Feedback Systems Steady-state error can be calculated from a system's closed-loop transfer function, T(s), or the open-loop transfer function, G(s), for unity feedback systems. Steady-State Error in Terms of T(s) Basil Hamed5

6. Example 7.1 P 345 Basil Hamed6 Since T(s) is stable and, subsequently, E(s) does not have right-half- plane poles or jw poles other than at the origin, we can apply the final value theorem

7. Steady-State Error in Terms of G(s) Basil Hamed7 E(s)= R(s)- C(s); C(s) =G(s) E(s) We now apply the final value theorem

8. Test Signals The three test signals we use to establish specifications for a control system's steady-state error characteristics are : 1. Step 2. Ramp 3. Parabola Let us take each input and evaluate its effect on the steady-state error by using Eq. Basil Hamed8

9. Test Signals Using Eq. With R(s) = 1/s, we find Basil Hamed9 Step Input

10. Test Signals Basil Hamed10

11. Example 7.2 P. 347 Basil Hamed11

12. Example 7.2 P. 347 Basil Hamed12

13. Example 7.3 P 348 Basil Hamed13

14. Example 7.3 P 348 Basil Hamed14

15. 7.3 Static Error Constants and System Type We continue our focus on unity negative feedback systems and define parameters that we can use as steady-state error performance specifications. These steady-state error performance specifications are called static error constants. Basil Hamed15

16. Static Error Constants For a step input, u(t), Basil Hamed16 For Ramp input, tu(t),

17. Static Error Constants Basil Hamed17

18. Example 7.4 P 350 PROBLEM: For each system in the Figure below, evaluate the static error constants and find the expected error for the standard step, ramp, and parabolic inputs. Basil Hamed18

19. Example 7.4 P 350 SOLUTION: First verify that all closed-loop systems shown are indeed stable. a) Basil Hamed 19

20. Example 7.4 P 350 b) Basil Hamed20

21. Example 7.4 P 350 C) Basil Hamed21

22. System Type The values of the static error constants,, depend upon the form of G(s), especially the number of pure integrations in the forward path. Since steady-state errors are dependent upon the number of integrations in the forward path. we define system type to be the value of n in the denominator or, equivalently, the number of pure integrations in the forward path. Basil Hamed22 Therefore, a system with n = 0 is a Type 0 system. If n = 1 or n = 2, the corresponding system is a Type 1 or Type 2 system, respectively.

23. Example Basil Hamed23 Type 1 Type 2 Type 3

24. Relationships between input and system type TypeStep InputRamp InputParabola Input 0 10 200 3000 Basil Hamed24

25. Example Basil Hamed25 Solution: The system is stable and of type 2 Zero SS error for step and ramp input For parabolic input -11.6063 -0.1968 + 0.6261i -0.1968 - 0.6261i The closed loop poles are

26. 7.4 Steady-State Error Specifications Static error constants can be used to specify the steady-state error characteristics of control systems. Just as damping ratio, ζ, settling time, Ts, peak time, Tp, and percent overshoot, % OS, are used as specifications for a control system's transient response, so the position constant, Kp, velocity constant, Kv, and acceleration constant, Ka, can be used as specifications for a control system's steady-state errors. Basil Hamed26 For example, if a control system has the specification Kv = 1000, we can draw several conclusions: 1. The system is stable 2. The system is of Type 1 3. A ramp input is the test signal.

27. Example 7.5 P. 354 PROBLEM: What information is contained in the specification K p = 1000? Basil Hamed27 SOLUTION: The system is stable. The system is Type 0, since only a Type 0 system has a finite K p. Type 1 and Type 2 systems have K p = ∞. The input test signal is a step, since K p is specified. Finally, the error per unit step is

28. Example 7.6 P 355 PROBLEM: Given the control system in Figure, find the value of K so that there is 10% error in the steady state. Basil Hamed28 SOLUTION: Since the system is Type 1, the error stated in the problem must apply to a ramp input; only a ramp yields a finite error in a Type 1 system. Thus, k= 672

29. 7.5 Steady-State Error for Disturbances Feedback control systems are used to compensate for disturbances or unwanted inputs that enter a system. The advantage of using feedback is that regardless of these disturbances, the system can be designed to follow the input with small or zero error. Basil Hamed29

30. Basil Hamed30 But Substituting Eq. (2) into Eq. (1) and solving for E(s), we obtain 1 2 To find the steady-state value of the error, we apply the final value theorem 3 to Eq. (3) and obtain 3

31. Example 7.7 P.357 Basil Hamed31 SOLUTION: The system is stable. The steady-state error component due to a step disturbance is found to be

32. 7.6 Steady-State Error for Nonunity Feedback Systems Control systems often do not have unity feedback because of the compensation used to improve performance or because of the physical model for the system. Basil Hamed32

33. Example 7.8 P. 359 PROBLEM: For the system shown, find the system type, the appropriate error constant associated with the system type, and the steady-state error for a unit step input. Assume input and output units are the same. Basil Hamed33 SOLUTION: After determining that the system is indeed stable, one may impulsively declare the system to be Type 1. This may not be the case, since there is a nonunity feedback element, The first step in solving the problem is to convert the system into an equivalent unity feedback system.

34. Example 7.8 P. 359 Basil Hamed34 we find Thus, the system is Type 0, since there are no pure integrations in above Eq. The appropriate static error constant is then K p, whose value is The steady-state error, e( ∞) is

35. Root Locus Techniques

36. Root Locus – What is it? W. R. Evans developed in 1948. Pole location characterizes the feedback system stability and transient properties. Consider a feedback system that has one parameter (gain) K > 0 to be designed. Root locus graphically shows how poles of CL system varies as K varies from 0 to infinity. Basil Hamed36 L(s): open-loop TF

37. Root Locus – A Simple Example Basil Hamed37 Characteristic eq. K = 0: s = 0,-2 K = 1: s = -1, -1 K > 1: complex numbers

38. Root Locus – A Complicated Example Basil Hamed38 Characteristic eq. It is hard to solve this analytically for each K. Is there some way to sketch a rough root locus by hand?

39. 8. 1 Introduction Root locus, a graphical presentation of the closed-loop poles as a system parameter is varied, is a powerful method of analysis and design for stability and transient response(Evans, 1948; 1950). Feedback control systems are difficult to comprehend from a qualitative point of view, and hence they rely heavily upon mathematics. The root locus covered in this chapter is a graphical technique that gives us the qualitative description of a control system's performance that we are looking for and also serves as a powerful quantitative tool that yields more information than the methods already discussed. Basil Hamed39

40. 8.2 Defining the Root Locus The root locus technique can be used to analyze and design the effect of loop gain upon the system's transient response and stability. Basil Hamed40 Assume the block diagram representation of a tracking system as shown, where the closed-loop poles of the system change location as the gain, K, is varied.

41. 8.2 Defining the Root Locus Basil Hamed41 Pole location as a function of gain for the system

42. 8.4 Sketching the Root Locus Basil Hamed42

43. 8.4 Sketching the Root Locus Basil Hamed43

44. Example Basil Hamed44 Find R-L

45. Example Sketch R-L Basil Hamed45 Solution: Indicate the direction with an arrowhead

46. Example Basil Hamed46 Asymptotes (Not root locus) Breakaway points are among roots of s  2.4656,  0.7672  0.7925 j

47. Example Basil Hamed47 Breakaway point -2.46 K=.4816

48. Root Locus – Matlab Command “rlocus.m” Basil Hamed48

49. Example Basil Hamed49 There are three finite poles, at s = 0, - 1, and - 2, and no finite zeros

50. Example Basil Hamed50

51. Example 8.2 P. 400 PROBLEM: Sketch the root locus for the system shown in Figure Basil Hamed51 SOLUTION: Let us begin by calculating the asymptotes α = n – m =4-1=3

52. Example 8.2 P. 400 Basil Hamed52

53. Example Basil Hamed53

54. Example Basil Hamed54 ii) a = 9. The breakaway point at s = -3. iii) a = 8. No breakaway point on RL

55. Example Basil Hamed55 iv) a = 3. v) a = b = 1. The pole at s = -a and the zero at -b cancel each other out, and the RL degenerate into a second-order case and lie entirely on the jw-axis.

56. Example Basil Hamed56

57. Example Basil Hamed57 Not valid

58. Example Basil Hamed58 For stability need b= (-1/4)(k-10)>0 k<10 C= k-6 k> 6 6<k<10

59. Example Basil Hamed59 Find R-L and find k for critical stability Solution Breakaway points are among roots of

60. Example Basil Hamed60

61. Example Characteristic equation Basil Hamed61 Routh array When K = 30

62. Example Basil Hamed62

63. Example Basil Hamed63 Solution >> n=[1 1]; >> d=[1 4 0 0]; >> rlocus(n,d) There is no Imj axes crossing

64. Example Basil Hamed64 >> n=[1]; >> d=[1 4 1 -6]; >> rlocus(n,d) Solution

65. Example Given check if the following poles are on R-L, if so, find the value of k; i) s=-1+j, ii) s=-2+j Solution: R-L is i) Select a point s=-1+j, we can see that s is on R-L, find value of k Basil Hamed 65 ii) Select a point s=-2+j, we can see that s is not on R-L there is no k value. s is NOT on root locus..

66. Example Basil Hamed66

67. Example Basil Hamed67

68. Example Basil Hamed68

69. Root Locus – Control Example Basil Hamed69 a) Set Kt = 0. Draw R-L for K > 0. b) Set K = 10. Draw R-L for K t > 0. c) Set K = 5. Draw R-L for K t > 0. Solution:Root Locus – (a) K t = 0 There is no stabilizing gain K!

70. Root Locus – Control Example Basil Hamed70 Root Locus – (b) K = 10 Characteristic eq. By increasing K t, we can stabilize the CL system..

71. Root Locus – Control Example Basil Hamed71 Characteristic equation R-H array When K t = 2

72. Root Locus – Control Example Basil Hamed72 Root Locus – (c) K = 5 Characteristic eq. >> n=[1 0]; >> d=[1 5 0 5]; >> rlocus(n,d)

73. Root Locus – Effect of Adding Poles Basil Hamed73 Pulling root locus to the RIGHT – Less stable – Slow down the settling

74. Root Locus – Effect of Adding Zeros Basil Hamed74 Pulling root locus to the LEFT – More stable – Speed up the settling Add a zero

75. Example Basil Hamed75 The Plant Feedback Control System

76. Example Basil Hamed76 >> n=[1]; >> d=[1 0 0]; >> rlocus(n,d) Marginal stable for all value of k P control is unacceptable

77. HW 3 7.22, 7.32, 7.39, 8.14, 8.22, 8.37 Due Next Class Basil Hamed77