Control Systems Lect 3 Steady State Error Basil
Control Systems Lect. 3 Steady State Error Basil Hamed
Definition and Test Inputs Basil Hamed 2
Application to Stable Systems Basil Hamed 3
SS Error for Unit Step Input Basil Hamed 4
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 Hamed 5
Example 7. 1 P 345 Since T(s) is stable and, subsequently, E(s) does not have right-halfplane poles or jw poles other than at the origin, we can apply the final value theorem Basil Hamed 6
Steady-State Error in Terms of G(s) E(s)= R(s)- C(s); C(s) =G(s) E(s) We now apply the final value theorem Basil Hamed 7
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 Hamed 8
Test Signals Step Input Using Eq. With R(s) = 1/s, we find Basil Hamed 9
Test Signals Basil Hamed 10
Example 7. 2 P. 347 Basil Hamed 11
Example 7. 2 P. 347 Basil Hamed 12
Example 7. 3 P 348 Basil Hamed 13
Example 7. 3 P 348 Basil Hamed 14
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 Hamed 15
Static Error Constants For a step input, u(t), For Ramp input, tu(t), Basil Hamed 16
Static Error Constants Basil Hamed 17
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 Hamed 18
Example 7. 4 P 350 SOLUTION: First verify that all closed-loop systems shown are indeed stable. a) Basil Hamed 19
Example 7. 4 P 350 b) Basil Hamed 20
Example 7. 4 P 350 C) Basil Hamed 21
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. 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. Basil Hamed 22
Example Type 1 Type 2 Type 3 Basil Hamed 23
Relationships between input and system type Type Step Input Ramp Input Parabola Input 0 1 0 2 0 0 3 0 0 Basil Hamed 0 24
Example The closed loop poles are -11. 6063 -0. 1968 + 0. 6261 i -0. 1968 - 0. 6261 i Solution: The system is stable and of type 2 Zero SS error for step and ramp input For parabolic input Basil Hamed 25
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. 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. Basil Hamed 26
Example 7. 5 P. 354 PROBLEM: What information is contained in the specification Kp = 1000? SOLUTION: The system is stable. The system is Type 0, since only a Type 0 system has a finite Kp. Type 1 and Type 2 systems have Kp = ∞. The input test signal is a step, since Kp is specified. Finally, the error per unit step is Basil Hamed 27
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. 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 Basil Hamed 28
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 Hamed 29
1 But 2 Substituting Eq. (2) into Eq. (1) and solving for E(s), we obtain 3 To find the steady-state value of the error, we apply the final value theorem 3 to Eq. (3) and obtain Basil Hamed 30
Example 7. 7 P. 357 SOLUTION: The system is stable. The steady-state error component due to a step disturbance is found to be Basil Hamed 31
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 Hamed 32
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. 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. Basil Hamed 33
Example 7. 8 P. 359 we find Thus, the system is Type 0, since there are no pure integrations in above Eq. The appropriate static error constant is then Kp, whose value is The steady-state error, e(∞) is Basil Hamed 34
Root Locus Techniques
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. L(s): open-loop TF • Root locus graphically shows how poles of CL system varies as K varies from 0 to infinity. Basil Hamed 36
Root Locus – A Simple Example Characteristic eq. K = 0: s = 0, -2 K = 1: s = -1, -1 K > 1: complex numbers Basil Hamed 37
Root Locus – A Complicated Example Characteristic eq. • It is hard to solve this analytically for each K. • Is there some way to sketch a rough root locus by hand? Basil Hamed 38
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 Hamed 39
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. 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. Basil Hamed 40
8. 2 Defining the Root Locus Pole location as a function of gain for the system Basil Hamed 41
8. 4 Sketching the Root Locus Basil Hamed 42
8. 4 Sketching the Root Locus Basil Hamed 43
Example Find R-L Basil Hamed 44
Example Sketch R-L Solution: Basil Hamed Indicate the direction with an arrowhead 45
Example Asymptotes (Not root locus) Breakaway points are among roots of s = -2. 4656, -0. 7672 ± 0. 7925 j Basil Hamed 46
Example Breakaway point -2. 46 K=. 4816 Basil Hamed 47
Root Locus – Matlab Command “rlocus. m” Basil Hamed 48
Example There are three finite poles, at s = 0, - 1, and - 2, and no finite zeros Basil Hamed 49
Example Basil Hamed 50
Example 8. 2 P. 400 PROBLEM: Sketch the root locus for the system shown in Figure SOLUTION: Let us begin by calculating the asymptotes α = n – m =4 -1=3 Basil Hamed 51
Example 8. 2 P. 400 Basil Hamed 52
Example Basil Hamed 53
Example ii) a = 9. The breakaway point at s = -3. iii) a = 8. No breakaway point on RL Basil Hamed 54
Example 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. Basil Hamed 55
Example Basil Hamed 56
Example Not valid Basil Hamed 57
Example For stability need b= (-1/4)(k-10)>0 C= k-6 k<10 k> 6 Basil Hamed 6<k<10 58
Example Find R-L and find k for critical stability Solution Breakaway points are among roots of Basil Hamed 59
Example Basil Hamed 60
Example Characteristic equation Routh array When K = 30 Basil Hamed 61
Example Basil Hamed 62
Example Solution >> n=[1 1]; >> d=[1 4 0 0]; >> rlocus(n, d) There is no Imj axes crossing Basil Hamed 63
Example Solution >> n=[1]; >> d=[1 4 1 -6]; >> rlocus(n, d) Basil Hamed 64
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 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. . Basil Hamed 65
Example Basil Hamed 66
Example Basil Hamed 67
Example Basil Hamed 68
Root Locus – Control Example a) Set Kt = 0. Draw R-L for K > 0. b) Set K = 10. Draw R-L for Kt > 0. c) Set K = 5. Draw R-L for Kt > 0. Solution: Root Locus – (a) Kt = 0 There is no stabilizing gain K! Basil Hamed 69
Root Locus – Control Example Root Locus – (b) K = 10 Characteristic eq. By increasing Kt, we can stabilize the CL system. . Basil Hamed 70
Root Locus – Control Example Characteristic equation R-H array When Kt = 2 Basil Hamed 71
Root Locus – Control Example Root Locus – (c) K = 5 Characteristic eq. >> n=[1 0]; >> d=[1 5 0 5]; >> rlocus(n, d) Basil Hamed 72
Root Locus – Effect of Adding Poles Pulling root locus to the RIGHT – Less stable – Slow down the settling Basil Hamed 73
Root Locus – Effect of Adding Zeros Pulling root locus to the LEFT – More stable – Speed up the settling Add a zero Basil Hamed 74
Example The Plant Feedback Control System Basil Hamed 75
Example >> n=[1]; >> d=[1 0 0]; >> rlocus(n, d) Marginal stable for all value of k P control is unacceptable Basil Hamed 76
HW 3 7. 22, 7. 39, 8. 14, 8. 22, 8. 37 Due Next Class Basil Hamed 77
- Slides: 77