Control System Design: A Case Study of Speed Control in MEEN 651 at Texas A&M University, Study notes of Mechanical Engineering

Download Control System Design: A Case Study of Speed Control in MEEN 651 at Texas A&M University and more Study notes Mechanical Engineering in PDF only on Docsity! Texas A & M University Department of Mechanical Engineering MEEN 651 Control System Design Dr. Alexander G. Parlos Fall 2003 Lecture 10A: Basic Properties of Feedback A Case Study of Speed Control The objective of this lecture is to present some of the most fundamental elements of control loops, including open-loop and closed-loop (or feedback) control. There are two basis structures for control of dynamic systems: (a) open-loop control and (b) feedback (or closed-loop) control. These two control structures are shown in Figures 1 and 2. Case Study of Motor Speed Control Let us write the equations of motion for a DC motor coupled to an inertial load. The electrical dynamics can be expressed as Keθ̇m(t) + La dia(t) dt + Raia(t) = va(t), (1) whereas the mechanical dynamics can be expressed as Jmθ̈m(t) + bθ̇m(t) = Ktia(t) + Tℓ. (2) Let us define the output to the motor speed y(t) = θ̇m(t) and let’s name the motor load as the disturbance w(t) = Tℓ. Taking the Laplace transforms of equations (1) and (2) and eliminating the current Ia(s) we obtain an equation of the form 1 ( JmLa bRa + KtKe s2 + JmRa + bLa bRa + KtKe s + 1)Y (s) = = Kt bRa + KtKe Va(s) + 1 bRa + KtKe W (s).(3) 1NOTE: Please, do the algebra to convince yourselves of this result. 1 Figure 1: Open-loop Control System. Figure 2: Closed-loop or Feedback Control System. Equation (3) can be simplified to the form (τ1s + 1)(τ2s + 1)Y (s) = AVa(s) + BW (s), (4) where the time constants τ1 and τ2, as well as the constants A and B are expressed in terms of the DC motor variables. Equation (4) can be written in a familiar transfer function form, as follows: Y (s) = A (τ1s + 1)(τ2s + 1) Va(s) + B (τ1s + 1)(τ2s + 1) W (s). (5) At steady-state, when both w(t) and va(t) are constant, we have the steady-state motor response as yss = Ava + Bw. (6) Figures 3 and 4 show the DC motor in open-loop and closed-loop speed control configu- ration. We now compare the various properties of the two control systems we considered, i.e. open and close-loop control systems. 2 So, the speed error is proportional to the load torque, where the constant of proportionality is B and fix for a given problem! For the closed-loop system we have the steady-state speed given by yss = AK 1 + AK r + B 1 + AK w. (16) If the controller is designed such that AK ≫ 1 and AK ≫ B, then there will be no significant error in the motor speed, despite the presence of any amount of load torque. Advantage of Feedback: Reduce Impact of Disturbances Output errors can be made less sensitive to disturbances with feedback than without feedback, by an amount of 1 + AK. Sensitivity to Gain Changes Another comparison that can be made between the two controllers is that of changing the controller gain value. Assume that temperature effects have resulted in the motor gain to shift from A to A + δA. In the open-loop case the controller gain K will still be 1 A , and the new overall system gain would be Tol + δTol = K(A + δA) = 1 A (A + δA) = 1 + δA A , (17) where Tol is the open-loop torque. So, the gain error is δA A . In terms of percent changes, defined as δT T , we have δTol Tol = δA A , (18) which means that a 10% error in A would translate in a 10% in Tol. The ratio of δT/T to δA/A is called the sensitivity of the gain from r to yss, with respect to A. For the open-loop case this ratio is 1. Applying the same change in A to the feedback case yields, Tcl + δTcl = (A + δA)K 1 + (A + δA)K . (19) One can compute the sensitivity using equation (19) and differential calculus. However, an easier approach is to first define the sensitivity as STcl A ≡ A Tcl dTcl dA , (20) 5 and then apply it to the close-loop transfer function Tcl = AK 1 + AK . (21) The result is STcl A = 1 1 + AK , (22) which reveals another major advantage of feedback control. Advantage of Feedback: Reduce Impact of Uncertainties When using feedback control, output errors can be made less sensitive to variations in the plant gain A by a factor of 1 + AK, as compared to open-loop control. Dynamic Tracking So far we have looked at steady-state system responses under constant disturbances and references. However, most control systems must track time-varying inputs. The differences between open-loop and closed-loop control with respect to dynamic tracking can best be seen through an example. Consider a servomechanism with τ1 = 1 60 , τ2 = 1 600 , A = 10 and B = 50 and with the open-loop controller gain set at 0.1. Determine the output for a step change in the load torque of −0.1 N · m. For the open-loop case, assuming that r = 0 the output is expressed as Y (s) = 50 ( 1 60 s + 1)( 1 600 s + 1) W (s). (23) The disturbance is given by W (s) = − 0.1 s . (24) So, Y (s) = −5 ( 1 60 s + 1)( 1 600 s + 1)s . (25) The system response is shown in Figure 5. Now, let’s assume that we wish to use feedback control in order to improve the system’s ability to reject steady-state disturbances by a factor of 100 with respect to the open-loop case. The output is given by Y (s) = 50 τ1τ2s2 + (τ1 + τ2)s + (1 + AK) W (s). (26) 6 Figure 5: Open-loop transient response to a disturbance input. Since we want the error reduction to be 100 we must have 1 + AK = 100. So, K = 9.9. Considering that W (s) = −0.1 s , from the final value theorem we have that yss = − 5 1 + AK = −0.05. (27) The transfer function from the reference to the output is given by T (s) = Y (s) R(s) = 10K τ1τ2s2 + (τ1 + τ2)s + (1 + AK) . (28) The system response to the given disturbance is shown in Figure 6(a), whereas the system response to a step refernce input shown in Figure 6(b). Notice the differences in the responses of the two control systems! Reading Assignment Read pages 200-215 of the textbook. Read the examples in Handout E.19 posted on the course web page. 7