1. سیستمهای کنترل خطی پاییز 1389 بسم ا... الرحمن الرحيم دکتر حسين بلندي- دکتر سید مجید اسما عیل زاده

2. Recap. Signal Flow Graph : –Algebra, –Mason Rule, Introduction to Time Domain Analysis 2

3. Time Domain Performance Specification 3

4. Outline Introduction Test Input Signals Performance of a first-order system Performance of a second-order system Effects of a Third Pole and a Zero on the Second- Order System Response Estimation of the Damping Ratio The s-plane Root Location and the Transient Response 4

5. Introduction/1 Objectives The ability to adjust the transient and steady-state response of a feedback control system is a beneficial outcome of the design of control systems. 5

6. Introduction/2 One of the first steps in the design process is to specify the measures of performance. We introduce the common time-domain specifications such as percent overshoot, settling time, time to peak, time to rise, and steady-state tracking error. 6

7. Introduction/3 We will use selected input signals such as the step and ramp to test the response of the control system. The correlation between the system performance and the location of the system transfer function poles and zeros in the s-plane is discussed. 7

8. Introduction/4 We will develop valuable relationships between the performance specifications and the natural frequency and damping ratio for second-order systems. Relying on the notion of dominant poles, we can extrapolate the ideas associated with second-order systems to those of higher order. 8

9. Test Input Signal Since the actual input signal of the system is usually unknown, a standard test input signal is normally chosen. Commonly used test signals include step input, ramp input, and the parabolic input. 9

10. For example in a radar tracking system for antiaircraft missiles, the position and speed of the target to be tracked may vary in an unpredictable manner. Mobile robot with collision avoidance, … From the step function to the parabolic function, they become progressively faster with respect to time. 10

11. For the purposes of analysis and design, it is necessary to assume some basic types of test inputs so that the performance of a system can be evaluated (CLASSFICATION OF INPUTS). The responses due to these inputs allow the prediction of the system’s performance to other more complex inputs. 11

12. General form of the standard test signals r(t) = t n R(s) = n!/s n+1 12

13. Test signals r(t) = A t n n = 0 n = 1 n = 2 r(t) = Ar(t) = Atr(t) = At 2 R(s) = 2A/s 3 R(s) = A/s R(s) = A/s 2 13

14. The ramp signal is the integral of the step input, and the parabola is the integral of the ramp input. The unit impulse function is also useful for test signal purposes. The responses due to these inputs allow the prediction of the system’s performance to other more complex inputs. 14

15. Step-for representation a stationary target Ramp- for track a constant angular position (first derivatives are constant) Parabolas- can be used to represent accelerating targets (second derivatives are constant) 15

16. Test Signal Inputs Test Signal r(t) R(s) Step position r(t) = A, t > 0 = 0, t < 0 R(s) = A/s Ramp velocity r(t) = At, t > 0 = 0, t < 0 R(s) = A/s 2 Parabolic acceleration r(t) = At 2, t > 0 = 0, t < 0 R(s) = 2A/s 3 16

17. Test inputs vary with target type parabola ramp step 17

18. Steady-state error Is a difference between input and the output for a prescribed test input as 18

19. Application to stable systems Unstable systems represent loss of control in the steady state and are not acceptable for use at all. 19

20. Steady-state error: a) step input, b) ramp input 20

21. Time response of systems c(t) = c t (t) + c ss (t) The time response of a control system is divided into two parts: c t (t) - transient response c ss (t) - steady state response 21

22. Transient response All real control systems exhibit transient phenomena to some extend before steady state is reached. lim ct(t) = 0 for t  22

23. Steady-state response The response that exists for a long time following any input signal initiation. 23

24. Poles and zeros of a first order system Css(t)Ct(t) 24

25. Poles and zeros A pole of the input function generates the form of the forced response ( that is the pole at the origin generated a step function at the output). The zeros and poles generate the amplitudes for both the transit and steady state responses 25

26. A pole on the real axis generate an exponential response of the form Exp[-  t] where -  is the pole location on real axis. The farther to the left a pole is on the negative real axis, the faster the exponential transit response will decay to zero. Effect of a real-axis pole upon transient response 26

27. Evaluating response using poles Css(t ) Ct(t) 27

28. First order system 28

29. First-order system response to a unit step Transient response specification: 1. Time-constant, 1/a 2. Rise time, Tr 3. Settling time, Ts 29

30. Transient response specification for a first-order system 1.Time-constant, 1/a Can be described as the time for (1 - Exp[- a t]) to rise to 63 % of final value. 1.Rise time, T r = 2.2/a The time for the waveform to go from 0.1 to 0.9 of its final value. 3.Settling time, T s = 4/a The time for response to reach, and stay within, 2% of its final value 30

31. Transfer function via laboratory testing 31

32. Identify K and a from testing The time for amplitude to reach 63% of its final value: 63 x 0.72 = 0.45, or about 0.13 sec, a = 1/0.13 = 7.7 From equation, we see that the forced response reaches a steady-state value of K/a =0.72. K= 0.72 x 7.7= 5.54 G(s) = 5.54/(s+7.7). 32

33. Exercise A system has a transfer function G(s)= 50/(s+50). Find the transit response specifications such as Tc, Tr, Ts. 33

34. 34

35. Steady-state response If the steady-state response of the output does not agree with the steady-state of the input exactly, the system is said to have a steady-state error. It is a measure of system accuracy when a specific type of input is applied to a control system. 35

36. Y(s) = R(s) G(s) 36

37. Steady-state error T(s) = 9/(s + 10) Y(s) = 9/s(s+10) y(t) = 0.9(1- e -10t ) y(∞) = 0.9 E(s) = R(s) - Y(s) ess = lim s E(s) = 0.1 37

38. 38

39. Performance of a second-order system 39

40. Numerical example of the second- order system 40

41. A second order system exhibits a wide range of response. The general case: which has two finite poles and no zeros. The term in the numerator is simple scale. Changing a and b we can show all possible transient responses. 41

42. Changes in the parameters of a second- order system can change the form of the response. Can display characteristic much like a first-order system or display damped or pure oscillation for its transient response. 42

43. The unit step response then can be found using C(s)= R(s) G(s), where R(s) = 1/s, followed by a partial- fraction expansion and inverse Laplace transform 43

44. Overdamped 44

45. Underdamped 45

46. Undamped 46

47. Critically damped 47

48. fig_04_11 48

49. Step response for second order system damping cases 49

50. Summary Overdamped Poles: Two real at -  1, -  2 Underdamped Poles: Two complex at -  d + j  d, -  d - j  d Undamped Poles: Two imaginary at + j  1, - j  1 Critically damped Poles: Two real at -  1, 50