Control chap4

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  • 1. CONTROL SYSTEMS THEORYTransient response Chapter 4

2. Objectives To find time response from transfer function To describe quantitatively the transient response of a 1st and 2nd order system To determine response of a control system using poles and zeros 3. Introduction In Chapter 1, we learned that the total response of a system, c(t) is given byc ( t ) = cforced ( t ) + cnatural ( t ) In order to qualitatively examine and describe this output response, the poles and zeros method is used. 4. Poles & zeros The poles of a TF are the values of the Laplace variable that cause the TF to become infinite (denominator) The zeros of a TF are the values of the Laplace variable that cause the TF to become zero (numerator) 5. Poles & zeros Example : Given the TF of G(s), find the poles and zeros Solution : G(s) = zero/pole Pole at s=-5 Zero at s=-2 6. Poles & zeros Zero (o), Pole (x) Transfer function = Numerator Denominator = Zeros Poles 7. Poles & zeros Example : Given G(s), obtain the pole-zero plot of the systemZero (o) Pole (x) 8. Poles & zeros Exercise : Obtain and plot the poles and zeros for the system given 9. First order system First order system with no zeros 10. First order system Performance specifications: Time constant, t 1/a, time taken for response to rise to 63% of its final value Rise time, Tr time taken for response to go from 10% to 90% of its final value Settling time, Ts time for response to reach and stay within 5% of final value 11. First order system System response 12. Second order system 13. Second order system 14. Second order system Exercise : Is this system under/over/critically damped? 15. Second order system Performance specifications damping ratio ln ( %OS / 100 ) = 2 + ln 2 ( %OS / 100 ) % Overshoot = cmax cfinalcfinalx 100 16. Second order system Settling time, Ts4 Ts = n Peak time, TpTp = n 1 2a = 2n 17. Second order system 2nd order underdamped response 18. Second order system Second-order response as a function of damping ratio 19. Second order system 20. Second order system Step responses of second-order under-damped systems as poles move: a. with constant real part b. with constant imaginary part c. with constant damping ratio (constant on the diagonal) 21. Second order system 22. Exercise Describe the damping of each system given the information below 23. Solution Find value of zeta 24. 2nd order general form 25. Exercise Given these 2nd order systems, find the value of and . Describe the damping 26. Solution 27. Example Given Find settling time, peak time, %OS Hint : 28. Solution 29. Block diagram: Analysis Finding transient response For the system shown below, find the peak time, percent overshoot and settling time. 30. Block diagram: Analysis Answers: n=10 =0.25 Tp=0.324 %OS=44.43 Ts=1.6 31. Block diagram: Analysis and design Gain design for transient response Design the value of gain, K, for the feedback control system of figure below so that the system will respond with a 10% overshoot 32. Block diagram: Analysis and design Solution: Closed-loop transfer function is2 n = 5 andn = K Thus, 5 = 2 KK T (s) = 2 s + 5s + K 33. Block diagram: Analysis and design Can be calculated using the %OS = ln ( %OS / 100 ) 2 + ln 2 ( %OS / 100 ) = 0.591 We substitute the value and calculate K, we get K=17.9 34. Higher order systems Systems with >2 poles and zeros can be approximated to 2nd order system with 2 dominant poles 35. Higher order systems Placement of third pole. Which most closely resembles a 2nd order system? 36. Higher order systems Case I : Non-dominant pole is near dominant second-order pair (=) Case II : Non-dominant pole is far from the pair (>>) Case III : Non-dominant pole is at infinity (=) How far away is infinity? 5 times farther away to the LEFT from dominant poles 37. Exercises Find , n, Ts, Tp and %OS a)T(s) =b) T(s) =16 s2 + 3s + 16 0.04 s2 + 0.02s + 0.04c) T(s) =1.05 x 107 s2 + (1.6 x 103)s + (1.5 x 107) 38. Solution part (a) n = 4 = 0.375 Ts =4s Tp = 0.8472 s %OS = 28.06 % 39. Solution part (b) n = 0.2 = 0.05 Ts =400s Tp = 15.73s %OS = 85.45 % 40. Solution part (c) n = 3240 = 0.247 Ts =0.005 s Tp = 0.001 s %OS = 44.92 %