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Chapter 9

Chapter 9. Design via Root Locus. Figure 9.1 a. Sample root locus, showing possible design point via gain adjustment ( A ) and desired design point that cannot be met via simple gain adjustment ( B ); b. responses from poles at A and B. Improving transient response.

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Chapter 9

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  1. Chapter 9 Design via Root Locus

  2. Figure 9.1a. Sample root locus,showing possibledesign point viagain adjustment (A)and desired designpoint that cannot bemet via simple gainadjustment (B);b. responses frompoles at A and B Improving transient response

  3. Compensationtechniques:a. cascade;b. feedbackIdeal compensators are implemented with active networks. Improving steady-state error

  4. Pole at A is:a. on the rootlocus without compensator;b. not on theroot locus withcompensatorpole added;(figure continues) Improving steady-state error via cascade compensation

  5. Ideal Integral compensation (PI) c. approximately on the root locus with compensator pole and zero added

  6. Closed-loop system for Example 9.1a. before compensation;b. after ideal integral compensation Problem: The given system operating with damping ratio of 0.174. Add an ideal integral compensator to reduce the ss error. Solution: We compensate the system by choosing a pole at the origin and a zero at -0.1

  7. Root locus for uncompensatedsystem of Figure 9.4(a) The gain K = 164.6 yields Kp=8.23 and

  8. Root locus for compensated system of Figure 9.4(b) Almost same transient response and gain, but with zero ss error since we have a type one system.

  9. Ideal integral compensated system response and theuncompensated system response of Example 9.1

  10. PI controller A method to implement an Ideal integral compensator is shown.

  11. Lag Compensatora. Type 1 uncompensated system;b. Type 1 compensated system;c. compensator pole-zero plot Using passive networks, the compensation pole and zero is moved to the left, close to the origin. The static error constant for uncompensated system is Assuming the compensator is used as in b & c the static error is

  12. Root locus:a. before lag compensation;b. after lag compensation Effect on transient response Almost no change on the transient response and same gain K. While the ss error is effected since

  13. Lag compensator design Example 9.2 Problem: Compensate the shown system to improve the ss error by a factor of 10 if the system is operating with a damping ratio of 0.174 Solution: the uncompensated system error from previous example is 0.108 with Kp= 8.23. a ten fold improvement means ss error = 0.0108 so Kp= 91.59. so the ratio arbitrarily selecting Pc=0.01 and Zc=11.13Pc 0.111

  14. Root locus for compensated system

  15. Predicted characteristics of uncompensated and lag-compensated systems for Example 9.2

  16. Step responses of uncompensated andlag-compensated systems forExample 9.2

  17. Step responses of the system for Example 9.2 using different lag compensators

  18. Ideal Derivative compensator is called PD controller When using passive network it’s called lead compensatorUsing ideal derivative compensation:a. uncompensated;b. compensator zero at –2; Improving Transient response via Cascade Compensation

  19. c. Compensator zero at –3;d. Compensator zero at – 4 Improving Transient response via Cascade Compensation

  20. Uncompensated system and ideal derivativecompensation solutions from Table 9.2

  21. Table 9.2 Predicted characteristics for the systems of previous slides

  22. Feedback control system for Example 9.3 Problem: Given the system in the figure, design an ideal derivative compensator to yield a 16% overshoot with a threefold reduction in settling time. Root locus for uncompensated system of Example 9.3

  23. Compensated dominant pole superimposed over the uncompensated root locus for Example 9.3 The settling time for the uncompensated system shown in next slide is In order to have a threefold reduction in the settling time, the settling time of the compensated system will be one third of 3.32 that is 1.107, so the real part of the compensated system’s dominant second order pole is And the imaginary part is The figure shows the designed dominant 2nd order poles.

  24. Evaluating the location of the compensating zero for Example 9.3 The sum of angles from all poles to the desired compensated pole -3.613+j6.193 is -275.6 The angle of the zero to be on the root locus is 275.6-180=95.6 The location of the compensator zero is calculated as

  25. Uncompensated and compensated system characteristics for Example 9.3

  26. Root locus for the compensated system of Example 9.3

  27. Uncompensated and compensated system step responses of Example 9.3

  28. PD controller implementation K2 is chosen to contribute to the required loop-gain value. And K1/K2 is chosen to equal the negative of the compensator zero.

  29. Geometry of lead compensation • Advantages of a passive lead network over an active PD controller: • no need for additional power supply • noise due to differentiation is reduced Three of the infinitepossible lead compensator solutions

  30. Lead compensator design, Example 9.4 Problem: Design 3 lead compensators for the system in figure that will reduce the settling time by a factor of 2 while maintaining 30% overshoot. Solution: The uncompensated settling time is To find the design point, new settling time is From which the real part of the desired pole location is And the imaginary part is

  31. S-plane picture used to calculate the location of the compensator pole for Example 9.4 Arbitrarily assume a compensator zero at -5 on the real axis as possible solution. Then we find the compensator pole location as shown in figure. Note sum of angles of compensator zero and all uncompensated poles and zeros is -172,69 so the angular contribution of the compensator pole is -7.31.

  32. Compensated system root locus

  33. Comparison of lead compensation designs for Example 9.4

  34. Uncompensated system and lead compensation responses for Example 9.4

  35. PID controller or using passive network it’s called lag-lad compensator Improving Steady-State Error and Transient Response

  36. PID controller design Design Steps: • Evaluate the performance of the uncompensated system to determine how much improvement is required in transient response • Design the PD controller to meet the transient response specifications. The design includes the zero location and the loop gain. • Simulate the system to be sure all requirements have been met. • Redesign if the simulation shows that requirements have not been met. • Design the PI controller to yield the required steady-sate error. • Determine the gains, K1, K2, and K3 shown in previous figure. • Simulate the system to be sure all requirements have been met. • Redesign if simulation shows that requirements have not been met.

  37. PID controller design Example 9.5 Problem: Using the system in the Figure, Design a PID controller so that the system can operate with a peak time that is 2/3 that of the uncompensated system at 20% overshoot and with zero steady-state error for a step input Solution: The uncompensated system operating at 20% overshoot has dominant poles at -5.415+j10.57 with gain 121.5, and a third pole at -8.169. The complete performance is shown in next table.

  38. Root locus for the uncompensated system of Example 9.5 To compensate the system to reduce the peak time to 2/3 of original, we must find the compensated system dominant pole location. The imaginary part of the dominant pole is Thus the real part is

  39. Predicted characteristics of uncompensated, PD- , and PID- compensated systems of Example 9.5

  40. Calculating the PD compensator zero for Example 9.5 To design the compensator, we find the sum of angles from the uncompensated system’s poles and zeros to the desired compensated dominant pole to be -198.37. Thus the contribution required from the compensator zero is 198.37-180=18.37. Then we calculate the location of the zero as: Thus the PD controller is GPD(s) = (s+55.92) The complete root locus sketch is shown in next slide. Using program the gain at the design point is 5.34

  41. Root locus for PD-compensated system of Example 9.5

  42. Step responses for uncompensated, PD-compensated, and PID-compensated systems of Example 9.5

  43. Root locus for PID-compensated system of Example 9.5 Choosing the ideal integral compensator to be And sketching the root locus for the PID-compensated system as shown. Searching the 0.456 damping ratio line, we find the dominant poles at -7.516+j14.67 The characteristics of the PID compensated system are shown in table.

  44. Predicted characteristics of uncompensated, PD- , and PID- compensated systems of Example 9.5

  45. Finally to implement the compensator and find the K’s, using the PD and PI compensatorsand compare towe find K1= 259.5, K2=128.6, and K3=4.6 Improving Steady-State Error and Transient Response

  46. Lag-Lead Compensator Design Example 9.6 Problem: Using the system in the Figure, Design a lag-lead compensator so that the system can operate with a twofold reduction in settling time, and 20% overshoot and a tenfold improvement in steady-state error for a ramp input Solution: The uncompensated system operating at 20% overshoot has dominant poles at -1.794+j3.501 with gain 192.1, and a third pole at -12.41. The complete performance is shown in next table.

  47. Root locus for uncompensated system of Example 9.6 To compensate the system to realize a twofold reduction in settling time, the real part of the dominant poles must be increased by a factor of 2, thus, And the imaginary part is

  48. Predicted characteristics of uncompensated, lead-compensated, and lag-lead- compensated systems of Example 9.6

  49. Evaluating the compensator pole for Example 9.6 Now to design the lead compensator, arbitrarily select a location for the lead compensator zero at -6, to cancel the pole. To find the location of the compensator pole. Using program sum the angles to get -164.65. and the contribution of the pole is -15.35 we find the location of the pole from the figure as Results are satisfactory see results in next slide

  50. Root locus for lead-compensated system of Example 9.6

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