Feedback Control System

1. Feedback Control System THE ROOT-LOCUS DESIGN METHOD Chapter 5 Dr.-Ing. Erwin Sitompul http://zitompul.wordpress.com

2. Example 3: Plotting a Root Locus Now, sketch the root locus for: RULE 1 • There are 3 branches of locus. • Two starting from s = 0 and one from s = –4. • There will be two zeros at infinity.

3. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 3: Plotting a Root Locus RULE 2

4. Example 3: Plotting a Root Locus Angles of Asymptotes Center of Asymptotes

5. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 3: Plotting a Root Locus 90° RULE 3 RULE 4 Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros. 270°

6. Example 3: Plotting a Root Locus Replacing s with jω0, ≡ 0 ≡ 0 Points of Cross-over

7. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 3: Plotting a Root Locus RULE 5

8. Example 3: Plotting a Root Locus The root locus must have a break-away point, which can be found by solving:

9. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 3: Plotting a Root Locus On the root locus The break-away point Not on the root locus RULE 6 0

10. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 3: Plotting a Root Locus After examining RULE 1 up to RULE 6, now there is enough information to draw the root locus plot. 0

11. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 3: Plotting a Root Locus The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as: Final Result

12. Example 4: Plotting a Root Locus Now, the characteristic equation is changed a little bit: RULE 1 • There are 3 branches of locus. • Two starting from s = 0 and one from s = –9. • There will be two zeros at infinity.

13. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –9 –1 –2 –3 Example 4: Plotting a Root Locus RULE 2

14. Example 4: Plotting a Root Locus Angles of Asymptotes Center of Asymptotes

15. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –9 –1 –2 –3 Example 4: Plotting a Root Locus 90° RULE 3 270° RULE 4 Not applicable.

16. Example 4: Plotting a Root Locus Replacing s with jω0, ≡ 0 ≡ 0 Points of Cross-over

17. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –9 –1 –2 –3 Example 4: Plotting a Root Locus RULE 5

18. Example 4: Plotting a Root Locus The root locus must have a break-away point, which can be found by solving:

19. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –9 –1 –2 –3 Example 4: Plotting a Root Locus On the root locus At the same time, the break-in and the break-away point RULE 6 On the root locus The break-away point –3 0

20. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –9 –1 –2 –3 Example 4: Plotting a Root Locus After examining RULE 1 up to RULE 6, now there is enough information to draw the root locus plot. –3 0

21. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –9 –1 –2 –3 Example 4: Plotting a Root Locus The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as: Final Result

22. Conclusion: Example 3 and Example 4 • The characteristic equations can be so similar, yet the resulting root locus plots are very different. • It is very important to examine each rule carefully.

23. The Effect of Adding Poles to a System α = –2 α = –3 If a pole is added to a system: • The root locus is pulled to the right. • The stability tends to decrease. • The settling time tends to increase (for the same value of ζ, the value of ωd decreases).

24. The Effect of Adding Zeros to a System If a zero is added to a system: • The root locus is pulled to the left. • The stability tends to increase. • The settling time tends to decrease (for the same value of ζ, the value of ωd increases). α = –1.5 α = –2.5 α = –3.5

25. Design Using Dynamic Compensation • If the process dynamics are of such a nature that a satisfactory design cannot be obtained by adjustment of K alone, then some modification or compensation of the process dynamics must be done. • Two compensation schemes have been found to be particularly simple and effective: • Lead compensation, approximates the function of PD control and acts mainly to speed up a response by lowering the rise time and decreasing the transient overshot. • Lag compensation, approximates the function of PI control and is usually used to improve the steady-state accuracy of the system. • The techniques to select the parameters of each compensation schemes will be discussed now.

26. Design Using Dynamic Compensation • A compensation scheme is written generally in the form of a transfer function: • If z < p, it is called lead compensation. • If z > p, it is called lag compensation. • The characteristic equation of the system is:

27. Design Using Lead Compensation • Consider a second-order position control system with normalized transfer function: • The root locus of the system will be compared for:

28. Design Using Lead Compensation α = –0.5 α = –4.5 α = –9.5 • Selecting the value of z and p is usually done by trial and error, which can be minimized with experience. • In general, the zero is placed near the closed-loop ωn, as determined by time domain specification, the pole is located at a distance 5 to 20 times the value of the zero location. • If the pole is too far to the right, the root locus moves back too far toward its uncompensated shape, while if the pole is too far the left, sensor noise will be amplified too much.

29. First Design Using Lead Compensation • Find a compensation for G(s) = 1/[s(s+1)] that will provide overshoot of no more than 20% and rise time of no more than 0.25 sec.

30. First Design Using Lead Compensation • Giving grid to the root locus plot of using D2(s)G(s), and using the calculation results: Area of Eligible Roots Design requirement fulfilled with:

31. Second Design Using Lead Compensation • The closed-loop system of G(s) = 1/[s(s+1)] is now required to have a pole at (corresponds to ωn = 7 and ζ = 0.5). Now using , the value of K and z must be determined. • There is exactly one location for the zero where the angle ψ1will fulfill the phase condition of r0. • If the zero is placed on that location, r0 will be on the root locus. 1 s = –z

32. Second Design Using Lead Compensation 1

33. Second Design Using Lead Compensation The position of the zero can now be calculated: Solving for K using the characteristic equation, 1 s = –z

34. Second Design Using Lead Compensation Thus, the compensation that will make to be on the root locus is:

35. Design Using Lag Compensation • Once satisfactory dynamic response has been obtained, perhaps by using one or more lead compensations, we may discover that the low-frequency gain (the value of the relevant steady-state error constant, such as Kv, Ka) is still too low. • In order to increase this constant, it is necessary to do so in a way that does not upset the already satisfactory dynamic response. • The new compensation D(s) should yield a significant gain at s = 0 to raise the steady-state error constant but is nearly unity at the higher frequency ωn. • The result is: • The value of z and p are small compared with ωn, yet D(0) = z / p can be adjusted to be big enough to adjust the steady-state gain.

36. Design Using Lag Compensation • To study the effects of lag compensation, we use again the result of the “Second Design Using Lead Compensation”. • The uncompensated closed-loop system is:G(s) = 1/[s(s+1)] • It is required to have a pole at (corresponds to ωn = 7 and ζ = 0.5). • The obtained lead compensation is D(s) = 127(s+5.402)/(s+20). • At the operating point, the velocity constant is given by: • Suppose not big enough • Required Kv= 100

37. Design Using Lag Compensation • To obtain Kv = 100, an additional lag compensation is designed, with: • z/p = 3  to increase the velocity constant by 3 at s = 0. • a pole at p = –0.01  to keep the values of both z and p very small so that the lag compensation would have little effect around ωn = 7, the dominant dynamics already obtained previously by the lead compensation. • The overall open-loop transfer function with lead-lag compensation is now given by: • Remark: The design using lag compensation is performed after adjusting the gain K and performing the design using lead compensation(s)

38. Design Using Lag Compensation • The root locus of the new L(s) is plotted below. Dominant root, hardly affected The root locus near the lag compensation • The transient response corresponding to the lag-compensation zero will be very slowly decaying, with small magnitude, and might seriously influence the settling time. • The lag pole-zero combination must be placed at the highest frequency possible without shifting the dominant roots.

39. Example: Compensation Design Problem 5.24 FPE Let Using root-locus techniques, find values for the parameters a, b, and K of the compensation D(s) that will produce closed-loop poles at s =–1 ± j for the system shown below. Unity Feedback System

40. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example: Compensation Design Before compensation desired closed-loop poles

41. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example: Compensation Design desired closed-loop poles The zero of D(s) cancels the pole of G(s) : roots of the compensation The pole of D(s) is placed in such a way that the desired closed loop poles are on the future root locus

42. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example: Compensation Design : roots of the compensation

43. Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example: Compensation Design K?

44. Example: Compensation Design Solving for K using the characteristic equation, The compensation D(s) can now completely be written as:

45. Homework 7 • No.1, FPE (5th Ed.), 5.23. • No.2, FPE (5th Ed.), 5.30. • No.3 • Consider the unity feedback system, with: • Show that the system cannot operate with a 2%-settling-time of 2/3 second and a percent overshoot of 1.5% with a simple gain adjustment. • Design a lead compensator so that the system meets the transient response characteristics of part (a). Specify the compensator’s pole, zero, and the required gain.