1 / 44

INC341 Design Using Graphical Tool (continue)

INC341 Design Using Graphical Tool (continue). Lecture 10. Improving Both Steady-State Error and Transient Response. PI, Lag improve steady-state error PD, Lead improve transient response PID, Lead-lag improve both (PID = Proportional plus Intergal plus Derivative controller).

floria
Download Presentation

INC341 Design Using Graphical Tool (continue)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. INC341Design Using Graphical Tool(continue) Lecture 10

  2. Improving Both Steady-State Error and Transient Response • PI, Lag improve steady-state error • PD, Lead improve transient response • PID, Lead-lag improve both (PID = Proportional plus Intergal plus Derivative controller)

  3. PID Controller

  4. PID controller design • Evaluate the performance of the uncompensated system • Design PD controller to meet transient response specifications • Simulate and Test, redesign if necessary • Design PI controller to get required steady-state error • Find K constant of PID • Simulate and Test, redesign if necessary

  5. Example Design PID controller so that the system can operate with apeak time that is2/3 of uncompensated system, at 20% OS, and steady-state error of0 for a step input

  6. Step 1 • %OS = 20%  damping ratio = 0.456  Ѳ= 62.87 • Search along ther line to find a point of 180 degree (-5.415±j10.57) • Find a correspoding K=121.51 • Then find the peak time

  7. Step 2 • Decrease peak time by a factor of2/3  get imaginary point of a compensator pole: • To keep adamping ratio constant, real part ofthe pole will be at • The compensatorpoles will be at -8.13±j15.867

  8. Sum of the angles from uncompensated poles and zeros to the test point (-8.13±j15.867) is -198.37 • The contribution angle for the compensator zerois then 180-198.371= 18.37 ดPD controller is(s+55.92)

  9. Step 3 • Simulate the PD compensated system to see if it reduces peak time and improves ss error

  10. Step 4 • design PI compensator (one pole at origin and a zero near origin; at -0.5 in this example) • Find a new point along the 0.456 damping ratio line (-7.516±j14.67), with an associate gain of 4.6

  11. Step 5 • Evaluate K1, K2, K3 ofPID controller • Compare to K1 = 259.5, K2 = 128.6, K3 = 4.6

  12. Step 6

  13. Lead-Lag Compensator Design Same procedures as in designing PID: • Begin withdesigning lead compensator to get the desired transient response • design lag compensator to improvesteady-state error

  14. Example Design lead-lag compensator so that the system can operate with 20% OS, twofold reduction in settling time, and tenfold improvement in steady-state error for a ramp input

  15. Step 1 • %OS = 20%  damping ratio = 0.456  Ѳ= 62.87 • Search along ther line to find a point of 180 degree (-1.794±j3.501) • Find a correspoding K=192.1 • Then find the settling time

  16. Step 2 • Decrease settling time by a factor of2  get a real part of a compensator pole: • To keep adamping ratio constant, imaginary part ofthe pole will be at • The compensatorpoles will be at -3.588±j7.003

  17. Select the compensator zero at -6 to coincide with the open-loop pole • Sum of the angles from uncompensated poles and zeros to the test point (-3.588±j7.003) is -164.65 • The contribution angle for the compensator zerois then 180-164.65= 15.35 Lead compensator is

  18. Then find a new K at the design point (K=1977)

  19. Step 3 Simulate the lead compensated system

  20. Step 4 • Originally the uncompensated system has the transfer function:

  21. After adding the lead compensator, the system has changed to • Static error constant, Kv, is then 6.794 (lead compensator has improved ss error by a factor of 6.794/3.201=2.122) • So the lag compensator must be designed to improve ss error by a factor of 10/2.122=4.713

  22. Step 5 • Pick a pole at 0.01, then the associated zero will be at 0.04713 • Lag-lead compensator • Lag-lead compensatated open loop system

  23. Step 6

  24. Conclusions

  25. Feedback Compensation Put a compensator in the feedback path

  26. Tachometer Popular feedback compensator, rate sensor Tachometer generates a voltage output proportional to input rational speed

  27. rate feedback

  28. Example Design a feedback compensator to decreasesettling time by afactor of 4 and keep a constant %OS of 20

  29. Step 1 %OS = 20%  damping ratio = 0.456  Ѳ= 62.87 Search along the daping ratio line to get a summation of angle of 180 degrees at -1.809±j3.531 Find the corresponding K from the magnitude rule settling time

  30. Step 2 Reduce the Settling time by a factor of 4 A new location of poles is at-7.236±j14.123

  31. Atdominant pole -7.236±j14.123, KG(s)H(s) has a net angle of= -277.33  needs an additional angle from zero of277.33-180= 97.33 Find the correspondingK to the pole at -7.236+j14.123 using the magnitude rule: K= 256.819

  32. Feedback block is0.185(s+5.42)

  33. Physical System Realization PI Compensator

  34. Lag Compensator

  35. PD Compensator

  36. Lead Compensator

  37. PID Compensator

More Related