1. INC341 Design 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

5. Example Design PID controller so that the system can operate
with a peak time that is 2/3 of uncompensated system,
at 20% OS, and steady-state error of 0 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

8. Step 2
Decrease peak time by a factor of 2/3  get imaginary point of a compensator pole:
To keep a damping ratio constant, real part of the pole will be at
The compensator poles will be at -8.13±j15.867

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

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

12. 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 damping ratio line (-7.516±j14.67), with an associate gain of 4.6

14. Step 5 Evaluate K1, K2, K3 of PID controller Compare to

15. Step 6

16. Lead-Lag Compensator Design
Same procedures as in designing PID:
Begin with designing lead compensator to get the desired transient response
design lag compensator to improve steady-state error

17. 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

18. 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

20. Step 2
Decrease settling time by a factor of 2  get a real part of a compensator pole:
To keep a damping ratio constant, imaginary part of the pole will be at
The compensator poles will be at ±j7.003

21. 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
The contribution angle for the compensator zero is then = 15.35
Lead compensator is

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

23. Step 3
Simulate the lead compensated system

24. Step 4
Originally the uncompensated system has the transfer function:

25. After adding the lead compensator, the system has changed to
Static error constant, Kv, is then (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

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

29. Step 6

30. Conclusions

32. Feedback Compensation
Put a compensator in the feedback path

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

34. rate feedback

35. Example Design a feedback compensator to decrease settling time
by a factor of 4 and
keep a constant %OS of 20

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

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

38. At dominant pole -7.236±j14.123, KG(s)H(s) has a net
angle of =  needs an additional angle from zero
of = 97.33
Find the corresponding K to the pole at j14.123
using the magnitude rule:
K =

39. Feedback block is 0.185(s+5.42)

40. Physical System Realization
PI Compensator

41. Lag Compensator

42. PD Compensator

43. Lead Compensator

44. PID Compensator