1. INC341 Design with Root Locus
Lecture 9

2. 2 objectives for desired response
Improving transient response
Percent overshoot, damping ratio, settling time, peak time
Improving steady-state error
Steady state error

3. Gain adjustment
Higher gain, smaller steady stead error, larger percent overshoot
Reducing gain, smaller percent overshoot, higher steady state error

4. Compensator Allows us to meet transient and steady state error.
Composed of poles and zeros.
Increased an order of the system.
The system can be approx. to 2nd order using some techniques.

5. Improving transient response
Point A and B have the same damping ratio.
Starting from point A, cannot reach a faster response at point B by adjusting K.
Compensator is preferred.

6. Compensator configulations
Cascade
Compensator
Feedback
Compensator
The added compensator can change a pattern of root locus

7. Types of compensator Active compensator Passive compensator
PI, PD, PID use of active components, i.e., OP-AMP
Require power source
ss error converge to zero
Expensive
Passive compensator
Lag, Lead use of passive components, i.e., R L C
No need of power source
ss error nearly reaches zero
Less expensive

8. Improving steady-state error
Placing a pole at the origin to increase system order; decreasing ss error as a result!!

9. The pole at origin affects the transeint response  adds a zero close to the pole to get an ideal integral compensator

10. Example Choose zero at -1
Damping ratio = in both uncompensated and PI cases

11. Draw root locus without compensator
Draw a straight line of damping ratio
Evaluate K from the intersection point
From K, find the last pole (at )
Calculate steady-state error

12. Finding an intersection between damping ratio line and root locus
Damping ratio line has an equation:
where a = real part, b = imaginary part of the intersection point,
Summation of angle from open-loop poles and zeros to the point is 180 degrees

13. Arctan formula

14. Magnitude of open loop system is 1
Use the formula to get the real and imaginary part of the intersection point and get
Magnitude of open loop system is 1
No open loop zero

15. Draw root locus with compensator (system order is up by 1--from 3rd to 4th)
Needs complex poles corresponding to damping ratio of (K=158.2)
From K, find the 3rd and 4th poles (at and )
Pole at can do phase cacellation with zero at -1 (3th order approx.)
Compensated system and uncompensated system have similar transient response (closed loop poles and K are aprrox. The same)

16. Comparason of step response of the 2 systems

17. PI Controller

18. Lag Compensator Build from passive elements
Improve ss error by a factor of Zc/Pc
To improve both transient and ss responses, put pole and zero close to the origin

19. Uncompensated system
With lag compensation
(root locus remains the same)

20. Example With damping ratio of 0.174, add lag
Compensator to improve steady-state error
by a factor of 10

21. Step I: find an intersection of root locus and
damping ratio line ( j3.926 with K=164.56)
Step II: find Kp = lim G(s) as s0 (Kp=8.228)
Step III: steady-state error = 1/(1+Kp)= 0.108
Step IV: want to decrease error down to
[Kp = (1 – )/ = ]
Step V: require a ratio of compensator zero to pole
as /8.228 =
Step VI: choose a pole at 0.01, the corresponding
Zero will be at *0.01 = 0.111

22. 3rd order approx. for lag compensator
(= uncompensated system)  making
Same transient response but 10 times
Improvement in ss response!!!

24. If we choose a compensator pole at 0.001 (10 times
closer to the origin), we’ll get a compensator zero
at (Kp=91.593)
New compensator:
4th pole is at -0.01
(compared to )
producing a longer
transient response.

25. SS response improvement conclusions
Can be done either by PI controller (pole at origin) or lag compensator (pole closed to origin).
Improving ss error without affecting the transient response.
Next step is to improve the transient response itself.

26. Improving Transient Response
Objective is to
Decrease settling time
Get a response with a desired %OS (damping ratio)
Techniques can be used:
PD controller (ideal derivative compensation)
Lead compensator

27. Ideal Derivative Compensator
So called PD controller
Compensator adds a zero to the system at –Zc to keep a damping ratio constant with a faster response

28. (a) Uncompensated system, (b) compensator zero at -2 (d) compensator zero at -3, (d) compensator zero at -4
Indicate peak time
Indicate settling time

29. Settling time & peak time: (b)<(c)<(d)<(a)
%OS: (b)=(c)=(d)=(a)
ss error: compensated systems has lower value than uncompensated one cause improvement in transient response always yields an improvement in ss error

31. Example
design a PD controller to yield 16% overshoot with a threefold reduction in settling time

32. Step I: calculate a corresponding damping ration (16% overshoot = 0
Step I: calculate a corresponding damping ration (16% overshoot = damping ratio)
Step II: search along the damping ratio line for an odd multiple of 180 (at ±j2.064) and corresponding K (43.35)
Step III: find the 3rd pole (at -7.59) which is far away from the dominant poles  2nd order approx. works!!!

33. More details in step II and III
Characteristic equation:

34. Step IV: evaluate a desired settling time:
Step V: get corresponding real and imagine number of the dominant poles
( and )

35. Location of poles
as desired is at
-3.613±j6.192

36. Step VI: summation of angles at the desired pole location, -275
Step VI: summation of angles at the desired pole location, , is not an odd multiple of 180 (not on the root locus) need to add a zero to make the sum of 180.
Step VII: the angular contribution for the point to be on root locus is =95.6  put a zero to create the desired angle

37. Compensator: (s+3.006)
Might not have a pole-zero cancellation for
compensated system

39. PD Compensator

40. Lead Compensation
Zeta2-zeta1=angular contribution

41. Can put pairs of poles/zeros to get a desired θc

42. Example Design three lead compensators for the system
that has 30% OS and will reduce settling time down
by a factor of 2.

43. Step I: %OS = 30% equaivalent to damping ratio = 0.358, Ѳ= 69.02
Step II: Search along the line to find a point that gives 180 degree (-1.007±j2.627)
Step III: Find a corresponding K ( )
Step IV: calculate settling time of uncompensated system
Step V: twofold reduction in settling time (Ts=3.972/2 = 1.986), correspoding real and imaginary parts are:

44. Step VI: let’s put a zero at -5 and find the net angle to the test point (-172.69)
Step VII: need a pole at the location giving 7.31 degree to the test point.

46. Note: check if the 2nd order approx
Note: check if the 2nd order approx. is valid for justify our estimates of percent overshoot and settling time
Search for 3rd and 4th closed-loop poles
(-43.8, )
-43.8 is more than 20 times the real part of the dominant pole
is close to the zero at -5
The approx. is then valid!!!