1. Chapter 9 Design via Root Locus <<<4.1>>>
###Continuous Time Signals###

2. Improving transient response
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

3. Improving steady-state error
Compensation techniques: a. cascade; b. feedback Ideal compensators are implemented with active networks.

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

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

6. Closed-loop system for Example 9. 1 a. before compensation; b
Closed-loop system for Example 9.1 a. before compensation; b. after ideal integral compensation
Problem: The given system operating with damping ratio of 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 uncompensated system of Figure 9.4(a)
The gain K = 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 the uncompensated system response of Example 9.1

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

11. Lag Compensator a. Type 1 uncompensated system; b
Lag Compensator a. 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
Root locus: a. before lag compensation; b. after lag compensation
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 with Kp= a ten fold improvement means ss error = so Kp= so the ratio arbitrarily
selecting Pc=0.01 and Zc=11.13Pc

14. Root locus for compensated system

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

16. Step responses of uncompensated and lag-compensated systems for Example 9.2

17. Step responses of the system for Example 9
Step responses of the system for Example 9.2 using different lag compensators

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

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

20. Uncompensated system and ideal derivative compensation 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 j6.193 is
The angle of the zero to be on the root locus is =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 infinite possible 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

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

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.
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 j10.57 with gain 121.5, and a third pole at 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 Thus the contribution required from the compensator zero is = 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 damping ratio line, we find the dominant poles at 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. Improving Steady-State Error and Transient Response
Finally to implement the compensator and find the K’s, using the PD and PI compensators and compare to we find K1= 259.5, K2=128.6, and K3=4.6

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 j3.501 with gain 192.1, and a third pole at 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 and the contribution of the pole is 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

51. Improvement in step response for lag-lead- compensated system of Example 9.6

52. Root locus for lag-lead- compensated system of Example 9.6
Since the uncompensated system’s open-loop transfer function is
The static error constant of the uncompensated system is 3.201
Since the open-loop transfer function of the lead-compensated system is
the static error constant of the lead-compensated system is 6.794, so we have improvement by a factor of
To improve the original system error by a factor of 10, the lag compensator must be designed to improve the error by a factor of 10/2.122 = 4.713

53. Root locus for lag-lead- compensated system of Example 9.6
We arbitrarily choose the lag compensator pole at 0.01, which then places the zero at yielding
as a lag compensator
and
as lag-lead-compensated system open-loop transfer function

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

55. Improvement in step response for lag-lead- compensated system of Example 9.6

56. Improvement in ramp response error for the system of Example 9.6:

57. a. Root locus before cascading notch filter; b
a. Root locus before cascading notch filter; b. typical closed-loop step response before cascading notch filter;
c. pole-zero plot of a notch filter; d. root locus after cascading notch filter; e. closed-loop step response after cascading notch filter.

58. Types of cascade compensators

59. Types of cascade compensators

60. Generic control system with feedback compensation
Equivalent block diagram

61. a. Transfer function of a tachometer
b. Tachometer feed-back compensation

62. Equivalent block diagram for the feedback compensator
The Figure shows that the loop
gain, G(s)H(s), is
G(s)H(s) = K1G1(s){KfHc(s)+KG2(s)}
Without feedback, KfHc(s), the loop gain is
G(s)H(s) = KK1G1(s)G2(s)
Thus, the effect of adding feedback is to
replace the poles and zeros of G2(s) with the
poles and zeros of [KfHc(s) + KG2(s)].
Hence, this method is similar to cascade
compensation in that we add new poles and zeros via H(s) to reshape the root locus to go through the design point. However, one must remember that zeros of the equivalent feedback shown in the Figure, H(s) = [KfHc(s) +KG2(s)]/KG2(s), are not closed-loop zeros.
For example, if G2(s) = 1 and the minor-loop feedback, KfHc(s), is a rate
sensor, = KfS, then the loop gain is G(s)H(s) = KfK1G1(s)(s+K/Kf)
Thus, a zero at -K/Kf is added to the existing open-loop poles and zeros. This zero reshapes the root locus to go through the desired design point. Again, this zero is not a closed-loop zero.

63. Example 9.7 Compensating Zero via Rate Feedback
PROBLEM: Given the system of Figure 9.49(a), design rate feedback compensation,
as shown in Figure 9.49(b), to reduce the settling time by a factor of 4 while continuing to operate the system with 20% overshoot.
SOLUTION: First design a PD compensator. For the uncompensated system, search
along the 20% overshoot line (ξ = 0.456) and find that the dominant poles are at
±j3.531, as shown in Figure in next slide.

64. Root locus for uncompensated system
The settling time is 2.21 seconds and must be reduced by a factor of 4 to 0.55 second.
Next determine the location of the dominant poles for the compensated
system. To achieve a fourfold decrease in the settling time, the real part of the pole
must be increased by a factor of 4. Thus, the compensated pole has a real part of
4(-1.809) = The imaginary part is then
ωd = tan ° = 14.12
where ° is the angle of the 20% overshoot line.

65. Step response for uncompensated system of Example

66. Predicted characteristics of uncompensated and compensated systems of Example

67. Finding the compensator zero in Example
Using the compensated dominant pole position of
±j14.12, we sum the angles from the uncompensated system's poles and obtain °. This angle requires a compensator zero contribution of ° to yield 180° at the design point. The geometry shown in the Figure leads to the calculation of the compensator's zero location. Hence, 14.12/ (7.236 – zc) = tan(180° °); from which zc = 5.42.

68. Root locus for the compensated system of Example
The root locus for the equivalent compensated system is shown in the Figure. The gain at the design point, which is K1Kf from Figure 9.49(c), is found to be Since Kf is the reciprocal of the compensator zero, Kf = Thus, K1 = 1388.
In order to evaluate the steady-state error characteristic, Kv is found from Figure 9.49(d) to be
Kv = K1 /(75 + K1Kf) = 4.18

69. Step response for the compensated system of Example
we see that the closed-loop transfer function is T(s) = G(s) / {1 + G(s)H(s)} =
K1 / (s3 + 20s2 + (75 + K1Kf)s + K1)
Thus, as predicted, the open-loop zero is not a closed-loop zero, and there is no pole-zero cancellation. Hence, the design must be checked by simulation.
The results of the simulation are shown in Figure and show an over-damped response with a settling time of 0.75 second, compared to the uncompensated system's settling time of approximately 2.2 seconds
Although not meeting the design requirements, the response still
represents an improvement over the uncompensated system.
Typically, less overshoot is acceptable. The system should be redesigned for further reduction in settling time.

70. Vo(s)/Vi(s) = Z2(s)/Z1(s)
Physical Realization of Compensation -- Active-Circuit Realization
Vo(s)/Vi(s) = Z2(s)/Z1(s)
Operational amplifier configured for transfer function realization

72. Lag-lead compensator implemented with operational amplifiers

73. Example 9.9 Implementing a PID Controller
PROBLEM: Implement the PID controller of Example 9.5.
SOLUTION: The transfer function of the PID controller is
Gc(s) = (s )(s + 0.5)/s
which can be put in the form Gc(s) = s /s
Comparing the PID controller in Table 9.10 with previous equation we obtain the following three relationships:
R2/R1+C1/C2 = 56.42
R2C1 = 1
1/R1C2 = 27.96
(Since there are four unknowns and three equations, we arbitrarily select a practical value for one of the elements. Selecting
C2 = 0.1μF, the remaining values are found to be R1 = kΩ, R2 = 178,891 kΩ, and C1 = 5.59 μF.
The complete circuit is shown in Figure 9.62, where the circuit element values have been rounded off.

75. Example 9.10 Realizing a Lead Compensator