1. Compensators

2. Assume that the desired transient response is represented by point B with a specified settling time.
For the current root locus, at the specified overshoot , we can only get the settling time represented by point A.
Our goal is to speed up the response A to B without affecting the percent overshoot.

3. Two ways to solve the problem:
Replace the system with a system whose root locus intersects the desired point B Expensive and counterproductive
Use a compensated system with additional poles and zeros so that compensated system has a root locus that goes through the desired pole location Poles and zero can be added at the low-power end and need not interfere the power output requirement s.
Compensating poles and zeros can be generated with a passive or an active network.
Possible disadvantage of compensating a system is the increase of system order with a subsequent effect on the desired response.
Compensators can be used to improve transient response and steady state error.

4. Both transient response and steady state error constant were related to the gain.
The higher the gain, the smaller the steady state error, but the higher the percent overshoot.
Reducing the gain reduce overshoot but increase the steady state error.
Transient response is improved with the addition of differentiation while the steady state error is improved with the addition of integration in the forward path.

5. Configuration
Two configurations: cascade compensation and feedback compensation.
With cascade compensation, the compensating network, G1(s) is placed at the low-power end of the forward path in cascade with the plant.
With feedback compensation, the compensator, H1(s) is placed in the feedback path.

6. Compensator
Compensators that use pure integration for improving steady-state error or pure differentiation for improving transient response are defined as ideal compensators.
Ideal compensators must be implemented with active networks. E.g. active amplifiers and possible additional power sources.
An advantage of ideal integral compensator is that steady state error is reduced to zero.
Other compensators that do not use pure integration or differentiation or active devices but adopt passive elements such as resistor and capacitor are not ideal compensators.
Advantages of passive networks are: less expensive, do not require additional power sources.
Disadvantage of passive networks: steady-state error not driven to zero.

7. Systems that feed the error forward to the plant are called proportional control systems.
Ideal integral compensator is called a proportional-plus-integral (PI) controller.
Non-ideal integral compensator is called lag compensator.
Ideal derivative compensator is called proportional-plus-derivative (PD) controller.
Non-ideal derivative compensator is called lead compensator.

8. Ideal Integral Compensation (PI)
Improve steady state error by placing an open-loop pole at the origin (this increase system type by one).
Type 0 system responding to a step input with a finite error will respond with zero error if the system is increased by one.
We will learn how to improve steady state error without affecting the transient response.
Figure 9.3(a) shows the original system with desired transient response generated by the close-loop poles at A.
To increase the system type, we add a pole to the origin. Thus the angular contribution of the open loop poles at point A is no longer 1800 and the root locus no longer goes through point A as shown in Figure 9.3(b).

9. Figure 9. 3 Pole at A is: a. on the root locus without compensator; b
Figure 9.3 Pole at A is: a. on the root locus without compensator; b. not on the root locus with compensator pole added; (figure continues)

10. Adding a zero close to the pole at the origin to cancel out the angular contribution of compensator zero and compensator pole.
Thus, point A is still at the root locus and the system type has been increased.
Furthermore, the required gain of the dominant pole is about the same as before compensation, because the ratio of lengths from the compensator pole and the compensator zero is approximately unity.

11. Thus, the steady state error is improved without affecting the transient response.
Example (9.1):
Given the system operates
with a damping ratio 0.174,
show that the PI controller
reduces the steady state error
to zero for a step input without
affecting the transient response.
Choose a pole at the origin and
a zero at -0.1, close to the
compensator pole.
Figure 9.4 Closed-loop system: a. before compensation; b. after ideal integral compensation

12. Then, evaluate the steady state error for a unit step input.
First analyze the uncompensated system and determine the location of the dominant, second order poles.
Then, evaluate the steady state error for a unit step input.
Uncompensated system

13. Searching along the line (damping ratio=0
Searching along the line (damping ratio=0.174) with the root locus program, the dominant poles are ±j3.926 for a gain, K= The third pole is approximately when K= The gain yields Kp=8.23, hence the steady state error is
Adding an ideal integral compensator with zero=-0.1, a new root locus is obtained. The dominant second order poles, the third pole beyond -10, and the gain are approximately the same as uncompensated system. The gain K is now 158.2, and the fourth close loop pole is found at (close enough to the zero allow zero pole cancellation).

15. Compensator with its pole at the origin is a type 1 system; unlike the uncompensated system, it will respond to a step input with zero error.

16. A method of implementing an ideal integral compensator is shown below.
Since both proportional and integral control exist, the controller is called PI Controller.

17. Lag Compensator
If passive networks (lag compensator) are used, the pole and zero are moved to the left, close to the origin (Figure below part c).
a. Type 1 uncompensated system; b. Type 1 compensated system; c. compensator pole-zero plot

18. Although this method does not increase the system type, it does yield improvement in static error constant.
Assume the uncompensated system, the static error constant Kvo is:
Assume the lag compensator, the new static error constant is:
After inserting the compensator, the required gain K is virtually the same since the lengths of the vectors are approximately equal.

19. The improvement in the compensated system’s Kv over the uncompensated system’s Kv is equal to the ratio of the magnitude of the compensator zero to the compensator pole.
In lag compensation, the compensator’s pole and zero must be close to each other to minimize the angular contribution (to keep the transient response unchanged).
The ratio of zc to pc can be large to yield appreciable improvement in steady state error.
Example:, if the pole is at , the zero is at -0.01, the ratio is 10.

20. Using Example (9.1), improve the steady state error by a factor 0f 10 if the system is operating with a damping ratio of
The uncompensated system error was with Kp=8.23 in Example (9.1).
A tenfold improvement means a steady state error of
The ratio,

21. Arbitrarily, select pc=0.01, then zc=11.13pc≈0.111.
Sketch the root locus, search along the ζ=0.174 line for multiple 180o. The second-order dominant pole s are ±j3.836 with a gain K of The third and fourth close loop pole are at and respectively.

25. Ideal Derivative Compensation (PD)
Poles and zeros can be added in the forward path to produce a new open loop function whose root locus goes through the design point on the s plane.
One way to speed up the original system is to add a single zero to the forward path.
Gc(s)=s+zc
Example below shows compensated system by adding compensating zero at -2,-3,-4.

26. Uncompensated

27. Each of the compensated cases has dominant poles with the same ζ, the percent overshoot will be the same for each case.
The compensated, dominant closed loop poles have more negative real parts, thus the settling time for the compensated cases will be shorter than the uncompensated system.
The larger the imaginary parts of the compensated system, the smaller the peak time is.

29. Given a system below, design ideal derivative compensator to yield a 16% overshoot with a threefold reduction in settling time.
16% overshoot is equivalent to ζ= Search along the damping ratio line for an odd multiple 180 degree and find the dominant poles at ±j2.064.
The settling time of uncompensated system is

30. In order to have threefold reduction in settling time, the new settling time will be Therefore the real part of the compensated dominant second order pole is
Imaginary part is

32. Using the open loop poles and the test point -3. 613+j6
Using the open loop poles and the test point j6.913, we obtain the sum of angles as o.
Hence, the angular contribution required from the compensator zero for test point to be on the root locus is =95.6o.
The location of the compensator zero is

36. PD controller.

37. Lead Compensation
Passive lead compensator can approximate an active ideal derivative compensator.
A compensator zero and a pole is result rather than a single zero.
If the pole is farther from the imaginary axis than zero, the angular contribution is still positive , thus it approximates an equivalent single zero.
Advantages: (1) no additional power supplies (2) reduce noise due to differentiation.

40. Example: Design a lead compensator of system below that will reduce the settling time by a factor of 2 while maintaining 30% overshoot.
overshoot 30% equivalent to ζ= Uncompensated settling time is Ts=4/1.007=3.972 seconds
Twofold reduction, Ts=3.972/2=1.986, the real part of desired pole is -4/Ts= The imaginary part is wd=-2.014tan(110.98)=5.252

41. Assume compensator zero at -5, the resulting angle is -172.69o.
Thus the angular contribution required from the compensator pole is -7.31o.

44. PID Controller

46. Lag-Lead Controller
First design with lead compensator to improve transient response then follow by lag
compensator to improve steady state error.

47. Transient Response via Gain Adjustment
If we desire a phase margin, фM, represented by CD,
we have to raise the magnitude curve by AB.
-- simple gain adjustment can be used to design phase margin to present overshoot

48. Design Procedure
(i) With a chosen gain, obtain the Bode plot and the gain crossover frequency,
(ii) Determine the required phase margin from the present overshoot
(iii)Find the frequency ωфM , on the Bode Plot that yield desired phase margin CD.
(iv) Change the gain by amount AB, forcing the magnitude curve to go through
0dB at ωфM . The amount of gain adjustment is the additional gain needed.

49. Find the preamplifier gain to yield 9
Find the preamplifier gain to yield 9.5% overshoot in the transient response for a step input. Use only frequency response method.
Choose K=3.6 to start magnitude plot at 0dB at ω=0.1.
9.5% overshoot implies ζ=0.6 for the closed loop dominance poles.
The following equation yields a 59.2o phase margin for a damping ratio of 0.6.

51. Locate on the phase plot the frequency that yields 59. 2o phase margin
Locate on the phase plot the frequency that yields 59.2o phase margin. -180o+59.2o=-120.8o. The value of the phase margin frequency is 14.8 rad/s.
At a frequency of 14.8 rad/s on the magnitude plot, the gain is found to be -44.2dB. The magnitude has to be raised to 0dB to yield the required phase margin.
20log M=44.2
M=162.2
As the log magnitude plot was drawn for K=3.6, , the adjusted K=3.6x162.2=583.9
The gain adjusted open loop transfer function is

52. Lag Compensation The function of lag compensator on Bode plot is to:
(1) IMPROVE STEADY STATE ERROR CONSTANT BY INCREASING ONLY LOW FREQUENCY GAIN
(2) INCREASE THE PHASE MARGIN OF THE SYSTEM TO YIELD DESIRED TRANSIENT RESPONSE
An uncompensated system is unstable if the gain at 180o is greater than 0dB.
The stability effect of lag compensator comes about as the gain at 180o is reduced below 0dB.

54. Frequency response plots of a lag compensator, Gc(s) = (s + 0
Frequency response plots of a lag compensator, Gc(s) = (s + 0.1)/(s )

55. Observation from the plot:
At low frequency, the compensator will not have influenced
on the open-loop
Design Procedure
Determine the gain K for the required steady state error
and draw the Bode plot
Obtained the required phase margin and add, compensator
phase angle, =5 - 12 o o
Compensator’s zero will be a decade below of the new gain
crossover frequency of (ii).
Compensator’s pole is obtained by drawing a slope of -20dB/dec
from the compensator’s zero until the line touched the 0dB.

56. The transfer function of the lag compensator is
Example: Use Bode Plot to design a lag compensator to yield a tenfold improvement in steady state error over the gain compensated system while keeping the percent overshoot at 9.5%.
From the previous example, a gain K of yields a 9.5% overshoot. For the system, Kv=16.22.
A tenfold improvement, Kv= Therefore, the value of K =5839, the open loop transfer function is

57. Bode Plot for K=5839

58. The phase margin required by 9. 5% overshoot is found to be 59. 2o
The phase margin required by 9.5% overshoot is found to be 59.2o. We increase this value of phase margin by 10o to 69.2o in order to compensate for the phase angle contribution.
Find the frequency where the phase margin is 69.2o. This frequency occurs at phase angle of -180o+69.2o =110.8oand is 9.8rad/s.
At 9.8rad/s, the magnitude plot must go through 0dB. The magnitude at 9.8rad/s is at +24dB. Thus the lag compensator must provide -24dB attenuation at 9.8rad/s.
First, draw the high frequency asymptote at -24dB. Arbitrarily select the higher break frequency to be about one decade below the phase margin frequency or 0.98rad/s.

59. Starting at the intersection with the lag compensator’s high frequency asymptote, draw a -20dB/decade line until 0dB is reached. The lower break frequency is found to be 0.062rad/s.
Thus, the lag compensator transfer function is
The gain is to yield a dc gain of unity.
The compensated system’s forward transfer function is thus