1. Control Systems (CS) Lecture-4-5-6 Lag Compensation &
Lag-Lead Compensation
Dr. Imtiaz Hussain
Associate Professor
Mehran University of Engineering & Technology Jamshoro, Pakistan
URL :

2. Lag Compensation 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛽 𝑇𝑠+1 𝛽𝑇𝑠+1 , (β>1)
Lag compensation is used to improve the steady state error of the system.
Generally Lag compensators are represented by following transfer function Or
Where 𝐾 𝑐is gain of lag compensator.
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛽 𝑇𝑠+1 𝛽𝑇𝑠+1 , (β>1)
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 , (β>1)

3. Lag Compensation
𝐺 𝑐 𝑠 =3 𝑠+10 𝑠+1 , (𝛽=10)

4. Lag Compensation
Consider the problem of finding a suitable compensation network for the case where the system exhibits satisfactory transient-response characteristics but unsatisfactory steady-state characteristics.
Compensation in this case essentially consists of increasing the open loop gain without appreciably changing the transient-response characteristics.
This means that the root locus in the neighborhood of the dominant closed-loop poles should not be changed appreciably, but the open-loop gain should be increased as much as needed.

5. Lag Compensation
To avoid an appreciable change in the root loci, the angle contribution of the lag network should be limited to a small amount, say less than 5°.
To assure this, we place the pole and zero of the lag network relatively close together and near the origin of the s plane.
Then the closed-loop poles of the compensated system will be shifted only slightly from their original locations. Hence, the transient-response characteristics will be changed only slightly.

6. Lag Compensation 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 , (β>1)
Consider a lag compensator Gc(s), where
If we place the zero and pole of the lag compensator very close to each other, then at s=s1 (where s1is one of the dominant closed loop poles then the magnitudes 𝑠 𝑇 and 𝑠 𝛽𝑇 are almost equal, or
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 , (β>1)
𝐺 𝑐 ( 𝑠 1 ) = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 ≅ 𝐾 𝑐

7. Lag Compensation −5 ° <𝑎𝑛𝑔𝑙𝑒 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 < 0 °
To make the angle contribution of the lag portion of the compensator small, we require
This implies that if gain 𝐾 𝑐 of the lag compensator is set equal to 1, the alteration in the transient-response characteristics will be very small, despite the fact that the overall gain of the open-loop transfer function is increased by a factor of 𝛽, where 𝛽>1.
−5 ° <𝑎𝑛𝑔𝑙𝑒 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 < 0 °

8. Lag Compensation
If the pole and zero are placed very close to the origin, then the value of 𝛽 can be made large.
A large value of 𝛽 may be used, provided physical realization of the lag compensator is possible.
It is noted that the value of T must be large, but its exact value is not critical.
However, it should not be too large in order to avoid difficulties in realizing the phase-lag compensator by physical components.

9. Lag Compensation 𝐾 𝑣 = lim 𝑠→0 𝑠𝐺(𝑠)
An increase in the gain means an increase in the static error constants.
If the open loop transfer function of the uncompensated system is G(s), then the static velocity error constant Kv of the uncompensated system is
Then for the compensated system with the open-loop transfer function Gc(s)G(s) the static velocity error constant 𝐾 𝑣 becomes
𝐾 𝑣 = lim 𝑠→0 𝑠𝐺(𝑠)
𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠) = 𝐾 𝑣 lim 𝑠→0 𝐺 𝑐 𝑠
𝐾 𝑣 = 𝐾 𝑣 lim 𝑠→0 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 = 𝐾 𝑣 𝐾 𝑐 𝛽

10. Lag Compensation
The main negative effect of the lag compensation is that the compensator zero that will be generated near the origin creates a closed-loop pole near the origin.
This closed loop pole and compensator zero will generate a long tail of small amplitude in the step response, thus increasing the settling time.

11. Electronic Lag Compensator
The configuration of the electronic lag compensator using operational amplifiers is the same as that for the lead compensator shown in following figure.
𝐸 𝑜 (𝑠) 𝐸 𝑖 (𝑠) = 𝑅 4 𝐶 1 𝑅 3 𝐶 2 𝑠+ 1 𝑅 1 𝐶 1 𝑠+ 1 𝑅 2 𝐶 2
𝐾 𝑐 = 𝑅 4 𝐶 1 𝑅 3 𝐶 2
𝑇= 𝑅 1 𝐶 1
𝑅 2 𝐶 2 > 𝑅 1 𝐶 1
𝛽𝑇= 𝑅 2 𝐶 2

12. Electronic Lag Compensator
Pole-zero Configuration of Lag Compensator
𝑅 2 𝐶 2 > 𝑅 1 𝐶 1

13. Electrical Lag Compensator
Following figure lag compensator realized by electrical network.
𝑅 2
𝑅 1 𝐶
𝐸 2 (𝑠) 𝐸 1 (𝑠) = 𝑅 2 𝐶𝑠+1 𝑅 1 + 𝑅 2 𝐶𝑠+1

14. Electrical Lag Compensator
𝐸 2 (𝑠) 𝐸 1 (𝑠) = 𝑅 2 𝐶𝑠+1 𝑅 1 + 𝑅 2 𝐶𝑠+1
Then the transfer function becomes
𝛽= 𝑅 1 + 𝑅 2 𝑅 2 >1
𝑇= 𝑅 2 𝐶
𝐸 2 (𝑠) 𝐸 1 (𝑠) = 𝑇𝑠+1 𝛽𝑇𝑠+1

15. Electrical Lag Compensator
𝐸 2 (𝑠) 𝐸 1 (𝑠) = 𝑇𝑠+1 𝛽𝑇𝑠+1
If an RC circuit is used as a lag compensator, then it is usually necessary to add an amplifier with an adjustable gain 𝐾 𝑐 𝛽 so that the transfer function of compensator is
𝐸 2 (𝑠) 𝐸 1 (𝑠) = 𝐾 𝑐 𝛽 𝑇𝑠+1 𝛽𝑇𝑠+1
𝐸 2 (𝑠) 𝐸 1 (𝑠) = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇

16. Mechanical Lag Compensator (Home Work)

17. Design Procedure
The procedure for designing lag compensators by the root-locus method may be stated as follows.
We will assume that the uncompensated system meets the transient-response specifications by simple gain adjustment.
If this is not the case then we need to design a lag-lead compensator which we will discuss in next few classes.

18. Design Procedure Step-1
Draw the root-locus plot for the uncompensated system whose open-loop transfer function is G(s).
Based on the transient-response specifications, locate the dominant closed-loop poles on the root locus.

19. Design Procedure 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛽 𝑇𝑠+1 𝛽𝑇𝑠+1 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 Step-2
Assume the transfer function of the lag compensator to be given by following equation
Then the open-loop transfer function of the compensated system becomes Gc(s)G(s).
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛽 𝑇𝑠+1 𝛽𝑇𝑠+1 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇

20. Design Procedure Step-3
Evaluate the particular static error constant specified in the problem.
Determine the amount of increase in the static error constant necessary to satisfy the specifications.

21. Design Procedure Step-4
Determine the pole and zero of the lag compensator that produce the necessary increase in the particular static error constant without appreciably altering the original root loci.
The ratio of the value of gain required in the specifications and the gain found in the uncompensated system is the required ratio between the distance of the zero from the origin and that of the pole from the origin.

22. Design Procedure Step-5
Draw a new root-locus plot for the compensated system.
Locate the desired dominant closed-loop poles on the root locus.
(If the angle contribution of the lag network is very small—that is, a few degrees—then the original and new root loci are almost identical.
Otherwise, there will be a slight discrepancy between them.
Then locate, on the new root locus, the desired dominant closed-loop poles based on the transient-response specifications.

23. Design Procedure Step-6
Adjust gain of the compensator from the magnitude condition so that the dominant closed-loop poles lie at the desired location.
𝐾 𝑐 will be approximately 1.

24. Example-1 Consider the system shown in following figure.
The damping ratio of the dominant closed-loop poles is The undamped natural frequency of the dominant closed-loop poles is rad/sec. The static velocity error constant is 0.53 sec–1.
It is desired to increase the static velocity error constant Kv to about 5 sec–1 without appreciably changing the location of the dominant closed-loop poles.

25. Example-1 (Step-1) The dominant closed-loop poles of given system are
s = ± j0.5864

26. Example-1 (Step-2) 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛽 𝑇𝑠+1 𝛽𝑇𝑠+1 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇
According to given conditions we need to add following compensator to fulfill the requirement.
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛽 𝑇𝑠+1 𝛽𝑇𝑠+1 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇

27. Example-1 (Step-3)
The static velocity error constant of the plant ( 𝐾 𝑣 ) is
The desired static velocity error constant ( 𝐾 𝑣 ) of the compensated system is 5 𝑠 −1 .
𝐾 𝑣 = lim 𝑠→0 𝑠𝐺(𝑠)= lim 𝑠→0 𝑠 𝑠 𝑠+1 𝑠+2 =0.53 𝑠 −1
𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠) = 𝐾 𝑣 lim 𝑠→0 𝐺 𝑐 𝑠
𝐾 𝑣 = 𝐾 𝑣 lim 𝑠→0 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 = 𝐾 𝑣 𝐾 𝑐 𝛽

28. Example-1 (Step-3) 𝐾 𝑣 = 𝐾 𝑣 lim 𝑠→0 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇 = 𝐾 𝑣 𝐾 𝑐 𝛽
𝐾 𝑣 = 𝐾 𝑣 𝐾 𝑐 𝛽
5=0.53𝛽
𝛽=10

29. Example-1 (Step-4) 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇
Place the pole and zero of the lag compensator
Since 𝛽=10, therefore
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛽𝑇
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 0.1 𝑇

30. Example-1 (Step-4) 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+0.05 𝑠+0.005
Solution-1
Place the zero and pole of the lag compensator at s=–0.05 and s=–0.005, respectively.
The transfer function of the lag compensator becomes
Open loop transfer function is given as
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+0.05 𝑠+0.005
𝐺 𝑐 𝑠 𝐺(𝑠)= 𝐾 𝑐 𝑠+0.05 𝑠 𝑠(𝑠+1)(𝑠+2)
𝐺 𝑐 𝑠 𝐺(𝑠)= 𝐾(𝑠+0.05) 𝑠(𝑠+0.005)(𝑠+1)(𝑠+2)
𝑤ℎ𝑒𝑟𝑒 𝐾=1.06 𝐾 𝑐

31. Example-1 (Step-5)
Solution-1
Root locus of uncompensated and compensated systems.
New Closed Loop poles are
𝑠=−0.31±𝑗0.55

32. Example-1 (Step-5)
Solution-1
Root locus of uncompensated and compensated systems.

33. 𝐾(𝑠+0.05) 𝑠(𝑠+0.005)(𝑠+1)(𝑠+2) 𝑠=−0.31+𝑗0.55 =1
Example-1 (Step-6)
Solution-1
The open-loop gain K is determined from the magnitude condition.
Then the compensator gain 𝐾 𝑐 is determined as
𝐾(𝑠+0.05) 𝑠(𝑠+0.005)(𝑠+1)(𝑠+2) 𝑠=−0.31+𝑗0.55 =1
𝐾=1.0235
𝐾=1.06 𝐾 𝑐
𝐾 𝑐 = 𝐾 1.06 =0.9656

34. Example-1 (Step-6) 𝐺 𝑐 𝑠 =0.9656 𝑠+0.05 𝑠+0.005
Solution-1
Then the compensator transfer function is given as
𝐺 𝑐 𝑠 = 𝑠+0.05 𝑠+0.005

35. Example-1 (Final Design Check)
Solution-1
The compensated system has following open loop transfer function.
Static velocity error constant is calculated as
𝐺 𝑐 𝑠 𝐺(𝑠)= (𝑠+0.05) 𝑠(𝑠+0.005)(𝑠+1)(𝑠+2)
𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠)
𝐾 𝑣 = lim 𝑠→0 𝑠[ 𝑠 𝑠 𝑠 𝑠+1 𝑠+2 ]
𝐾 𝑣 = =5.12 𝑠 −1

36. Example-1 (Step-4) 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+0.01 𝑠+0.001
Solution-2
Place the zero and pole of the lag compensator at s=–0.01 and s=–0.001, respectively.
The transfer function of the lag compensator becomes
Open loop transfer function is given as
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+0.01 𝑠+0.001
𝐺 𝑐 𝑠 𝐺(𝑠)= 𝐾 𝑐 𝑠+0.01 𝑠 𝑠(𝑠+1)(𝑠+2)
𝐺 𝑐 𝑠 𝐺(𝑠)= 𝐾(𝑠+0.01) 𝑠(𝑠+0.001)(𝑠+1)(𝑠+2)
𝑤ℎ𝑒𝑟𝑒 𝐾=1.06 𝐾 𝑐

37. Example-1 (Step-5)
Solution-2
Root locus of uncompensated and compensated systems.
New Closed Loop poles are
𝑠=−0.33±𝑗0.55

38. Example-2
Design a lag compensator for following unity feedback system such that the static velocity error constant is 50 sec-1 without appreciably changing the closed loop poles, which are at 𝑠=−2±𝑗 6 .

39. Lag-Lead Compensation
Lead compensation basically speeds up the response and increases the stability of the system.
Lag compensation improves the steady-state accuracy of the system, but reduces the speed of the response.
If improvements in both transient response and steady-state response are desired, then both a lead compensator and a lag compensator may be used simultaneously.
Rather than introducing both a lead compensator and a lag compensator as separate units, however, it is economical to use a single lag–lead compensator.

40. Lag-Lead Compensation
Lag-Lead compensators are represented by following transfer function
Where Kc belongs to lead portion of the compensator.
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 1 𝑠+ 𝛾 𝑇 𝑠+ 1 𝑇 2 𝑠+ 1 𝛽𝑇 2 , (γ>1 𝑎𝑛𝑑 β>1)

41. Design Procedure
In designing lag–lead compensators, we consider two cases where
Case-1: γ≠𝛽
Case-2: γ=𝛽
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 1 𝑠+ 𝛾 𝑇 𝑠+ 1 𝑇 2 𝑠+ 1 𝛽𝑇 2 , (γ>1 𝑎𝑛𝑑 β>1)
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 1 𝑠+ 𝛽 𝑇 𝑠+ 1 𝑇 2 𝑠+ 1 𝛽𝑇 2 , (β>1)

42. Design Procedure (Case-1)
Step-1: Design Lead part using given specifications.
Step-1: Design lag part according to given values of static error constant.
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 1 𝑠+ 𝛾 𝑇 𝑠+ 1 𝑇 2 𝑠+ 1 𝛽𝑇 2 , (γ>1 𝑎𝑛𝑑 β>1)

43. Example-1 (Case-1)
Consider the control system shown in following figure
The damping ratio is 0.125, the undamped natural frequency is 2 rad/sec, and the static velocity error constant is 8 sec–1.
It is desired to make the damping ratio of the dominant closed-loop poles equal to 0.5 and to increase the undamped natural frequency to 5 rad/sec and the static velocity error constant to 80 sec–1.
Design an appropriate compensator to meet all the performance specifications.

44. Example-1 (Case-1)
From the performance specifications, the dominant closed-loop poles must be at
Since
Therefore the phase-lead portion of the lag–lead compensator must contribute 55° so that the root locus passes through the desired location of the dominant closed-loop poles.
𝑠=−2.50±𝑗4.33

45. Example-1 (Case-1) 𝐾 𝑐 𝑠+ 1 𝑇 1 𝑠+ 𝛾 𝑇 1 = 𝐾 𝑐 𝑠+0.5 𝑠+5.02
The phase-lead portion of the lag–lead compensator becomes
Thus 𝑇 1 =2 and 𝛾=10.04.
Next we determine the value of Kc from the magnitude condition:
𝐾 𝑐 𝑠+ 1 𝑇 1 𝑠+ 𝛾 𝑇 1 = 𝐾 𝑐 𝑠+0.5 𝑠+5.02
𝐾 𝑐 (𝑠+0.5) 𝑠 𝑠(𝑠+0.5) 𝑠=−2.5+𝑗4.33 =1
𝐾 𝑐 = 𝑠(𝑠+5.02) 4 𝑠=−2.5+𝑗4.33 =5.26

46. Example-1 (Case-1)
The phase-lag portion of the compensator can be designed as follows.
First the value of 𝛽is determined to satisfy the requirement on the static velocity error constant
𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠)
80= lim 𝑠→0 𝑠 𝑠+ 1 𝑇 2 𝑠 𝑠 𝑠+ 1 𝛽𝑇 2
80=4.988𝛽
𝛽=16.04

47. Example-1 (Case-1)
Finally, we choose the value of 𝑇 2 such that the following two conditions are satisfied:

48. Example-1 (Case-1) 𝐺 𝑐 𝑠 =6.26 𝑠+0.5 𝑠+5.02 𝑠+0.2 𝑠+0.0127
Now the transfer function of the designed lag–lead compensator is given by
𝐺 𝑐 𝑠 =6.26 𝑠+0.5 𝑠 𝑠+0.2 𝑠

49. Example-1 (Case-2)
Home Work

50. Home Work Electronic Lag-Lead Compensator
Electrical Lag-Lead Compensator
Mechanical Lag-Lead Compensator

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