1. Modern Control Systems (MCS)
Lecture
Lead Compensation
Dr. Imtiaz Hussain
Assistant Professor
URL :

2. Lecture Outline Introduction to Lead Compensation
Electronic Lead Compensator
Electrical Lead Compensator
Mechanical Lead Compensator

3. Lead Compensation
Lead Compensation essentially yields an appreciable improvement in transient response and a small change in steady state accuracy.
There are many ways to realize lead compensators and lag compensators, such as electronic networks using operational amplifiers, electrical RC networks, and mechanical spring-dashpot systems.

4. Lead Compensation 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛼 𝑇𝑠+1 𝛼𝑇𝑠+1 , (0<𝛼<1)
Generally Lead compensators are represented by following transfer function or
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛼 𝑇𝑠+1 𝛼𝑇𝑠+1 , (0<𝛼<1)
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 , (0<𝛼<1)

5. Lead Compensation
𝐺 𝑐 𝑠 =3 𝑠+1 𝑠+10 , (𝛼=0.1)

6. Electronic Lead Compensator
Following figure shows an electronic lead compensator using operational amplifiers.
𝐸 𝑜 (𝑠) 𝐸 𝑖 (𝑠) = 𝑅 2 𝑅 4 𝑅 1 𝑅 𝑅 1 𝐶 1 𝑠+1 𝑅 2 𝐶 2 𝑠+1

7. Electronic Lead Compensator
𝐸 𝑜 (𝑠) 𝐸 𝑖 (𝑠) = 𝑅 2 𝑅 4 𝑅 1 𝑅 𝑅 1 𝐶 1 𝑠+1 𝑅 2 𝐶 2 𝑠+1
This can be represented as
Where,
Then,
Notice that
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
𝑎𝑇= 𝑅 2 𝐶 2
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 , (0<𝛼<1)
𝑅 1 𝐶 1 > 𝑅 2 𝐶 2

8. Electronic Lead Compensator
Pole-zero Configuration of Lead Compensator
𝑅 1 𝐶 1 > 𝑅 2 𝐶 2

9. Lead Compensation Techniques Based on the Root-Locus Approach.
The root-locus approach to design is very powerful when the specifications are given in terms of time-domain quantities, such as
damping ratio
undamped natural frequency
desired dominant closed-loop poles
maximum overshoot
rise time
settling time.

10. Lead Compensation Techniques Based on the Root-Locus Approach.
The procedures for designing a lead compensator by the root-locus method may be stated as follows:
Step-1: Analyze the given system via root locus.

11. Step-2
From the performance specifications, determine the desired location for the dominant closed-loop poles.

12. Step-3
From the root-locus plot of the uncompensated system (original system), ascertain whether or not the gain adjustment alone can yield the desired closed loop poles.
If not, calculate the angle deficiency.
This angle must be contributed by the lead compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles.

13. Step-4 Assume the Lead Compensator to be:
Where α and T are determined from the angle deficiency.
Kc is determined from the requirement of the open-loop gain.

14. Step-5
If static error constants are not specified, determine the location of the pole and zero of the lead compensator so that the lead compensator will contribute the necessary angle.
If no other requirements are imposed on the system, try to make the value of α as large as possible.
A larger value of α generally results in a larger value of Kv, which is desirable.
Larger value of α will produce a larger value of Kv and in most cases, the larger the Kv is, the better the system performance.

15. Step-6
Determine the value of Kc of the lead compensator from the magnitude condition.

16. Final Design check
Once a compensator has been designed, check to see whether all performance specifications have been met.
If the compensated system does not meet the performance specifications, then repeat the design procedure by adjusting the compensator pole and zero until all such specifications are met.

17. Final Design check
If the selected dominant closed-loop poles are not really dominant, or if the selected dominant closed-loop poles do not yield the desired result, it will be necessary to modify the location of the pair of such selected dominant closed-loop poles.

18. Example-1
Consider the position control system shown in following figure.
It is desired to design an Electronic lead compensator Gc(s) so that the dominant closed poles have the damping ratio 0.5 and undamped natural frequency 3 rad/sec.

19. Step-1 (Example-1) Draw the root Locus plot of the given system.
The closed loop transfer function of the given system is:
The closed loop poles are

20. Step-1 (Example-1)
Determine the characteristics of given system using root loci.
The damping ratio of the closed-loop poles is
The undamped natural frequency of the closed-loop poles is rad/sec.
Because the damping ratio is small, this system will have a large overshoot in the step response and is not desirable.

21. Step-2 (Example-1)
From the performance specifications, determine the desired location for the dominant closed-loop poles.
Desired performance Specifications are:
It is desired to have damping ratio 0.5 and undamped natural frequency 3 rad/sec.

22. Desired Closed Loop Pole
Step-2 (Example-1)
Alternatively desired location of closed loop poles can also be determined graphically
Desired ωn= 3 rad/sec
Desired damping ratio= 0.5
Desired Closed Loop Pole

23. Desired Closed Loop Pole
Step-3 (Exampl-1)
From the root-locus plot of the uncompensated system ascertain whether or not the gain adjustment alone can yield the desired closed loop poles.
Desired Closed Loop Pole

24. Desired Closed Loop Pole
Step-3 (Exampl-1)
If not, calculate the angle deficiency.
To calculate the angle of deficiency apply Angle Condition at desired closed loop pole. -1
Desired Closed Loop Pole -2
120o
100.8o

25. Step-3 (Exampl-1)
Alternatively angle of deficiency can be calculated as.
Where are desired closed loop poles

26. Step-4 (Exampl-1)
This angle must be contributed by the lead compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles.
Note that the solution to such a problem is not unique. There are infinitely many solutions.

27. Step-5 (Exampl-1) Solution-1
If we choose the zero of the lead compensator at s = -1 so that it will cancel the plant pole at s =-1, then the compensator pole must be located at s =-3.

28. Step-5 (Example-1)
Solution-1
If static error constants are not specified, determine the location of the pole and zero of the lead compensator so that the lead compensator will contribute the necessary angle.

29. Step-5 (Example-1) 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1 𝑠+3
Solution-1
The pole and zero of compensator are determined as
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1 𝑠+3
The Value of 𝛼 can be determined as
1 𝑇 =1 yields 𝑇=1
1 𝛼𝑇 =3 yields 𝛼=0.333

30. Step-6 (Example-1) 𝐺 𝑐 𝑠 =0.9 𝑠+1 𝑠+3
Solution-1
The Value of Kc can be determined using magnitude condition.
𝐾 𝑐 (𝑠+1) 𝑠 𝑠(𝑠+1) 𝑠=−1.5+𝑗 =1
𝐾 𝑐 10 𝑠(𝑠+3) 𝑠=−1.5+𝑗 =1
𝐾 𝑐 = 𝑠(𝑠+3) 10 𝑠=−1.5+𝑗 =0.9
𝐺 𝑐 𝑠 =0.9 𝑠+1 𝑠+3

31. Final Design Check
Solution-1
The open loop transfer function of the designed system then becomes
The closed loop transfer function of compensated system becomes.
𝐺 𝑐 𝑠 𝐺(𝑠)= 9 𝑠(𝑠+3)
𝐶(𝑠) 𝑅(𝑠) = 9 𝑠 2 +3𝑠+9

32. Final Design Check
Solution-1
𝐺(𝑠)= 10 𝑠(𝑠+1)
𝐺 𝑐 𝑠 𝐺(𝑠)= 9 𝑠(𝑠+3)

33. Final Design Check
Solution-1
The static velocity error constant for original system is obtained as follows.
The steady state error is then calculated as
𝐾 𝑣 = lim 𝑠→0 𝑠𝐺(𝑠)
𝐾 𝑣 = lim 𝑠→0 𝑠 10 𝑠(𝑠+1) =10
𝑒𝑠𝑠= 1 𝐾 𝑣 = 1 10 =0.1

34. Final Design Check
Solution-1

35. Final Design Check
Solution-1
The static velocity error constant for the compensated system can be calculated as
The steady state error is then calculated as
𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠)
𝐾 𝑣 = lim 𝑠→0 𝑠 9 𝑠(𝑠+3) =3
𝑒𝑠𝑠= 1 𝐾 𝑣 = 1 3 =0.333

36. Step-5 (Exampl-1)
Solution-2
Solution-2 -1 -2
90o
49.2o -3

37. Step-5 (Exampl-1) 𝐺 𝑐 𝑠 =1.03 𝑠+1.5 𝑠+3.6 Solution-2 Solution-2 -1 -2-3
𝐺 𝑐 𝑠 =1.03 𝑠+1.5 𝑠+3.6

38. Step-5 (Example-1)
Solution-3
If no other requirements are imposed on the system, try to make the value of α as large as possible. A larger value of α generally results in a larger value of Kv, which is desirable.
Procedure to obtain a largest possible value for α.
First, draw a horizontal line passing through point P, the desired location for one of the dominant closed-loop poles. This is shown as line PA in following figure.
Draw also a line connecting point P and the origin O. P A -1 -2 -3 O

39. Step-5 (Example-1)
Solution-3
Bisect the angle between the lines PA and PO, as shown in following figure. O P A -2 -1 -3 -2 -1

40. Step-5 (Example-1)
Solution-3
Draw two lines PC and PD that make angles ± 𝜃 𝑑 2 with the the bisector PB.
The intersections of PC and PD with the negative real axis give the necessary locations for the pole and zero of the lead network. O P A C D -2 -1 -3 -2 -1 B

41. 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1.9432 𝑠+4.6458
Step-5 (Example-1)
Solution-3
The lead compensator has zero at s=– and pole at s=–
Thus, Gc(s) can be given as -1 -2 B -3 O P A C D
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠 𝑠

42. 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1.9432 𝑠+4.6458
Step-5 (Example-1)
Solution-3
For this compensator value of 𝛼 is
Also
𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠 𝑠
1 𝑇 = yields 𝑇=0.514
1 𝛼𝑇 = yields 𝛼=0.418

43. Step-6 (Example-1)
Solution-3
Determine the value of Kc of the lead compensator from the magnitude condition.
𝐺 𝑠 𝐺 𝑐 𝑠 𝐻(𝑠)= 10 𝐾 𝑐 (𝑠 ) 𝑠(𝑠+1)(𝑠 )
10 𝐾 𝑐 (𝑠 ) 𝑠(𝑠+1)(𝑠 ) 𝑠=−1.5+𝑗 =1

44. Step-6 (Example-1) The Kc is calculated as
Solution-3
The Kc is calculated as
Hence, the lead compensator Gc(s) just designed is given by
𝐾 𝑐 =1.2287
𝐺 𝑐 𝑠 = 𝑠 𝑠

45. Final Design Check Solution-3 Compensated System Uncompensated System
Desired Closed Loop Pole
Uncompensated System
Desired Closed Loop Pole
Compensated System

46. Final Design Check
Solution-3
It is worthwhile to check the static velocity error constant Kv for the system just designed.
Steady state error is
𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠)
𝐾 𝑣 = lim 𝑠→0 𝑠 𝑠 𝑠 𝑠(𝑠+1) =5.139
𝑒𝑠𝑠= 1 𝐾 𝑣 = =0.194

47. Final Design Check
Solution-3

48. Final Design Check
Solution-1
Solution-3

49. Mechanical Lead Compensator
Figure shows the mechanical lead compensator.
Equations are obtained as
Taking Laplace transform of these equations assuming zero initial conditions and eliminating Y(s), we obtain

50. Mechanical Lead Compensator
By defining
We obtain

51. Exampl-2 Design a mechanical lead compensator for following system.
The damping ratio of closed loop poles is 0.5 and natural undamped frequency 2 rad/sec. It is desired to modify the closed loop poles so that natural undamped frequency becomes 4 rad/sec without changing the damping ratio.

52. Electrical Lead Compensator
𝑉𝑜(𝑠)
𝑉𝑖(𝑠)
𝑉𝑜(𝑠) 𝑉𝑖(𝑠) 𝑐 = 𝑅 2 𝑅 1 + 𝑅 2 𝑅 1 𝐶𝑠+1 𝑅 1 𝑅 2 𝑅 1 + 𝑅 2 𝐶𝑠+1
𝑎𝑇= 𝑅 1 𝑅 2 𝐶 𝑅 1 + 𝑅 2
𝑇= 𝑅 1 C
𝑎= 𝑅 2 𝑅 1 + 𝑅 2
𝐾 𝑐 =1

53. Example-3
Consider the model of space vehicle control system depicted in following figure.
Design an Electrical lead compensator such that the damping ratio and natural undamped frequency of dominant closed loop poles are 0.5 and 2 rad/sec.

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