1. Modern Control Systems (MCS)
Lecture-19-20
Lag-Lead Compensation
Dr. Imtiaz Hussain
Assistant Professor
URL :

2. Lecture Outline Introduction to lag-lead compensation
Design Procedure of Lag-lead Compensator
Case-1
Case-2
Electronic Lag-lead Compensator
Mechanical Lag-lead Compensator
Electrical Lag-lead Compensator

3. Introduction
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.

4. 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)

5. 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)

6. 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)

7. 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.

8. 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

9. 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

10. 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

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

12. 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

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

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

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End of Lecture-19-20