Modern Control Systems (MCS)

Dr. Imtiaz HussainAssistant Professor

email: imtiaz.hussain@faculty.muet.edu.pkURL :http://imtiazhussainkalwar.weebly.com/

Lecture-19-20Lag-Lead Compensation

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

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.

Lag-Lead Compensation• Lag-Lead compensators are represented by following transfer

function

• Where Kc belongs to lead portion of the compensator.

, ()

Design Procedure• In designing lag–lead compensators, we consider two

cases where

• Case-1:

• Case-2:

, ()

, ()

Design Procedure (Case-1)• Case-1:

• Step-1: Design Lead part using given specifications.

• Step-1: Design lag part according to given values of static error constant.

, ()

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.

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± 𝑗 4.33

235)5.0(

4

33.450.2 jsss

Example-1 (Case-1)• The phase-lead portion of the lag–lead compensator becomes

• Thus and .

• Next we determine the value of Kc from the magnitude condition:

=

|𝐾 𝑐

(𝑠+0.5)𝑠+5.02

4𝑠(𝑠+0.5)|𝑠=−2.5+ 𝑗 4 . 33

=1

𝐾 𝑐=|𝑠(𝑠+5.02)4 |

𝑠=− 2.5+ 𝑗4 .33

=5 .26

Example-1 (Case-1)• The phase-lag portion of the compensator can be designed as

follows.• First the value of is determined to satisfy the requirement

on the static velocity error constant

�̂� 𝑣=lim𝑠→ 0

𝑠𝐺𝑐 (𝑠 )𝐺(𝑠)

80=lim𝑠→ 0

𝑠 [ 25.04 (𝑠+ 1𝑇 2

)𝑠 (𝑠+5.02 )(𝑠+ 1

𝛽𝑇 2) ]

80=4.988 𝛽

𝛽=16.04

Example-1 (Case-1)• Finally, we choose the value of such that the following two

conditions are satisfied:

Example-1 (Case-1)• Now the transfer function of the designed lag–lead

compensator is given by

Example-1 (Case-2)

Home Work

Home Work

• Electronic Lag-Lead Compensator• Electrical Lag-Lead Compensator• Mechanical Lag-Lead Compensator

END OF LECTURE-19-20

To download this lecture visithttp://imtiazhussainkalwar.weebly.com/