Modern Control Systems (MCS) Lecture-19-20 Lag-Lead Compensation Dr. Imtiaz Hussain Assistant Professor email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/

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. 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 1 𝑠+ 𝛾 𝑇 1 𝑠+ 1 𝑇 2 𝑠+ 1 𝛽𝑇 2 , (γ>1 𝑎𝑛𝑑 β>1)

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 1 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 1 𝑠+ 1 𝑇 2 𝑠+ 1 𝛽𝑇 2 , (β>1)

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 1 𝑠+ 1 𝑇 2 𝑠+ 1 𝛽𝑇 2 , (γ>1 𝑎𝑛𝑑 β>1)

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±𝑗4.33

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+5.02 4 𝑠(𝑠+0.5) 𝑠=−2.5+𝑗4.33 =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 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 25.04 𝑠+ 1 𝑇 2 𝑠 𝑠+5.02 𝑠+ 1 𝛽𝑇 2 80=4.988𝛽 𝛽=16.04

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

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+5.02 𝑠+0.2 𝑠+0.0127

Example-1 (Case-2) Home Work

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

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