Transcript Modern Control Systems (MCS)

Modern Control Systems (MCS)
Lecture-19-20
Lag-Lead Compensation
Dr. Imtiaz Hussain
Assistant Professor
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture Outline
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
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑇1
𝛾
𝑠+
𝑇1
𝑠+
1
𝑇2
1
𝑠+
𝛽𝑇2
𝑠+
, (γ > 1 𝑎𝑛𝑑 β > 1)
• Where Kc belongs to lead portion of the compensator.
Design Procedure
• In designing lag–lead compensators, we consider two
cases where
• Case-1: γ ≠ 𝛽
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑠+
𝑇1
𝛾
𝑠+
𝑇1
1
𝑠+
𝑇2
1
𝑠+
𝛽𝑇2
, (γ > 1 𝑎𝑛𝑑 β > 1)
• Case-2: γ = 𝛽
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑠+
𝑇1
𝛽
𝑠+
𝑇1
1
𝑠+
𝑇2
1
𝑠+
𝛽𝑇2
, (β > 1)
Design Procedure (Case-1)
• Case-1: γ ≠ 𝛽
𝐺𝑐 𝑠 = 𝐾𝑐
1
𝑇1
𝛾
𝑠+
𝑇1
𝑠+
• Step-1: Design
specifications.
1
𝑇2
1
𝑠+
𝛽𝑇2
𝑠+
Lead
, (γ > 1 𝑎𝑛𝑑 β > 1)
part
using
given
• 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
𝑠 = −2.50 ± 𝑗4.33
• Since
4

 235
s ( s  0.5) s  2.50 j 4.33
• 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.
Example-1 (Case-1)
• The phase-lead portion of the lag–lead compensator becomes
𝐾𝑐
1
𝑠+
𝑇1
𝛾
𝑠+
𝑇1
𝑠+0.5
= 𝐾𝑐
𝑠+5.02
• Thus 𝑇1 = 2 and 𝛾 = 10.04.
• Next we determine the value of Kc from the magnitude
condition:
(𝑠 + 0.5)
4
𝐾𝑐
𝑠 + 5.02 𝑠(𝑠 + 0.5)
𝑠(𝑠 + 5.02)
𝐾𝑐 =
4
=1
𝑠=−2.5+𝑗4.33
= 5.26
𝑠=−2.5+𝑗4.33
Example-1 (Case-1)
• The phase-lag portion of the compensator can be designed as
follows.
• First the value of 𝛽 is determined to satisfy the
requirement on the static velocity error constant
𝐾𝑣 = lim 𝑠𝐺𝑐 𝑠 𝐺(𝑠)
𝑠→0
80 = lim 𝑠
𝑠→0
1
25.04 𝑠 +
𝑇2
𝑠 𝑠 + 5.02
1
𝑠+
𝛽𝑇2
80 = 4.988𝛽
𝛽 = 16.04
Example-1 (Case-1)
• Finally, we choose the value of 𝑇2 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
𝐺𝑐 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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END OF LECTURE-19-20