Ch.09 Design via Root Locus
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24/03/2016
System Dynamics and Control
9.01
Design via Root Locus
09. Design via Root Locus
HCM City Univ. of Technology, Faculty of Mechanical Engineering
System Dynamics and Control
9.03
Nguyen Tan Tien
Design via Root Locus
§1.Introduction
- The root locus typically allows us to choose the proper loop
gain to meet a transient response specification
- Setting the gain at a particular value yields the transient
response dictated by the poles at that point on the root locus
Improving Transient Response
System Dynamics and Control
9.02
Design via Root Locus
Learning Outcome
After completing this chapter, the student will be able to
• Use the root locus to design cascade compensators to improve
the steady-state error
• Use the root locus to design cascade compensators to improve
the transient response
• Use the root locus to design cascade compensators to improve
both the steady-state error and the transient response
• Use the root locus to design feedback compensators to
improve the transient response
• Realize the designed compensators physically
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Design via Root Locus
§1.Introduction
Improving Steady-State Error
Compensators can be used independently to improve the
steady-state error characteristics
Configurations
• Cascade compensation
• Feedback compensation
Sample root locus, showing possible
design point via gain adjustment 𝐴 and
desired design point that cannot be met via
simple gain adjustment 𝐵
Responses from poles at 𝐴 and 𝐵
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Design via Root Locus
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Design via Root Locus
§1.Introduction
Compensators
- Ideal compensators
• pure integration for improving steady-state error
• pure differentiation for improving transient response
- Ideal compensators must be implemented with active networks,
which, in the case of electric networks, require the use of active
amplifiers and possible additional power source
- An advantage of ideal integral compensators is that steadystate error is reduced to zero. Electromechanical ideal
compensators, such as tachometers, are often used to improve
transient response, since they can be conveniently interfaced
with the plant
§2.Improving Steady-State Error via Cascade Compensation
Ideal Integral Compensation (PI)
HCM City Univ. of Technology, Faculty of Mechanical Engineering
HCM City Univ. of Technology, Faculty of Mechanical Engineering
Nguyen Tan Tien
The system operating with a desirable Add a pole at the origin to increase the
transient response generated by the system type, the angular contribution of the
closed-loop poles at 𝐴
open-loop poles at point 𝐴 is no longer
1800 , and the root locus no longer goes
through point 𝐴
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System Dynamics and Control
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
A compensator with a pole at the
origin and a zero close to the pole is
called an ideal integral compensator
The system operating with a desirable Add a zero close to the pole at the origin,
transient response generated by the the
angular
contribution
of
the
closed-loop poles at 𝐴
compensator zero and compensator pole
cancel out, point 𝐴 is still on the root locus,
and the system type has been increased
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
Solution
The dominant poles −0.694 ± 𝑗3.926 for a gain, 𝐾 = 164.6
The third pole
−11.61
1
1
𝐾𝑝 = lim 𝐾𝐺 𝑠 = 8.23 ⟹ 𝑒 ∞ =
=
= 0.108
𝑠→0
1 + 𝐾𝑝 1 + 8.23
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
The step response of
• the ideal integral compensated system 𝑐(∞) → 1.000
• the uncompensated system
𝑐(∞) → 0.892
The compensated system reaches the uncompensated
system’s final value in about the same time. The remaining time
is used to improve the steady-state error over that of the
uncompensated system
HCM City Univ. of Technology, Faculty of Mechanical Engineering
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System Dynamics and Control
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
- Ex.9.1
Effect of an Ideal Integral Compensator
Given the system, operating with 𝜁 = 0.174, show that the
addition of the ideal
integral
compensator
reduces the steady-state
error to zero for a step
input without appreciably
affecting transient response.
The compensating network
network is chosen with a pole at the origin to increase the
system type and a zero at −0.1, close to the compensator pole,
so that the angular contribution of the compensator evaluated
at the original, dominant, 2nd-order poles is approximately zero.
Thus, the original, dominant, 2nd-order closed-loop poles are
still approximately on the new root locus
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
Adding an ideal integral compensator with a zero at −0.1
The dominant poles −0.678 ± 𝑗3.837 for a gain, 𝐾 = 158.2
The third pole
−11.55
The fourth pole
−0.0902
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
The ideal integral compensator that improved steady-state error
was implemented with a proportional-plus-integral (PI) controller
𝐾1
𝐾2 𝐾1 𝑠 + 𝐾2
(9.2)
=
𝑠
𝑠
The value of the zero can be adjusted by varying 𝐾2 /𝐾1. In this
implementation, the error and the integral of the error are fed
forward to the plant, 𝐺(𝑠)
𝐺𝑐 𝑠 = 𝐾1 +
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
Lag Compensation
Type 1 uncompensated system
Type 1 compensated system
Root locus before lag compensation
Root locus after lag compensation
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
The improvement in 𝐾𝑝 from the uncompensated system to the
compensated system is the required ratio of the compensator
zero to the compensator pole, or
𝑧𝑐 𝐾𝑝𝑁 91.59
=
=
= 11.13
𝑝𝑐 𝐾𝑝𝑂
8.23
Arbitrarily selecting
𝑝𝑐 = 0.01
to get
𝑧𝑐 = 11.13𝑝𝑐 = 11.13 × 0.01 ≈ 0.111
Compensated system
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
- Ex.9.2
Lag Compensator Design
Compensate the system, whose root locus is shown, to
improve the steady-state
error by a factor of 10 if
the system is operating
with 𝜁 = 0.174
Solution
The uncompensated system error (Ex.9.1) is 0.108 with 𝐾𝑝 = 8.23
A tenfold improvement means a steady-state error of 𝑒 ∞ =
0.108/10 = 0.0108
1
𝑒 ∞ =
= 0.0108
1 + 𝐾𝑝
1−𝑒 ∞
1 − 0.0108
⟹ 𝐾𝑝 =
=
= 91.59
𝑒 ∞
0.0108
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
Sketch the root locus of the compensated system
The dominant poles −0.678 ± 𝑗3.836 for a gain, 𝐾 = 158.1
The third pole
−11.55
The fourth pole
−0.101
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
All transient and steady-state results for both the uncompensated
and the compensated systems
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
Step responses of uncompensated and lag compensated
systems
The fourth pole of the compensated system cancels its zero.
The three closed-loop poles of the compensated system ≅ the
uncompensated system ⟹ the transient response of both
systems is approximately the same, as is the system gain
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
TryIt 9.1
Use the following MATLAB
and Control System Toolbox
statements to reproduce
Fig.9.13.
Gu=zpk([ ],...
[-1 -2 -10],164.6);
Gc=zpk([-0.111],[-0.01],1);
Gce=Gu*Gc;
Tu=feedback(Gu,1);
Tc=feedback(Gce,1);
step(Tu)
hold
step(Tc)
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
Solution a.Evaluate the steady-state error for a unit ramp input
The point on root locus
−3.5 + 𝑗5.8,𝐾 = 45.84
For uncompensated system
𝐾𝑣 = lim 𝑠𝐺(𝑠)
𝑠→0
𝐾 45.84
=
= 6.55
7
7
1
𝑒ቚ
∞ =
= 0.1527
𝐾𝑣
𝑟𝑎𝑚𝑝
b.Design a lag compensator to improve the steadystate error by a factor of 20
Compensator zero should be 20 × further to the left
than the compensator pole. Arbitrarily select
𝐺𝑐 𝑠 = (𝑠 + 2)/(𝑠 + 0.01)
=
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Ideal Derivative Compensation (PD)
Compensator zero at −2
Uncompensated system
Ideal derivative, or PD controller
𝐺𝑐 𝑠 = 𝑠 + 𝑧𝑐
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§2.Improving Steady-State Error via Cascade Compensation
Skill-Assessment Ex.9.1
Problem A unity feedback system with the forward TF
𝐾
𝐺 𝑠 =
𝑠(𝑠 + 7)
is operating with a closed-loop step response that has
15% overshoot. Do the following
a.Evaluate the steady-state error for a unit ramp input.
b.Design a lag compensator to improve the steadystate error by a factor of 20
c.Evaluate the steady-state error for a unit ramp input
to your compensated system
d.Evaluate how much improvement in steady-state
error was realized
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Design via Root Locus
§2.Improving Steady-State Error via Cascade Compensation
c. Evaluate the steady-state error for a unit ramp input
The point on root locus
−3.4 + 𝑗5.63,𝐾 = 44.64
For compensated system
𝐾𝑣 = lim 𝑠𝐺(𝑠)
𝑠→0
𝐾 × 0.2
= 127. 5
7 × 0.01
1
𝑒ቚ
∞ =
= 0.0078
𝐾𝑣
𝑟𝑎𝑚𝑝
d.Evaluate how much improvement in steady-state error
𝑐𝑜𝑚𝑝𝑒𝑛𝑠𝑎𝑡𝑒𝑑
𝑒ȁ𝑟𝑎𝑚𝑝
0.1527
𝑢𝑛𝑐𝑜𝑚𝑝𝑒𝑛𝑠𝑎𝑡𝑒𝑑 = 0.0078 = 19.58
𝑒ȁ𝑟𝑎𝑚𝑝
=
⟹ 19.58 times improvement
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Uncompensated system
Compensator zero at −3
(9.12)
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
9.26
Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Summarizes the results obtained from the root locus of each of
the design cases
Compensator zero at −4
Uncompensated system
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Uncompensated system and ideal derivative compensation
solutions
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
- Ex.9.3
Ideal Derivative Compensator Design
Given the system, design an ideal derivative compensator to
yield a 16% overshoot, with a
threefold reduction in settling time
Solution
𝜁=−
𝑙𝑛 %𝑂𝑆/100
𝜋 2 + 𝑙𝑛2 %𝑂𝑆/100
=−
𝑙𝑛 16%
𝜋2 + 𝑙𝑛2 16%
= 0.504
For the uncompensated system
• dominant, 2nd-order poles −1.205 ± 𝑗2.064
• settling time, 𝑇𝑠𝑢
4
4
𝑇𝑠𝑢 =
=
𝜁𝜔𝑛 1.205
= 3.320
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
• the transient and steady-state error characteristics
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
For the compensated system
1
1
ã settling time = ì = × 3.320 = 1.107
3
3
nd
• dominant, 2 -order poles
4
4
- real part
𝜎= =
= 3.613
𝑇𝑠 1.107
- imaginary part 𝜔𝑑 = −3.613 𝑡𝑎𝑛 120.260
= 6.193
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Design the location of the compensator zero
Using the open-loop poles and the desired dominant 2nd-order
pole, −3.613 ± 𝑗6.193
𝜃𝑧𝑐 − 𝜃𝑝1 − 𝜃𝑝2 − 𝜃𝑝3 = 1800
⟹
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Summarizes the results for both the uncompensated system
and the compensated system
𝜃𝑝1 = 𝑎𝑛𝑔𝑙𝑒((−3.613 + 6.193 ∗ 𝑗) − 0) = 120.260
𝜃𝑝2 = 𝑎𝑛𝑔𝑙𝑒((−3.613 + 6.193 ∗ 𝑗) − (−4)) = 86.420
𝜃𝑝3 = 𝑎𝑛𝑔𝑙𝑒((−3.613 + 6.193 ∗ 𝑗) − (−6)) = 68.920
𝜃𝑧𝑐 = 1800 + 𝜃𝑝1 + 𝜃𝑝2 + 𝜃𝑝3 = 455.60
= 95.60
The location of the compensator zero
6.193
= 𝑡𝑎𝑛 95.60
−3.613 − (−𝜎)
⟹ 𝜎 = 3.006
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Root locus for the compensated system
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Run ch9p1 in Appendix B
Learn how to use MATLAB to
• design a PD controller
• solve Ex.9.3
Uncompensated and compensated system
step responses
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
TryIt 9.2
Use MATLAB, the Control System Toolbox, and the following steps to use SISOTOOL
to perform the design of Ex.9.3
1.
2.
3.
4.
Type SISOTOOL in the MATLAB Command Window
Select Import in the File menu of the SISO Design for SISO Design Task Window
In the Data field for G, type zpk([ ], [0, −4, −6], 1) and hit ENTER on the keyboard. Click OK
On the Edit menu choose SISO Tool Preferences … and select Zero/pole/gain: under the Options
tab. Click OK
5. Right-click on the root locus white space and choose Design Requirements/New ...
6. Choose Percent overshoot and type in 16. Click OK
7. Right-click on the root locus white space and choose Design Requirements/New ...
8. Choose Settling time and click OK
9. Drag the settling time vertical line to the intersection of the root locus and 16% overshoot radial line
10.Read the settling time at the bottom of the window
11.Drag the settling time vertical line to a settling time that is 1/3 of the value found in Step 9
12.Click on a red zero icon in the menu bar. Place the zero on the root locus real axis by clicking again
on the real axis
13.Left-click on the real-axis zero and drag it along the real axis until the root locus intersects the
settling time and percent overshoot lines
14.Drag a red square along the root locus until it is at the intersection of the root locus, settling time
line, and the percent overshoot line
15.Click the Compensator Editor tab of the Control and Estimation Tools Manager window to see the
resulting compensator, including the gain
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
- The ideal derivative compensator used to improve the transient
response is implemented with a proportional-plus-derivative
(PD) controller
𝐺𝑐 𝑠 = 𝐾1 + 𝐾2 𝑠 = 𝐾2 𝑠 +
𝐾1
𝐾2
(9.17)
• 𝐾1 /𝐾2 is chosen to equal the negative of the compensator zero
• 𝐾2 is chosen to contribute to the required loop-gain value
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Design via Root Locus
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
- While the ideal derivative compensator can improve the
transient response of the system, it has two drawbacks
• First, it requires an active circuit to perform the differentiation
• Second, differentiation is a noisy process
The level of the noise is low, but the frequency of the noise is
high compared to the signal. Differentiation of high
frequencies can lead to large unwanted signals or saturation
of amplifiers and other components
⟹ To overcome the disadvantages of ideal differentiation and still
retain the ability to improve the transient response: using the
passive network lead compensator
§3.Improving Transient Response via Cascade Compensation
Lead Compensation
- An active ideal derivative compensator can be approximated
with a passive lead compensator
- Advantages
• no additional power supplies are required, and
• noise due to differentiation is reduced
- Disadvantage
• the additional pole does not reduce the number of branches of
the root locus that cross the imaginary axis into the right halfplane, or
• the addition of the single zero of the PD controller tends to
reduce the number of branches of the root locus that cross
into the right half-plane
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
- Ex.9.4
Lead Compensator Design
Design three lead compensators for the system that will reduce
the settling time by a factor of 2 while
maintaining 30% overshoot. Compare
the system characteristics between
the three designs
Solution
Determine the characteristics of the
uncompensated system operating at
30% overshoot
𝑂𝑆% = 30% ⟹ 𝜁 = 0.358
The uncompensated settling time
4
4
𝑇𝑠𝑢 =
=
= 3.972
𝜁𝜔𝑛 1.007
§3.Improving Transient Response via Cascade Compensation
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§3.Improving Transient Response via Cascade Compensation
Find the design point
1
1
𝑇𝑠 = × 𝑇𝑠𝑢 = × 3.972 = 1.986
The settling time
2
2
The desired pole location
• real part
−𝜁𝜔𝑛 = −4/𝑇𝑠 = −2.014
• imaginary part
𝜔𝑑 = −2.014𝑡𝑎𝑛 110.980 = 5.252
Designing the lead compensator
Arbitrarily assume a compensator zero at
− 5 on the real axis as a possible solution
𝜃𝑧𝑐 − 𝜃𝑝𝑐 − 𝜃1 − 𝜃2 − 𝜃3 = 1800
⟹
𝜃1 = 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − 0) = 110.980
𝜃2 = 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − (−4)) = 69.290
𝜃3 = 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − (−6)) = 52.800
𝜃𝑧𝑐 = 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − (−5)) = 60.380
𝜃𝑝𝑐 = 𝜃𝑧𝑐 − 1800 − 𝜃1 − 𝜃2 − 𝜃3
= −352.690 = 7.310
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The summarized characteristics of the uncompensated system
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Find the design point
The location of the compensator pole
5.252
= 𝑡𝑎𝑛7.310 ⟹ 𝑝𝑐 = 42.96
−2.014 − (−𝑝𝑐 )
Justify the estimates of percent overshoot and settling time
The third and fourth poles are at −43.8 and −5.134. Since
− 43.8 is more than 20 times the real part of the dominant pole,
the effect of the third closed-loop pole is negligible. Since the
closed-loop pole at −5.134 is close to the zero at −5, we have
pole-zero cancellation, and the 2nd-order approximation is valid
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Comparison of lead compensation designs
§3.Improving Transient Response via Cascade Compensation
Uncompensated system and lead compensation responses
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Design via Root Locus
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Run ch9p2 in Appendix B
Learn how to use MATLAB to
• design a lead compensator
ã solve Ex.9.4
Đ3.Improving Transient Response via Cascade Compensation
Skill-Assessment Ex.9.2
Problem A unity feedback system with the forward TF
𝐾
𝐺 𝑠 =
𝑠(𝑠 + 7)
is operating with a closed-loop step response that has
15% overshoot. Do the following
a.Evaluate the settling time
b.Design a lead compensator to decrease the settling
time by three times. Choose the compensator’s zero
to be at −10
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Design via Root Locus
§3.Improving Transient Response via Cascade Compensation
Solution a.Evaluate the settling time
𝜁=
=
− 𝑙𝑛 %𝑂𝑆/100
𝜋2 +𝑙𝑛2 %𝑂𝑆/100
o
− 𝑙𝑛 15%
𝜋 2 + 𝑙𝑛2 15%
= 0.517
The desired point
𝑝 = −3.5 ± 𝑗5.8
𝐾 = 45.84
Setting time for uncompensated system
4
4
𝑇𝑠𝑢 =
=
= 1.143
ȁ𝑅𝑒ȁ 3.5
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§3.Improving Transient Response via Cascade Compensation
b.Design a lead compensator
For the compensated system
ã settling time
1
1
= ì = ì 1.143 = 0.381
3
3
• dominant, 2nd-order poles
4
4
𝜎= =
= 10.50
𝑇𝑠 0.381
𝜃0 = 𝑎𝑛𝑔𝑙𝑒((−3.5 + 5.8 ∗ 𝑗) − 0) = 121.110
𝜔𝑑 = −10.499 𝑡𝑎𝑛 121.110 = 17.40
other way, the real part of the design point must be
three times larger than the uncompensated pole’s
real part 3 −3.5 + 𝑗5.8 = −10.50 + 𝑗17.40
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§3.Improving Transient Response via Cascade Compensation
Designing the lead compensator
Compensator zero at −10 on the real axis
𝜃𝑧𝑐 − 𝜃𝑝𝑐 − 𝜃1 − 𝜃2 = 1800
⟹
𝜃1 = 𝑎𝑛𝑔𝑙𝑒 −10.50 + 17.40 ∗ 𝑗 − 0
= 121.110
𝜃2 = 𝑎𝑛𝑔𝑙𝑒 −10.50 + 17.40 ∗ 𝑗 − −7
= 101.370
𝜃𝑧𝑐 = 𝑎𝑛𝑔𝑙𝑒 −10.50 + 17.40 ∗ 𝑗 − −10
= 91.650
𝜃𝑝𝑐 = 𝜃𝑧𝑐 − 1800 − 𝜃1 − 𝜃2
= 91.650 − 1800 − 121.110 − 101.370
= −310.840
= 49.20
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§4.Improving Steady-State Error and Transient Response
PID Controller Design
𝐾2
𝐺𝑐 𝑠 = 𝐾1 +
+ 𝐾2 𝑠 (9.21)
𝑠
𝐾
𝐾
2
𝐾3 𝑠 + 1 𝑠 + 2
𝐾3
𝐾3
=
𝑠
A PID controller has two zeros plus a pole at the origin
• One zero and the pole at the origin can be designed as the
ideal integral compensator
• The other zero can be designed as the ideal derivative
compensator
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§4.Improving Steady-State Error and Transient Response
- Ex.9.5
PID Controller Design
Given the system, design a PID controller so that the system
can operate with a peak time that is 2/3 that of the
uncompensated system at 20% overshoot and with zero
steady-state error for a step input
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§3.Improving Transient Response via Cascade Compensation
dominant, 2nd-order poles
𝜎 = 10.50
𝜔𝑑 = 17.40
𝜔𝑝𝑐 = 49.20
The location of the compensator pole
17.40
= 𝑡𝑎𝑛 49.20
−10.50 − (−𝑝𝑐)
⟹ 𝑝𝑐 = 25.52
with this pole, the gain 𝐾 = 476.3. A higher-order
closed-loop pole is found to be at −11.54. This pole
may not be close enough to the closed-loop zero at
− 10. Thus, we should simulate the system to be
sure the design requirements have been met
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§4.Improving Steady-State Error and Transient Response
The design technique consists of the following steps
1.Evaluate the performance of the uncompensated system to
determine how much improvement in transient response is required
2.Design the PD controller to meet the transient response
specifications. The design includes the zero location and the
loop gain
3.Simulate the system to be sure all requirements have been met
4.Redesign if the simulation shows that requirements have not
been met
5.Design the PI controller to yield the required steady-state error
6.Determine the gains, 𝐾1 , 𝐾2 , and 𝐾3
7.Simulate the system to be sure all requirements have been met
8.Redesign if simulation shows that requirements have not been met
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§4.Improving Steady-State Error and Transient Response
Solution
Step 1
Evaluate the performance of the
uncompensated system operating
at 20% overshoot
• the dominant poles
−5.415 ± 𝑗10.57, 𝐾 = 121.5
• The third pole
−8.169, 𝐾 = 121.5
The complete performance of
the uncompensated system is
shown in the first column of
Table 9.5, where we
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§4.Improving Steady-State Error and Transient Response
The complete performance of
the uncompensated system
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§4.Improving Steady-State Error and Transient Response
Step 2
Design the PD controller to meet
the transient response specifications
The imaginary part of the
compensated dominant pole
𝜋
𝜔𝑑 =
𝑇𝑝
𝜋
=
2
× 0.297
3
= 15.87
𝜔𝑑
𝑡𝑎𝑛117.130
= −8.13
𝜎=
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§4.Improving Steady-State Error and Transient Response
The compensating zero’s location
𝜃𝑧𝑐 + 𝜃−8 − 𝜃−3 − 𝜃−6 − 𝜃−10 = 1800
⟹
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§4.Improving Steady-State Error and Transient Response
The complete root locus for the PD-compensated system
𝜃−8 = 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−8))
= 90.470
𝜃−3 = 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−3))
= 107.910
𝜃−6 = 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−6))
= 97.640
𝜃−10 = 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−10))
= 83.280
𝜃𝑧𝑐 = 1800 − 𝜃−8 + 𝜃−3 + 𝜃−6 + 𝜃−10
0
= 18.37
15.87
= 𝑡𝑎𝑛18.370 → 𝑧𝑐 = 55.92
−8.13 − (−𝑧𝑐)
The PD controller
𝐺𝑃𝐷 𝑠 = 𝑠 + 55.92
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§4.Improving Steady-State Error and Transient Response
Complete specifications for ideal derivative compensation
§4.Improving Steady-State Error and Transient Response
Step 3&4
Simulate the system to be sure all requirements have been met
Step responses for uncompensated, PD-compensated, and
PID-compensated systems
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§4.Improving Steady-State Error and Transient Response
Step 5
Design the PI controller to yield the required steady-state error
𝑠 + 0.5
Choosing the ideal integral compensator to be 𝐺𝑃𝐼 𝑠 =
𝑠
§4.Improving Steady-State Error and Transient Response
Complete specifications for PID compensated system
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§4.Improving Steady-State Error and Transient Response
Step 6
Determine the gains, 𝐾1 , 𝐾2 , and 𝐾3
𝐾(𝑠 + 55.92)(𝑠 + 0.5)
𝐺𝑃𝐼𝐷 𝑠 =
𝑠
4.6(𝑠 + 55.92)(𝑠 + 0.5)
=
𝑠
4.6(𝑠 2 + 56.42𝑠 + 27.96)
=
𝑠
Matching Eqs.(9.21) to get 𝐾1 = 259.2, 𝐾2 = 128.6, 𝐾3 = 4.6
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§4.Improving Steady-State Error and Transient Response
Step 7&8
- PD controller improved the transient response by decreasing the time required to
reach the first peak as well as yielding some improvement in the steady-state error
- PID controller further improved the steady-state error without appreciably changing
the transient response designed with the PD controller
𝐾
𝐺𝑐 𝑠 = 𝐾1 +
𝐾2
𝑠
+ 𝐾2 𝑠 =
𝐾
𝐾3 𝑠2 +𝐾1 𝑠+𝐾2
3
3
(9.21)
𝑠
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- PID controller exhibits a slower response, reaching the final value of unity at
approximately 3𝑠. If this is undesirable, the speed of the system must be increased
by redesigning the ideal derivative compensator or moving the PI controller zero
farther from the origin
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§4.Improving Steady-State Error and Transient Response
Lag-Lead Compensator Design
Used for improving both transient response and the steady-state
error
• First design the lead compensator to improve the transient
response
• Next evaluate the improvement in steady-state error still
required
• Finally, design the lag compensator to meet the steady-state
error requirement
§4.Improving Steady-State Error and Transient Response
The design technique consists of the following steps
1.Evaluate the performance of the uncompensated system to
determine how much improvement in transient response is required
2.Design the lead compensator to meet the transient response
specifications. The design includes the zero location, pole
location, and the loop gain
3.Simulate the system to be sure all requirements have been met
4.Redesign if the simulation shows that requirements have not been met
5.Evaluate the steady-state error performance for the leadcompensated system to determine how much more
improvement in steady-state error is required
6.Design the lag compensator to yield the required steady-state error
7.Simulate the system to be sure all requirements have been met
8.Redesign if the simulation shows that requirements have not been met
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§4.Improving Steady-State Error and Transient Response
- Ex.9.6
Lag-Lead Compensator Design
Design a lag-lead compensator for the system so that the system
will operate with 20% overshoot and
a twofold reduction in settling time.
Further, the compensated system will
exhibit a tenfold improvement in steady-state error for a ramp input
Solution
Step 1
Evaluate the performance of
the uncompensated system
𝑂𝑆% = 20% ~ 𝜁 = 0.456
§4.Improving Steady-State Error and Transient Response
The performance of the uncompensated system
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§4.Improving Steady-State Error and Transient Response
Step 2
Design the lead compensator
1
𝑇𝑠 = 𝑇𝑠𝑢 → −𝜁𝜔𝑛 = −2 × 1.794 = −3.588
2
𝜃𝑧−6 − 𝜃𝑝𝑐 − 𝜃𝑝0 − 𝜃𝑝−6 − 𝜃𝑝−10 = 1800
⟹
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§4.Improving Steady-State Error and Transient Response
Step 3&4
Check the design with a simulation
𝜃𝑝0 = 𝑎𝑛𝑔𝑙𝑒( −3.588 + 7.003 ∗ 𝑗 − 0) = 117.130
𝜃𝑝−6 = 𝜃𝑧−6 = 𝑎𝑛𝑔𝑙𝑒( −3.588 + 7.003 ∗ 𝑗 − (−6)) = 71.000
𝜃𝑝−10 = 𝑎𝑛𝑔𝑙𝑒 −3.588 + 7.003 ∗ 𝑗 − −10 = 47.520
𝜃𝑝𝑐 = 𝜃𝑧−6 − 1800 − 𝜃𝑝0 − 𝜃𝑝−6 − 𝜃𝑝−10 = 15.350
7.003
= 𝑡𝑎𝑛 15.350
−3.588 − (−𝑝𝑐 )
⟹ 𝑝𝑐 = 29.1
The result for the lead-compensated system is shown in the
figure and is satisfactory
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§4.Improving Steady-State Error and Transient Response
Step 5
Design the lag compensator to improve the steady-state error
The uncompensated system’s open-loop TF
192.1
1
𝐺 𝑠 =
𝐾𝑣 = lim 𝑠𝐺(𝑠) =
= 3.202
𝑠→0
𝑠(𝑠 + 6)(𝑠 + 10)
𝑒 ∞
The open-loop TF of the lead-compensated system
1977
1
𝐺𝐿𝐶 𝑠 =
𝐾 = lim 𝑠𝐺(𝑠) =
= 6.794
𝑠(𝑠 + 10)(𝑠 + 29.1) 𝑣 𝑠→0
𝑒 ∞
The addition of lead compensation has improved the steadystate error by a factor of6.794/3.202 = 2.122
The requirements of the problem specified a tenfold
improvement, the lag compensator must be designed to
improve the steady-state error by a factor of 10/2.122 = 4.713
over the lead-compensated system
§4.Improving Steady-State Error and Transient Response
Step 6
Arbitrarily choose the lag compensator pole at 0.01, which then
places the lag compensator zero at 0.04713, yielding
𝑠 + 0.04731
𝐺𝑙𝑎𝑔 𝑠 =
𝑠 + 0.01
The lag-lead compensated system’s open-loop TF
𝐾(𝑠 + 0.04731)
𝐺𝐿𝐿𝐶 𝑠 =
𝑠(𝑠 + 10)(𝑠 + 29.1)(𝑠 + 0.01)
The lag-lead-compensated root locus
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§4.Improving Steady-State Error and Transient Response
The characteristics of uncompensated, lead-compensated, and
lag-lead-compensated systems
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§4.Improving Steady-State Error and Transient Response
Step 7
(a)
(b)
- Improvement in the transient response
• the peak time occurring sooner in the lag-lead-compensated system
- Improvement in the steady-state error for a ramp input for
• the lead-compensated system is shown in (a)
• the final improvement due to the addition of the lag is shown in (b)
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§4.Improving Steady-State Error and Transient Response
Notch Filter
Root locus before cascading notch filter
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§4.Improving Steady-State Error and Transient Response
One way of eliminating the high-frequency oscillations is to
cascade a notch filter with the plant (Kuo, 1995). The notch filter
has zeros close to the low-damping-ratio poles of the plant as well
as two real poles. The root locus branch from the high-frequency
poles now goes a short distance from the high frequency pole to
the notch filter’s zero. The high-frequency response will now be
negligible because of the pole-zero cancellation
Typical closed-loop step response before
cascading notch filter
High-frequency vibration modes can be modeled as part of the
plant’s TF by pairs of complex poles near the imaginary axis. In a
closed-loop configuration, these poles can move closer to the
imaginary axis or even cross into the RHP. Instability or highfrequency oscillations superimposed over the desired response can result
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§4.Improving Steady-State Error and Transient Response
Closed-loop step response after cascading notch filter
§4.Improving Steady-State Error and Transient Response
Skill-Assessment Ex.9.3
Problem A unity feedback system with the forward TF
𝐾
𝐺 𝑠 =
𝑠(𝑠 + 7)
is operating with a closed-loop step response that has
20% overshoot. Do the following
a.Evaluate the settling time
b.Evaluate the steady-state error for a unit ramp input
c.Design a lag-lead compensator to decrease the
settling time by 2 times and decrease the steadystate error for a unit ramp input by 10 times. Place
the lead zero at −3
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§4.Improving Steady-State Error and Transient Response
Solution a.Evaluate the settling time
𝑂𝑆% = 20% ~ 𝜁 = 0.456
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§4.Improving Steady-State Error and Transient Response
b.Evaluate the steady-state error for a unit ramp input
For the uncompensated system
𝐾 58.9
𝐾𝑣𝑢 = lim 𝑠𝐺(𝑠) = =
= 8.41
𝑠→0
7
7
1
1
𝑒𝑟𝑎𝑚𝑝 ∞ =
=
= 0.1189
𝐾𝑣 8.41
c.Design a lag-lead compensator
1
𝑇𝑠 = 𝑇𝑠𝑢 → −𝜁𝜔𝑛 = −2 × 6.83 = −13.66
2
The desired poles −7 ± 13.66
For the uncompensated system
4
4
𝑇𝑠𝑢 =
=
= 1.143𝑠
ȁ𝑅𝑒 ȁ 3.5
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§4.Improving Steady-State Error and Transient Response
𝜃𝑧−3 − 𝜃𝑝𝑐 − 𝜃𝑝0 − 𝜃𝑝−7 = 1800
⟹
𝜃𝑧−3 = 𝑎𝑛𝑔𝑙𝑒( −7 + 13.66 ∗ 𝑗 − (−3)) = 106.320
𝜃𝑝0 = 𝑎𝑛𝑔𝑙𝑒( −7 + 13.66 ∗ 𝑗 − 0) = 117.130
𝜃𝑝−6 = 𝑎𝑛𝑔𝑙𝑒( −7 + 13.66 ∗ 𝑗 − (−7)) = 90.000
𝜃𝑝𝑐 = 𝜃𝑧−3 − 1800 − 𝜃𝑝0 − 𝜃𝑝−7 = 79.190
13.66
= 𝑡𝑎 𝑛 79.190 ⟹ 𝑝𝑐 = 9.61, 𝐾 = 204.9
−7 − (−𝑝𝑐 )
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§4.Improving Steady-State Error and Transient Response
For the uncompensated system 𝐾𝑣𝑢 = 8.41
For the lead-compensated system
𝐾×3
205.9 × 3
𝐾𝑣𝑙𝑒𝑎𝑑 = lim 𝑠𝐺(𝑠)𝐺𝑙𝑒𝑎𝑑(𝑠) =
=
= 9.138
𝑠→0
7 × 9.61 7 × 9.61
The desired system
𝑙𝑎𝑔
𝐾𝑣 = 10𝐾𝑣𝑢 = 10 × 8.41 = 84.1 = 𝐾𝑣𝑙𝑒𝑎𝑑 𝐾𝑣
⟹ for the lag-compensated system
𝐾𝑣
84.1
𝑙𝑎𝑔
𝐾𝑣 = 𝑙𝑒𝑎𝑑 =
= 9.2
9.138
𝐾𝑣
The lag compensator zero should be 9.2 × further to
the left than the compensator pole, arbitrarily select
𝑠 + 0.092
𝐺𝑐 𝑠 =
𝑠 + 0.01
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§4.Improving Steady-State Error and Transient Response
Using all the compensator poles, we find the gain at
the point to be 𝐾 = 205.4. Summarizing the forward
path with plant, compensator, and gain yields
205.4(𝑠 + 3)(𝑠 + 0.092)
𝐺𝑒 𝑠 =
𝑠(𝑠 + 7)(9.61)(𝑠 + 0.01)
Higher-other poles are found at −0.928 and −2.6. It
would be advisable to simulate the system to see if
there is indeed pole-zero cancellation
§4.Improving Steady-State Error and Transient Response
Types of cascade compensators
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§4.Improving Steady-State Error and Transient Response
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§5.Feedback Compensation
Generic control system with feedback compensation
The design procedures for feedback compensation can be more
complicated than for cascade compensation
⟹ Refer the textbook
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§6.Physical Realization of Compensation
Active-Circuit Realization
𝑉𝑜 (𝑠)
𝑍2 (𝑠)
=−
𝑉𝑖 (𝑠)
𝑍1 (𝑠)
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§6.Physical Realization of Compensation
Active-Circuit Realization
𝑉𝑜 (𝑠)
𝑍2 (𝑠)
=−
𝑉𝑖 (𝑠)
𝑍1 (𝑠)
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§6.Physical Realization of Compensation
Other compensators can be realized by cascading compensators.
Ex., a lag-lead compensator can be formed by cascading the lag
compensator with the lead compensator
Lag-lead compensator implemented with operational amplifiers
HCM City Univ. of Technology, Faculty of Mechanical Engineering
HCM City Univ. of Technology, Faculty of Mechanical Engineering
Nguyen Tan Tien
HCM City Univ. of Technology, Faculty of Mechanical Engineering
System Dynamics and Control
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Design via Root Locus
§6.Physical Realization of Compensation
- Ex.9.9
Implementing a PID Controller
(𝑠 + 55.92)(𝑠 + 0.5)
Implement the PID controller 𝐺𝑐 𝑠 =
𝑠
Solution
The TF of the PID controller
(𝑠 + 55.92)(𝑠 + 0.5)
27.96
𝐺𝑐 𝑠 =
= 𝑠 + 56.42 +
𝑠
𝑠
From the table
𝑅2 𝐶1
+ = 56.42,
𝑅1 𝐶2
𝑅2 𝐶1 = 1,
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1
= 27.96
𝑅1 𝐶2
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Design via Root Locus
System Dynamics and Control
9.92
§6.Physical Realization of Compensation
𝑅2 𝐶1
1
+ = 56.42,
𝑅2 𝐶1 = 1,
= 27.96
𝑅1 𝐶2
𝑅1 𝐶2
Arbitrarily select a practical value 𝐶2 = 0.1𝜇𝐹
⟹ 𝑅1 = 357.65𝑘Ω,
𝑅1 = 178.89𝑘Ω,
𝐶1 = 5.59𝜇𝐹
The complete circuit with the circuit element values have been
rounded off
§6.Physical Realization of Compensation
𝐺𝑐 𝑠 =
Passive-Circuit Realization
HCM City Univ. of Technology, Faculty of Mechanical Engineering
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System Dynamics and Control
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§6.Physical Realization of Compensation
A lag-lead compensator can be realized with passive networks by
cascading the lead and lag networks. A possible realization using
the passive networks uses an operational amplifier to provide
isolation
Lag-lead compensator implemented with cascaded lag and lead networks with isolation
Note: The two networks must be isolated to ensure that one
network does not load the other. If the networks load each
other, the TF will not be the product of the individual TFs
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Design via Root Locus
1
𝑠+𝑇
1
𝑠+
1
𝛼𝑇1
9.94
2
𝑠+
𝛼
𝑇1
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Design via Root Locus
§6.Physical Realization of Compensation
- Ex.9.10
Realizing a Lead Compensator
𝑠+4
Realize the lead compensator 𝐺𝑐 𝑠 =
𝑠 + 20.09
Solution
Comparing the TF of a lead network to obtain the relationships
1
1
1
= 4,
+
= 20.09
𝑅1 𝐶
𝑅1 𝐶 𝑅2 𝐶
Letting 𝐶 = 1𝜇𝐹, then 𝑅1 = 250𝑘Ω and 𝑅2 = 62.2𝑘Ω
HCM City Univ. of Technology, Faculty of Mechanical Engineering
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§6.Physical Realization of Compensation
Skill-Assessment Ex.9.5
Problem Implement the compensators shown in a. and b. below.
a.
(𝑠 + 0.1)(𝑠 + 5)
𝐺𝑐 𝑠 =
𝑠
b.
(𝑠 + 0.1)(𝑠 + 2)
𝐺𝑐 𝑠 =
(𝑠 + 0.01)(𝑠 + 20)
Choose a passive realization if possible
§6.Physical Realization of Compensation
Solution a.An active PID controller must be used
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𝑠+𝑇
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Matching the given TF with the TF of the PID controller
(𝑠 + 0.1)(𝑠 + 5) 𝑠2 + 5.1𝑠 + 0.5
0.5
𝐺𝑐 𝑠 =
=
= 𝑠 + 5.1 +
𝑠
𝑠
𝑠
𝑅2 𝐶2
1/(𝑅1𝐶2)
𝐺𝑐 𝑠 = −
+
+ 𝑅2𝐶1𝑠 +
𝑅1 𝐶1
𝑠
1
𝑅2 𝐶2
⟹
= 0.5,
𝑅2 𝐶1 = 1,
+ = 5.1
𝑅1 𝐶2
𝑅1 𝐶1
−5
Choose 𝐶1 = 10 𝜇𝐹 → 𝑅1 = 20𝑘Ω, 𝑅2 = 105Ω,𝐶2 = 100𝜇𝐹
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§6.Physical Realization of Compensation
b.The lag-lead compensator can be implemented with
the following passive network, since the ratio of
the lead pole-to-zero is the inverse of the ratio of
the lag pole-to zero
Matching the given TF with the TF of the passive laglead compensator yields
(𝑠 + 0.1)(𝑠 + 2)
(𝑠 + 0.1)(𝑠 + 2)
𝐺𝑐 𝑠 =
=
(𝑠 + 0.01)(𝑠 + 20) 𝑠2 + 20.01𝑠 + 0.2
1
1
𝑠+
𝑠+
𝑅1𝐶1
𝑅2𝐶2
𝐺𝑐 𝑠 =
1
1
1
1
𝑠2 +
+
+
𝑠+
𝑅1𝐶1 𝑅2𝐶2 𝑅2𝐶1
𝑅1𝑅2𝐶1𝐶2
1
1
1
1
1
⟹
= 0.1,
= 0.1,
+
+
= 20.01
𝑅1 𝐶1
𝑅2 𝐶2
𝑅1𝐶1 𝑅2𝐶2 𝑅2𝐶1
Choose 𝐶1 = 100𝜇𝐹 → 𝑅1 = 100𝑘Ω,𝑅2 = 558𝑘Ω,𝐶2 = 900𝜇𝐹
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