Ch.09 Design via Root Locus

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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.04 Nguyen Tan Tien 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.05 Nguyen Tan Tien Design via Root Locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.06 Nguyen Tan Tien 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 Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.07 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.09 Nguyen Tan Tien 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 𝐾𝑝 = lim 𝐾𝐺 𝑠 = 8.23 ⟹ 𝑒 ∞ = = = 0.108 𝑠→0 + 𝐾𝑝 + 8.23 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.11 Nguyen Tan Tien 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 Nguyen Tan Tien System Dynamics and Control 9.08 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.10 Nguyen Tan Tien 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.12 Nguyen Tan Tien 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 + HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.13 Design via Root Locus §2.Improving Steady-State Error via Cascade Compensation Lag Compensation Type uncompensated system Type compensated system Root locus before lag compensation Root locus after lag compensation HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.15 Nguyen Tan Tien 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 System Dynamics and Control 9.14 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 𝑒 ∞ = = 0.0108 + 𝐾𝑝 1−𝑒 ∞ − 0.0108 ⟹ 𝐾𝑝 = = = 91.59 𝑒 ∞ 0.0108 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.16 Nguyen Tan Tien 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.17 Nguyen Tan Tien 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.18 Nguyen Tan Tien 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 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.19 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) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 9.21 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 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) = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 9.23 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.20 Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.22 Nguyen Tan Tien 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 × 0.01 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.24 Nguyen Tan Tien Design via Root Locus §3.Improving Transient Response via Cascade Compensation Uncompensated system Compensator zero at −3 (9.12) Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.25 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control 9.27 Nguyen Tan Tien Design via Root Locus §3.Improving Transient Response via Cascade Compensation Uncompensated system and ideal derivative compensation solutions HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 9.28 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 + 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 𝑇𝑠𝑢 = = 𝜁𝜔𝑛 1.205 = 3.320 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.29 Nguyen Tan Tien Design via Root Locus Đ3.Improving Transient Response via Cascade Compensation ã the transient and steady-state error characteristics 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 9.30 Nguyen Tan Tien Design via Root Locus §3.Improving Transient Response via Cascade Compensation For the compensated system 1 • settling time 𝑇𝑠 = × 𝑇𝑠𝑢 = × 3.320 = 1.107 3 nd • dominant, -order poles 4 - real part 𝜎= = = 3.613 𝑇𝑠 1.107 - imaginary part 𝜔𝑑 = −3.613 𝑡𝑎𝑛 120.260 = 6.193 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.31 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 ⟹ System Dynamics and Control 9.32 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.33 Nguyen Tan Tien Design via Root Locus §3.Improving Transient Response via Cascade Compensation Root locus for the compensated system System Dynamics and Control Nguyen Tan Tien 9.34 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control HCM City Univ of Technology, Faculty of Mechanical Engineering 9.35 Nguyen Tan Tien 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 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 Right-click on the root locus white space and choose Design Requirements/New Choose Percent overshoot and type in 16 Click OK Right-click on the root locus white space and choose Design Requirements/New Choose Settling time and click OK 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 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 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 Nguyen Tan Tien 9.36 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 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.37 Design via Root Locus System Dynamics and Control 9.38 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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.39 Nguyen Tan Tien Design via Root Locus System Dynamics and Control 9.40 Nguyen Tan Tien 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 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 𝑇𝑠𝑢 = = = 3.972 𝜁𝜔𝑛 1.007 §3.Improving Transient Response via Cascade Compensation HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.41 Nguyen Tan Tien Design via Root Locus §3.Improving Transient Response via Cascade Compensation Find the design point 1 𝑇𝑠 = × 𝑇𝑠𝑢 = × 3.972 = 1.986 The settling time 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 − 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 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien The summarized characteristics of the uncompensated system System Dynamics and Control 9.42 Nguyen Tan Tien 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 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.43 Design via Root Locus System Dynamics and Control 9.44 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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.45 Nguyen Tan Tien Design via Root Locus System Dynamics and Control 9.46 Nguyen Tan Tien 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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.47 Nguyen Tan Tien 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 + 𝑙𝑛2 15% = 0.517 The desired point 𝑝 = −3.5 ± 𝑗5.8 𝐾 = 45.84 Setting time for uncompensated system 4 𝑇𝑠𝑢 = = = 1.143 ȁ𝑅𝑒ȁ 3.5 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.48 Nguyen Tan Tien Design via Root Locus §3.Improving Transient Response via Cascade Compensation b.Design a lead compensator For the compensated system • settling time 1 𝑇𝑠 = × 𝑇𝑠𝑢 = × 1.143 = 0.381 3 • dominant, 2nd-order poles 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 −3.5 + 𝑗5.8 = −10.50 + 𝑗17.40 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.49 Design via Root Locus §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 − = 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.51 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response PID Controller Design 𝐾2 𝐺𝑐 𝑠 = 𝐾1 + + 𝐾2 𝑠 (9.21) 𝑠 𝐾 𝐾 𝐾3 𝑠 + 𝑠 + 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.53 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.50 Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.52 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.54 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response Solution Step 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 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24/03/2016 System Dynamics and Control 9.55 Design via Root Locus §4.Improving Steady-State Error and Transient Response The complete performance of the uncompensated system System Dynamics and Control 9.56 Design via Root Locus §4.Improving Steady-State Error and Transient Response Step Design the PD controller to meet the transient response specifications The imaginary part of the compensated dominant pole 𝜋 𝜔𝑑 = 𝑇𝑝 𝜋 = × 0.297 = 15.87 𝜔𝑑 𝑡𝑎𝑛117.130 = −8.13 𝜎= HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.57 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response The compensating zero’s location 𝜃𝑧𝑐 + 𝜃−8 − 𝜃−3 − 𝜃−6 − 𝜃−10 = 1800 ⟹ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.58 Nguyen Tan Tien Design via Root Locus §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 = 18.37 15.87 = 𝑡𝑎𝑛18.370 → 𝑧𝑐 = 55.92 −8.13 − (−𝑧𝑐) The PD controller 𝐺𝑃𝐷 𝑠 = 𝑠 + 55.92 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.59 Nguyen Tan Tien Design via Root Locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.60 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 10 24/03/2016 System Dynamics and Control 9.61 Design via Root Locus System Dynamics and Control 9.62 Design via Root Locus §4.Improving Steady-State Error and Transient Response Step 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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.63 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response Step 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 + 56.42𝑠 + 27.96) = 𝑠 Matching Eqs.(9.21) to get 𝐾1 = 259.2, 𝐾2 = 128.6, 𝐾3 = 4.6 System Dynamics and Control 9.64 Nguyen Tan Tien Design via Root Locus §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 (9.21) 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.65 Nguyen Tan Tien Design via Root Locus - 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.66 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 11 24/03/2016 System Dynamics and Control 9.67 Design via Root Locus System Dynamics and Control 9.68 Design via Root Locus §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 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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.69 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response Step Design the lead compensator 𝑇𝑠 = 𝑇𝑠𝑢 → −𝜁𝜔𝑛 = −2 × 1.794 = −3.588 𝜃𝑧−6 − 𝜃𝑝𝑐 − 𝜃𝑝0 − 𝜃𝑝−6 − 𝜃𝑝−10 = 1800 ⟹ System Dynamics and Control 9.70 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.71 Nguyen Tan Tien Design via Root Locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.72 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response Step Design the lag compensator to improve the steady-state error The uncompensated system’s open-loop TF 192.1 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 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 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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 12 24/03/2016 System Dynamics and Control 9.73 Design via Root Locus §4.Improving Steady-State Error and Transient Response The characteristics of uncompensated, lead-compensated, and lag-lead-compensated systems System Dynamics and Control 9.74 Design via Root Locus §4.Improving Steady-State Error and Transient Response Step (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) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.75 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response Notch Filter Root locus before cascading notch filter HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.76 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.77 Nguyen Tan Tien Design via Root Locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.78 Nguyen Tan Tien Design via Root Locus §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 times and decrease the steadystate error for a unit ramp input by 10 times Place the lead zero at −3 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 13 24/03/2016 System Dynamics and Control 9.79 Design via Root Locus §4.Improving Steady-State Error and Transient Response Solution a.Evaluate the settling time 𝑂𝑆% = 20% ~ 𝜁 = 0.456 System Dynamics and Control 9.80 Design via Root Locus §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 1 𝑒𝑟𝑎𝑚𝑝 ∞ = = = 0.1189 𝐾𝑣 8.41 c.Design a lag-lead compensator 𝑇𝑠 = 𝑇𝑠𝑢 → −𝜁𝜔𝑛 = −2 × 6.83 = −13.66 The desired poles −7 ± 13.66 For the uncompensated system 4 𝑇𝑠𝑢 = = = 1.143𝑠 ȁ𝑅𝑒 ȁ 3.5 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.81 Nguyen Tan Tien Design via Root Locus §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 ) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.83 Nguyen Tan Tien Design via Root Locus HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.82 Nguyen Tan Tien Design via Root Locus §4.Improving Steady-State Error and Transient Response For the uncompensated system 𝐾𝑣𝑢 = 8.41 For the lead-compensated system 𝐾×3 205.9 × 𝐾𝑣𝑙𝑒𝑎𝑑 = lim 𝑠𝐺(𝑠)𝐺𝑙𝑒𝑎𝑑(𝑠) = = = 9.138 𝑠→0 × 9.61 × 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.84 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 14 24/03/2016 System Dynamics and Control 9.85 Design via Root Locus §4.Improving Steady-State Error and Transient Response System Dynamics and Control 9.86 Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.87 Nguyen Tan Tien Design via Root Locus §6.Physical Realization of Compensation Active-Circuit Realization 𝑉𝑜 (𝑠) 𝑍2 (𝑠) =− 𝑉𝑖 (𝑠) 𝑍1 (𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.89 System Dynamics and Control 9.88 Nguyen Tan Tien Design via Root Locus §6.Physical Realization of Compensation Active-Circuit Realization 𝑉𝑜 (𝑠) 𝑍2 (𝑠) =− 𝑉𝑖 (𝑠) 𝑍1 (𝑠) Nguyen Tan Tien Design via Root Locus §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 9.90 Nguyen Tan Tien 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, HCM City Univ of Technology, Faculty of Mechanical Engineering = 27.96 𝑅1 𝐶2 Nguyen Tan Tien 15 24/03/2016 System Dynamics and Control 9.91 Design via Root Locus System Dynamics and Control 9.92 §6.Physical Realization of Compensation 𝑅2 𝐶1 + = 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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.93 Nguyen Tan Tien Design via Root Locus §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 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 9.95 Nguyen Tan Tien Design via Root Locus System Dynamics and Control Design via Root Locus 𝑠+𝑇 𝑠+ 𝛼𝑇1 9.94 𝑠+ 𝛼 𝑇1 Nguyen Tan Tien 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 = 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 System Dynamics and Control 9.96 §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 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 𝑠+𝑇 Nguyen Tan Tien Design via Root Locus 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 𝑅2 𝐶2 ⟹ = 0.5, 𝑅2 𝐶1 = 1, + = 5.1 𝑅1 𝐶2 𝑅1 𝐶1 −5 Choose 𝐶1 = 10 𝜇𝐹 → 𝑅1 = 20𝑘Ω, 𝑅2 = 105Ω,𝐶2 = 100𝜇𝐹 Nguyen Tan Tien 16 24/03/2016 System Dynamics and Control 9.97 Design via Root Locus §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 𝑠+ 𝑠+ 𝑅1𝐶1 𝑅2𝐶2 𝐺𝑐 𝑠 = 1 1 𝑠2 + + + 𝑠+ 𝑅1𝐶1 𝑅2𝐶2 𝑅2𝐶1 𝑅1𝑅2𝐶1𝐶2 1 1 ⟹ = 0.1, = 0.1, + + = 20.01 𝑅1 𝐶1 𝑅2 𝐶2 𝑅1𝐶1 𝑅2𝐶2 𝑅2𝐶1 Choose 𝐶1 = 100𝜇𝐹 → 𝑅1 = 100𝑘Ω,𝑅2 = 558𝑘Ω,𝐶2 = 900𝜇𝐹 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 17 ... and Control 9.95 Nguyen Tan Tien Design via Root Locus System Dynamics and Control Design via Root Locus