1              Design via State Space•   How to design a state-feedback controller using pole placement    to meet transie...
2                       Introduction•   State space techniques can be applied to a wider class of systems    than transfor...
3                          Controller Design•     Let us consider an n-th order control system with an n-th-order      clo...
4Now, we introduce feedback             , wherein order to set the poles to the desired location                          ...
5Phase-variable representation for plantPlant with state-variable feedback                                                ...
6Pole Placement for Plants in Phase-Variable Form      (Method of Matching the Coefficients)1.   Represent the plant in ph...
7Phase-variable representation of the plant is given by: The characteristic equation of the plant is      Feedback:       ...
8        Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise  Copyright © 2004 by John Wiley...
9The characteristic equation can be found by inspection as: Now if the desired characteristic equation for proper pole pla...
10         Problem         Design the phase-variable feedback gains to yield 9.5% overshoot and a settling         time of...
11•   The closed-loop systems characteristic equation is•   Equating coefficients with the desired characteristic equation...
12                              11.5% overshoot and a settling time                              of 0.8 second            ...
13  Pole Placement for Plants NOT in Phase-Variable Form           (Method of Matching the Coefficients)- consists of matc...
14                           ⎛ ⎡1 0⎤ ⎛ ⎡− 2 1 ⎤ ⎡0⎤          ⎞⎞       ⎛ ⎡1 0⎤ ⎡ − 2                          1 ⎤⎞det( sI −...
15                         Controllability •    If any one of the state variables cannot be controlled by the control u,  ...
16The Controllability Matrix•     enables to determine controllability for a plant under any representation      or choice...
17                   Observer Design•      In some applications, some of the state variables may not be available at      ...
18       Plant                               Observer•   the speed of convergence between the actual state and the estimat...
19Closed loop observer with feedback    •   when designing an observer, the observer canonical form yields the easy       ...
20LetThe design then consists of solving for the values of Lc to yield a desiredcharacteristic equation. The characteristi...
21Let us consider nth-order plant represented in observer canonical form: The characteristic equation for (A-LC) is det[sI...
22If the desired characteristic equation of the observer is Then, we can find li’s as:Observer Design for Plants in Observ...
23SolutionThe state equations for the estimated plant are The observer error is Characteristic polynomial is              ...
24Characteristic polynomial isThe closed loop controlled system has poles at -1+/-j2 and -10.We choose observer poles 10 t...
25Observer Design for Plants NOT in Observer-Canonical Form          (Method of Matching the Coefficients)  •   match the ...
26Comparing the coefficients we can find the values of l1 and l2                                           l1 = −38.397   ...
27        Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise  Copyright © 2004 by John Wile...
28                       Observability•   If any state variable has no effect upon the output, then we cannot evaluate    ...
29The Observability Matrix- enables to determine observability for systems under any representation orchoice of state vari...
30Steady-State Error Design via Integral Control•   enables to design a system for zero steady-state error for a step inpu...
31•   we now have an additional    pole to place.                          Dr Branislav Hredzak                  Control S...
32Problema. Design a controller with integral control to yield a 10% overshoot and asettling time of 0.5 second. Solution ...
33Using the requirements for settling time and percent overshoot, we find that thedesired second order poles are s1,2= -8+...
34Now we match the coefficients of the desired 3rd order characteristic equationwith the characteristic polynomial for the...
35The steady-state error for a unit step input:             e(∞) = 1 + CA −1B                                             ...
36                                                                    Name: _______________________                       ...
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12 elec3114

  1. 1. 1 Design via State Space• How to design a state-feedback controller using pole placement to meet transient response specifications• How to design an observer for systems where the states are not available to the controller• How to design steady-state error characteristics for systems represented in state space Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  2. 2. 2 Introduction• State space techniques can be applied to a wider class of systems than transform methods, for example, to systems with nonlinearities, multi-input multi-output (MIMO) systems• Frequency domain methods of design – cannot be used to specify all closed-loop poles of the higher-order system• State space techniques allow to place all poles of the closed-loop system• Frequency domain methods – allow placement of zero through zero of the lead compensator• State space techniques – do not allow to specify zero locations• State space techniques – more sensitive to parameter variations Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  3. 3. 3 Controller Design• Let us consider an n-th order control system with an n-th-order closed-loop characteristic equation • we need n-adjustable parameters in order to be able to set the poles to any desired location Topology for Pole Placement Let us consider a plant: Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  4. 4. 4Now, we introduce feedback , wherein order to set the poles to the desired location Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  5. 5. 5Phase-variable representation for plantPlant with state-variable feedback Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  6. 6. 6Pole Placement for Plants in Phase-Variable Form (Method of Matching the Coefficients)1. Represent the plant in phase-variable form.2. Feed back each phase variable to the input of the plant through a gain, ki3. Find the characteristic equation for the closed-loop system represented in step 2.4. Decide upon all closed-loop pole locations and determine an equivalent characteristic equation.5. Equate like coefficients of the characteristic equations from steps 3 and 4 and solve for ki Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  7. 7. 7Phase-variable representation of the plant is given by: The characteristic equation of the plant is Feedback: Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  8. 8. 8 Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  9. 9. 9The characteristic equation can be found by inspection as: Now if the desired characteristic equation for proper pole placement is: Then we can find the feedback gains ki as: Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  10. 10. 10 Problem Design the phase-variable feedback gains to yield 9.5% overshoot and a settling time of 0.74 second for the given plant 20 s + 100 = s 3 + 5s 2 + 4 s Solution ⎡0 1 0 ⎤ ⎛ ⎡0 1 0 ⎤ ⎡0 ⎤ ⎞ ⎜ ⎟A = ⎢0 0 1⎥, ( A - BK ) = ⎜ ⎢0 0 1⎥ − ⎢0⎥[k1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ k2 k 3 ]⎟, ⎜⎢ ⎟ ⎣0 4 5 ⎥ ⎢ ⎦ ⎝ ⎣0 4 5⎥ ⎢1⎥ ⎦ ⎣ ⎦ ⎠ • Based on the desired response, we choose two closed loop poles as p1,2 = -5.4 +/- j7.2 • We choose the third closed-loop pole as p3 = -5.1 • Then the desired characteristic equation is ( s − p1 )( s − p 2 )( s − p3 ) = 0 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  11. 11. 11• The closed-loop systems characteristic equation is• Equating coefficients with the desired characteristic equation we obtain• Hence T ( s ) = C( sI − A) −1 B + D Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  12. 12. 12 11.5% overshoot and a settling time of 0.8 second - does not meet the desiredNote, that there is a large specifications because the zero at -5steady-state error ! was not cancelled - if the third pole is chosen at -5 then the design will meet the desired specifications Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  13. 13. 13 Pole Placement for Plants NOT in Phase-Variable Form (Method of Matching the Coefficients)- consists of matching the coefficients of det(sI - (A - BK)) with the coefficientsof the desired characteristic equation – (can result in difficult calculations)Problem Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  14. 14. 14 ⎛ ⎡1 0⎤ ⎛ ⎡− 2 1 ⎤ ⎡0⎤ ⎞⎞ ⎛ ⎡1 0⎤ ⎡ − 2 1 ⎤⎞det( sI − ( A − BK )) = det⎜ s ⎢ ⎥ −⎜ − [k k 2 ]⎟ ⎟ = det⎜ s ⎢ ⎜ 0 1⎥ − ⎢− k ⎟= ⎜ ⎣0 1⎦ ⎜ ⎢ 0 − 1⎥ ⎢1⎥ 1 ⎝ ⎝⎣ ⎦ ⎣ ⎦ ⎟⎟ ⎠⎠ ⎝ ⎣ ⎦ ⎣ 1 − (k 2 + 1)⎥ ⎟ ⎦⎠ Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  15. 15. 15 Controllability • If any one of the state variables cannot be controlled by the control u, then we cannot place the poles of the system where we desire.controllable uncontrollablesystem system If an input to a system can be found that takes every state variable from a desired initial state to a desired final state, the system is said to be controllable; otherwise, the system is uncontrollable. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  16. 16. 16The Controllability Matrix• enables to determine controllability for a plant under any representation or choice of state variables An nth-order plant whose state equation is is completely controllable if the controllability matrix CM is of rank n. The rank of CM equals the number of linearly independent rows or columns. The rank of CM equals n if the determinant of CM is non-zero. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  17. 17. 17 Observer Design• In some applications, some of the state variables may not be available at all, or it is too costly to measure them or send them to the controller• In that case we can estimate states and the estimated states, rather than actual states, are then fed to the controller Observer (Estimator) is used to calculate state variables that are not accessible from the plant.Open-loop observer Closed-loop observer Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  18. 18. 18 Plant Observer• the speed of convergence between the actual state and the estimated state is the same as the transient response of the plant since the characteristic equation is the same. Hence we cannot use the estimated states for the controller.• to increase the speed of convergence between the actual and estimated states, we use feedback Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  19. 19. 19Closed loop observer with feedback • when designing an observer, the observer canonical form yields the easy solution for the observer gains • the observer has to be faster than the response of the controlled loop in order to yield a rapidly updated estimate of the state vector • the design of the observer is separate from the design of the controller Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  20. 20. 20LetThe design then consists of solving for the values of Lc to yield a desiredcharacteristic equation. The characteristic equation is det[sI - (A - LC)]. Then, we select the eigenvalues of the observer to yield stability and a desired transient response that is faster ( about 10 times) than the controlled closed-loop response. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  21. 21. 21Let us consider nth-order plant represented in observer canonical form: The characteristic equation for (A-LC) is det[sI - (A - LC)] : Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  22. 22. 22If the desired characteristic equation of the observer is Then, we can find li’s as:Observer Design for Plants in Observer-Canonical Form (Method of Matching the Coefficients) Problem Design an observer for the plant represented in observer canonical form. The observer will respond 10 times faster than the closed loop control system with poles at -1+/-j2 and -10. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  23. 23. 23SolutionThe state equations for the estimated plant are The observer error is Characteristic polynomial is Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  24. 24. 24Characteristic polynomial isThe closed loop controlled system has poles at -1+/-j2 and -10.We choose observer poles 10 times faster, -10+/-j20 and 100, then the desiredpolynomial isAfter equating coefficients, we can find l1 = 112, l2 = 2483, l3 = 49990 … simulation response to r(t) = 100t Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  25. 25. 25Observer Design for Plants NOT in Observer-Canonical Form (Method of Matching the Coefficients) • match the coefficients of det[sI - (A - LC)] with the coefficients of the desired characteristic polynomial (can yield difficult calculations for higher-order systems) Problem Design an observer for the phase variables with a transient response described by ζ= 0.7 and ωn = 100. Solution Plant in phase variable form will be Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  26. 26. 26Comparing the coefficients we can find the values of l1 and l2 l1 = −38.397 l2 = 35.506 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  27. 27. 27 Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  28. 28. 28 Observability• If any state variable has no effect upon the output, then we cannot evaluate this state variable by observing the output Observable Unobservable Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  29. 29. 29The Observability Matrix- enables to determine observability for systems under any representation orchoice of state variables Plantis completely observable if the observability matrix OM, is of rank n (i.e., the determinant of OM is non-zero) Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  30. 30. 30Steady-State Error Design via Integral Control• enables to design a system for zero steady-state error for a step input as well as design the desired transient response Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  31. 31. 31• we now have an additional pole to place. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  32. 32. 32Problema. Design a controller with integral control to yield a 10% overshoot and asettling time of 0.5 second. Solution Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  33. 33. 33Using the requirements for settling time and percent overshoot, we find that thedesired second order poles are s1,2= -8+/-j10.9 and desired characteristicpolynomial isWe choose the third pole at -100 (real part more than 5 times greater than thedesired second order poles s1,2= -8+/-j10.9).Hence, the desired 3rd order characteristic equation is The characteristic polynomial for the system with integral action is det(sI - )= Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  34. 34. 34Now we match the coefficients of the desired 3rd order characteristic equationwith the characteristic polynomial for the system with integral actionand we can findThen, T ( s ) = C( sI − A) −1 B + D = Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  35. 35. 35The steady-state error for a unit step input: e(∞) = 1 + CA −1B Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
  36. 36. 36 Name: _______________________ Student ID: ___________________ Signature: ____________________ THE UNIVERSITY OF NEW SOUTH WALES School of Electrical Engineering & Telecommunications FINAL EXAMINATION Session 2 2010 ELEC3114 Control Systems TIME ALLOWED: 3 hours TOTAL MARKS: 100 TOTAL NUMBER OF QUESTIONS: 4THIS EXAM CONTRIBUTES 65% TO THE TOTAL COURSE ASSESSMENT.Reading Time: 10 minutes.This paper contains 9 pages .Candidates must ATTEMPT ALL 4 questions.Answer each question in a separate answer book.Marks for each question are indicated beside the question.This paper MAY be retained by the candidate.Print your name, student ID and question number on the front page of each answer book.Authorised examination materials: Drawing instruments may be brought into the examination room. Candidates should use their own UNSW-approved electronic calculators. This is a closed book examination.Assumptions made in answering the questions should be stated explicitly.All answers must be written in ink. Except where they are expressly required, pencils mayonly be used for drawing, sketching or graphical work. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.

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