MODERN CONTROL DESIGN

Similar documents
Australian Journal of Basic and Applied Sciences, 3(4): , 2009 ISSN Modern Control Design of Power System

Course Outline. Higher Order Poles: Example. Higher Order Poles. Amme 3500 : System Dynamics & Control. State Space Design. 1 G(s) = s(s + 2)(s +10)

State Regulator. Advanced Control. design of controllers using pole placement and LQ design rules

Topic # Feedback Control Systems

Suppose that we have a specific single stage dynamic system governed by the following equation:

Optimal Control. Quadratic Functions. Single variable quadratic function: Multi-variable quadratic function:

Linear State Feedback Controller Design

1 Steady State Error (30 pts)

OPTIMAL CONTROL. Sadegh Bolouki. Lecture slides for ECE 515. University of Illinois, Urbana-Champaign. Fall S. Bolouki (UIUC) 1 / 28

Control Systems Design

Lecture 4 Continuous time linear quadratic regulator

MATH4406 (Control Theory) Unit 6: The Linear Quadratic Regulator (LQR) and Model Predictive Control (MPC) Prepared by Yoni Nazarathy, Artem

CONTROL DESIGN FOR SET POINT TRACKING

5. Observer-based Controller Design

SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015

X 2 3. Derive state transition matrix and its properties [10M] 4. (a) Derive a state space representation of the following system [5M] 1

EE221A Linear System Theory Final Exam

Dynamic Compensation using root locus method

EE C128 / ME C134 Final Exam Fall 2014

LINEAR QUADRATIC GAUSSIAN

Full State Feedback for State Space Approach

Control Systems. State Estimation.

Quadratic Stability of Dynamical Systems. Raktim Bhattacharya Aerospace Engineering, Texas A&M University

Control Systems. Design of State Feedback Control.

Module 07 Controllability and Controller Design of Dynamical LTI Systems

Homework Solution # 3

MS-E2133 Systems Analysis Laboratory II Assignment 2 Control of thermal power plant

Steady State Kalman Filter


Topic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback

Time-Invariant Linear Quadratic Regulators Robert Stengel Optimal Control and Estimation MAE 546 Princeton University, 2015

Chapter 3. LQ, LQG and Control System Design. Dutch Institute of Systems and Control

Extensions and applications of LQ

Time-Invariant Linear Quadratic Regulators!

Autonomous Mobile Robot Design

Exam. 135 minutes, 15 minutes reading time

Control Systems Lab - SC4070 Control techniques

LQR, Kalman Filter, and LQG. Postgraduate Course, M.Sc. Electrical Engineering Department College of Engineering University of Salahaddin

Topic # Feedback Control Systems

1. Find the solution of the following uncontrolled linear system. 2 α 1 1

State Space Control D R. T A R E K A. T U T U N J I

Intro. Computer Control Systems: F8

Subject: Optimal Control Assignment-1 (Related to Lecture notes 1-10)

Chapter 9 Observers, Model-based Controllers 9. Introduction In here we deal with the general case where only a subset of the states, or linear combin

Pole placement control: state space and polynomial approaches Lecture 2

Controls Problems for Qualifying Exam - Spring 2014

Pitch Rate CAS Design Project

Intermediate Process Control CHE576 Lecture Notes # 2

Topic # Feedback Control

EECS C128/ ME C134 Final Wed. Dec. 14, am. Closed book. One page, 2 sides of formula sheets. No calculators.

The output voltage is given by,

State Feedback and State Estimators Linear System Theory and Design, Chapter 8.

Problem 1 Cost of an Infinite Horizon LQR

EE363 homework 2 solutions

Chap 4. State-Space Solutions and

Module 03 Linear Systems Theory: Necessary Background

6. Linear Quadratic Regulator Control

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.

Chapter 8 Stabilization: State Feedback 8. Introduction: Stabilization One reason feedback control systems are designed is to stabilize systems that m

Solution to HW State-feedback control of the motor with load (from text problem 1.3) (t) I a (t) V a (t) J L.

Topic # /31 Feedback Control Systems

Identification Methods for Structural Systems

Topic # Feedback Control Systems

Control, Stabilization and Numerics for Partial Differential Equations

Extreme Values and Positive/ Negative Definite Matrix Conditions

EL2520 Control Theory and Practice

ESC794: Special Topics: Model Predictive Control

SUCCESSIVE POLE SHIFTING USING SAMPLED-DATA LQ REGULATORS. Sigeru Omatu

Here represents the impulse (or delta) function. is an diagonal matrix of intensities, and is an diagonal matrix of intensities.

Linear Quadratic Optimal Control Topics

ẋ n = f n (x 1,...,x n,u 1,...,u m ) (5) y 1 = g 1 (x 1,...,x n,u 1,...,u m ) (6) y p = g p (x 1,...,x n,u 1,...,u m ) (7)

Integral action in state feedback control

Chap 8. State Feedback and State Estimators

Inverted Pendulum: State-Space Methods for Controller Design

DESIGN OF LINEAR STATE FEEDBACK CONTROL LAWS

STATE VARIABLE (SV) SYSTEMS

4F3 - Predictive Control

6 OUTPUT FEEDBACK DESIGN

Design Methods for Control Systems

Lec 6: State Feedback, Controllability, Integral Action

16.31 Fall 2005 Lecture Presentation Mon 31-Oct-05 ver 1.1

Didier HENRION henrion

1. LQR formulation 2. Selection of weighting matrices 3. Matlab implementation. Regulator Problem mm3,4. u=-kx

Topic # /31 Feedback Control Systems. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback

Synthesis via State Space Methods

Advanced Control Theory

Topic # Feedback Control Systems

Lecture 2 and 3: Controllability of DT-LTI systems

Control Systems Design

An Introduction to Control Systems

Test 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010

Control Design. Lecture 9: State Feedback and Observers. Two Classes of Control Problems. State Feedback: Problem Formulation

The norms can also be characterized in terms of Riccati inequalities.

EECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.

Robust Control 5 Nominal Controller Design Continued

6.241 Dynamic Systems and Control

Lecture 9: Discrete-Time Linear Quadratic Regulator Finite-Horizon Case

Appendix 3B MATLAB Functions for Modeling and Time-domain analysis

Control Systems. University Questions

Linear Quadratic Regulator (LQR) Design I

Transcription:

CHAPTER 8 MODERN CONTROL DESIGN The classical design techniques of Chapters 6 and 7 are based on the root-locus and frequency response that utilize only the plant output for feedback with a dynamic controller In this final chapter on design, we employ modern control designs that require the use of all state variables to form a linear static controller Modern control design is especially useful in multivariable systems; however, in this chapter the ideas of state-space design are illustrated using single-input, single-output systems One approach in modern control systems accomplished by the use of state feedback is known as pole-placement design The pole-placement design allows all roots of the system characteristic equation to be placed in desired locations This results in a regulator with constant gain vector K The state-variable feedback concept requires that all states be accessible in a physical system, but for most systems this requirement is not met; ie, some of the states are inaccessible For systems in which all states are not available for feedback, a state estimator (observer) may be designed to implement the pole-placement design The other approach to the design of regulator systems is the optimal control problem where a specified mathematical performance criterion is minimized 7

8 Pole-Placement Design 7 8 Pole-Placement Design The control is achieved by feeding back the state variables through a regulator with constant gains Consider the control system presented in the state-variable form _x(t) =Ax(t) +Bu(t) (8) y(t) =Cx(t) Consider the block diagram of the system shown in Figure 8 with the following state feedback control u(t) = Kx(t) (8) where K is a n vector of constant feedback gains The control system input r(t) is assumed to be zero The purpose of this system is to return all state variables to values of zero when the states have been perturbed r(t) = ffο ρß k n u(t) y(t) Plant x n (t) x (t) x (t) k k FIGURE 8 Control system design via pole placement is Substituting (8) into (8), the closed-loop system state-variable representation The closed-loop system characteristic equation is _x(t) =(A BK)x(t) =A f x(t) (8) jsi A + BKj = (8) Assume the system is represented in the phase variable canonical form as follows 6 _x _x _x n _x n 7 5 = 6 ::: ::: ::: a a a ::: a n 7 5 6 x x x n x n 7 5 + 6 7 5 (85) u(t)

7 8 Modern Control Design Substituting for A and B into (8), the closed-loop characteristic equation for the control system is found jsi A + BKj = s n +(a n + k n )s n + +(a + k )s +(a + k )= (86) For the specified closed-loop pole locations ;:::; n, the desired characteristic equation is ff c (s) =(s + ) (s + n )=s n + ff n s n + + ff s + ff = (87) The design objective is to find the gain matrix K such that the characteristic equation for the controlled system is identical to the desired characteristic equation Thus, the gain vector K is obtained by equating coefficients of equations (86) and (87) k i = ff i a i (88) If the state model is not in the phase-variable canonical form, we can use the transformation technique of Chapter Section 5 to transform the given state model to the phase-variable canonical form The gain factor is obtained for this model and then transformed back to confirm with the original model This procedure results in the following formula, known as Ackermann s formula where the matrix S is given by and the notation ff c (A) is given by K = Λ S ff c (A) (89) S = B AB A B ::: A n B Λ (8) ff c (A) =A n + ff n A n + + ff A + ff I (8) The function [K, A f ]=placepol(a,b,c,p)is developed for the pole-placement design A, B, C are system matrices and p is a row vector containing the desired closed-loop poles This function returns the gain vector K and the closed-loop system matrix A f Also, the MATLAB Control System Toolbox contains two functions for pole-placement design Function K = acker(a, B, p) is for single input systems, and function K = place(a, B, p), which uses a more reliable algorithm, is for multi-input systems The condition that must exist to place the closed-loop poles at the desired location is to be able to transform the given state model into phase-variable canonical form That is, the controllability matrix S, given in (8), must have a nonzero determinant This characteristic is known as controllability Example 8 For the plant _x _x _x 5 = 5 x x x 5 + 5 u

8 Pole-Placement Design 7 y = Λ x design a state feedback control to place the closed-loop pole at ± j and 8 Obtain the initial condition response, given x () =, x () = and x () = The following commands A=[- ; - - ; ]; B=[; ; ]; C=[ ]; D=; j=sqrt(-); P=[-+j* --j* -8]; % desired closed-loop poles [K,Af]=placepol(A,B,C,P); % returns gain K and closed-loop % system matrix % initial condition response t=::; r=zeros(,length(t)); % generates a row of zero input x=[ -]; % initial state [y,x]=lsim(af, B, C, D, r, t, x); % initial state response subplot(,,), plot(t, x(:,)),title('x_(t)'), grid subplot(,,), plot(t, x(:,)),title('x_(t)'), grid subplot(,,), plot(t, x(:,)),title('x_(t)'), grid subplot(,,), plot(t, y),title('y(t)'), grid subplot() result in Feedback gain vector K 5 Uncompensated Plant s^ + s ----------------- s^ + s^ + s Compensated system closed-loop s^ + s ------------------------- s^ + s^ + 7 s + Compensated system matrix A - B*K - -5 - - - The results are given in Figure 8

7 8 Modern Control Design x (t) x (t) 5 5 5 x (t) 5 5 5 5 5 5 5 y(t) 5 5 FIGURE 8 Initial condition response for Example 8 Example 8 A single-input single-output plant has the transfer function G(s) = s(s +)(s +) Obtain a state model for the plant and design a state feedback that will place the closed-loop pole at ± j and 5 Also, obtain the closed-loop transfer function for the controlled system The following commands num=; den=[ 5 ]; [A, B, C, D]=tfss(num, den) % converts transfer function % to state model j=sqrt(-); P=[-+j* --j* -5]; % desired closed-loop poles [K,Af]=placepol(A,B,C,P); % returns gain K & closed-loop % system matrix result in A = -5 -

8 Controllability 75 B = C = D = Feedback gain vector K Uncompensated Plant ----------------- s^ + 5 s^ + s Compensated system closed-loop ----------------------- s^ + 9 s^ + 8 s + Compensated system matrix A - B*K -9-8 - From the above results the closed-loop transfer function for the controlled plant is T (s) = C(s) R(s) = s +9s +8s + 8 Controllability A system is said to be controllable when the plant input u can be used to transfer the plant from any initial state to any arbitrary state in a finite time The plant described by (8) with the system matrix having dimension n n is completely state controllable if and only if the controllability matrix S in (8) has a rank of n The function S=cntrable(A, B) is developed which returns the controllability matrix S and determines whether or not the system is state controllable 8 Observer Design In the pole-placement approach to the design of control systems, it was assumed that all state variables are available for feedback However, in practice it is impractical to install all the transducers which would be necessary to measure all of the states If the state variables are not available because of system configuration or cost, an observer

76 8 Modern Control Design r(t) = ffο ρß u(t) y(t) Plant Observer Controller Estimated State _x FIGURE 8 State feedback design with an observer or estimator may be necessary The observer is an estimator algorithm based on the mathematical model of the system The observer creates an estimate ^x(t) of the states from the measurements of the output y(t) The estimated states, rather than the actual states, are then fed to the controller One scheme is shown in Figure 8 Consider a system represented by the state and output equations _x(t) =Ax(t) +Bu(t) (8) y(t) =Cx(t) (8) Assume that the state x(t) is to be approximated by the state ^x(t) of the dynamic model _^x(t) =A^x(t) +Bu(t) +G(y(t) ^y(t)) (8) ^y(t) =C^x(t) (85) Subtracting (8) from (8), and (85) from (8), we have ( _x(t) _^x(t)) = A(x(t) ^x(t)) G(y(t) ^y(t)) (86) (y(t) ^y(t)) = C(x(t) ^x(t)) (87) where x(t) ^x(t) is the error between the actual state vector and the estimated vector, and y(t) ^y(t) is the error between the actual output and the estimated output Substituting the output equation into the state equation, we obtain the equation for the error between the estimated state vector and the actual state vector where _e(t) =(A GC)e(t) =A e e(t) (88) e(t) =x(t) ^x(t) (89) If G is chosen such that eigenvalues of matrix A GC all have negative real parts, then the steady-state value of the estimated state vector error e(t) for any initial condition will tend to zero That is, ^x(t) will converge to x(t) The design of the observer is similar to the design of the controller However, the eigenvalues of A GC must

8 Observer Design 77 be selected to the left of the eigenvalues of A This ensures that the observer dynamic is faster than the controller dynamic for providing a rapid updated estimate of the state vector The estimator characteristic equation is given by jsi A + GCj = (8) For a specified speed of response, the desired characteristic equation for the estimator is ff e (s) =s n + ff n s n + + ff s + ff = (8) Thus, the estimator gain G is obtained by equating coefficients of (8) and (8) This is identical to the pole-placement technique, and G is found by the application of Ackermann s formula G = ff e (A) and the notation ff e (A) is given by 6 C CA CA n 7 5 6 7 5 (8) ff e (A) =A n + ff n A n + + ff A + ff I (8) The function [G, A e ] = observer(a, B, C, p e ) is developed for the estimator p e is the desired estimator eigenvalues This function returns the gain vector G and the closed-loop system matrix A f The necessary condition for the design of an observer is that all the states can be observed from the measurements of the output This characteristic is known as observability Example 8 For the plant» _x = _x» 6 y =» x x + Λ x design a full-state observer such that the observer is critically damped with eigenvalues at 8 and 8 The following commands» A=[ ; 6 ]; B=[; ]; C=[ ]; D=; Pe=[-8-8]; % desired observer eigenvalues [K,Ae]= observer(a,b,c,pe); % returns gain K and closed-loop result in u

78 8 Modern Control Design Estimator gain vector G 6 8 Open-loop Plant -------- s^ - 6 Error matrix A - G*C -6-6 8 Observability A system is said to be observable if the initial vector x(t) can be found from the measurement of u(t) and y(t) The plant described by (8) is completely state observable if the inverse matrix in (8) exists The function V=obsvable(A, C) returns the observability matrix V and determines whether or not the system is state observable 85 Combined Controller-Observer Design Consider the system represented by the state and output equations (8) and (8) with the state feedback control based on the observed state ^x(t) given by Substituting in (8), the state equation becomes u(t) = K^x(t) (8) _x(t) =(A BK)x(t) +BKe(t) (85) Combining the above equation with the error equation given by (89), we have»»» _x(t) A BK BK x(t) = (86) _e(t) A GC e(t) The function [K, G, A c ] = placeobs(a, B, C, p, p e ) is developed for the combined controller-observer design A is the system matrix, B is the input column vector, and C is the output row vector p is a row vector containing the desired closed-loop poles and p e is the desired observer eigenvalues The function displays the gain vectors K and G, open-loop plant transfer function, and the controlled system closed-loop transfer function Also, the function returns the gain vector K, and the combined system matrix in (86)

85 Combined Controller-Observer Design 79 Example 8 For the system of Example 8, design a controller-observer system such that the desired closed-loop poles for the system are at ± j Choose the eigenvalues of the observer gain matrix to be p e = p e = 8 The following commands A=[ ; 6 ]; B=[; ]; C=[ ]; D=; j=sqrt(-); P=[-+j* --j*]; % desired regulator roots Pe=[-8-8]; % desired observer roots [K,G,Af]= placeobs(a,b,c,p,pe); % returns gain K,G & closed-loop % system matrix result in Feedback gain vector K Estimator gain vector G: 6 8 Open-loop Plant -------- s^ - 6 Controller-estimator 96 s + 9 ---------------- s^ + 8 s + 7 Controlled system closed-loop 96 s + 9 ------------------------------------ s^ + 8 s^ + s^ + 8 s + Combined controller observer system matrix -5 - -6-6 From the above results the controller-observer transfer function is 96s +9 G ce = s +8s + 7 The closed-loop transfer function for the controlled plant is T (s) = C(s) R(s) = 96s + 9 s +8s +s + 8s +

8 8 Modern Control Design 86 Optimal Regulator Design The object of the optimal regulator design is to determine the optimal control law u Λ (x;t) which can transfer the system from its initial state to the final state (with zero system input) such that a given performance index is minimized The performance index is selected to give the best trade-off between performance and cost of control The performance index that is widely used in optimal control design is known as the quadratic performance index and is based on minimum-error and minimum-energy criteria Consider the plant described by The problem is to find the vector K(t) of the control law _x(t) =Ax(t) +Bu(t) (87) u(t) = K(t)x(t) (88) which minimizes the value of a quadratic performance index J of the form Z tf J = (x Qx + u Ru)dt (89) t subject to the dynamic plant equation in (87) In (89) Q is a positive semidefinite matrix and R is a real symmetric matrix Q is positive semidefinite if all its principal minors are nonnegative The choice of the elements of Q and R allows the relative weighting of individual state variables and individual control inputs To obtain a formal solution, we can use the method of Lagrange multipliers The constraint problem is solved by augmenting (87) into (89) using an n-vector of Lagrange multipliers, The problem reduces to the minimization of the following unconstrained function L(x; ;u;t)=[x Qx + u Ru] + [Ax + Bu _x] (8) The optimal values (denoted by the subscript Λ) are found by equating the partial derivatives to zero @L @ = AXΛ + Bu Λ _x Λ = ) _x Λ = AX Λ + Bu Λ (8) @L @u =RuΛ + B = ) u Λ = R B (8) @L @x =x Λ Q+ _ + A = ) _ = Qx Λ A (8) Assume that there exists a symmetric, time varying positive definite matrix p(t) satisfying =p(t)x Λ (8) Substituting (8) into (8) gives the optimal closed-loop control law u Λ (t) = R B p(t)x Λ (85)

86 Optimal Regulator Design 8 Obtaining the derivative of (8), we have Finally equating (8) with (86), we obtain _ =(_px Λ + p _x Λ ) (86) _p(t) = p(t)a A p(t) Q + p(t)br B p(t) (87) The above equation is referred to as the matrix Riccati equation The boundary condition for (87) is p(t f ) = Therefore, (87) must be integrated backward in time Since a numerical solution is performed forward in time, a dummy time variable fi = t f t is replaced for time t Once the solution to (87) is obtained the solution of the state equation (8) in conjunction with the optimum control equation (85) is obtained The function [fi,p,k,t,x]=riccatiis developed for the time-domain solution of the Riccati equation The function returns the solution of the matrix Riccati equation, p(fi ), the optimal feedback gain vector k(fi ), and the initial state response x(t)inorder to use this function, the user must declare the function [A; B; Q; R; t ; t f ; x ]=system (A; B; Q; R; t ; t f ; x ) containing system matrices and the performance index matrices in an M-file named systemm For linear time-invariant systems, since _p =, when the process is of infinite duration, that is t f =, (87) reduces to the algebraic Riccati equation pa + A p + Q pbr B p = (88) The MATLAB Control System Toolbox function [k, p]=lqr(a, B, Q, R) can be used for the solution of the algebraic Riccati equation Example 85 Design a state feedback system for the plant described by» _x = 5 x x _x 5 x y = Λ x 5 + Find the optimal control law to minimize the performance index Z J = x (t) +x +x + u dt (89) The admissible states and control values are unconstrained The states are initially at x () =, x () = and x () = For this system we have A = 5 5 ; B = 5 u 5 ; Q = First an M-file named systemm is created and the function [A,B,Q,R,t ;t f ;x )= system (A, B, Q, R, t ;t f ;x ) is defined as follows 5 ; and R =

8 8 Modern Control Design Vector K(t) of the control law u(t)= k(t)x(t) 5 5 Initial state response x, x, x 5 5 FIGURE 8 Optimal feedback gain vector K(t) and initial condition response x(t) function [A,B,Q,R,t,tf,x]=system(A,B,Q,R,t,tf,x) A=[ ; ; - -5]; B=[;; ]; Q=[ ; ; ]; R=5; t=; tf=5; x=[ -]; The above function is saved in an M-file named systemm Then, the following commands [tt,p,k,t,x]=riccati subplot(,,), plot(tt,k), title('vector K(t) of the control law u(t)=-k(t)x(t)'), grid, subplot(,,), plot(t,x), title('initial state response x_, x_, x_'), grid subplot() result in the control law and the response is given in Figure 8 Example 86 For Example 85 use the MATLAB Control System Toolbox lqr to obtain the solution to the algebraic Riccati equation The following commands

86 Optimal Regulator Design 8 A=[ ; ; - -5]; B=[;; ]; Q=[ ; ; ];R=5; [K, p]=lqr(a,b,q,r) result in K = 88 78 996 p = 5755 88 88 6 79 79 98