State Feedback. State Feedback Block Diagram. State Feedback. Closed-Loop State-Space Equations

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1 State Feedback M. S. Fadali Professor of Electrical Engineering University of Nevada, Reno State Feedback The control is a function of the state vector. State Feedback Law: Linear State Feedback Law: state gain matrix (select for good transient response). Affine State Feedback Law: reference input 1 2 State Feedback Block Diagram v(k) u(k) x(k+1) x(k) + B + T C y(k) Closed-Loop State-Space Equations State-Space Equations for LTI System Affine State Feedback Law A K 3 4

2 Output Feedback Output measurements y must be used to obtain the control u x( k 1) v(k) Ax( k) B K ycx( k) A BK C x( k) Bv( k) u(k) + B + y x(k+1) T A x(k) u( k) K v( k) C = K A y(k) y y y y( k) v( k) Cx( k) v( k) A BK y C Control Systems Regulator The control law drives the system to the zero state (zero output, ) or to a specified constant state (constant output, ). Design: Select to provide i. A good transient zero-input response. ii. Good disturbance rejection. Tracker The control law makes the system output track a specified time function. K 5 6 Pole Placement (Assignment, Allocation) Choose the gain matrix or to assign the system eigenvalues to an arbitrary set Easier with state feedback but output feedback is more realistic. Poles can be arbitrarily placed if and only if the system is controllable. Pole placement is easier if the system is given in controllable form. Stability Closed-loop stability: asymptotically stable Select so that the eigenvalues of the closedloop state matrix are inside the unit circle (open LHP for continuous-time systems). 7 8 Lim Lim

3 Theorem 9.1: State Feedback There exists a state feedback gain matrix that arbitrarily places the eigenvalues of the closedloop matrix if and only if the pair is controllable. Furthermore, if the pair is stabilizable, then the controllable modes can all be arbitrarily assigned. Sufficiency Proof (if) Assume w. l. o. g. that the system is in controllable form (any controllable system can be transformed to controllable form) Find the matrix that arbitrarily places the eigenvalues of the closed-loop system with open-loop pair in controllable form (shown later) Proof of Necessity Controllability Invariance: Uncontrollable modes are invariant under state feedback. Let be an uncontrollable mode then Then is an eigenvalue and is a left eigenvector of the closed-loop system for any state feedback gain matrix. Eigenvalues cannot be arbitrarily assigned for uncontrollable. 11 Remarks on Theorem For any controllable single-input pair, and any set of real or complex conjugate values, there exists a unique 1 by vector such that the eigenvalues of the closed-loop matrix are placed at 2. For controllable -input systems, an matrix exists but is not unique. 3. For stabilizable systems, the controllable modes can be arbitrarily selected but the uncontrollable modes are invariant. 12

4 Selection of Eigenvalues Use the time response criteria based on second order approximation. Inequality constraints on design criteria. Poles at, for CT and at ±, for DT systems. For higher order systems, place additional poles at the origin for DT systems (far in the LHP for CT) or to cancel with system zeros. Procedure 9.1: Pole Placement by Equating Coefficients 1. Evaluate the desired characteristic polynomial from the specified eigenvalues using the expression 2. Evaluate the closed-loop characteristic polynomial using the expression 3. Equate the coefficients of the two polynomials to obtain equations to be solved for the entries of the matrix Equating Coefficients Example C The system is controllable The c.l. eigenvalues can be chosen arbitrarily 15 Numerical Values Desired eigenvalues:, Feedback Gain: 16

5 Limitations of Equating Coefficients Difficult to obtain closed-loop characteristic equation in terms of the gains. Equating coefficients may result in complex equations. Suitable for second order systems but difficult in general. What guarantees that the eigenvalues can be arbitrarily assigned? 17 Phase Variable (Controllable) Form, Gain vector in terms of coefficients vectors. 18 Controller (Controllable) Form, Gain vector in terms of coefficients vectors. 19 Procedure: Either Controllable Form 1. Obtain the open-loop characteristic polynomial and the vector or. 2. Obtain the desired closed-loop characteristic equation and vector or. 3. Obtain the gain vector,or (same values: in ascending order, and in descending order). Note: Identical procedure for continuous-time systems but poles selected in LHP (change ) 20

6 Example: Phase Variable Form Example: Controller Form Desired eigenvalues: Desired characteristic polynomial: Open-loop characteristic polynomial: Gain vector: Desired eigenvalues: Desired characteristic polynomial: Open-loop characteristic polynomial: Gain vector: Example: Phase Variable Form Example: Controller Form Desired eigenvalues: Desired characteristic polynomial: s 2 +6 s +18 Open-loop characteristic polynomial: s 2 +4 s +3 Gain vector: Desired poles Desired characteristic polynomial: Open-loop characteristic polynomial: Gain vector: Check closed-loop state matrix with MATLAB 23 24

7 MATLAB Commands >> a=[ ; ; ]; >>polya= poly(a) % Characteristic polynomial ans = >> polyd=poly([-10,-3+3*j,-3-j*3]) % Desired polynomial ans = >> polyd-polya; kc=ans(2:4) % For the controller form kc = MATLAB: fliplr needed for phase variable form. 25 Lemma Any pair can be transformed to controllable form if and only if it is controllable. Note: 1. Transform to phase variable form or to controller form. 2. Use the characteristic polynomial to form 3. There exists a nonsingular by matrix such that,where is in controllable form. 26 Proof of Sufficiency Assume the system is controllable, then C is invertible If the system is controllable then CC and C C exist (invertible matrices). C The similarity transformation 27 Proof of Necessity Controllable form is controllable. Similarity transformation does not change controllability. If the system is not controllable we cannot transform it to a controllable form. Transformation CC C C does not exist if the system is not controllable (C is singular). 28

8 State Feedback (Bass-Gura) For any controllable pair exists. For controllable form CC transformation to controllable form CC C C Write the control in terms of (for any ) C C C C Calculate (invert C only). Do not calculate 29 Example Desired eigenvalues: C C C C 30 Example: DT Pole placement for the DT system with Desired eigenvalues: (±)(.) Discretize the system with sampling period Example: DT (cont.) Desired eigenvalues: Characteristic polynomial of DT system C C C 31 32

9 MATLAB Pole Placement >> a=[ ; ; ]; % Earlier Ex. >> b=[0; 0; 1]; >> k=place(a, b, [-10,-3+3*j,-3-j*3]) k = >> eig(a-b*k) % Check closed-loop eigenvalues ans = i i Uncontrollable Modes If some modes are uncontrollable >> b= [1; 3; 1]; >> rank(ctrb(a,b)) % Not even stabilizable! 1 >> k=place(a,b,[-10,-3+3*j,-3-j*3])??? Error using ==> place at 179 Can't place eigenvalues there Ackermann s Formula to place the eigenvalues at C C = row of C 35 Proof For phase var. (controllable) form Cayley-Hamilton (similarity transf. invariant) Subtract then use 36

10 Proof: Elementary Vectors entry only is unity Similarly show: 37 Proof: Controllable Form only entry is unity 38 Proof (Cont.) Similarity transformation Bass-Gura formula and C C 39 Proof (Cont.) Transformation to phase var. form C C C Substitute in the formula derived earlier C C C 40

11 Example: Ackermann Example: Ackermann CT C C Desired poles C C = = , C = Desired poles C As obtained earlier MATLAB: Basic Commands >> pd=poly([-3+j*3,-3-j*3]) % Desired polynomial ans = >> con=ctrb(a,[0;1]) % Controllability matrix con = >> k=[0,1]/con* polyvalm(pd,a) % Gain vector k = Desired eigenvalues: DT Example Discretize the system with sampling period >> poldes=poly(pold) % pold =poles for DT system >> k=[0 1]/cond*polyvalm(poldes,pd.A) k =

12 MATLAB: acker >> A=[0 1;-3-4];B=[0;1]; % Controllable pair >> acker(a,b,[-3+j*3,-3-j*3]) % Gain vector ans = 15 2 Theorem: C.L. Controllability A LTI system with the state feedback is controllable with the external input if and only if is controllable. Closed-loop system controllable controllable Proof of Sufficiency (if): For controllable (reachable), there exists an input such that and for any state. Let =state associated with. Since, define the external input takes the closed-loop system to arbitrary in finite time, i.e. the closed-loop system is controllable. 47 Proof of Necessity (only if): If is not controllable (reachable), then there are some states that are not reachable from the origin regardless of how the input is generated (including the state feedback ). All point in the uncontrollable subspace are not reachable. If is not controllable, the closed-loop system is also not controllable. 48

13 Transmission Zero Solve for and equations in unknowns Can determine and direction of. for..... Transmission Zeros :, 49 Transmission Zero Condition A complex number is a transmission zero if it satisfies Rewrite as two equations Matrix is not full rank if the two equations hold. 50 Theorem: Transmission Zeros The transmission zeros of a LTI system are invariant under the state feedback Proof Let be a transmission zero of the LTI system. Then 51 Closed-Loop System Zero transmission zero of the open-loop system is also a transmission zero for the closed-loop system. Cannot change zeros with state feedback 52

14 Loss of Observability State feedback can place a pole at the location of a zero: pole-zero cancellation. If the original system is controllable, the closed-loop system is controllable. Pole-zero cancellation: closed-loop system can have unobservable modes even if the original system is completely observable. 53

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