Final Exam Review Final Exam Study Guide Important Topics to Review Final Exam Dates 1 Final Exam Guidelines Review the lecture notes and text sections corresponding to the final exam topics to be discussed. Review Homework problems and solutions Review the posted sample final exam The final exam will be closed-book but you are allowed two(2) 8.5x11 sheets (both sides) of notes 2 1 Final Exam Topics State-Space Realizations (L3) Time Response Analysis (L4) Modal Decomposition (L9-10) Stability (L11) Controllability & Observability (L12-14) State Feedback and Observers (L15-18) Linear Quadratic Optimal Control (L19-20) 3 Review of Important Topics Concepts & Examples 4 2 State & Transfer Function Models State Space Representation of an LTI system: x Ax Bu y Cx Du Transfer Function L(output) L(input) zero initial conditions G(s) C sI A B D 1 5 General Linear State Transformation State and output equations State transformation xˆ Tx x Ax Bu y Cx Du x Ax Bu xˆ Tx Tx TAx TBu xˆ TAT-1xˆ TBu y Cx Du y CT-1xˆ Du y Cx Du Transformation Formulas A,B,C,DAˆ ,Bˆ , Cˆ ,D T ˆ CT 1 ˆ TAT1, B ˆ TB, C A 6 3 Controllable Realization General n-th order transfer function: G(s) Y(s) b s n 1 b2 s n 2 bn 1s bn 1 n U(s) s a1s n 1 a n 1s a n Controllable Canonical form (CCF): x Ac x Bcu y Cc x 1 0 0 0 0 0 0 0 1 0 Ac , Bc , Cc bn bn1 b2 b1 0 1 0 0 0 an an1 a2 a1 1 7 Observable Realization General n-th order transfer function: G(s) Y(s) b s n 1 b2 s n 2 bn 1s bn 1 n U(s) s a1s n 1 a n 1s a n Observable Canonical form (OCF): a1 a 2 Ao an1 an x Ao x Bou y Co x 1 0 0 b1 b 0 1 0 2 , Bo , Co 1 0 0 0 0 0 1 bn bn1 0 0 0 8 4 Observable Realization General n-th order transfer function: G(s) Y(s) b1 s n 1 b2 s n 2 bn 1 s bn U(s) s n a1 s n 1 a n 1 s a n Canonical observable form: a1 a 2 Ao an1 an 1 0 0 0 0 1 0 0 x Aox Bou y Cox 0 b1 b 0 2 , Bo , Co 1 0 0 0 1 bn bn1 0 9 Time Response of LTI Systems x Ax B u LTI System: t A t-t Time Response: xt e xt 0 e A t- Bud t The matrix exponential (eAt) 0 0 is key in computing response can be computed using Laplace Transform: e At L 1 s I - A 1 10 5 Jordan Decomposition Theorem (JDT) Let A be a real nxn matrix with eigenvalues i of multiplicity mi. Then there exits an nxn invertible complex matrix M such that J1 0 0 J 2 1 M AM 0 0 i 1i 0 2 0 i i , Ji mi 1 0 i 0 Jk i 0 0 where Jordan Blocks Ji are mi x mi matrices with super diagonal elements ij=1 or 0. 11 Comments on JDT A defective matrix can only be block diagonalized The columns of modal matrix consists of regular and generalized eigenvectors Multiplicity of eigenvalues determines size of each Jordan block Number of super-diagonal 1’s of block i is equal to number of generalized eigenvectors for i 12 6 State Transition Matrix: eAt State Transition Matrix: Let M be the modal matrix of A. Then (t , t 0 ) e A (t t0 ) A n MJ n M e A t M e J t M 1 e At e J1t M M 1 e Jm t 0 e J2t 0 13 Exponential of Jordan Block Given mxm Jordan block Laplace Transform: 0 J 0 0 1 0 0 1 0 0 1 0 0 e Jt L1[( s I J ) 1 ] e t 0 Jt e 0 0 te t t 2e t 2! e t te t 0 e t 0 0 t m 1e t (m 1)! t m 2e t (m 2)! t te t e 14 7 Exponential of Complex Jordan Block Using Laplace Transformation J e t cos t e t sin t Jt e t t e sin t e cos t 15 Example 0 0 A 0 0 Find exp(At) of Solution: 0 0 A 0 0 e At 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 1 t 0 e 0 0 0 At e 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 t 1 0 0 cos t 0 sin t 0 0 0 0 0 1 t e 1 0 0 0 sin t cos t 16 8 Stability of Continuous-Time LTI Systems Theorem: The LTI system x Ax is stable iff all the eigenvalues of A have non-positive real parts imaginary eigenvalues are distinct It is asymptotically stable iff all eigenvalues of A have negative real parts 17 Illustration of Stability Criteria Im X X X X X Re X X X X X X: Unstable, X: Stable, X: Asy. Stable 18 9 Controllability & Observability Controllability: Any initial state can be transferred to any final state in finite time by means of a control input Observability: The initial state can be uniquely determined from the output and input measurements for any input including 0. 19 Rank Test Consider the LTI system (A,B) is controllable iff rank B AB x Ax Bu y Cx Du A n 1B n (A,C) is observable iff C CA rank n CAn1 20 10 PBH Rank Test Consider the LTI system (A,B) is controllable iff x Ax Bu y Cx Du ranksI A B n, for all complex s (A,C) is observable iff sI A rank n, for all complex s C 21 Transfer Function Test If the system x Ax Bu y Cx is uncontrllable and/or unobservable then its transfer G( s ) C( sI A) B function has pole-zero cancellations. 1 22 11 State Feedback Control Block Diagram r State Feedback Control: Block Diagram u Kx r y x u B C A -K 23 Closed-loop System Plant: x Ax Bu y Cx u Kx r Control: Closed-loop System: x A BK x Br y Cx Open-loop numerator Closed Loop TF: YR((ss )) C(sI A BK ) 1 B N (s ) des (s ) 24 12 Integral Control t Control law: u K s x K I e(t ) dt 0 yd Ki y x u B Integral controller C A plant -Ks Automatically generates reference input r! 25 Closed-Loop Integral Control System Plant: x Ax Bu w y Cx t u K s x K I e(t ) dt , e y y d Control: 0 t x I e( ) d Integral state: 0 Closed-loop system d dt x A 0 x B x C 0 x 0 K s I I x w K I x I y d 26 13 Steady State Error Observation: If the control gains are chosen such that the closed-loop system is asymptotically stable, then all the state variable must reach a constant value since the inputs w and yd are constant. Thus lim t e (t ) 0 27 Pole-Placement with Integral Control Need to find K=[Ks KI] to place the eigenvalues of A B K where A 0 B A C 0 ,B 0 Pole placement technique can be used to find K. Matlab commands: SISO Systems: MIMO Systems: K acker A , B , P K place A , B , P d d 28 14 Pole Placement Problem Choose the state feedback gain to place the poles of the closed-loop system, i.e., Eigenvalue s of A : A BK At specified locations Computation of K: Direct method Bass-Gura Formula: des , , des 1 n 1 0 K ( d a )T P 0 a1 1 0 a n 1 a n 2 1 1 29 Luenberger Observer x Ax Bu Plant y Cx Observer: xˆ Axˆ Bu Ly Cxˆ Closed-loop System: ~x A LC ~x, ~x x xˆ If eigenvalues of A-LC are in LHP then lim ~ x(t ) 0 xˆ (t ) x(t ) as t t Observer pole placement problem is equivalent to state feedback pole placement of (AT,CT) 30 15 Observer Block Diagram u y System L x B A C y Observer 31 Observer Transfer Function Observer State Equation: xˆ Axˆ Bu Ly Cxˆ x u y Observer Sate-Space Realization: ss( A, [B L], I,0 ) Transfer Function: Xˆ (s ) Gu (s )U(s ) G y (s )Y(s ) Gu (s ) sI A B, 1 A A LC G y (s ) sI A L 1 32 16 State Equation Form for Reduced Order Observer (ROO) System State Equation Form: x 1 A 11 A 12 x 1 x A 2 21 A 22 x 2 x y Iq q 0 1 x 1 x 2 d dt ROO Summary: z A r z Buu B y y xˆ 2 z Ly B 1 B u 2 A r A 22 LA12 Bu B2 LB1 B y A r L A 21 LA11 ROO State Realization: ss(Ar,[Bu By],I,[0 L]) 33 Combined State Feedback Controller and Observer Combined State Feedback Controller and Observer Formulation Separation Principle Transfer Function Representation Combined State Feedback Controller and Reduced Order Observer Illustrative Examples 34 17 Motivation In most control applications all state variables are not measurable A full or reduced order observer may be used to estimate needed states Separation principle allows independent controller and observer design 35 Combined Controller-Observer Formulation Plant: x Ax Bu Controller: u Kxˆ r Observer: xˆ Axˆ Bu Ly Cxˆ y Cx Open-loop numerator Closed-Loop TF: Y (s ) N (s ) R ( s ) des ( s ) 36 18 State Feedback Observer Block Diagram u r y System L B A x C y Observer K 37 Closed-Loop System Closed-Loop Control Subsystem: Closed-Loop Observer Subsystem: Overall Closed-Loop System: x Ax B Kxˆ r ~ Br x A BK x BKx xˆ Axˆ Bu L y Cxˆ ~ A LC x ~ x d x A BK BK x B ~ 0 ~ 0 r A LC x dt x 38 19 Separation Principle Eigenvalues of the closed loop systems are the union of eigenvalues of Closed-loop poles, i.e., eigenvalues of ABK and Observer poles A-LC This is known as the separation principle 39 Block Digaram Representation Block Diagram Y R P F1 F2 Feedforward and Feedback Blocks: 1 F1(s ) I K(sI A )1B , F2 (s ) K(sI A )1L Closed-Loop TF Gclosed (s ) I PF1F2 PF1 CsI A BK B 1 1 N (s ) des (s ) 40 20 State Realizations of F1 and F2 F2 (s ) K(sI A )1L ss( A,L,K,0 ) F2: What is the state space realization of F1 1 F1(s ) I F1(s ) , F1(s ) K(sI A )1B ss( A,B,K,0 ) F1(s ) : v ss( A BK,B,K, I) z F’1 u 41 Combined Controller Reduced Order Observer (CCRO) Plant State Equation: x 1 A 11 A 12 x 1 B 1 x A u 2 21 A 22 x 2 B 2 x y I q q 0 1 x 1 x 2 d dt Reduced Order Observer z A r z B u u B y y xˆ 2 z Ly A r A 22 LA 12 B u B 2 LB 1 B y A r L A 21 LA 11 42 21 Transfer Function Block Digaram Representation of CCRO Block Diagram Y R P F1 F2 Block and Closed-Loop TF: F1 ( s ) 1 K 2 ( s I A r ) 1B u 1 F2 ( s ) K 2 ( s I A r ) 1B y K 1 K 2L r Gclosed (s ) I PF1F2 PF1 CsI A BK B 1 1 F1 and F2 are of order n-q (lower than FOO) 43 Alternative Output Feedback Control: Polynomial Design The block diagram representation of the combined observer and state feedback controller raises the possibility of a direct s-plane design approach: Y R P=N(s)/M(s) F1 F2 F1(s ) ( s ) s n 1 n 2s n 1 (s ) 0s n 1 n 2s n 1 F2 (s ) (s ) 0s n 1 n 2s n 1 n 1 ( s ) s n 2s n 1 Indeed for SISO systems, we can find the transfer functions F1 and F2 directly. 44 22 Determination of Compensation Coefficients Choose the reduced observer characteristic polynomial (z). Solve the Diophantine equation (s)A(s)+(s)B(s)=D(s)= (s)des(s) for (s) and (s), where des(s) is the desired closed-loop characteristic equation. The Bezout equation may be rearranged as shown next. 45 Linear Quadratic (LQ) Optimal Control Given continuous-time state equation x Ax Bu Find the control function u(t) to minimize tf J ( x, t0 ) x (tf )Sx(tf ) xT (t )Qx(t ) uT (t )Ru(t ) dt T 0 S, Q 0, R 0 and symmetric 46 23 Summary of LQR Solution x Ax Bu Plant: Control law: ut K t x t , Ricatti Equation: K t R 1 BT P t P PA A T P Q PBR 1B T P 0, P ( t f ) S J x , t x T P ( t ) x Cost-to-go: 47 Infinite Horizon LQ Given continuous-time state equation x Ax Bu Find the control function u(t) to minimize J ( x, t0 ) xT (t )Qx(t ) uT (t )Ru(t ) dt 0 Q 0, R 0 and symmetric Solution is obtained as the limiting case of Ricatti Eq. 48 24 Summary of Steady-State LQR Solution x Ax Bu Plant: u Kx , K R 1 BT P Control law: Algebraic Ricatti Equation (ARE): PA A T P Q PBR 1B T P 0 Cost-to-go: J x x T Px 49 Existence Conditions Theorem: If (A,B) is stabilizable and (A,Q1/2) is detectable (note: Q=QT/2Q1/2) then the ARE has a unique positive definite solution. Reason: (A,B) stabilizable implies J* (hence P) is finite (A,C) detectable implies J* >0 for nonzero x(0) 50 25