Kalman Decomposition B 2. z = T 1 x, where C = ( C. z + u (7) T 1, and. where B = T, and

Transcription

1 Kalman Decomposition Controllable / uncontrollable decomposition Suppose that the controllability matrix C R n n of a system has rank n 1 <n. Then there exists an invertible transformation, T R n n such that: z = T 1 x, (Ã11 à ż = 12 z + à 22 ( B where B = T, and A = T T 1. à 22 (Ã11 à 12 ( B u (7 Observable / unobservable decomposition Hence if the observability matrix is not full rank, then using basis for its null space as the last k basis vectors of R n, the system can be represented as: where C = ( C ( (Ã11 B1 ż = z + u à 21 à 22 B 2 y = ( C z T 1, and A = T T 1. à 22 (Ã11 à 12 and the dim of Ã22 is non-zero (and is the dim of the null space of the observability matrix. and the dim of Ã22 is n n 1. M.E. University of Minnesota 64 M.E. University of Minnesota 65

2 Theorem There exists a coordinate transformation z = T 1 x R n such that à 11 à 13 B 1 à ż = 21 à 22 A 23 A 24 à 33 z + B 2 u à 43 à 44 y = ( C1 C2 z Let T = ( t 1 t 2 t 3 t 4 (with compatible block sizes, then t 2 (C Ō: t 2 = basis for Range(C Null(O. t 1 (C O: t 1 t 2 = basis for Range(C t 4 ( C Ō: t 2 t 4 = basis for Null(O. t 3 ( C O: t 1 t 2 t 3 t 4 = basis for R n. Note: Onlyt 2 is uniquely defined. Stabilizability and Detectability If the uncontrollable modes Ā33, Ā 44 are stable (have eigenvalues on the Left Half Plane, then, system is called stabilizable. One can use feedback to make the system stable; Uncontrollable modes decay, so not an issue. If the unobservable modes Ā22, Ā 44 are stable (have eigenvalues on the Left Half Plane, then, system is called detectable. The states z 2, z 4 decay to Eventually, they will have no effect on the output State can be reconstructed by ignoring the unobservable states (eventually. M.E. University of Minnesota 66 M.E. University of Minnesota 67

3 Relation to Transfer Function Poles For the system they are values of s C s.t. G(s becomes infinite. ẋ = Ax + Bu y = Cx + Du The transfer function from U(s Y (s is given by: G(s =C(sI A 1 B + D = Cadj(sI AB det(si A + D. Note that transfer function is not affected by similarity transform. From Kalman decomposition, it is obvious that G(s = C 1 (si Ã11 1 B1 + D where Ã11, B1, C1 correspond to the controllable and observable component. poles of G(s are clearly the eigenvalues of Ã11. Zeros For the SISO case, z C(complex numbers is a zero if it makes G(z =. Thus, zeros satisfy Cadj(zI AB + Ddet(zI A = assuming system is controllable and observable, otherwise, need to apply Kalman decomposition first. In the general MIMO case, a zero implies that there can be non-zero inputs U(s that produce output Y (s the is identically zero. Therefore, there exists X(z, U(z such that: zi AX(z =BU(z =CX(z+DU(z M.E. University of Minnesota 68 M.E. University of Minnesota 69

4 This will be true if and only if at s = z ( si A B rank C D is less than the normal rank of the matrix at other s. We shall return to the multivariable zeros when we discuss the effect of state feedback on the zero of the system. Pole-zero cancellation Cascade system with input/output (u 1,y 2 : System 1 with input/output (u 1,y 1 has a zero at α and a pole at β. ẋ 1 = A 1 x 1 + B 1 u 1 y 1 = C 1 x 1 System 2 with input/output (u 2,y 2 has a pole at α and a zero at β. ẋ 2 = A 2 x 2 + B 2 u 2 y 2 = C 2 x 2 Cascade interconnection: u 2 = y 2. Then 1. The system pole at β is not observable from y 2 2. The system pole at α is not controllable from u 1. M.E. University of Minnesota 7 M.E. University of Minnesota 71

5 The cascade system is given by: A = ( A1 ; B = B 2 C 1 A 2 Consider x = ( x1 x 2. ( B1 ; C = ( C 2 Use PBH test with λ = β. Does there exist x 1 and x 2 s.t. βi A 1 B 2 C 1 βi A 2 ( x1 =??? x C 2 2 Let x 1 be eigenvector associated with β for A 1. and β is a zero of system 2, PHB test shows that the mode β is not observable. Similar case for α mode in system 2 not controllable by u: PHB test for controllability for α mode: ( ( x T 1 x T αi A 1 B 1 2 = (??? B 2 C 1 αi A 2 Let x T 2 be the left eigenvector of A 1 s.t. x T 2 A 1 = αx T 2 Solve for x T 1 in the 1st column: x T 1 (αi A 1 =x T 2 B 2 C 1 x T 1 = x T 2 B 2 C 1 (αi A 1 1 Let x 2 =(βi A 2 1 B 2 C 2 x 1 (i.e. row in PBH test. Then since solve second With this choice of x 1 and x 2, x T 1 B 1 = x T 2 B 2 C 1 (αi A 1 1 B }{{} 1 = G 1 (s=α= Cx = C 2 x 2 =(C 2 (βi A 2 1 B 2 C }{{} 2 x 1 G 2 (s=β= Thus, by the PHB test, the α mode is not controllable. DO NOT CANCEL OUT UNSTABLE MODES!!! M.E. University of Minnesota 72 M.E. University of Minnesota 73

6 Degree of controllability / observability Formal definitions are black and white. Similarly, if A is stable, as T, =A T W o + W o A + C T C Degree of controllability and observability can be evaluated by the size of W r and W o with infinite time horizon: W r (, = lim = T W o (, = lim T T e A(T τ BB T e AT (T τ dτ e At BB T e AT t dt T e AT τ C T Ce Aτ dτ W r satisfies differential equation: d dt W r(,t=aw r (,t+w r (,ta T + BB T If all eigenvalues of A have strictly negative real parts (i.e. A is stable, then, as T, =AW r + W r A T + BB T Minimum norm control: To reach x( = x from x( =in infinite time, u = L T r W 1 c x min u( J(u =min u( u T (τu(τdτ = x T W 1 r x. If state x is difficult to reach, then x T Wr 1 x is large practically uncontrollable. Signal in an initial state, if x( = x, then, with u(τ y T (τy(τdτ = x T W o x. If x T W o x is small, the signal of x in the output y( is small, thus, it is hard to observe practically unobservable. M.E. University of Minnesota 74 M.E. University of Minnesota 75

7 Generally, one can look at the smallest eigenvalues W r and W o, the span of the associated eigenvectors will be difficult to control, or difficult to observe. Balanced Realization Reduce the number of states while minimizing effect on I/O response (transfer function State equations from distributed parameters system, P.D.E. via finite elements methods, etc. generate many states. If unobservable and uncontrollable remove. What about lightly controllable, lightly observable states? Issues: States can be lightly controllable, observable. but heavily States can be lightly observable, but heavily controllable. Both contribute to significant component to input/output (transfer function response. M.E. University of Minnesota 76 M.E. University of Minnesota 77

8 Idea: Exhibit states so that they are simultaneously lightly (heavily controllable and observable. Consider a stable system: ẋ = Ax + Bu y = Cx + Du If A is stable, controllability and observability grammians can be computed by solving the Lyapunov equations: =AW c + W c A T + BB T =A T W o + W o A + C T C If state x is difficult to reach, then x T Wc 1 x is large practically uncontrollable. If x T W o x is small, the signal of x in the output y( is small, thus, it is hard to observe practically unobservable. Generally, one can look at the smallest eigenvalues W c and W o, the span of the associated eigenvectors will be difficult to control, or difficult to observe. Transformation of grammians: (beware of which side T goes!: z = Tx Minimum norm control should be invariant to coordinate transformation: Therefore x T Wc 1 x = z T Wc,z 1 z = x T T T Wc,z 1 Tx W 1 c = T T W 1 c,z T W c,z = T W c T T. Similarly, the energy in a state transmitted to output should be invariant: x T W o x = zw o,z z W o,z = T T W o T 1. Theorem: (Balanced realization If the linear time invariant system is controllable and observerable, then, there exists an invertible transformation T R n n s.t. W o,z = W c,z. M.E. University of Minnesota 78 M.E. University of Minnesota 79

9 The balanced realization algorithm: 5. This gives W r,z = W o,z =Σ. 1. Compute controllability (reachability and observability grammians: >> W r = gram(sys, c ; >> W o = gram(sys, o 2. Write W o = R T R [One can use SVD] >> [U,S,V]=svd(W o ; >> R = sqrtm(s*v ; 3. Diagonalize RW r R T [via SVD again]: RW r R T = UΣ 2 U T, where UU T = I and Σ is diagonal. >> [U 1,S 1,V 1 ]=svd(r*w r *R ; >> Σ = sqrtm(s 1 ; 4. Take T =Σ 1 2U T R >> T = inv(sqrtm(σ*u 1 *R; >> W r,z = T W r T, >> W o,z = inv(t W o inv(t Proof: Direct substitution! Once W r,z and W o,z are diagonal, and equal, one can decide to eliminate states that correspond to the smallest few entries. Do not remove unstsable states from the realization since these states need to be controlled (stabilized It is also possible to maintain some D.C. performance. Consider the system in the balanced realization form: ( ( ( ( d z1 A11 A = 12 z1 B1 + u dt z 2 A 21 A 22 z 2 B 2 y = C 1 z 2 + C 2 z 2 + Du where z 1 R n 1 and z 2 R n 2, and n 1 + n 2 = n. Suppose that the z 1 are associated with M.E. University of Minnesota 8 M.E. University of Minnesota 81

10 the simultaneously lightly controllable and lightly observable mode. Naive trunction - Remove z 2 : ż 1 = A 11 z 1 + B 1 u y = C 1 z 1 + Du However, this does not retain the D.C. (steady state gain from u to y (which is important from a regulation application point of view. To maintain the steady state response, instead of eliminating z 2 completely, replace z 2 by its steady state value: In steady state: ż 2 ==A 21 z 1 + A 22 z2 + B 2 u z2 = A 1 22 (A 21z 1 + B 2 u Truncation that maintains steady state gain: ż 1 = A 11 z 1 + A 12 z2 + B 1 u y = C 1 z 1 + C 2 z2 + Du ż 1 = [ A 11 A 12 A 1 22 A ] 21 }{{} A r + [ B1 A 12 A 1 22 B ] 2 u }{{} y = [ C 1 C 2 A 1 22 A ] [ 21 z }{{} 1 + D C2 A 1 22 B ] 2 u }{{} C r D r So that the truncated system z 1 R n 1, ż 1 = A r z 1 + B r u y = C r z 1 + D r u with have the same steady state response as the original system. B r This is feasible if A 22 eigenvalue. is invertible ( is not an M.E. University of Minnesota 82 M.E. University of Minnesota 83