Controllability and Observability

Transcription

1 Chapter 3 Controllability and Observability In this chapter, we study the controllability and observability concepts. Controllability is concerned with whether one can design control input to steer the state to arbitrarily values. Observability is concerned with whether without knowing the initial state, one can determine the state of a system given the input and the output. 3.1 Some linear mapping concepts The study of controllability and observability for linear systems essentially boils down to studying two linear maps: the reachability map L r which maps, for zero initial state, the control input u( to the final state x(t 1 ; and the observability map, L o which maps, for zero input, from the initial state x(t to the output trajectory y(. The range and the null spaces of these maps respectively are critical for their studies. Let L : X(= R p Y(= R q be a linear map: i.e. R(L Y denotes the range of L, given by: N(L X denotes the null space of L, given by: L(αx 1 + βx 2 = α(lx 1 + β (Lx 2 R(L = {y Y y = Lx for some x X } N(L = {x X Lx = } Note: Both R(L and N(L are linear subspaces, i.e. for scalar α, β, y 1, y 2 R(L (αy 1 + βy 2 R(L x 1, x 2 N(L (αx 1 + βx 2 R(L L is called onto or surjective if R(L = Y (everything is within range. L is called into or one-to-one (cf. many-to-one if N(L = {}. Thus, Lx 1 = Lx 2 x 1 = x 2 27

2 28 c Perry Y.Li The Finite Rank Theorem Proposition : Let L R n m, so that L : u R m y R n R(L = R(LL T N(L = N(L T L 3. Any point u R n can be written as u = L T y 1 + u n R n ;where y 1 R n, Lu n =. In other words, u can always be written as a sum of u 1 R(L T and u n N(L. 4. Similarly, any point u R n can be written as Note: y = Lu 1 + y n R n ; where u 1 R m, L T y n =. In other words, u can always be written as a sum of y 1 R(L and y n N(L T. If L R n m and m > n (short and fat, then LL T R n n is both short and thin. To check whether the system is controllable we need to check if LL T is full rank - e.g. check its determinant. If m < n (tall and thin, then L T L R n n is also both short and thin. Checking N(L T L will be useful when we talk about observerability. 3.2 Controllability Consider a state determined dynamical system with a transition map s(t 1, t, x, u( such that This can be continuous time or discrete time. x(t 1 = s(t 1, t, x(t, u(. Definition (Controllability : The dynamical system is controllable on [t, t 1 ] if initial and final states x, x 1, u( so that s(t 1, t, x, u = x 1. It is said to controllable at t if x, x 1, t 1 t and u( U so that s(t 1, t, x, u = x 1 A system that is not controllable probably means the followings: If state space system is realized from input-output data, then the realization is redundant. i.e. some states have no relationship to either the input or the output. If states are meaningful (physical variables that need to be controlled, then the design of the actuators are deficient. The effect of control is limited. There is also a possibililty of instability

3 ME 8281:Advanced Control Systems, F1, F2, S6, S8 29 Remarks: The followings are equivalent: 1. A system is controllable at t = t. 2. The system can be transferred from any state at t = t to any other state in finite time. 3. For each x Σ, s(, t, x, : [t, u( s(t 1, t, x, u( is surjective (onto. Controllability does not say that x(t remains at x 1. e.g. for ẋ = Ax + Bu to make x(t = x 1 t t, it is necessary that Ax 1 Range(B. This is generally not true. Example - An Uncontrollable System An linear actuator pushing 2 masses on each end of the actuator in space. m 1 ẍ 1 = u; m 2 ẍ 2 = u State: X = [x 1, ẋ 1, x 2, ẋ 2 ] T. Action and reaction are equal and opposite, and act on different bodies. Momentum is conserved: m 1 ẋ 1 + m 2 ẋ 2 = constant = m 1 ẋ 1 (t + m 2 ẋ 2 (t E.g. if both masses are initially at rest: it is not possible to choose u( to make ẋ 1 =, ẋ 2 = Effects of feedback on controllability Consider a system ẋ = f(t, x(t, u(t. Suppose that the system is under memoryless (static state feedback: with v(t being the new input. u(t = v(t g(t, x(t A system is controllable if and only if the system under memoryless feedback with v(t as the input is controllable.

4 3 c Perry Y.Li Proof: Suppose that the original system is controllable. Given a pair of initial and final states, x and x 1, suppose that the control u( transfers the state from x at t to x 1 at t 1. For the new system, which has also x(t as the state variables, we can use the control v(t = u(t + g(t, x(t to transfer the state from x at t to x 1 at t 1. Therefore, the new system is also controllable. Suppose that the feedback system is controllable, and v( is the control that can achieve the desired state transfer. Then for the original system, we can use the control u(t = v(t g(t, x(t to achieve the same state transfer. Thus, the original system is also controllable. Consider now the system under dynamic state feedback: ż(t = α(t, x(t, z(t u(t = v(t g(t, x(t, z(t where v(t is the new input and z(t is the state of the controller. If a system under dynamic feedback is controllable then the original system is. The converse is not true, i.e. the original system is controllable does not necessarily imply that the system under dynamic state feedback is. Proof: The same proof for the memoryless state feedback works for the first statement. For the second statement, we give an example of the controller: ż = z; u(t = v(t g(t, x(t, z(t. The state of the new system is (x, z but no control can affect z(t. Thus, the new system is not controllable. These results show that feedback control cannot make a uncontrollable system become controllable, but can even make a controllable system, uncontrollable. To make a system controllable, one must redesign the actuators. 3.3 Controllability of Linear Systems Continuous time system: ẋ(t = A(tx(t + B(tu(t x(t 1 = Φ(t 1, t x + The reachability map on [t, t 1 ] is defined to be: L r,[t,t 1 ](u( = t t x R n Φ(t 1, tb(tu(tdt. Φ(t 1, tb(tu(tdt Thus, it is controllable on [t, t 1 ] if and only if L r,[t,t 1 ](u( is surjective (onto. Notice that L r,[t,t 1 ] determines the set of states that can be reached from the origin at t = t 1. The study of the range space of the linear map: is central to the study of controllability. L r,[t,t 1 ] : {u( } R n

5 ME 8281:Advanced Control Systems, F1, F2, S6, S8 31 Proposition If a continuous time, possibly time varying, linear system is controllable on [t, t 1 ] then it is controllable on [t, t 2 ] where t 2 t 1. Proof: To transfer from x(t = x to x(t 2 = x 2, define x 1 = Φ(t 1, t 2 x 2 Suppose that u [t,t 1 ] transfers x(t = x to x(t 1 = x 1. Choose { u (t t [t, t 1 ] u(t = t (t 1, t 2 ] This result does not hold for t < t 2 < t 1. e.g. The system { t [t, t 2 ] ẋ(t = B(tu(t B(t = I n t (t 2, t 1 ]. is controllable on [t, t 1 ] but x(t 2 = x(t for any control u( Discrete time system. x(k + 1 = A(kx(k + B(ku(k, x(k 1 = Φ(k 1, k x(k + k 1 1 x R n, u R m k=k Φ(k 1, k + 1B(ku(k The transition function of the discrete time system is given by: x(k 1 = s(k, k 1, x(k, u( = Φ(k 1, k x(k + k 1 1 The reachability map of the discrete time system can be written as: where L r (k, k 1 R n (k 1 k m, U = k=k Φ(k 1, k + 1B(ku(k (3.1 Φ(k f, k i = Π k f 1 k=k i A(k (3.2 k 1 1 L r,[k,k 1 ](u(k,..., u(k 1 1 = Φ(k 1, k + 1B(ku(k = L r (k, k 1 U k=k u(k.. u(k 1 1 Thus, the system is controllable on [k, k 1 ] if and only if L r,[k,k 1 ] is surjective, which is true if and only if L r (k, k 1 has rank n Proposition Suppose that A(k is nonsingular for each k 1 k < k 2, then the discrete time linear system is controllable on [k, k 1 ] implies that it is controllable on [k, k 2 ] where k 2 k 1. Proof: : The proof is similar to the continuous case. However for a discrete time system, we need A(k nonsingular for k 1 k < k 2 to ensure that Φ(k 2, k 1 is invertible. Q.E.D.

6 32 c Perry Y.Li Example Consider a unit point mass under control of a force: ( ( ( d x 1 x = + dt ẋ ẋ ( u 1 This is the same as ẍ = u. Suppose that x( = ẋ( =, we would like to translate the mass to x(t = 1, ẋ(t =. The reachability map L : u( x(t is x(t = T Φ(T, tb(tu(tdt Let use try to solve this problem using piecewise constant control: u t < T/1 u 1 T/1 t < 2T/1 u(t =. 9T/1 t T and find U = [u 1, u 2,...,u 1 ] T. The reachability map becomes: u 1 u 1 u 2 u 1 x(t = ( L 1 L 2... L 1 } {{ }. L r where [ ] T/1 L 1 = Φ(T, tb(tdt = T ( T 1 1 [ ] 2T/1 L 2 = Φ(T, tb(tdt = T 1 T/1, (.9T 1,. L 1 = T 1 (.1T 1 Whether the matrix denoted by L r R 2 1 is surjective (onto (i.e. R(L r = R 2 determines whether we can achieve our task using piecewise constant control. In our case, let T = 1 for convenience, ( L r = and L r L T r = (

7 ME 8281:Advanced Control Systems, F1, F2, S6, S8 33 which is non-singular (full rank. Thus, from the finite rank theorem R(L r = R(L r L T r, the particle transfer problem is solvable using piecewise constant control. If we make the number of pieces in the piecewise constant control larger and larger, then the multiplication of L r by L T r becomes an integral: L r L T r t Φ(t 1, tb(tb(t T Φ(t 1, t T dt. Therefore for the continuous time linear system is controllable over the interval [t, t 1 ] if and only if the matrix, t Φ(t 1, tb(tb(t T Φ(t 1, t T dt (3.3 is full rank. 3.4 Reachability Grammian The matrix W r = L r L T r R n n is called the Reachability Grammian for the interval that L r is defined. For the continuous time system ẋ = A(tx + B(tu the Reachability Grammian on the time interval [t, t 1 ] is defined to be: W r,[t,t 1 ] = For time invariant systems, t W r ([t 1, t ] = L r L r = = For the discrete time system, Φ(t 1, tb(tb(t T Φ(t 1, t T dt t t 1 t e Aτ BB T e AT τ dτ. e A(t 1 t BB T e AT (t 1 t dt x(k + 1 = A(kx(k + B(ku(k, the Reachability Grammian on the interval [k, k 1 ] is k 1 1 k=k Φ(k 1, k + 1B(kB(k T Φ(k 1, k + 1 T. From the finite rank theorem, we know that the controllability of a system on [t, t 1 ] (or on [k, k 1 ] is equivalent to Reachability Grammian on the same time interval being full rank.

8 34 c Perry Y.Li Reduction theorem: Controllability in terms of Controllability to Zero Thus far, we have been deciding controllability by considering the reachability map L r which concerns what states we are able to steer to from x(t =. The reduction theorem allows us to formulate the question in terms of whether a state can be steered to x(t f = in finite time. Theorem The followings are equivalent for a linear time varying differential system: 1. It is controllable on [t, t 1 ]. ẋ = A(tx + B(tu 2. It is controllable to on [t, t 1 ], i.e. x(t = x, there exists u(τ, τ [t, t 1 such that final state is x(t 1 =. 3. It is reachable from x(t =, i.e. x(t 1 = x f, there exists u(τ, τ [t, t 1 such that for x(t =, the final state is x(t 1 = x f. Proof: Clearly, (1 (2 and (3. To prove (2 (1 and (3 (1, use x(t 1 = Φ(t 1, t x(t + Φ(t 1, τb(τu(τdτ. t and Φ(t 1, t is invertible, to construct the appropriate x(t and x(t f. Controllability map: L c,[t,t 1 ] The controllability (to zero map on [t, t 1 ] is the map between u( to the initial state x such that x(t 1 =. L c,[t,t 1 ](u( = Φ(t, tb(tu(tdt t Proof: Notice that = x(t 1 = Φ(t 1, t x + Φ(t 1, τb(τu(τdτ t x = Φ(t 1, t 1 Φ(t 1, τb(τu(τdτ x = x = t t t Φ(t, t 1 Φ(t 1, τb(τu(τdτ Φ(t, τb(τu(τdτ R(L c,[t,t 1 ] = Φ(t 1, t R(L r,[t,t 1 ] These two ranges are typically not the same for time varying systems. However, one is full rank if and only if the other is, since Φ(t 1, t is invertible.

9 ME 8281:Advanced Control Systems, F1, F2, S6, S8 35 For time invariant continuous time systems, R(L c,[t,t 1 ] = R(L r,[t,t 1 ]. The above two statements are not true for time invariant discrete time systems. Can you find counter-examples? Because of the reduction theorem, the linear continuous time system is controllable on the interval [t, t f ] if R(L c,[t,t 1 ] = R n. From the finite rank theorem, R(L c,[t,t 1 ] = R(L c,[t,t 1 ]L c,[t,t 1 ] = R n. The controllability-to-zero grammian on the time interval [t, t 1 ] is defined to be: For time invariant system, W c = L c L c = W c = L c L c = t Φ(t, tb(tb(t T Φ(t, t T dt t e A(t t BB T e AT (t t dt = e Aτ BB T e AT τ dτ. t 1 t Since controllability of a system on [t, t 1 ] is equivalent to controllability to which is determined by the controllability map being surjective, it is also equivalent to controllability grammian on the same time interval [t, t 1 ] being full rank. 3.5 Minimum norm control Formulation The Controllability Grammian is related to the cost of control. Given x(t = x, find a control function or sequence u( so that x(t 1 = x 1. Let x d = x 1 Φ(t 1, t x. Then we must have x d = L r,[t,t 1 ](u where L r,[t,t 1 ] : {u( } R is the reachability map which, for continuous time system is: and for discrete time system: where C R n (k 1 k m, U = L r,[t,t 1 ](u = k 1 1 t Φ(t 1, tb(tu(tdt. L r,[k,k 1 ](u( = Φ(k 1, k + 1B(ku(k = C(k, k 1 U k=k u(k.. u(k 1 1 Let us focus on the discrete time system. Since x d Range(L r,[k,k 1 ] for solutions to exist, we assumerange(l r,[k,k 1 ] = R n. This implies that W r,[k,k 1 ] is invertible. Generally there are multiple solutions. To resolve the non-uniqueness, we find the solution so that k 1 1 J(U = 1 u(k T u(k = U T U 2 k=k is minimized. Here U = [u(k ; u(k + 1;...;u(k 1 1].

10 36 c Perry Y.Li Least norm control solution This is a constrained optimization problem (with J(U as the cost to be minimized, and L r U x d = as the constraint. It can be solved using the Lagrange Multiplier method of converting a constrained optimization problem into an unconstrained optimization. Define an augmented cost function, with the Lagrange multipliers λ R n The optimal given by (U, λ must satisfy: J (U, λ = J(U + λ T (L r U x d J U (U, λ = ; The second condition is just the constraint: L r U = x d. This means that L T r λ + U = ; Solving, we get: The optimal cost of control is: J λ (U, λ = L r U = x d λ = (L r L T r 1 x d = Wr 1 x d U = L T r λ = L T r Wr 1 x d. J(U = x T d W 1 r L r L T r W 1 r x d = x T d W r 1 x d Thus, the inverse of the Reachability Grammian tells us how difficult it is to perform a state transfer from x = to x d. In particular, if W r is not invertible, for some x d, the cost is infinite Geometric interpretation of least norm solution Geometrically, we can think of the cost as J = U T U, i.e. the inner product of U with itself. In notation of inner product, J =< U, U > R The advantage of the notation is that we can change the definition of inner product, e.g. < U, V > R = U T RV where R is a positive definite matrix. The usual inner (dot product has R = I. We say that U and V are normal to each other if < U, V > R =. Any solution that satisfies the constraint must be of the form where L r U p = x f is any particular solution. (U U p Null(L r Claim: Let U be the optimal solution, and U is any solution that satisfies the constraint. Then, (U U U, i.e. < (U U, U > R = which is the normal equation for the least norm solution problem. Proof: Direct substitution. (U U T R U R 1 U LT pc ( Lr R 1 1 U LT r xf = (L r (U U T ( L r R 1 1 U LT r xf =.

11 ME 8281:Advanced Control Systems, F1, F2, S6, S Open-loop least norm control Applying the above result to linear continuous time system using standard norms, i.e. J = t u T (τu(τdτ. (L r,[t,t 1 ] (t = B T (tφ(t, t 1 T [ ] W r,[t,t 1 ] = L r L r = Φ(t 1, τb(τb T (τφ(t 1, τ T dτ t u opt (t = B(t T Φ(t, t 1 T W 1 r,[t,t 1 ] (x 1 Φ(t 1, t x (3.4 Eq, (3.4 solves u opt ( in a batch form given x, i.e. it is openloop Recursive formulation Openloop control does not make use of feedback. It is not robust to disturbances or model unceratinty. Suppose now that x(t is measured. Conceptually, we can solve (3.4 for u opt (t and let t = t. Computing W r,[t,t1 ] is expensive. Try recursion: u opt (t = B(t T Φ(t, t 1 T W 1 r,[t,t 1 ] (x 1 Φ(t 1, tx(t ( ( = B(t T Φ(t, t 1 T W 1 r,[t,t 1 ] x 1 B(t T Φ(t, t 1 T W 1 r,[t,t 1 ] Φ(t 1, t x(t For any invertible matrix Q(t R n n, Q(tQ 1 (t = I, therefore Since W r,[t,t1 ] = d dt Q 1 (t = Q 1 (t Q(tQ 1 (t t Φ(t 1, τb(τb(τ Φ(t 1, τ dτ d dt W r,[t,t 1 ] = Φ(t 1, tb(tb(t Φ(t 1, t Thus we can solve for W 1 r,[t,t 1 ] dynamically by integrating on-line d dt W 1 1 r,[t,t 1 ] = Wr,[t,t 1 ] Φ(t 1, tb(tb(t Φ(t 1, t W 1 r,[t,t 1 ] Issue: Does not work well when t t 1 because W r,[t,t1 ] can be close to singular.

12 38 c Perry Y.Li Optimal cost Optimal cost as a function of initial time t Since W r,[t,t f ] = tf t Φ(t f, τb(τb(τ Φ(t f, τ dτ and the integrand is a positive semi-definite term, for each v R n, v T W 1 r,[t,t f ] v decreases as t decreases, i.e. given more time, the cost of control decreases. To see this, fix the final time t f, and consider the minimum cost function, E min (x f, t = x T f W 1 r,[t,t f ] x f and consider Since d dt f W 1 r,[t,t f ] d dt E min (x f, t f = v T d dt f W 1 r,[t,t f ] x f 1 d = Wr,[t,t f ] W dt r,[t,t f ]W 1 r,[t,t f f ] { } = W 1 r,[t,t f ] Φ(t f, t B(t B(t Φ(t f, t W 1 r,[t,t f ] is negative semi-definite, d dt E min (v, t f. Hence, the optimal control increases as the initial time decreases, increasing the allowable time interval. In other words, the set of states reachable with a cost less than a given value grows as t decreases. If the system is unstable, in that, Φ(t f, t becomes unbounded as t, then one can see that W r,[t,t f ] also becomes unbounded. This means that for some x f, E min (x f, t and t f. This means that some directions do not require any cost of control. Which direction do these correspond to? For time invariant system, decreasing t is equivalent to increasing t f. Thus for time-invariant system, increasing the time interval increases W r, thus decreasing cost. Optimal cost as a function final state x f Since W r,[t,t f ] is symmetric and positive semi-definite, has n orthogonal eigenvectors, V = [v 1,...,v n ], and associated eigenvalues Λ = diag(λ 1,...,λ n, such that V V T = I. Then W r,[t,t f ] = V ΛV T W 1 r,[t,t f ] = V diag(λ 1 1,...,λ 1 n V T This shows that most expensive direction to transfer to is the eigenvector associated with the minimum λ i, and the least expensive direction is the eigenvector associated with the maximum λ i 3.6 Linear Time Invariant (LTI Continuous Time Systems Consider first the LTI continuous time system: ẋ = Ax + Bu y = Cx (3.5

13 ME 8281:Advanced Control Systems, F1, F2, S6, S8 39 Note: Because of time invariance, the system is controllable (observable at t means that it is controllable (observable at any time. We will develop tests for controllability and observability by using directly the matrices A and B. The three tests for controllability of system (3.5 is given by the following theorem. Theorem For the continuous time system (3.5, the followings are equivalent: 1. The system is controllable over the interval [, T], for some T >. 2. The system is controllable over any interval [t, t 1 ] with t 1 > t. 3. The reachability grammian, W r,t := = T T Φ(T, tb(tb(t T Φ(T, t T dt e At BB e A t dt is full rank n. (This test works for time varying systems also 4. The controllability matrix C := ( B AB A 2 B A n 1 B has rank n. Notice that the controllability matrix has dimension n nm where m is the dimension of the control vector u. 5. (Belovich-Popov-Hautus test For each s C, the matrix, ( si A B has rank n. Note that rank of (si A, B is less than n only if s is an eigenvalue of A. We need the Cayley-Hamilton Theorm to prove this result: Theorem (Cayley Hamilton Let A R n n. Consider the polynomial (s = det(si A Then (A =. Notice that if λ i is an eigenvalue of A, then (λ i =. Proof: This is true for any A, but is easy to see using the eigen decomposition of A when A is semi-simple. A = TΛT 1 where Λ is diagonal with eigenvalues on its diagonal. Since ψ(λ i = by definition, (A = Tψ(ΛT 1 =.

14 4 c Perry Y.Li Proof: Controllability test (1 (3: We have already seen before that controllability over the interval [, T] means that the range space of the reachability map, L r (u = T must be the whole state space, R n. Notice that W r,t := T e A(T τ Bu(τdτ e At BB e A t dt = L r L r where L r is the adjoint (think transpose of L r. From the finite rank linear map theorem, R(L r = R(L r L r but, L r L r is nothing but W r,t. (3 (4: We will show this by showing not (4 not (3. Suppose that (4 is not true. Then, there exists a 1 n vector v T so that, Consider now, Since v T B = v T AB = v T A 2 B = v T A n 1 B =. v T W r,t v = T e At = I + At + A2 t 2 2 v T e At B 2 2dt. + An 1 t n 1 n 1! + and by the Cayley Hamilton Theorem, A k is a linear combination of I, A,... A n 1, therefore, v T e At B = for all t. Hence, v T W r,t v = or W r,t is not full rank. (3 (2: We will show this by showing not (2 not (3. If (2 is not true, then there exists 1 n vector v T so that v T W r,t v = T v T e At B 2 2dt =. Because e At is continuous in t, this implies that v T e At B = for all t [, T]. Hence, the all time derivatives of v T e At B = for all t [, T]. In particular, at t =, v T e At B t= = v T B = v T d dt eat B t= = v T AB = v T d2 dt 2eAt B t= = v T A 2 B =. v T dn 1 dt n 1eAt B t= = v T A n 1 B =

15 ME 8281:Advanced Control Systems, F1, F2, S6, S8 41 Hence, v T( B AB A 2 B A n 1 B = ( i.e. the controllability matrix of (A, B is not full rank. (4 (5: We will show not (5 implies not (4. Suppose (5 is not true so that there exists a 1 n vector v, and λ C, v T (λi A =, v T B =. for some λ. Then v T A = λv T, v T A 2 = λ 2 v T etc. Because v T B = by assumption, we have v T AB = λv T B =, v T A 2 B = λ 2 v T B = etc. Hence, Therefore the controllability is not full rank. v T B = v T AB = v T A 2 B = v T A n 1 B =. The proof of (4 (3 requires the 2nd representation theorem, and we shall leave it afterward. Note that (2 (1 is obvious. To see that (1 (2, notice that the controllability matrix C does not depend on T, so controllability does not depend on duration of time interval. Since the system is time invariant, it does not depend on the initial time (t. Proof of (4 (3 - Optional To prove (4 (3, we need the so called 2nd Representation Theorem. It is included here for completeness. We will not go through this part. Definition Let A : R n R n and W R n a subspace with the property that for each w W, Aw W. We say that W is A invariant. Theorem (2nd Representation theorem Let A : R n R n is a linear map (i.e. a matrix W R n be an A invariant k dimensional subspace of R n. There exists a nonsingular T = ( e 1 e 2 e n s.t. e 1,, e k form a basis of W, in the basis of e 1, e 2,, e n, A is represented by: Ã 11 R k k where k = dim W. T 1 AT = (Ã11 Ã 12 Ã 22 For any vector B W, the representation of B is given by B = T ( B.

16 42 c Perry Y.Li The point of the theorem is that A can be put in a form with in strategic locations (Ã21 are zeros, similarly for any vector B W ( B 2 are zeros. Proof: Take e 1,, e k to be a basis of W, and choose e k+1, e n to complete the basis of R n. Because Ae i W for each i = 1, k, Ae i = k α li e l. l=1 Therefore, α li are the coefficients of Ã11 and the coefficients of Ã21 are all zero. Let B W, then k B = β l e l so that β l are simply the coefficients in B 1 and the coefficients of B 2 are zeros. l=1 Proof of (4 (3 in controllability test: Notice that A ( B AB A 2 B A n 1 B = ( AB A 2 B A n 1 B A n B therefore, by the Cayley Hamilton Theorem, the range space of the controllability matrix is A invariant. Suppose that (3 is not true, so that the rank of ( B AB A 2 B A n 1 B is less than n. Since the controllability matrix is A-invariant, by the 2nd representation theorem, there is an invertible matrix T so that in the new coordinates, z = T 1 x, ( (Ã11 Ã ż = 12 B z + u (3.6 Ã 22 where B = T ( B, and A = T (Ã11 Ã 12 Ã 22 and the dim of Ã22 is non-zero (it is n - rank(c. Let v T 2 be a left eigenvector of Ã22 so that T 1. v2 T (λi Ã22 = for some λ C. Define v T := ( v2 T T 1. Evaluating v T B, v T (λi A gives v T B = ( v2 T B T T( 1 = v T (λi A = λ ( v2 T T 1 (λtt 1 TÃT 1 = ( v2 T (λi ÃT 1 =. This shows that [λi A, B] has rank less than n. Remarks:

17 ME 8281:Advanced Control Systems, F1, F2, S6, S Notice that the controllability matrix is not a function of the time interval. Hence, if a LTI system is controllable over some interval, it is controllable over any (non-zero interval. c.f. with result of linear time varying system. 2. Because of the above fact, we often say that the pair (A, B is controllable. 3. Controllability test can be done by just examining A and B without computing the grammian. The test in (4 is attractive in that it enumerates the vectors in the controllability subspace. However, numerically, since it involves power of A, numerical stability needs to be considered. 4. The PBH Test in (5, or the Hautus test for short, involves simply checking the condition at the eigenvalues. It is because for (si A, B to have rank less than n, s must be an eigenvalue. 5. The range space of the controllability matrix is of special interests. It is called the controllable subspace and is the set of all states that can be reached from zero-initial condition. This is A invariant. 6. Using the basis for the controllable subspace as part of the basis for R n, the controllability property can be easily seen in ( Linear Time Invariant (LTI discrete time system For the discrete time system (3.7, x(k + 1 = Ax(k + Bu(k y(k = Cx(k (3.7 where x R n, u R m, the conditions for controllability is fairly similar except that we need to consider controllability over period of length greater than n. First, recall that the reachability map for a linear discrete time system L r,[k,k1 ] : u( s(k, k, x =, u( is given by: L r,[k,k 1 ][u( ] = k 1 Πj=i+1 k 1 i=k Theorem For the discrete time system (3.7, the followings are equivalent: A(jB(iu(i ( The system is controllable over the interval [, T], with T n (i.e. it can transfer from any arbitrary state at k = to any other arbitrary state at k = T. 2. The system is controllable over the interval [k, k 1 ], with k 1 k n 3. The controllability grammian, is full rank n. 4. The controllability matrix T 1 W r,t := A l BB A l l= ( B AB A 2 B A n 1 B has rank n. Notice that the controllability matrix has dimension n nm where m is the dimension of the control vector u.

18 44 c Perry Y.Li 5. For each z C, the matrix, ( zi A B has rank n. Proof: The proof of this theorem is exactly analogous to the continuous time case. However, in the case of the equivalence of (2 and (3, we can notice that u(k 1 x(k = A k x( + [B, AB, A 2 B, A n 1 u(k 2 B]. u( so clearly one can transfer to arbitrary states at k = n iff the controllability matrix is full rank. Controllability does not improve for T > n is again the consequence of Cayley-Hamilton Theorem. 3.8 Observability The observability question deals with the question of while not knowing the initial state, whether one can determine the state from the input and output. This is equivalent to the question of whether one can determine the initial state x( when given the input u(t, t [, T] and the output y(t, t [, T]. Definition A state determined dynamical system is called observable on [t, t 1 ] if for all inputs, u(τ, and outputs y(τ, τ [t, t 1 ], the state x(t = x can be uniquely determined. For observability, the null space of the observability map on the interval [t, t 1 ], given by: L o : x y( = C( Φ(, t x, y(τ = C(τΦ(τ, t x Φ(τ, x, τ [t, t 1 ] must be trivial (i.e. only contains. Otherwise, if L o (x n (t = y(t =, for all t [, T], then for any α R, y(t = C(tΦ(t, x = Φ(t, (x + αx n. Recall from the finite rank theorem that if the observability map L o is a matrix, then the null space of L o, N(L o = N(L T o L o. For continuous time system, L o can be thought of as an infinitely long matrix, then the matrix product operation in L T o L o becomes an integral, and is written as L ol o where L denotes the adjoint of the L. The observability Grammian is given by: W o,[t,t 1 ] = L ol o = t Φ(t, t T C T (tc(tφ(t, t R n n where L o is the adjoint (think transpose of the observability map.

19 ME 8281:Advanced Control Systems, F1, F2, S6, S Least square estimation Formulation and solution Consider the continuous time varying system: ẋ(t = A(tx(t y(t = C(tx(t. The observability map on [, t 1 ] is given by: τ [, t 1 ] L o : x( y(τ = C(τΦ(τ, x( (3.9 The so-called Least squares estimate of x( given the output y(τ, τ [, t 1 ] is: ˆx( t 1 = arg min x e T (τe(τdτ. The least squares estimation problem can be cast into a set of linear equations. For example, we can stack all the outputs y(t, t [, t 1 ] into a vector Y = [Y (,..., Y (.1,...,Y (.2,..., Y (t 1 ] T R pp, and represent L o as a matrix R pp n, then, Y = L o x + E; x o = argmin x E T E = (L T o L o 1 L T o Y Here, E = [E(,...,E(t 1 ] T = R pp is the modeling error or noise. The optimal estimate x uses the minimum amount of noise E to explain the data Y. Notice that the resulting E satisfies E T L o =, i.e. E is normal to Range(L o. Using the observability map for the LTI continuous time system (3.9 as L o, ˆx( t 1 = Wo 1 (t 1 W o (t 1 = L ol o = Φ T (τ, C T (τy(τdτ Φ T (τ, C T (τc(τφ(τ, dτ. One can also find estimates of the states at time t 2 given data up to time t 1 : ˆx(t 2 t 1 = Φ(t 2, ˆx( t 1 This refers to three classes of esimation problems: t 2 < t 1 : Filtering problem t 2 > t 1 : Prediction problem t 2 = t 1 : Observer problem The estimate ˆx(t t is especially useful for control systems as it allows us to effective do state feedback control.

20 46 c Perry Y.Li Recursive Least Squares Observer The least squares solution computed above is expensive to compute: ˆx(t t = Φ(t, Wo 1 (t W o (t = L ol o = t t Φ T (τ, C T (τy(τdτ Φ T (τ, C T (τc(τφ(τ, dτ. It is desirable if the estimate ˆx(t t can be obtained recursively via a differential equation. where d dt t d 1 ˆx(t t = A(tΦ(t, Wo (t Φ T (τ, C T (τy(τdτ dt } {{ } + Φ(t, d dt W o 1 (t + Φ(t, W 1 o t ˆx(t t Φ T (τ, C T (τy(τdτ (tφ T (t, C T (ty(t Now by differentiating W o (two 1 (t = I, we have: d dt W o 1 ˆx(t t = A(tˆx(t t Φ(t, W 1 o P(t satisfies: (t = W 1 o (t dw o dt 1 (two (t d dt W o(t = Φ T (t, C T (tc(tφ(t, (tφ T (t, C T (tc(tˆx(t t + Φ(t, Wo 1 (tφ T (t, C T (ty(t = A(tˆx(t t Φ(t, Wo 1 (tφ T (t, C T (t[c(tˆx(t t y(t] (3.1 } {{ } } {{ } P(t ŷ(t t P(t = A(tP(t + P(tA T (t P(tC T (tc(tp(t Typically, we would initialize P( to be a large non-singular matrix, e.g. P( = α I, where α is large. In the recursive least squares observer in (3.1, the first term is an openloop prediction term, and the second term is called an output injection where an output prediction error is used to correct the estimation error. We can think of the observer feedback gain as: Covariance Windup L(t = P(tC T (t Let us rewrite P(t as follows: [ t 1 P(t = Φ(t, Φ T (τ, C T (τc(τφ(τ, dτ] Φ T (t, [ t 1 = Φ T (τ, tc T (τc(τφ(τ, tdτ] =: K 1 (t

21 ME 8281:Advanced Control Systems, F1, F2, S6, S8 47 The matrix K(t := P 1 (t is known as the Information Matrix and satisfies: K = C T (tc(t A T (tk(t K(tA(t Notice that W o (t increases with time. Thus, W o 1(t decreases with time, as would K 1 (t. This means that the observer gain L(t will decrease. The observer will asymptotically rely more and more on open loop estimation: Forgetting factor d ˆx(t t = A(tˆx(t t dt Forget old information (τ far away from current time: Choose λ > : K(t = t Φ T (τ, tc T (τc(τφ(τ, te λ(t τ dτ; K = C T (tc(t A T (tk(t K(tA(t λk Covariance reset If K(t becomes large, reset it to small number. Check maximum singular value of K(t, σk(t. If σk(t λ max, K(t = ǫi, i.e. P(t = I/ǫ. 3.1 Observability Tests The observability question deals with the question of whether one can determine what the initial state x( is, given the input u(t, t [, T] and the output y(t, t [, T]. For observability, the null space of the observability map, L o : x y( = Φ(, x, τ [, T], must be trivial (i.e. only contains. Otherwise, if L o (x n (t = y(t =, for all t [, T], then for any α R, y(t = Φ(t, x = Φ(t, (x + αx n. Hence, one cannot distinguish the various initial conditions from the output. Just like for controllability, it is inconvenient to check the rank (and the null space of the L o which is tall and thin. We can instead check the rank and the null space of the observability grammian given by: W o,t = L ol o where L o is the adjoint (think transpose of the observability map. Proposition Null(L o = Null(L ol o. Proof:

22 48 c Perry Y.Li Let x Nul(L o so, L o x =. Then, clearly, L T o L o x = L T. Therefore, Null(L o Null(L ol o. Let x Nul(L ol o. Then, x T L ol o x =. Or, (L o x T (L o x =. This can only be true if L o x =. Therefore, Null(L ol o Null(L o Observability Tests for Continuous time LTI systems Theorem For the LTI continuous time system (3.5, the followings are equivalent: 1. The system is observable over the interval [, T] 2. The observability grammian, is full rank n. 3. The observability matrix C CA CA 2. CA n 1 W o,t := T e A t C Ce At dt has rank n. Notice that the observability matrix has dimension np n where p is the dimension of the output vector y. 4. For each s C, the matrix, ( si A C has rank n. Proof: The proof is similar to the ones as in the controllable case and will not be repeated here. Some differences are that instead of considering 1 n vector v T multiplying on the left hand sides of the controllability matrix and the grammians, we consider n 1 vector multiplying on the RHS of the observability matrix and the grammian etc. Also instead of considering the range space of the controllability matrix, we consider the NULL space of the observability matrix. Its null space is also A-invariant. 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: ( (Ã11 B1 ż = z + u à 21 à 22 B 2 y = ( C z where C = ( C T 1, and A = T (Ã11 à 12 à 22 T 1.

23 ME 8281:Advanced Control Systems, F1, F2, S6, S8 49 and the dim of Ã22 is non-zero (and is the dim of the null space of the observability matrix. Remark 1. Again, observability of a LTI system does not depend on the time interval. So, theoretically speaking, if observing the output and input for an arbitrary amount of time will be sufficient to figure out x(. In reality, when more data is available, one can do more averaging to eliminate effects of noise (e.g. using the Least square ie. Kalman Filter approach. 2. The subspace of particular interest is the null space of the controllability matrix. An initial state lying in this set will generate identically zero-input response. This subspace is called the unobservable subspace. 3. Using the basis of the unobservable subspace as part of the basis of R n, the observability property can be easily seen Observability Tests for Discrete time LTI systems The tests for observability of the discrete time system (3.7 is given similarly by the following theorem. Theorem For the discrete time system (3.7, the followings are equivalent: 1. The system is observable over the interval [, T], for some T n. 2. The observability grammian, is full rank n. 3. The observability matrix C CA CA 2. CA n 1 T 1 W o,t := A k C CA k k= has rank n. Notice that the observability matrix has dimension np n where p is the dimension of the output vector y. 4. For each z C, the matrix, ( zi A C has rank n. The controllability matrix can have the following interpretation: zero-input response is given by: y( C y(1 CA y(2 = CA 2 x(.. y(n 1. CA n 1

24 5 c Perry Y.Li Thus, clearly if the observability matrix is full rank, one can reconstruct x( from measurements of y(,, y(n 1. One might think that by increasing the number of output measurement times (e.g. y(n the system can become observable. However, because of Cayley-Hamilton theorem, y(n = CA n x( can be expressed as n 1 i= a ica i x( consisting of rows already in the observability matrix PHB test and Eigen/Jordan Forms Consider ẋ = Ax + Bu, y = Cx Suppose first that A is semi-simple, then A = TDT 1 and D = diag λ 1,...,λ n where λ i might be repeating. The PBH test in the transformed eigen-coordinates is to check rank(λ i D, T 1 B for each i if it is n. This shows that T 1 B must be non-zero in each row. Also, if λ i = λ j for some i j, then a single input system cannot be controllable. Now suppose that A = TJT 1 and J is in Jordan form, J = λ 1 1 λ 1 λ 1 λ 2 This Jordan form has 3 generalized eigenvector chains. Controllability From homework, we saw that if the entries of B R 4 at the beginning of some chain is, i.e.. b 1 b 1 b 1 T 1 B = b 2 b3, T 1 B = b 2, T 1 B = b3 b 4 then the system is uncontrollable. Thus, T 1 B being nonzero at the beginning of each chain is a necessary condition for controllability. PBH test shows that it is in fact a sufficient condition. Observability Similarly, if C T R 1 n is a zero entry at the end of any chain (these correspond to the coordinates for the eigenvector, i.e. b 4 CT = ( c 2 c 3 c 4, CT = ( c1 c 2 c 4, or, CT = ( c 1 c 2 c 3 then we saw from the homework that the system is not observable.

25 ME 8281:Advanced Control Systems, F1, F2, S6, S8 51 Hence, CT having non-zero entries at the end of each chain is necessary for observability. PHB test shows that it is also sufficient. Note If the system is either uncontrollable or unobservable, then the order of the transfer function will be less than the order of A. However, for a multi-input-multi-output system, the fact that the order of each transfer function in the transfer function matrix has order less than the order of A, does NOT imply that the system is uncontrollable or unobservable. e.g. A = ( 1 2 B = C = ( 1 1 gives, Y (s = ( 1 s+1 1 U(s. s 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: where B = T ( B, and ż = z = T 1 x, (Ã11 Ã 12 A = T Ã 22 z + (Ã11 Ã 12 Ã 22 ( B u (3.11 T 1. and the dim of Ã22 is n n 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: ż = (Ã11 Ã 21 Ã 22 y = ( C z ( B1 z + u B 2 where C = ( C T 1, and A = T (Ã11 Ã 12 Ã 22 T 1. and the dim of Ã22 is non-zero (and is the dim of the null space of the observability matrix.

26 52 c Perry Y.Li Kalman decomposition The Kalamn decomposition is just the combination of the observable/unobservable, and the controllable/uncontrollable decomposition. Theorem There exists a coordinate transformation z = T 1 x R n such that à 11 à 13 B 1 ż = à 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: Only T 2 is uniquely defined. Proof: This is just combination of the obs/unobs and cont/uncont decomposition 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. The PBH tests can be modified to check if the system is stabilizable or detectable, namely, check rank ( si A B ( si A rank C for all s with non-negative real parts if they lose rank. Specifically, one needs to check only s which are the unstable eigenvalues of A.

27 ME 8281:Advanced Control Systems, F1, F2, S6, S Realization Degree of controllability / observability Formal definitions for controllability and observability are black and white. In reality, some states are very costly to control to, or have little effect on the outputs. They are therefore, effectively uncontrollable or unobservable. Degree of controllability and observability can be evaluated by the sizes of W r and W o with infinite time horizon: W r (, = lim = T W o (, = lim T Notice that W r satisfies differential equation: 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 d dt W r(, t = AW r (, t + W r (, ta T + BB T Thus, if all the eigenvalues of A have strictly negative real parts (i.e. A is stable, then, as T, = AW r + W r A T + BB T which is a set of linear equations in the entries of W r, and can be solved effectively. The equation is called a Lyapunov equation. Matlab can solve for W r using gram. Similarly, if A is stable, as T, W o satisfies the Lyapunov equation = A T W o + W o A + C T C which can be solved using linear method (or using Matlab via gram. W r is related to the minimum norm control. To reach x( = x from x( = in infinite time, u = L T r Wc 1 x min u( J(u = min u( u T (τu(τdτ = x T W 1 c x. If state x is difficult to reach, then x T W c 1 x is large meaning that x is practically uncontrollable. Similarly, if the initial state is x( = x, then, with u(τ y T (τy(τdτ = x T W o x. Thus, x T W ox measures the effect of the initial condition on the output. If it is small, the effect of x in the output y( is small, thus, it is hard to observe, or 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.

28 54 c Perry Y.Li Relation to Transfer Function For the system 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 they are values of s C s.t. G(s becomes infinite. 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 that is identically zero. Therefore, there exists X(z, U(z such that: This will be true if and only if at s = z zi A X(z = B U(z = C X(z + D U(z ( si A B rank C D is less than the normal rank of the matrix at other s. We shall return to the multi-variable zeros when we discuss the effect of state feedback on the zero of the system.

29 ME 8281:Advanced Control Systems, F1, F2, S6, S 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 Then Cascade interconnection: u 2 = y The system pole at β is not observable from y 2 2. The system pole at α is not controllable from u 1. Combined system: Consider x = ( x1 x 2. ( A1 A = B 2 C 1 A 2 ; B = ( B1 ; C = ( C 2 Use PBH test with λ = β. 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. Let x 2 = (βi A 2 1 B 2 C 2 x 1 (i.e. solve second row in PBH test. Then since Cx = C 2 x 2 = (C 2 (βi A 2 1 B 2 C 2 x 1 and β is a zero of system 2, PHB test shows that the mode β is not observable. Similar for the case when α mode in system 2 not controllable by u. These example shows that when cascading systems, it is important not to do unstable pole/zero cancellation.

30 56 c Perry Y.Li 3.16 Balanced Realization The objective is to reduce the number of states while minimizing effect on I/O response (transfer function. This is especially useful when the state equations are obtained from distributed parameters system, P.D.E. via finite elements methods, etc. which generate many states. If states are unobservable or uncontrollable, they can be removed. The issue at hand is if some states are lightly controllable while others are lightly observable. Do we cancel them? In particular, Some states can be lightly controllable, but heavily observable. Some states can be lightly observable, but heavily controllable. Both contribute to significant component to input/output (transfer function response. 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, reachability and observability grammians can be computed by solving the Lyapunov equations: = AW r + W r A T + BB T = A T W o + W o A + C T C If state x is difficult to reach, then x T W r 1 x is large practically uncontrollable. If x T W ox 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 r 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 Wr 1 x = z T Wr,z 1 z = x T T T Wr,z 1 Tx W 1 r = T T W 1 r,z T W r,z = T W r 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 observable, then, there exists an invertible transformation T R n n s.t. W o,z = W r,z.

31 ME 8281:Advanced Control Systems, F1, F2, S6, S8 57 The balanced realization algorithm: 1. Compute controllability 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]: where UU T = I and Σ is diagonal. RW r R T = UΣ 2 U T, >> [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; 5. This gives W r,z = W o,z = Σ. >> 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 unstable 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 + dt z 2 A 21 A 22 z 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 the simultaneously lightly controllable and light observable mode. The naive truncation approach is to remove z 2 to form the reduced system: ż 1 = A 11 z 1 + B 1 u y = C 1 z 1 + Du ( B1 B 2 u

32 58 c Perry Y.Li 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 z 2 + B 2 u z 2 = A 1 22 (A 21z 1 + B 2 u This is feasible if A 22 is invertible ( is not an eigenvalue. A truncation approach that maintains the steady state gain is to replace z 2 by its steady state value: ż 1 = A 11 z 1 + A 12 z 2 + B 1 u y = C 1 z 1 + C 2 z 2 + Du so that, 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 B r The truncated system with state z 1 R n 1 is ż 1 = A r z 1 + B r u y = C r z 1 + D r u which with have the same steady state response as the original system.