Topic #11 16.31 Feedback Control State-Space Systems State-space model features Observability Controllability Minimal Realizations Copyright 2001 by Jonathan How. 1 Fall 2001 16.31 11—1 State-Space Model Features There are some key characteristics of a state-space model that we need to identify. — Will see that these are very closely associated with the concepts of pole/zero cancellation in transfer functions. Example: Consider a simple system G(s)= 6 s +2 for which we develop the state-space model Model # 1 ˙x = ?2x+2u y =3x But now consider the new state space model ˉx =[xx 2 ] T Model # 2 ˙ ˉx = ? ? ?20 0 ?1 ? ? ˉx + ? 2 1 ? u y = ￡ 30 ¤ ˉx which is clearly di?erent than the first model. But let’s looks at the transfer function of the new model: ˉ G(s)=C(sI ?A) ?1 B + D Fall 2001 16.31 11—2 This is a bit strange, because previously our figure of merit when comparing one state-space model to another (page 8-8) was whether they reproduced the same same transfer function — Nowwehavetwoverydi?erentmodelsthatresultinthesame transfer function — Note that I showed the second model as having 1 extra state, but I could easily have done it with 99 extra states!! So what is going on? — The clue is that the dynamics associated with the second state of the model x 2 were eliminated when we formed the product ˉ G(s)= ￡ 30 ¤ " 2 s+2 1 s+1 # because the A is decoupled and there is a zero in the C matrix — Which is exactly the same as saying that there is a pole-zero cancellation in the transfer function ? G(s) 6 s +2 = 6(s +1) (s +2)(s +1) , ? G(s) — Note that model #2 is one possible state-space model of ? G(s) (has 2 poles) For this system we say that the dynamics associated with the second state are unobservable using this sensor (defines the C matrix). — There could be a lot “motion” associated with x 2 , but we would be unware of it using this sensor. Fall 2001 16.31 11—3 There is an analogous problem on the input side as well. Consider: Model # 1 ˙x = ?2x+2u y =3x with ˉx =[xx 2 ] T Model # 3 ˙ ˉx = ? ? ?20 0 ?1 ? ? ˉx + ? 2 0 ? u y = ￡ 32 ¤ ˉx which is also clearly di?erent than model #1, and has a di?erent form from the second model. ? G(s)= ￡ 32 ¤ ? ? sI ? ? ? ?20 0 ?1 ? ? ? ? ?1 ? 2 0 ? = ￡ 3 s+2 2 s+1 ¤ ? 2 0 ? = 6 s +2 !! Once again the dynamics associated with the pole at s = ?1are cancelled out of the transfer function. — But in this case it occurred because there is a 0 in the B matrix So in this case we can “see” the state x 2 in the output C = ￡ 32 ¤ , but we cannot “influence” that state with the input since B = ? 2 0 ? So we say that the dynamics associated with the second state are uncontrollable using this actuator (defines the B matrix). Fall 2001 16.31 11—4 Of course it can get even worse because we could have ˙ ˉx = ? ? ?20 0 ?1 ? ? ˉx + ? 2 0 ? u y = ￡ 30 ¤ ˉx So now we have ] G(s)= ￡ 30 ¤ ? ? sI ? ? ? ?20 0 ?1 ? ? ? ? ?1 ? 2 0 ? = ￡ 3 s+2 0 s+1 ¤ ? 2 0 ? = 6 s +2 !! Get same result for the transfer function, but now the dynamics associated with x 2 are both unobservable and uncontrollable. Summary: Dynamics in the state-space model that are uncontrollable, un- observable,orboth do not show up in the transfer function. Would like to develop models that only have dynamics that are both controllable and observable V called a minimal realization — It is has the lowest possible order for the given transfer function. But first need to develop tests to determine if the models are ob- servable and/or controllable Fall 2001 16.31 11—5 Observability Definition: An LTI system is observable if the initial state x(0) can be uniquely deduced from the knowledge of the input u(t) and output y(t) for all t between 0 and any T>0. — If x(0) can be deduced, then we can reconstruct x(t)exactly becauseweknowu(t) V we can find x(t) ? t. — Thus we need only consider the zero-input (homogeneous) solu- tion to study observability. y(t)=Ce At x(0) This definition of observability is consistent with the notion we used before of being able to “see” all the states in the output of the decoupled examples — ROT: For those decoupled examples, if part of the state cannot be “seen” in y(t), then it would be impossible to deduce that part of x(0) from the outputs y(t). Fall 2001 16.31 11—6 Definition: A state x ? 6= 0 is said to be unobservable if the zero-input solution y(t), with x(0) = x ? , is zero for all t ≥ 0 — Equivalent to saying that x ? is an unobservable state if Ce At x ? =0? t ≥ 0 For the problem we were just looking at, consider Model #2 with x ? =[0 1] T 6=0,then ˙ ˉx = ? ? ?20 0 ?1 ? ? ˉx + ? 2 1 ? u y = ￡ 30 ¤ ˉx so Ce At x ? = ￡ 30 ¤ ? e ?2t 0 0 e ?t ?? 0 1 ? = ￡ 3e ?2t 0 ¤ ? 0 1 ? =0? t So, x ? =[0 1] T is an unobservable state for this system. But that is as expected, because we knew there was a problem with the state x 2 from the previous analysis Fall 2001 16.31 11—7 Theorem: An LTI system is observable i? it has no unobservable states. — We normally just say that the pair (A,C) is observable. Pseudo-Proof: Let x ? 6= 0 be an unobservable state and compute the outputs from the initial conditions x 1 (0) and x 2 (0) = x 1 (0)+x ? — Then y 1 (t)=Ce At x 1 (0) and y 2 (t)=Ce At x 2 (0) but — Thus 2 di?erent initial conditions give the same output y(t), so it would be impossible for us to deduce the actual initial condition of the system x 1 (t)orx 2 (t)giveny 1 (t) Testing system observability by searching for a vector x(0) such that Ce At x(0) = 0 ? t is feasible, but very hard in general. — Better tests are available. Fall 2001 16.31 11—8 Theorem: The vector x ? is an unobservable state if ? ? ? ? ? ? ? C CA CA 2 . . . CA n?1 ? ? ? ? ? ? ? x ? =0 Pseudo-Proof: If x ? is an unobservable state, then by definition, Ce At x ? =0 ? t ≥ 0 But all the derivatives of Ce At exist and for this condition to hold, all derivatives must be zero at t =0.Then Ce At x ? ˉ ˉ t=0 =0 ? Cx ? =0 d dt Ce At x ? ˉ ˉ ˉ ˉ t=0 =0 ? d 2 dt 2 Ce At x ? ˉ ˉ ˉ ˉ t=0 =0 ? . . . d k dt k Ce At x ? ˉ ˉ ˉ ˉ t=0 =0 ? We only need retain up to the n ? 1 th derivative because of the Cayley-Hamilton theorem. Fall 2001 16.31 11—9 Simple test: Necessary and su?cient condition for observability is that rank M o ,rank ? ? ? ? ? ? ? C CA CA 2 . . . CA n?1 ? ? ? ? ? ? ? = n Whydoesthismakesense? — The requirement for an unobservable state is that for x ? 6=0 M o x ? =0 — Which is equivalent to saying that x ? is orthogonal to each row of M o . — But if the rows of M o are considered to be vectors and these span the full n-dimensional space, then it is not possible to find an n-vector x ? that is orthogonal to each of these. — To determine if the n rows of M o span the full n-dimensional space, we need to test their linear independence,whichis equivalent to the rank test 1 . 1 Let M be a m×p matrix, then the rank of M satisfies: 1. rank M ≡ number of linearly independent columns of M 2. rank M ≡ number of linearly independent rows of M 3. rank M ≤ min{m,p}