Topic #12 16.31 Feedback Control State-Space Systems ? State-space model features ? Controllability Copyright 2001 by Jonathan How. 1 Fall 2001 16.31 12—1 Controllability ? Definition: An LTI system is controllable if, for every x ? (t) and every T> 0, there exists an input function u(t), 0 <t ≤ T , such that the system state goes from x(0) = 0 to x(T )= x ? . — Starting at 0 is not a special case — if we can get to any state in finite time from the origin, then we can get from any initial condition to that state in finite time as well. — Need only consider the forced solution to study controllability. x(t)= Z t 0 e A(t?τ ) Bu(τ )dτ — Change of variables τ 2 = t ? τ , dτ = ?dτ 2 gives x(t)= Z t 0 e Aτ 2 Bu(t ? τ 2 )dτ 2 ? This definition of observability is consistent with the notion we used before of being able to “influence” all the states in the system in the decoupled examples we looked at before. — ROT: For those decoupled examples, if part of the state cannot be “influenced” by u(t), then it would be impossible to move that part of the state from 0 to x ? Fall 2001 16.31 12—2 ? Definition: A state x ? 6= 0 is said to be uncontrollable if the forced state response x(t) is orthogonal to x ? ? t> 0 and all input functions. — “You cannot get there from here” ? This is equivalent to saying that x ? is an uncontrollable state if (x ? ) T Z t 0 e Aτ 2 Bu(t ? τ 2 )dτ 2 = Z t 0 (x ? ) T e Aτ 2 Bu(t ? τ 2 )dτ 2 =0 ? Since this identity must hold for all input functions u(t ? τ 2 ), this can only be true if (x ? ) T e At B ≡ 0 ? t ≥ 0 ˙ Fall 2001 16.31 12—3 ? For the problem we were just looking at, consider Model #3 with x ? =[0 1 ] T 6=0, then ? ? ? ?2 0 Model # 3 xˉ = ? ? xˉ + 0 2 ? u 0 ?1 ￡ 32 ¤ xˉy = so ? e ?2t 0 0 e ?t ? ? 0 ? 2 ￡ 01 ¤ e ?t ¤ (x ? ) T e At B = ? 2 0 ? =0 ? t ￡ 0 = So x ? =[0 1 ] T is an uncontrollable 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 12—4 ? Theorem: An LTI system is controllable i? it has no uncontrollable states. — We normally just say that the pair (A,B) is controllable. Pseudo-Proof: The theorem essentially follows by the definition of an uncontrollable state. — If you had an uncontrollable state x ? , then it is orthogonal to the forced response state x(t), which means that the system cannot reach it in finite time ; the system would be uncontrollable. ? Theorem: The vector x ? is an uncontrollable state i? (x ? ) T ￡ BABA 2 B ··· A n?1 B ¤ =0 — See page 81. ? Simple test: Necessary and su?cient condition for controllability is that rank M c , rank ￡ BABA 2 B ··· A n?1 B ¤ = n ˙ ˙ Fall 2001 Examples 16.31 12—5 ? ? ? ?2 0 With Model # 2: xˉ = ? ? xˉ + 1 2 ? u 0 ?1 ￡ y = 30 ¤ xˉ ? ? C 30 M 0 = CA ? = ?60 ? ? M c = ￡ BAB ¤ = 2 ?4 1 ?1 ? — rank M 0 =1 and rank M c =2 — So this model of the system is controllable, but not observable. ? ? ? ?2 0 With Model # 3: xˉ = ? ? xˉ + 0 2 ? u 0 ?1 ￡ y = 32 ¤ xˉ ? ? C 3 2 M 0 = CA ? = ?6 ?2 ? M c = ￡ BAB ¤ = ? 2 0 ?4 0 ? — rank M 0 =2 and rank M c =1 — So this model of the system is observable, but not controllable. ? Note that controllability/observability are not intrinsic properties of a system. Whether the model has them or not depends on the representation that you choose. — But they indicate that something else more fundamental is wrong... Fall 2001 16.31 12—6 Example: Loss of Observability ? Typical scenario: consider system G(s)ofthe form 1 s + a x 1 s + a s +1 x 2 u → → → y so that s + a s +1 · 1 s + a G(s)= ? Clearly a pole-zero cancelation in this system (pole s = ?a) ? The state space model for the system is: x˙ 1 = ?ax 1 + u x˙ 2 = ?x 2 +(a ? 1)x 2 y = x 1 + x 2 ?? ? ?a 0 a ? 1 ?1 ? ,B = ? 1 0 ? ,C = ￡ 11 ¤ ,D =0? A = ? The Observability/Controllability tests are (a =2): ? ? C CA ? = rank 1 ?1 1 ? =1 <n =2 ?1 rank ? 1 ?2 1 ? =2 0 ￡ BAB ¤ = rankrank ? System controllable, but unobservable. Consistent with the picture: — Both states can be influenced by u — But e ?at mode dynamics canceled out of the output by the zero. Fall 2001 16.31 12—7 Example: Loss of Controllability ? Repeat the process, but now use the system G(s)ofthe form s + a s +1 x 2 1 s + a x 1 u → → → y so that 1 s + a · s + a s +1 G(s)= ? Still a pole-zero cancelation in this system (pole s = ?a) ? The state space model for the system is: x˙ 1 = ?ax 1 + x 2 + u x˙ 2 = ?x 2 +(a ? 1)u y = x 1 ?? ? ? ?a 1 0 ?1 ? ,B 2 = 1 a ? 1 ? ,C 2 = ￡ 10 ¤ ,D 2 =0? A 2 = ? The Observability/Controllability tests are (a =2): ? ? C 2 C 2 A 2 ? = rank 10 ?21 ? =2rank ? 1 ?1 ?1 ? =1 <n =2 1 ￡ B 2 A 2 B 2 ¤ = rankrank ? System observable, but uncontrollable. Consistent with the picture: — u can influence state x 2 ,but e?ect on x 1 canceled by zero — Both states can be seen in the output (x 1 directly, and x 2 because it drives the dynamics associated with x 1 ) ˙ Fall 2001 16.31 12—8 Modal Tests ? Earlier examples showed the relative simplicity of testing observ ability/controllability for system with a decoupled A matrix. ? There is, of course, a very special decoupled form for the state-space model: the Modal Form (8—5) ? Assuming that we are given the model x˙ = Ax + Bu y = Cx + Du and the A is diagonalizable (A = T ΛT ?1 ) using the transformation T = ? | | ? v 1 ··· v n | | ? ? based on the eigenvalues of A. Note that we wrote: T ?1 = ? ? ? w T ? 1 . . . ? w T ? n ? ? which is a column of rows. ? Then define anew statesothat x = Tz,then z = T ?1 x˙ = T ?1 (Ax + Bu)=(T ?1 AT )z + T ?1 Bu = Λz + T ?1 Bu y = Cx + Du = CTz + Du Fall 2001 16.31 12—9 ? The new model in the state z is diagonal. There is no coupling in the dynamics matrix Λ. ? But by definition, ? ? w T 1 . T ?1 B = ? . . ? B w T n and ￡ ¤ CT = C v 1 ··· v n ? Thus if it turned out that T w i B ≡ 0 then that element of the state vector z i would be uncontrollable by the input u. ? Also, if Cv j ≡ 0 then that element of the state vector z j would be unobservable with this sensor. ? Thus, all modes of the system are controllable and ob- servable if it can be shown that w i T B 6=0 ? i and Cv j 6=0 ? j Fall 2001 16.31 12—10 Cancelation ? Examples show the close connection between pole-zero cancelation and loss of observability and controllability. Can be strengthened. ? Theorem: The mode (λ i ,v i ) of a system (A,B,C,D) is unob ? ? .servable i? the system has a zero at λ i with direction v i 0 ? Proof: If the system is unobservable at λ i , then we know (λ i I ? A)v i =0 It is a mode Cv i = 0 That mode is unobservable Combine to get: ? (λ i I ? A) C ? v i =0 Or ? (λ i I ? A) ?B C D ? ? v i 0 ? =0 which implies that the system has a zero at that frequency as well, ? with direction v i 0 ? . ? Can repeat the process looking for los ¤ s of controllability, but now ￡ using zeros with left direction w T 0 i . Fall 2001 16.31 12—11 ? Combined Definition: when aMIMOzerocauseslossof ei ther observability or controllability we say that there is a pole/zero cancelation. — MIMO pole-zero (right direction generalized eigenvector) cance lation ? mode is unobservable — MIMO pole-zero (left direction generalized eigenvector) cancela tion ? mode is uncontrollable ? Note: This cancelation requires an agreement of both the fre quency and the directionality of the system mode (eigenvector) and ? zero v i 0 ? or ￡ w T 0 i ¤ . Fall 2001 16.31 12—12 Weaker Conditions ? Often it is too much to assume that we will have full observability and controllability. Often have to make do with the following: ? A system is called detectable if all unstable modes are observable ? A system is called stabilizable if all unstable modes are controllable ? So if you had a stabilizable and detectable system, there could be dynamics that you are not aware of and cannot influence, but you know that they are at least stable.