Topic #12 16.31 Feedback Control State-Space Systems e State-space model features o Controllability Copyright 2001 by Jonathan How
Topic #12 16.31 Feedback Control State-Space Systems • State-space model features • Controllability Copyright 2001 by Jonathan How. 1
Fall 2001 16.3112-1 Controllability Definition: An LTI system is controllable if, for every a*(t d every T>0, there exists an input function u(t),0<t<T, such that the system state goes from (0)=0 to a T)=I' 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 Bu(r)dr Change of variables T2=t-T, dr =-dr2 gives a(t)= eAry Bu(t-T2)dT2 0 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 ult), then it would be impossible to move that part of the state from 0 to *k
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 eA(t−τ ) Bu(τ )dτ — Change of variables τ2 = t − τ , dτ = −dτ2 gives x(t) = Z t 0 eAτ2Bu(t − τ2)dτ2 • 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 16.31122 Definition: A state o*+0 is said to be uncontrollable if the forced state response a(t)is orthogonal to x*vt>0 and all input functions You cannot get there from here This is equivalent to saying that a* is an uncontrollable state if 2 Bu(t-T2)di 2B(t-2)d2=0 Since this identity must hold for all input functions u(t-r2), this can only be true if B≡0t>0
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 eAτ2Bu(t − τ2)dτ2 = Z t 0 (x? ) T eAτ2Bu(t − τ2)dτ2 = 0 • Since this identity must hold for all input functions u(t − τ2), this can only be true if (x? ) T eAtB ≡ 0 ∀ t ≥ 0
Fall 2001 16.3112-3 For the problem we were just looking at, consider Model #3 with x=[01]≠0,then 20 2 del #3 a C+ 0 32 (x)e4B=[01 01「 0e-t||0 2 0 t Sor*=[0 1 is an uncontrollable state for this system. But that is as expected, because we knew there was a problem with the state 2 from the previous analysis
˙ 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 x¯ = x¯ + 0 2 ¸ u 0 −1 £ 3 2 ¤ y = x¯ so ∙ e−2t 0 0 e−t ¸ ∙ 0 ¸ £ 2 0 1 ¤ e−t ¤ (x? ) T eAtB = ∙ 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 x2 from the previous analysis
Fall 2001 16.3112-4 Theorem: An Lti system is controllable iff 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 a(t), which means that the system cannot reach it in finite time the system would be uncontrollable o Theorem The vector is an uncontrollable state iff (x)[BABA2B…A-B] See page 81 Simple test: Necessary and sufficient condition for controllability is that ankM会rank[BABA2B…A2-B]=n
Fall 2001 16.31 12—4 • Theorem: An LTI system is controllable iff 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 iff (x? ) T £ B AB A2B ··· An−1B ¤ = 0 — See page 81. • Simple test: Necessary and sufficient condition for controllability is that rank Mc , rank £ B AB A2B · · · An−1B ¤ = n
Fall 2001 Examples 6.3112-5 With Model #2: 30 CA 60 [ B AB I rank Mo= l and rank Mc=2 So this model of the system is controllable, but not observable 2 With Model #3: C CA 6-2 C B AB 00 rank Mo=2 and rank Mc=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 Examples 16.31 12—5 − ∙ 2 0 With Model # 2: x¯ = x¯ + 1 2 ¸ u 0 −1 £ y = 3 0 ¤ x¯ ∙ ∙ C 3 0 M0 = CA ¸ = −6 0 ¸ ∙ Mc = £ B AB ¤ = 2 −4 1 −1 ¸ — rank M0 = 1 and rank Mc = 2 — So this model of the system is controllable, but not observable. − ∙ 2 0 With Model # 3: x¯ = x¯ + 0 2 ¸ u 0 −1 £ y = 3 2 ¤ x¯ ∙ ∙ C 3 2 M0 = CA ¸ = −6 −2 ¸ Mc = £ B AB ¤ = ∙ 2 0 −4 0 ¸ — rank M0 = 2 and rank Mc = 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 16.3112-6 Example: Loss of observability Typical scenario: consider system G(s)of the form a s+a y so that s+a 1 S +1s+ Clearly a pole-zero cancelation in this system(pole s The state space model for the system is ac+ +(a-1 y=x1+2 A B C=11,D=0 a The Observability/ Controllability tests are(a= 2) 「11 rank rank CA rank[B AB rank /I System controllable, but unobservable. Consistent with the picture Both states can be influenced by But e -at mode dynamics canceled out of the output by the zero
Fall 2001 16.31 12—6 Example: Loss of Observability • Typical scenario: consider system G(s) of the form 1 s + a x1 s + a s + 1 x2 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 = −ax1 + u x˙ 2 = −x2 + (a − 1)x2 y = x1 + x2 −a 0 a − 1 −1 , B = ∙ 1 0 ¸ , C = £ 1 1 ¤ ⇒ A = , D = 0 • The Observability/Controllability tests are (a = 2): ∙ ∙ C CA ¸ = rank 1 −1 1 ¸ = 1 < n = 2 −1 rank ∙ 1 −2 1 ¸ = 2 0 £ B AB ¤ rank = rank • 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 16.3112-7 Example: Loss of Controllability Repeat the process, but now use the system G(s) of the form y a that G(s) s+a s+ Still a pole-zero cancelation in this system (pole s==a The state space model for the system is a1+a2+ +(a-1) C B D2=0 The Observability/Controllability tests are(a=2) rank rank 2 rank [B2AB2]=an/1-1=1<n=2 o System observable, but uncontrollable. Consistent with the picture u can influence state a 2, but effect on 1 canceled by zero Both states can be seen in the output(ai directly, and 2 because it drives the dynamics associated with 1)
Fall 2001 16.31 12—7 Example: Loss of Controllability • Repeat the process, but now use the system G(s) of the form s + a s + 1 x2 1 s + a x1 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 = −ax1 + x2 + u x˙ 2 = −x2 + (a − 1)u y = x1 − ∙ a 1 0 −1 , B2 = 1 a − 1 ¸ , C2 = £ 1 0 ¤ ⇒ A2 = , D2 = 0 • The Observability/Controllability tests are (a = 2): ∙ ∙ C2 C2A2 ¸ = rank 1 0 −2 1 ¸ rank = 2 ∙ 1 −1 −1 ¸ = 1 < n = 2 1 £ B2 A2B2 ¤ rank = rank • System observable, but uncontrollable. Consistent with the picture: — u can influence state x2, but effect on x1 canceled by zero — Both states can be seen in the output (x1 directly, and x2 because it drives the dynamics associated with x1)
Fall 2001 16.3112-8 Modal ests 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 Ar+B Ca+ du and the A is diagonalizable(A- TAT-)using the transformation 1 based on the eigenvalues of a. note that we wrote which is a column of rows Then define a new state so that =tz. then i=t-i=T(Ac+ Bu)=(T- AT)z+r- bu Az+T- Bu y= C+ Du= CTz+ du
˙ Fall 2001 16.31 12—8 Modal Tests • Earlier examples showed the relative simplicity of testing observability/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 = TΛT −1) using the transformation T = | | v1 · · · vn | | based on the eigenvalues of A. Note that we wrote: T −1 = − w1 T − . . . − wn T − which is a column of rows. • Then define a new state so that x = T z, 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 163112-9 The new model in the state z is diagonal. There is no coupling in the dynamics matrix A But by definition TB and CT . Thus if it turned out that B≡0 then that element of the state vector zi would be uncontrollable by the input u ●Also.if then that element of the state vector z; would be unobservable with this sensor Thus, all modes of the system are controllable and ob servable if it can be shown that B≠0 an Cv;≠0V
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, wT 1 . T −1 B = . . B wT n and £ ¤ CT = C v1 · · · vn • Thus if it turned out that T wi B ≡ 0 then that element of the state vector zi would be uncontrollable by the input u. • Also, if Cvj ≡ 0 then that element of the state vector zj would be unobservable with this sensor. • Thus, all modes of the system are controllable and observable if it can be shown that wi TB 6= 0 ∀ i and Cvj 6= 0 ∀ j
