Topic #11 16.31 Feedback Control State-Space Systems State-space model features . Observability o Controllability e Minimal realizations Copyright[2001by JOnathan dHow.D
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 16.3111-1 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 6 +2 for which we develop the state-space model Model#1 i==2c+2u . But now consider the new state space model t=[ a2T Model #2 a C 0-1 30x which is clearly different than the first model e But let's looks at the transfer function of the new model (s)=C(sI-A)B+D
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 ¯x = [x x2] T Model # 2 x¯˙ = −2 0 0 −1 x¯ + ∙ 2 1 ¸ u y = £ 3 0 ¤ x¯ which is clearly different than the first model. • But let’s looks at the transfer function of the new model: G¯(s) = C(sI − A) −1 B + D
Fall 2001 16.3111-2 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 Now we have two very different models that result in the same 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 2 were eliminated when we formed the product G(s)=[30] 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 6(s+1) s+2(s+2)(s+1 会G( 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 o. but we would be unware of it using this sensor
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 — Now we have two very different models that result in the same 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 x2 were eliminated when we formed the product G¯(s) = £ 3 0 ¤ " 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 x2, but we would be unware of it using this sensor
Fall 2001 6.3111-3 There is an analogous problem on the input side as well. Consider Model #1 =2a + 2u with a=a a2 2 #3 0-1 32 which is also clearly different than model #1, and has a different form from the second model 20 [32]sr 0-1 2 6 0 s+2 Once again the dynamics associated with the pole at s=-1 are cancelled out of the transfer function But in this case it occurred because there is a o in the b matrix So in this case we can"see"the state a2 in the output C=3 2 but we cannot "influence that state with the input since b 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—3 • There is an analogous problem on the input side as well. Consider: Model # 1 x˙ = −2x + 2u y = 3x with ¯x = [ x x2] T Model # 3 x¯˙ = −2 0 0 −1 x¯ + ∙ 2 0 ¸ u y = £ 3 2 ¤ x¯ which is also clearly different than model #1, and has a different form from the second model. Gˆ(s) = £ 3 2 ¤ sI − −2 0 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 = −1 are 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 x2 in the output C = £ 3 2 ¤ , 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 6.3111-4 Of course it can get even worse because we could have C+ 0-1 ●5 o now we have 20 30 0-1 s+2s+1 +2 e Get same result for the transfer function, but now the dynamics associated with a are both unobservable and uncontrollable unnary Dynamics in the state-space model that are uncontrollable, un observable, or both do not show up in the transfer function Would like to develop models that only have dynamics that are both controllable and observable 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—4 • Of course it can get even worse because we could have x¯˙ = −2 0 0 −1 x¯ + ∙ 2 0 ¸ u y = £ 3 0 ¤ x¯ • So now we have G ](s) = £ 3 0 ¤ sI − −2 0 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 x2 are both unobservable and uncontrollable. • Summary: Dynamics in the state-space model that are uncontrollable, unobservable, or both 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 observable and/or controllable
Fall 2001 16.3111-5 Observability Definition: An lti system is observable if the initial state (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 a(0) can be deduced, then we can reconstruct a(t) exactly because we know a (t)= we can find a(t)vt Thus we need only consider the zero-input(homogeneous) solu tion to study observability y(t)=CeAtr(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 e state canno be"seen '"in y(t), then it would be impossible to deduce that part of a(0)from the outputs y(t)
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 because we know u(t) V we can find x(t) ∀ t. — Thus we need only consider the zero-input (homogeneous) solution to study observability. y(t) = CeAtx(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 16.3111-6 Definition: A state a*f0 is said to be unobservable if the zero-input solution y(t), with a(0)=a*, is zero for all t20 equivalent to saying that k is an unobservable state if Ce 0t>0 For the problem we were just looking at, consider Model #2 with [01]7≠0,then 20 C+ 0-1 30 SO 00 e^- 0 3 =0t So, a*=[0 1 is an unobservable 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 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 CeAtx? = 0 ∀ t ≥ 0 • For the problem we were just looking at, consider Model #2 with x? =[0 1]T 6= 0, then x¯˙ = −2 0 0 −1 x¯ + ∙ 2 1 ¸ u y = £ 3 0 ¤ x¯ so CeAtx? = £ 3 0 ¤ ∙ 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 x2 from the previous analysis
Fall 2001 16.3111-7 Theorem: An lTi system is observable iff it has no unobservable states We normally just say that the pair(A, C) is observable Pseudo-Proof: Let a*fo be an unobservable state and compute the outputs from the initial conditions a10) and x2( 0)=31(0)+o* Ther g1(t)=CeAta1(0) and 92(t)=CeAtx20) but Thus 2 different initial conditions give the same output y(t), so it would be impossible for us to deduce the actual initial condition of the system cI(t)or 2(t) given g(t Testing system observability by searching for a vector x(0) such that CeAtx(0)=0v t is feasible, but very hard in general Better tests are available
Fall 2001 16.31 11—7 • Theorem: An LTI system is observable iff 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 x1(0) and x2(0) = x1(0)+x? — Then y1(t) = CeAtx1(0) and y2(t) = CeAtx2(0) but — Thus 2 different initial conditions give the same output y(t), so it would be impossible for us to deduce the actual initial condition of the system x1(t) or x2(t) given y1(t) • Testing system observability by searching for a vector x(0) such that CeAtx(0) = 0 ∀ t is feasible, but very hard in general. — Better tests are available
Fall 2001 16.3111-8 Theorem: The vector k is an unobservable state if CA CA CAT Pseudo-Proof: If a* is an unobservable state, then by definition Cex*=0yt≥0 But all the derivatives of CeAt exist and for this condition to hold all derivatives must be zero at t=0. Then Ce=0=0→Cx=0 C dt 0→ dt t=0 At dth 0 t=0 We only need retain up to the n- 1tn derivative because of the Cayley-Hamilton theorem
Fall 2001 16.31 11—8 • Theorem: The vector x? is an unobservable state if C CA CA2 . . . CAn−1 x? = 0 • Pseudo-Proof: If x? is an unobservable state, then by definition, CeAtx? = 0 ∀ t ≥ 0 But all the derivatives of CeAt exist and for this condition to hold, all derivatives must be zero at t = 0. Then CeAtx? ¯ ¯ t=0 = 0 ⇒ Cx? = 0 d dtCeAtx? ¯ ¯ ¯ ¯ t=0 = 0 ⇒ d2 dt2CeAtx? ¯ ¯ ¯ ¯ t=0 = 0 ⇒ . . . dk dtkCeAtx? ¯ ¯ ¯ ¯ t=0 = 0 ⇒ • We only need retain up to the n − 1th derivative because of the Cayley-Hamilton theorem
Fall 2001 163111-9 Simple test: Necessary and sufficient condition for observability is that CA rank Mo A rank CA2 CA Why does this make sense? he requirement for an unobservable state is that for *+0 Which is equivalent to saying that is orthogonal to each row of M But if the rows of Mo are considered to be vectors and these span the full n-dimensional space, then it is not possible to find an n-vector a* that is orthogonal to each of these To determine if the n rows of Mo span the full n-dimensional space, we need to test their linear independence, which is equivalent to the rank test ILet m be a m x 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}
Fall 2001 16.31 11—9 • Simple test: Necessary and sufficient condition for observability is that rank Mo , rank C CA CA2 . . . CAn−1 = n • Why does this make sense? — The requirement for an unobservable state is that for x? 6= 0 Mox? = 0 — Which is equivalent to saying that x? is orthogonal to each row of Mo. — But if the rows of Mo 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 Mo span the full n-dimensional space, we need to test their linear independence, which is equivalent to the rank test1. 1Let 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}
