1 Controllability and Observability

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1 1 Controllability and Observability 1.1 Linear Time-Invariant (LTI) Systems State-space: Dimensions: Notation Transfer function: ẋ = Ax+Bu, x() = x, y = Cx+Du. x R n, u R m, y R p. Note that H(s) is always proper! [ ] A B C D H(s) = C(sI A) 1 B +D Similarity transformation: [ T 1 AT T 1 ] B CT D Similarity does not change transfer function [ ] A B C D H(s) = CT(sI T 1 AT) 1 T 1 B +D = C(sI A) 1 B +D System response: Y(s) = H(s)U(s) }{{} + C(sI A) 1 x() }{{} Input Initial conditions MIMO comes for free! MAE 28 B 5 Maurício de Oliveira

2 1.2 Concepts from MAE 28 A Controllability Matrix: C(A,B) = [ B AB A n 1 B ]. Controllability Gramian: X(t) = t e Aξ BB T e ATξ dξ. Observability Matrix: Observability Gramian: O(A,C) = C CA. CA n 1. Y(t) = t e ATξ C T Ce Aξ dξ. MAE 28 B 6 Maurício de Oliveira

3 1.3 Controllability Problem: Given x() = and any x, can one compute u(t) such that x( t) = x for some t >? Theorem: The following are equivalent a) The pair (A,B) is controllable; b) The Controllability Matrix C(A, B) has full-row rank; c) There exists no z such that z A = λz, z B = ; d) The Controllability Gramian X(t) is positive definite for some t. 1.4 Observability Problem: Given y(t) over t [, t] with t > can one compute x(t) for all t [, t]? Theorem: The following are equivalent a) The pair (A,C) is observable; b) The Observability Matrix O(A, C) has full-column rank; c) There exists no x such that Ax = λx, Cx = ; d) The Observability Gramian Y = Y(t) is positive definite for some t. 1.5 Things you should already know 1. Why a) and b) are equivalent. 2. Why can we stop C(A,B) at A n 1 B and O(A,C) at CA n 1? 3. Kalman canonical forms. E.g. if (A,C) is not observable then [ ] A A B o B o Aōo Aō Bō C C o where A o R r r and (A o,c o ) is observable. MAE 28 B 7 Maurício de Oliveira

4 1.6 The Popov-Belevitch-Hautus Test Theorem: The pair (A,C) is observable if and only if there exists no x such that Ax = λx, Cx =. (1) Proof: Sufficiency: Assume there exists x such that (1) holds. Then CAx = λcx =, CA 2 x = λcax =,. CA n 1 x = λca n 2 x = so that O(A,C)x =, which implies that the pair (A,C) is not observable. Necessity: Assume that (A, C) is not observable. Then transform it into the equivalent non observable realization where [ ] Ao [ Ā =, C = Co ]. Aōo Aō Chose x such that Aōx = λx. Then [ Ao Aōo Aō ]( ) = λ x ( ), x [ Co ]( ) =. x MAE 28 B 8 Maurício de Oliveira

5 1.7 Controllability Gramian Problem: Given x() = and any x, compute u(t) such that x( t) = x for some t >. Solution: We know that x = x( t) = t e A( t τ) Bu(τ)dτ. If we limit our search to controls u of the form u(t) = B T e ( t t) z AT we have x = = = t ( t e A( t τ) BB T e ( t τ) zdτ, AT ) e A( t τ) BB T e AT ( t τ) dτ ( ) t e Aξ BB T e ATξ dξ z, z, ξ = t τ and ( 1 t z = e Aξ BB T e dξ) ATξ x, ( 1 t u(t) = B T e AT ( t t) e Aξ BB T e dξ) ATξ x The symmetric matrix X(t) := is the Controllability Gramian. t e Aξ BB T e ATξ dξ MAE 28 B 9 Maurício de Oliveira

6 1.8 Stabilizability Problem: Given any x() = x can one compute u(t) such that x( t) = for some t >? Theorem: The following are equivalent a) The pair (A,B) is stabilizable; b) There exists no z and λ such that z A = λz, z B = with λ+λ. 1.9 Detectability Problem: Given y(t) over t [, t] with t > can one compute x( t)? Theorem: The following are equivalent a) The pair (A,C) is detectable; b) There exists no x and λ such that Ax = λx, Cx = with λ+λ. MAE 28 B 1 Maurício de Oliveira

7 1.1 Example: satellite in circular orbit u r θ m u 1 e Satellite of mass m with thrust in the radial direction u 1 and in the tangential direction u 2. From Skelton, DSC, p. 11. Newton s law where f g is the gravitational force m r = u 1 + u 2 + f g, f g = km r 2 r r. Using cylindrical coordinates ( cosθ e 1 = sinθ we have ), e 2 = ( ) sinθ, cosθ r = r e 1, u 1 = u 1 e 1, u 2 = u 2 e 2, fg = km r 2 e 1. MAE 28 B 11 Maurício de Oliveira

8 We need to compute where That is r = d2 dt 2(r e 1) = d dt (ṙ e 1 +r e 1 ) = r e 1 +2ṙ e 1 +r e 1, θ( ) sinθ e 1 = = cosθ θ e 2, θ( ) ( ) sinθ e 1 = + cosθ θ cosθ 2 sinθ r = ( r r θ 2 ) e 1 +(2ṙ θ +r θ) e 2, = θ e 2 θ 2 e 1. so that Newton s law can be rewritten as m( r r θ 2 ) e 1 +m(2ṙ θ +r θ) e 2 = u 1 e 1 +u 2 e 2 km r 2 e 1, or, equivalently, as the two scalar differential equations m( r r θ 2 ) = u 1 km r 2, m(2ṙ θ+r θ) = u 2. In state space r x = θ ṙ, ẋ = θ ṙ θ r = θ x 3 x 4 x 1 x 2 4 k/x2 1 2x 3 x 4 /x 1 + 1/m 1/(mx 1 ) ( u1 u 2 ). This is a nonlinear system and we look for equilibrium ( r = θ = ) when u 1 = u 2 =. This can be stated as x 1 x 2 4 k/x 2 1 =, 2x 3 x 4 /x 1 =. The second condition implies x 3 = ṙ and/or x 4 = θ must be zero. We choose x 3 = ṙ = which implies x 1 = r = r constant and x 4 = θ k k = = r = ω k = r3 ω 2. 3 Note also that x 2 = θ = ωt. x 3 1 MAE 28 B 12 Maurício de Oliveira

9 This nonlinear system is in the form ẋ = f(x,t)+g(x)u. We will linearize f(x,t) and g(x)u around the equilibrium point ( x,ū) to obtain the linearized system ẋ = ( f x ) T [x(t) x(t)]+( g x ) T [x(t) x(t)]ū+g( x)u. For this problem r x(t) = ωt, ū =, f(x,t) = ω and ( f x ) T = ẋ = 1 1 x 3 x 4 x 1 x 2 4 k/x 2 1 2x 3 x 4 /x 1 2k/ r 3 + ω 2 2 r ω 2 ω/ r This produces the linearized system ω 2 2 r ω 2 ω/ r = x+, g(x) = 1/m 1/(mx 1 ) ω 2 2 r ω 2 ω/ r 1/m 1/(m r) ( u1 u 2 ). If we looking at the satellite (from the earth) we can say that we can observe r and θ (how?), that is [ ] 1 y = x. 1 Questions: 1) Can we estimate the state of the satellite by measuring only r? 2) Can we estimate the state of the satellite by measuring only θ? 3) Can we estimate the state of the satellite by measuring r and θ? 4) Can the system be controlled to remain in circular orbit using radial thrusting (u 1 ) alone? 5) Can the system be controlled using tangential thrusting (u 2 ) alone?., MAE 28 B 13 Maurício de Oliveira

10 Question: Can we estimate the state of the satellite by measuring only r? Answer: Is the system observable when C = [ 1 ]? Compute the observability matrix C O(A,C) = CA CA 2, CA 3 1 = 1 3 ω 2 2 r ω ω 2 Physical interpretation: measuring r does not give any information on θ or θ! Note that if we that know the satellite is in equilibrium and measure k then k θ = ω = r 3. But we still do not know θ since and we do not know θ()! θ(t) = θ()+ωt, MAE 28 B 14 Maurício de Oliveira

11 Question: Can we estimate the state of the satellite by observing θ only? Answer: Is the system observable when C = [ 1 ]? Compute the observability matrix C O(A,C) = CA CA 2, CA 3 1 = 2 ω/ r 6 ω 3 / r 4 ω 2 2 ω 3 / r Physical interpretation: again, if we try to reconstruct θ from θ we still need to know θ at some t! From that point on θ = θ( t)+ t t θ(τ)dτ. MAE 28 B 15 Maurício de Oliveira

12 Question: Can we estimate the state of the satellite by measuring r and θ? [ ] 1 Answer: Is the system observable when C =? Compute the 1 observability matrix C O(A,C) = CA CA 2, CA ω/ r = 3 ω 2 2 r ω 6 ω 3 / r 4 ω 2 ω 2 2 ω 3 / r Physical interpretation: can we estimate θ at all? MAE 28 B 16 Maurício de Oliveira

13 Question: Can the system be controlled to remain in circular orbit using radial thrusting (u 1 ) alone? Answer: Is the system controllable when B = controllability matrix Note that 1/m C(A,B) = [ B AB A 2 B A 3 B ], = 1 m ω 2 1 ω 2 2 ω/ r 1 ω 2 2 ω/ r 2 ω 3 / r 1 2 ω/ r = ω 2 2 ω 3 / r which implies that the system is not controllable from u 1!? Compute the Physical interpretation: there must be a change in the angular velocity θ if one changes the radius! MAE 28 B 17 Maurício de Oliveira

14 Question: Can the system be controlled using tangential thrusting (u 2 ) alone? Answer: Is the system controllable when B =? Compute the 1/(m r) controllability matrix C(A,B) = [ B AB A 2 B A 3 B ], = 1 m r 2 r ω 1 4 ω 2 2 r ω 2 r ω ω 2 The above matrix has full rank, so the system is controllable from u 2! Physical interpretation: we can change the radius by changing the angular velocity! MAE 28 B 18 Maurício de Oliveira

15 1.11 More on Gramians Theorem: The Controllability Gramian X(t) = t is the solution to the differential equation If X = lim t X(t) exists then e Aξ BB T e ATξ dξ, d dt X(t) = AX(t)+X(t)AT +BB T. AX +XA T +BB T =. Theorem: The Observability Gramian Y(t) = t is the solution to the differential equation If Y = lim t X(t) exists then e ATξ C T Ce Aξ dξ, d dt Y(t) = AT Y(t)+Y(t)A+C T C. A T Y +YA+C T C =. MAE 28 B 19 Maurício de Oliveira

16 Proof (Controllability): For the first part, compute t d dt X(t) = d e Aξ BB T e ATξ dξ = d e A(t τ) BB T e AT (t τ) dτ, dt dt t d = dt ea(t τ) BB T e AT (t τ) + e A(t τ) BB T e AT (t τ), τ=t ( t ) = A e A(t τ) BB T e AT (t τ) dτ ( t ) + e A(t τ) BB T e AT (t τ) dτ A T +BB T, = AX(t)+X(t)A T +BB T. For the second part, use the fact that X(t) is smooth and therefore t d lim X(t) = X lim X(t) =. t t dt MAE 28 B 2 Maurício de Oliveira

17 Lemma: Consider the Lyapunov Equation where A C n n and C C m n. A T X +XA+C T C = 1. A solution X C n n exists and is unique if and only if λ j (A)+λ i (A) for all i,j = 1,...,n. Furthermore X is symmetric. 2. If A is Hurwitz then X is positive semidefinite. 3. If (A,C) is detectable and X is positive semidefinite then A is Hurwitz. 4. If (A,C) is observable and A is Hurwitz then X is positive definite. Proof: Item 1. The Lyapunov Equation is a linear equation and it has a unique solution if and only if the homogeneous equation associated with the Lyapunov equation admits only the trivial solution. Assume it does not, that is, there X such that A T X + XA = Then, multiplication of the above on the right by x i, the ith eigenvector of A and on the right by x j yields = x i AT Xxj +x i XAx j = [λ j (A)+λ i (A)]x i Xx j. Since λ i (A)+λ j (B) by hypothesis we must have x i Xx j = for all i,j. One can show that this indeed implies X =, establishing a contradiction. That X is symmetric follows from uniqueness since so that X T X =. = (A T X +XA+C T C) T (A T X +XA+C T C) = A T (X T X)+(X T X)A Item 2. If A is Hurwitz then lim t e At =. But and X = A T X +XA = lim t e ATt C T Ce At dt d dt eatt C T Ce At dt = e ATt C T Ce At = CT C. MAE 28 B 21 Maurício de Oliveira

18 Item 3. Assume (A,C) is detectable, X and that A is not Hurwitz. Then there exists λ and x such that Ax = λx with λ+λ. But if X solves the Lyapunov equation x C T Cx = x A T Xx+x XAx = (λ+λ )x Xx which implies Cx =, hence (A,C) not detectable. Item 4. Assume that (A,C) is observable and A is Hurwitz. From Item 2. X. Assume X is not positive definite, that is, there exists x such that X x =. It follows that = x X x = x e ATt C T Ce At xdt = y (t)y(t)dt which implies that response y(t) = Ce At x = to a non null initial condition x() = x is null, which contradicts the hypothesis that (A,C) is observable. MAE 28 B 22 Maurício de Oliveira