1. The Transition Matrix (Hint: Recall that the solution to the linear equation ẋ = Ax + Bu is

Transcription

1 ECE 55, Fall 2007 Problem Set #4 Solution The Transition Matrix (Hint: Recall that the solution to the linear equation ẋ Ax + Bu is x(t) e A(t ) x( ) + e A(t τ) Bu(τ)dτ () This formula is extremely important for linear systems and should be memorized) First solve for x 2 x 2 tx 2 dx 2 /x 2 tdt ln x 2 (t) ln x 2 ( ) 2 (t2 t 2 0) x 2 (t) e 2 (t2 t 2 0 ) x 2 ( ) Substituting this into equation for x : ẋ x + e t 2 t2 x 2 x + e t 2 t2 0 x2 ( ) Since x 2 is not a function of x, we can regard it as the input u(t) So with A, B e 2 t2 0 x2 (, u(t) e t, we apply () to obtain: x (t) e (t t0) x ( ) + e (t τ) e τ 2 t2 0 x2 ( )dτ e (t ) x ( ) + e t 2 t2 0 x2 ( )(t ) We have ( ) [ x (t) x 2 (t) e (t ) (t )e t 2 t2 0 0 e 2 (t2 t 2 0 ) ] (x ) ( ) x 2 ( ) and so, by definition, the state transition matrix is [ ] e Φ(t, ) (t ) (t )e t 2 t2 0 0 e 2 (t2 t 2 0 ) You can check that this matrix indeed satisfies t Φ(t, ) A(t)Φ(t, ); Φ(, ) I, and thus, the solution of the system is x(t) Φ(t, )x( ) 2 Matrix Linear Differential Equation (0 points) (Hint: The Leibnitz Integral Rule: d dt a f(t, τ)dτ f(t, t) + a f(t, τ) dτ for a constant a t

2 See First we check the initial condition X( ) I X 0 I + 0 X 0 If Φ i is the transition matrix of ẋ A i x, then Φ i (t, ) t A i Φ i (t, ) Then using the Leibniz rule: [ ] [ ] Ẋ(t) t Φ (t, ) X 0 Φ T 2 (t, ) + Φ (t, )]X 0 t ΦT 2 (t, ) [ ] t [ ] + t Φ (t, τ) F (τ)φ T 2 (t, τ) dτ + Φ (t, τ)f (τ) t ΦT 2 (t, τ) dτ + F (t) A (t)φ (t, )X 0 Φ T 2 (t, ) + A (t) Φ (t, τ)f (τ)φ T 2 (t, τ) dτ + Φ (t, )X 0 Φ T 2 (t, )A T 2 (t) + Φ (t, τ)f (τ)φ T 2 (t, τ) dτa T 2 (t) + F (t) A (t)x(t) + X(t)A T 2 (t) + F (t) Hence X(t) as given is a solution to the original system 3 Convergence of LTI systems (5 points) (a) We use the Jordan from (see Homework 3 problem 4; notice that here we have e At instead of e A so you need to include the variable t accordingly) You can also calculate by other methods (eg using the Cayley-Hamilton theorem or inverse Laplace transform) e A t e A 2t e A 3t 0 At ( ) ( ) ( ) ( ) 0 0 P e P 0 t 0 t t 0 t + t ( ) ( ) ( ) ( 3 e t 0 3e 2 0 e t t 2e t 3e t + 3e t ) 2 3 2e t 2e t 2e t + 3e t C A t P e 0 0 P 0 0 e 2t e t te t e 0 e te t 0 e 2t 3e 2t 3e t 0 0 e t 2

3 (b) ( ) A : x(t) e At t(x2 (0) x x(0) (0)) + x (0) t(x 2 (0) x (0)) + x 2 (0) ( A 2 : x(t) e A2t e x(0) [3x (0) 3x 2 (0)] e t ) [2x (0) 3x 2 (0)] e t [2x (0) 2x 2 (0)] e t [2x (0) 3x 2 (0)] e t [x (0) + tx 3 (0)] A 3 : x(t) e A3t x(0) e [x 2 (0) + 3x 3 (0)] 3e t x 3 (0) x t x 3 (0) (c) A system: the solution stays bounded if x (0) x 2 (0), or goes to infinity otherwisethe subspace span{(, ) T } is stable in the sense of Lyapunov, but it is not a stable subspace (just check the definition of stability subspaces) A 2 system: if x (0) x 2 (0), the solution decays to 0, and goes to infinity otherwise The span{(, ) T } is its stable subspace corresponding to eigenvalue A 3 system: asymptotically decays to 0 for all initial conditions (d) Let λ i be the eigenvalues of the matrix A i The solution will decay to zero for arbitrary initial condition (ie the system is asymptotically stable) if and only if Re(λ i ) < 0 i ii If Re(λ i ) 0, and the eigenvalues with zero real parts have linearly independent eigenvectors (note: not generalized eigenvectors Generalized eigenvectors are not eigenvectors), the system will be bounded in the sense of Lyapunov iii Otherwise, the system is unstable (it can go to infinity for some initial states) If there exists some Re(λ i ) < 0, their corresponding eigenvectors will span an asymptotically stable subspace If Re(λ i ) > 0 for all i, the system is unbounded for all nonzero initial states To sum up, these are things one need to remember: i Re(λ) < 0 x 0 span(eigenvectors, gen eigenvectors) asymptotic stability ii Re(λ) 0 + have only eigenvectors, not gen eigenvectors x 0 span(eigenvectors) stability (but not asymptotic) iii Re(λ) 0 + have gen eigenvectors x 0 span(eigenvectors, gen eigenvectors) instability iv Re(λ) > 0 x 0 span(eigenvectors, gen eigenvectors) instability 4 Linear Time-Varying System with Negative Eigenvalues (0 points) For the given LTV system: (a) Similar to problem, we solve for x(t) ẋ 2 x 2 x 2 (t) e (t ) x 2 ( ) 3

4 Then solve for x (t): ẋ x + e 2t x 2 x (t) e (t ) x ( ) + e (t τ) e 2τ e (τ ) x 2 ( )dτ e (t ) x ( ) + 2 (et+ e t+3 )x 2 ( ) And thus, ( ) [ x (t) e (t ) 2 (et+ e t+3t ] ( ) 0 ) x ( ) x 2 (t) 0 e (t ) x 2 ( ) and hence, the state transition matrix is [ e (t ) Φ(t, ) 2 (et+ e (t 3t0) ) 0 e (t ) (b) The solution to the differential equation with x(0) [0, ] T is: x(t) Φ(t, 0)x(0) [ 2 (et e t ) e t (c) The eigenvalues of A(t) are, for all t 0 However, since A is time varying, this does not necessarily imply that the system is stable (d) As t, the first component of x above becomes unbounded, whereas the second component goes to zero This may at first seem surprising since both eigenvalues of A are negative, but not on a second thought since A is time-varying Also, this is inline with part (c) 5 Solution of Linear Systems (0 points) [ ] 0 We first find the solution of the system Eigenvalues of are, 4 and the corresponding eigenvectors are (, ) T, (, 4) T 4 5 Then t [ ] 4 0 5t [ ] x(t) e x(0) e x(0) 4 4 [ 4e t + e 4t e t + e 4t ] 3 4e t 4e 4t e t 4e 4t x(0) ( ( 4x0 x 02 )e t + (x 0 + x 02 )e 4t ) 3 (4x 0 + x 02 )e t 4(x 0 + x 02 )e 4t and hence, y(t) [ ] x(t) (x 0 + x 02 )e 4t ( ) Therefore, if x(0) span{ }, the mode corresponding to λ 4 does not appear at the output ie it s guaranteed that only the mode corresponding to the eigenvalue is excited at the output (which is zero in this case) 3 ] ] 4

5 Another solution: Eigenvalues λ, λ 2 4 and eigenvectors v (, ) T, v 2 ( 4) T, correspondingly Since the eigenvectors are linearly independent, for any x 0, we have x 0 αv + βv 2 for some α, β Then x(t) e At x 0 e At (αv + βv 2 ) αe At v + βe At v 2 Using the fact that Av λv and using Taylor expansion of e At, we can show that e At v e λ t v Similarly, e At v 2 e λ 2t v 2 Hence and so, x(t) αe λ t v + βe λ 2t v 2, y(t) Cx(t) αe λ t Cv + βe λ 2t Cv 2, We have Cv 2 0 so in order for e λ2t not appearing at the output, ( ) we must have β 0 Thus, x 0 is of the form αv, which means x 0 span{v } span{ } 6 Linear Time-Varying Systems with (Skew) Symmetric Matrix A(t) (0 points) (a) Choose a Lyapunov function candidate V (x, t) x T x It s clear that V is positive definite ie V (x) 0 for all x and V (x) 0 x 0 Then: V (x, t) x T A T x + x T Ax x T (A T + A)x 0 The last equation uses the fact that A is skew symmetric Since V (x, t) 0, the system is stable It is uniformly stable because V does not depend explicitly on t (b) Again choose a Lyapunov function candidate V (x, t) x T x It s clear that V is positive definite ie V (x) 0 for all x and V (x) 0 x 0 Then: V (x, t) x T A T x + x T Ax 2x T Ax 2x T (ɛi)x 2ɛx T x < 0 x 0 Hence the system is uniformly asymptotically stable (it is exponentially stable in this case) Uniformity follows from the fact that V does not depend explicitly on t (it s a function of x only) 7 Periodic linear systems (0 points) (a) We have P (t)e R(t t0) P ( ) Φ(t, 0)e Rt e R(t ) e R Φ (, 0) Φ(t, 0)Φ (, 0) Φ(t, ) Notice that for matrices, (AB) B A (b) Since x(t) Φ(t, 0)x(0), x(t) P (t)x(t) e Rt Φ (t, 0)Φ(t, 0)x(0) e Rt x(0) Therefore, x Re Rt x(0) R x 5

6 (c) By the definition of the transition matrix, Φ(t, ) A(t)Φ(t, ) Letting 0 and T, we have the following two differential equations: { Φ(t, 0) A(t)Φ(t, 0) Φ(T + t, T ) A(t + T )Φ(T + t, T ) Since A(t) to be periodic, A(t) A(t + T ) t, and we then have { Φ(t, 0) A(t)Φ(t, 0) Φ(T + t, T ) A(t)Φ(T + t, T ) Since Φ(T, T ) Φ(0, 0) I, which means the boundary conditions for t 0 are also the same, we have that Φ(T + t, T ) and Φ(t, 0) are the state transition matrices of the same equation ẋ A(t)x By the uniqueness of solutions of differential equations, we must have Φ(T + t, T ) Φ(t, 0) Note: We do not have a close form of Φ(t, ) for a time-varying A(t) in general, but only the Peano-Baker series expansion of it (page 56, ECE55 course note) (d) Property of the transition matrix: Φ(t+T, 0) Φ(T +t, T )Φ(T, 0) and so, Φ (t+t, 0) Φ (T, 0)Φ (T + t, T ) We then have P (t + T ) e R(t+T ) Φ (t + T, 0) e Rt e RT Φ (T, 0)Φ (t + T, T ) e Rt Φ (t + T, T ), since Φ(T, 0) e RT is given by the problem Since Φ(t, ) is periodic with period T, Φ (t, ) is also periodic with period T Thus, P (t + T ) e Rt Φ (t, 0) P (t) 6