EE263: Introduction to Linear Dynamical Systems Review Session 6

Transcription

1 EE263: Introduction to Linear Dynamical Systems Review Session 6 Outline diagonalizability eigen decomposition theorem applications (modal forms, asymptotic growth rate) EE263 RS6 1

2 Diagonalizability consider square matrix A R n n. Assume that λ C is an eigenvalue of A then, A is diagonalizable if and only if the multiplicity of λ i equals dim(n(λ i I A)) = n rank(λ i I A), for all i EE263 RS6 2

3 Diagonalizability to see this, note that we can write the characteristic polynomial of A as χ(λ) = (λ λ 1 ) n 1 (λ λ 2 ) n 2...(λ λ k ) n k the eigenvectors corresponding to λ i are given by the linear system or (A λ i I)X = 0 AX = λ i X so if dimn(λ i I A) = n i, then we can find n i independent eigenvectors corresponding to λ i if this condition holds for all i, then the matrix A is diagonalizable (and vice versa) EE263 RS6 3

4 Example is this matrix diagonalizable? A = R 3 3 Solution. let s write the characteristic polynomial for A: χ(a) = det(λi A) = (λ 4)(λ 3) 2 eigenvalues are λ 1 = 4 with multiplicity 1, and λ 2 = 3 with multiplicity 2 the condition is satisfied for λ 1, since λ 1 has multiplicity 1 EE263 RS6 4

5 for λ 2, since λ 2 I A = has rank 2, we have n rank(λ 2 I A) = 3 2 = 1, which is not the multiplicity of λ 2 this means that we cannot find two independent eigenvectors corresponding to λ 2 thus A is not diagonalizable (in fact, A is in Jordan canonical form) EE263 RS6 5

6 Example show that the following statement is true: if A C n n has distinct eigenvalues, i.e., λ i λ j for i j, then A is diagonalizable (Lecture 11-22). Solution. the characteristic polynomial of A has order of n. since its roots are distinct, we have n different eigenvalues with multiplicity of 1. without loss of generality, let s show dim(n(λ 1 I A)) = 1 clearly, v i αv 1 for i = 2,...,n since v 1 / span{v 2,...,v n }, {v 1,...,v n } is linearly independent also v i / N(λ 1 I A) for i = 2,...,n which implies dim(n(λ 1 I A)) = 1 therefore, dim(n(λ i I A)) = 1 for i = 1,...,n EE263 RS6 6

7 Example show that the following statement is true: It is possible for A to have repeated eigenvalues, but still be diagonalizable (Lecture 11-22) Solution. as an example, assume A = with X(λ) = (λ 1) 2 (λ 2) has multiplicity of 2 for λ 1 = 1 but still is diagonalizable here, e 1 and e 2 are eigenvectors for λ 1, and e 3 is for λ 2, so the condition still holds EE263 RS6 7

8 Eigen decomposition theorem if a matrix A C n n is diagonalizable, A can be written as an eigen decomposition A = TΛT 1 n = λ i v i wi T i=1 where T = [v 1 v n ], Λ = diag(λ 1,...,λ n ), and v i is the eigenvector corresponding to eigenvalue λ i if A is symmetric, i.e., A = A T, then A is diagonalizable and can be written as A = QΛQ T n = λ i q i qi T i=1 EE263 RS6 8

9 where Q = [q 1 q n ] is an orthogonal matrix you may see that this is useful for positive definiteness soon! EE263 RS6 9

10 Application: Modal form suppose A C n n is diagonalizable, i.e., A = TΛT 1 then we can rewrite ẋ = Ax as x = Λ x where x = T 1 x EE263 RS6 10

11 Application: Real modal form when eigenvalues (hence T) are complex, system can be put in real modal form assume A R n n is diagonalizable, i.e., A = TΛT 1, and has a complex eigenvalue λ i = σ i +jω i with corresponding eigenvector v i, i.e., Av i = λ i v i taking the conjugate, we have A v i = λ i v i, which implies v i is also an eigenvector corresponding to the eigenvalue λ i we can select λ i+1 = λ i = σ i jω i from this argument, we can show: Av i = λ i v i = (σ i +jω i )(R(v i )+ji(v i )) EE263 RS6 11

12 = σ i R(v i ) ω i I(v i )+j(ω i R(v i )+σ i I(v i )) and Av i+1 = A v i = λ i v i = (σ i jω i )(R(v i ) ji(v i )) = σ i R(v i ) ω i I(v i ) j(ω i R(v i )+σ i I(v i )) and thus, AR(v i ) = σ i R(v i ) ω i I(v i ) AI(v i ) = ω i R(v i )+σ i I(v i ) EE263 RS6 12

13 which can be rewritten in a matrix form: [ ] σi ω [R(v i ) I(v i )] i = A[R(v ω i σ i ) I(v i )] i now, let s bundle the real and complex eigenvalues together. assume λ i R for i = 1,...,r and λ i C\R for i = r +1,...,n and let and Λ r = [λ 1 λ r ], M i = [ σi ω i ω i σ i S = [ v 1 v r R(v r+1 ) I(v r+1 ) R(v r+3 ) I(v r+3 ) ] since S is invertible, we finally have ], S 1 AS = diag(λ r,m r+1,m r+3,...,m n 1 ) EE263 RS6 13

14 Example: Standard form for LDS given LDS as ẋ = Ax+Bu and y = Cx suppose A is diagonalizable, i.e., A = TΛT 1 let s change coordinates with x = T x then, x = T 1 ẋ = T 1 AT x+t 1 Bu = Λ x+ Bu y = CT x = C x where B = T 1 B and C = CT EE263 RS6 14

15 Example suppose that A R n n + is diagonalizable, and its eigenvalues are nonnegative. consider a discrete-time linear dynamic system, x(t+1) = Ax(t), and a given nonnegative initial state, i.e., x i (0) 0 for i = 1,...,n how can we find the asymptotic growth rate of sum, i.e., 1T x(t+1) 1 T x(t) t? as Solution. since A is diagonalizable, we can rewrite A = n i=1 λ iv i w T i as above EE263 RS6 15

16 without loss of generality let s assume that λ 1... λ n 0, and let s take m as the index of the largest eigenvalue for which x(0 is not in the associated nullspace, i.e., wmx(0) T 0, and wi T x(0) = 0 for i = 1,...,m 1 then, 1 T x(t) = 1 T A t x(0) = n i=1 λt i (1T v i )(w T i x(0)) as t, we get 1 T x(t) λ t m(1 T v m )(w T mx(0)) therefore, 1T x(t+1) 1 T x(t) λ m as t EE263 RS6 16

17 Transfer matrix and impulse matrix assume LDS is given as ẋ(t) = Ax(t)+Bu(t) and y(t) = Cx(t)+Du(t) taking Laplace transform, we can get: sx(s) x(0) = AX(s)+BU(s) X(s) = (si A) 1 x(0)+(si A) 1 BU(s) Y(s) = C(sI A) 1 x(0)+(c(si A) 1 B +D) U(s) }{{} H(s) since (si A) 1 Le ta, we can get: x(t) = e ta x(0)+e ta Bu(t) and y(t) = Ce ta x(0)+(ce ta B +Dδ(t)) u(t) }{{} h(t) where H(s) and h(t) are transfer matrix and impulse matrix, respectively EE263 RS6 17

18 discretization: if we sample the states and outputs with interval h as x d (k) = x(kh) and y d (k) = y(kh), we can get: and x d (k +1) = x(kh+h) = e (kh+h)a x(0)+e ta Bu(t) t=kh+h = e ha( e kha x(0)+e ta ) Bu(t) t=kh + h t=0 = e ha x d (k)+ e (h τ)a Bu(τ +kh)dτ h t=0 e (h τ)a Bu(τ +kh)dτ y d (k) = Ce kha x(0)+h(t) u(t) t=kh = Cx d (k)+du(kh) EE263 RS6 18

19 Example when inputs are piecewise constant,i.e., u(t) = u d (k) for kh t < (k+1)h, rewrite the linear dynamic system in the discretized form Solution. with the piecewise constant inputs, the above system of equations can be rewritten as: x d (k +1) = e ha x d (k)+ = e ha }{{} A d x d (k)+ h t=0 ( h e (h τ)a Bu(τ +kh)dτ t=0 e τa dτ ) } {{ } B d Bu d (k) and y d (k) = }{{} C x d (k)+ }{{} D u d (k) C d D d EE263 RS6 19