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EECE 568: Control Systems Handout #2: State transition for LTI systems

Dr. Meeko Oishi http://courses.ece.ubc.ca/568

October 15, 2009

Similarity Transformations

For diagonalizable A, every eigenvalue λi solves Aλi = viλi with eigenvector vi. With n independent eigenvectors as the columns of the n × n matrix V , and the eigenvalues λ1, · · · , λn the elements of the diagonal matrix Λ, λ1 0 · · · 0   0 λ 0 −1 2 A = V ΛV , Λ =  . . .  (1)  . .. .     0 0 · · · λn  i For non-diagonalizable (defective) A, generalized eigenvectors vj corresponding to eigenvalue λi i i solve Avj+1 = λivj, j ∈ {1, · · · , ni−1}. The generalized eigenvectors form the columns of the matrix T . The block-diagonal Jordan matrix J consists of bidiagonal Jordan blocks Ji, i ∈ {1, · · · , q}, q with ni = n. Pi=1 λ J1 i 1 · · · 0    . .  J2 . . −1 0 λi . . ni×ni A = T JT , J = , with Ji =   ∈ R (2)  ..   . . .   .   . .. .. 1       Jq   0 · · · 0 λi 

Continuous-Time Systems For the multi-input, multi-output, linear time-invariant system

x˙(t) = Ax(t) + Bu(t), x ∈ Rn, u ∈ Rm, with x(t ) = x 0 0 (3) y(t) = Cx(t) + Du(t), y ∈ Rp with real-valued matrices A, B, C, D of the appropriate dimensions, the state transition matrix k A(t−t0) ∞ k (t−t0) Φ(t, t0) = e = A is the . Assume t0 = 0 for convenience in Pk=0 k! the following.

1 For diagonalizable A, the state transition matrix is eAt = V eΛtV −1, with

eλ1t 0 · · · 0   0 eλ2t · · · 0 Λt e =  . . .  (4)  . .. .     0 0 · · · eλnt  and the resolvent is (sI − A)−1 = V (sI − Λ)−1V −1, with

1 0 · · · 0  s+λ1  0 1 · · · 0 −1 s+λ2 (sI − Λ) =  . . .  (5)  . .. .   1   0 0 · · · s+λn 

For defective A, the state-transition matrix is eAt = T eJtT −1, with

n λit λit t i λit J1t e te e e 0 · · · 0 · · · ni!     J2t λit . . 0 e · · · 0 0 e .. . × eJt = , and eJit =   ∈ Rni ni (6)  . .. .   . . .   . . .   . .. .. teλit   Jnt     0 0 · · · e   0 · · · 0 eλit  and the resolvent is (sI − A)−1 = V (sI − Λ)−1V −1, with

−1 (sI − J1) 0 · · · 0  −  0 (sI − J ) 1 · · · 0 −1 2 (sI − J) =  . . .  , and  . .. .   −1   0 0 · · · (sI − Jn)  1 1 1 2 · · · n (7)  s+λi (s+λi) (s+λi)i  . . 0 1 .. . −1  s+λi  Rni×ni (sI − Ji) =   ∈ . .. .. 1  . . . 2   (s+λi)   1   0 · · · 0 s+λi 

Discrete-Time Systems For the multi-input, multi-output, linear time-invariant system

x[k + 1] = Ax[k] + Bu[k], x ∈ Rn, u ∈ Rm, with x(k ) = x 0 0 (8) y[k] = Cx[k] + Du[k], y ∈ Rp with real-valued matrices A, B, C, D of the appropriate dimensions, the state transition matrix is (k−k0) Φ(k, k0) = A . Assume k0 = 0 for convenience in the following.

2 For diagonalizable A, the state transition matrix is Ak = V ΛkV −1, with

k (λ1) 0 · · · 0  λ k  k 0 ( 2) · · · 0 Λ =  . . .  (9)  . .. .   k   0 0 · · · (λn) 

For defective A, the state transition matrix is Ak = T J kT −1, with

− k (λ )k k(λ )k 1 · · · k! (λ )1 (J1) 0 · · · 0 i i 1!(k−1)! i     k k . . 0 (J2) · · · 0 0 (λ ) .. . × J k , J k i Rni ni =  . . .  and ( i) =  .  ∈  . .. .   . .. .. k−1     . . . k(λi)  J k  k   0 0 · · · ( n)   0 · · · 0 (λi)  (10) k k! k−j The (j + 1)th column in the first row of (Ji) has the value (k−j)!j!(λi) . The resolvent (used in calculation of transfer functions in the z-domain) is the same as was calculated for the continuous-time system, for A diagonalizable and for A defective, respectively.

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