Homogeneous and particular LTI solutions Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome Tor Vergata Fondamenti di Automatica e Controlli Automatici A.A. 2014-2015 1 / 11
Change of coordinates The general solution for a continuous-time LTI system is given by and in discrete-time it is As an example, assume that t x(t) = e A(t t0) x 0 + e A(t τ) Bu(τ) dτ t 0 (1) x(t) = k 1 x(k) = A k x 0 + A k h 1 Bu(h). (2) [ ] x1 (2) = x 2 (t) [ h=0 x 1 (0)e 2t ] x 1 (0)e 2t + x 2 (0)e t, and consider the change of coordinates [ ] [ 1 0 x1 (0)e z(t) = T x(t) = x(t) = 2t ] 1 1 x 2 (0)e t, (3) that allows to see in a simpler way the system modes: it is a change of coordinates operated via the matrix (invertible) T. 2 / 11
Change of coordinates: general approac Assume that the matrix A R n n is diagonalizable. A sufficient condition is when all the eigenvalues of A are distinct, i.e. σ{a} = n. Then define a new state variable z R n as z = T x, (4) where T is invertible and can be taken such that T 1 [ w 1, w 2,..., w n ] (5) where w i is i-th left-eigenvector (column) of the i-th eigenvalue λ i σ{a}. The time derivative of the new variable z is ż = T ẋ = T (Ax + Bu) = T A(T 1 z) + T Bu = T AT 1 z + T Bu, = Âz + ˆBz,  T AT 1, ˆB T B. (6) It is possible to show (do it...) that if T is picked as in (5), then  is diagonal, i.e. λ 1 0... 0 0 λ 2... 0  =........ = diag{λ 1, λ 2,..., λ n}. (7) 0... 0 λ n 3 / 11
Change of coordinates: general approac cont d In the previous case, although all the left-eigenvalues w i have to be evaluated, the exponential matrix of Â, that is diagonal, can be easily computed as (prove it) e λ 1 t 0... 0 0 e eâ t λ 2 t... 0 =......... (8) 0... 0 e λn t Note also that if T is chosen as in (5), then (prove it) v 1 v 2 T =..., (9) v n where v i is the i-th right-eigenvalue (column) of A of the i-th eigenvalue λ i σ{a}. Since  is diagonal, it would be convenient to evaluate the exponential matrix eât and then write back x(t) as (eât t ) x(t) = T 1 z(t) = T 1 z 0 + eâ(t τ) T Bu(τ), 0 t = T 1 eât T x 0 + T 1 eâ(t τ) T B u(τ) (10) 0 4 / 11
Change of coordinates: general approac cont d The previous formula allows to conclude that e At = T 1 eât T. (11) Everything holds in discrete-time as well (prove it), i.e. z + = T x + = T (Ax + Bu) = T A(T 1 z) + T Bu = T AT 1 z + T Bu, = Âz + ˆBz, Â T AT 1, ˆB T B. (12) and then ) k 1 x(k) = T 1 z(k) = T (Â 1 k z 0 + Â k h 1 T Bu(h), h=0 k 1 = T 1 Â k T x 0 + T 1 Â k h 1 T Bu(h), (13) h=0 yielding A k = T 1 Â k T. (14) 5 / 11
The modes of the system in the new coordinates z, when A is diagonalizable, are then written as + diag{λ 1 t, λ 2 t,..., λ nt} ˆΦ(t) h = eât = = e diag{λ 1t, λ 2 t,...,λ nt}, h! h=0 The same in discrete-time = diag{e λ 1t, e λ 2t,..., e λnt }. (15) ˆΦ(k) = Âk = diag{λ 1, λ 2,..., λ n} k = diag{λ k 1, λk 2,..., λk n }. (16) The spectral decomposition of A, when  is diagonal, is. λ. 1 0... 0 v 1. 0 λ A=T 1 2... 0 v 2 ÂT = w 1 w 2... w n.......... = n λ i w i v i,.. 0... 0 λ n v n (17) yielding e At = e λit w i v i, and Ak = λ k i w iv i. (18) 6 / 11
and the homogenous solution Diagonalizable dynamic matrix Recall that the solution x of a LTI system can be written as x = x l + x f, then by spectral decomposition of A it holds x l (t) = Φ(t)x 0 = e At x 0 = e λit w i v i x 0 = }{{} c i e λit w i c i, (19) and when only the free response x l has to be retrieved, then c i = T x 0. The same holds in discrete time x l (k) = Φ(k)x 0 = A k x 0 = λ k i w i v i x 0 = }{{} c i λ k i w ic i. (20) However, note that the general solution when there is not resonance can be written as x = x l + x f = L 1 { (si A) 1 x 0 + (si A) 1 Bu(s) }, { ν m i = L 1 1 ν (s λ h=1 i ) h R m i i,h x 0 + h=1 ξ i,h r (s λ i ) h + r i h=1 } χ i,h (s α i ) h, (21) 7 / 11
The homogenous x l and particular x u solution Diagonalizable dynamic matrix Then, without resonance, it is then possible to write in continuous time { ν } m i x o(t) L 1 R i,h x 0 + ξ i,h (s λ h=1 i ) h = e }{{} λit w i c i, (22) m i =1 { r } r i x u(t) L 1 χ i,h r r i t h 1 (s α h=1 i ) h = χ i,h (h 1)! eαt, (23) h=1 and in discrete time { ν m i x o(k) Z 1 z R } i,hx 0 + ξ i,h (z λ h=1 i ) h = λ }{{} k i w i c i, (24) m i =1 { r } r i x u(k) 1 χ i,h r r i ( ) k z (z α h=1 i ) h = χ i,h λ k h+1 h 1 i α k, (25) h=1 8 / 11
Diagonalizable dynamic matrix Diagonalizable dynamic matrix The following statement are equivalent: A matrix A is diagonalizable iff n i = µ i, where n i is the degree of the root λ i of the matrix A polynomial characteristic p A (λ) and µ i = dim{ker{a λ i I} the denominator of rational function (si A) 1 or (zi A) 1 has simple roots (simple poles) m i = 1 in eq. (22)-(24) Then, if there exists an initial condition x 0 such that x l ( ) has polynomial modes, the matrix A is not diagonalizable (Jordan form). See on the text-book the Jordan forms. 9 / 11
System discretizion Approximating the derivative of x(t) as x(t + T ) x(t) + T ẋ(t), a continuous time dynamical system described by ẋ = f(x, u) sampled at times kt, where T > 0 is the sampling time and k N, can be approximated by a discrete time LTI system via Backward Euler such as x(kt ) = x((k 1)T ) + T (f(x((k + 1)T ), u((k + 1)T )), (26) or simply, with an abuse of notation, as that for LTI systems becomes x(k + 1) = x(k) + T f(x(k), u(k)), x(k) = x(k 1) + T (Ax(k 1) + Bu(k 1)). (27) Selection and requirements on the sampling time T are not discussed here. 10 / 11
Exact discretizion for LTI Assume that the input u is constant within intersample times, then the exact discretization of a continuous time LTI system is given by x((k + 1)T ) = A d x(kt ) + B d u(kt ), or x(k + 1) = A d x(k) + B d u(k) (28) where T A d = e AT, B d = e A(t τ) B dτ. 0 11 / 11