Signal Processing and Linear Systems-B.P.Lathi copy

2 a s ystematic p rocedure for d etermining s tate e

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Unformatted text preview: cteristic values) of t he m atrix A . A t this point, t he r eader will recall t hat if Al, A2, . .. , A n a re t he poles of t he t ransfer function, t hen t he z ero-input response is o f t he form Yo(t) = (13.49) a nd Hence A l=-1 0 I + 3s + 2 :] t hen At = [ I: :1-1:2 ~31 -1 s = \ 2 s +3 805 T hus, if + ao (s - AIl(s - A2)··· (s - An) = 0 For t he s ystem i n E xample 13.6, 13.3 S olution o f S tate E quations Also n ote t hat from t he d efinition (13.48a), i t follows t hat We now show t hat t he s olution of t he vector differential E quation (13.46) is x (t) = e Atx(O) + l e A(t-T)Bf(T) dT (13.47) (13.53a) where Before proceeding further, we m ust define t he e xponential of t he m atrix a ppearing i n Eq. (13.47). A n e xponential o f a m atrix is defined by a n infinite series identical t o t hat u sed in defining an exponential of a scalar. We shall define A 2t 2 A 3t3 A ntn 2! 3! n! e At = I +At+ - - + - - + . .. + - - + . .. + (13.48a) I= [~ ~] I f we p remultiply o r p ostmultiply t he i nfinite series for e At [Eq. (13.48a)] b y a n i nfinite series for e - At , we find t hat ( e-At)(e At ) = ( eAt)(e- At ) (13.48b) I n Sec. B.6-3, we show t hat =I (13.53b) a 13 S tate-Space Analysis 806 dP d dQ x (t) = eAtx(O) d t Q + P dt ;U(PQ) = Using this relationship, we o bserve t hat ~[e-Atx] = (~e-At)x + e -Atx dt 807 13.3 Solution of S tate E quations dt (13.54) + e At * B f(t) (13.57b) Note t hat t he l imits of t he c onvolution integral [Eq. (13.57a)] are from 0 t o t. Hence, all t he e lements o f e At i n t he c onvolution t erm o f Eq. (13.57b) are implicitly assumed t o b e m ultiplied b y u (t). T he r esult of Eq. (13.57) can b e easily generalized for a ny i nitial value of t. I t is left as a n exercise for t he r eader t o show t hat t he s olution of t he s tate e quation c an b e expressed as We now premultiply b oth s ides of Eq. (13.46) b ye-At t o yield x (t) = e A(t-to)x(to) + t eA(t-r)Bf(r)dr (13.58) i to e - Atx = e - At A x + e - AtBf (13.55a) D etermining e At or (13.55b) A glance a t E q. (13.54) shows t hat t he l eft-hand side of Eq. (13.55b) is Hence ~[e-At] 1tfe - At ]. = e -AtBf T he e xponential e At r equired in Eq. (13.57) c an b e c omputed from t he definition in Eq. (13.51a). Unfortunately, t his is a n infinite series, a nd i ts c omputation c an b e q uite l aborious. Moreover, we m ay n ot b e a ble t o recognize t he closed-form : xpression for t he answer. T here a re several efficient m ethods o f determining e At I II closed form. I t is shown in Sec. B.6-5 t hat for a n n x n m atrix A , dt T he i ntegration of b oth sides of t his e quation from 0 t o t yields t o io e -ArBf(r)dr e -Atxl t = or (13.59a) (13.56a) where e -Atx(t) - x(O) = I t e -ArBf(r)dr (13.56b) (31 Hence Al A2 1 An - 1 1 A2 1 (30 A~ An 2 1 -1 e>.,t e>'2t ...................... e - Atx = x(O) + I t e - ArBf( r ) d r (13.56c) eAtx(O) ' -v--' z ero-input c omponent + t eA(t-r)Bf(r) dr io " -'----,.-------" T his is t he desired s...
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