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 eroinput 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: :11:2 ~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(tT)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
( eAt)(e At ) = ( eAt)(e At ) (13.48b)
I n Sec. B.63, we show t hat =I (13.53b) a
13 S tateSpace Analysis 806
dP d dQ x (t) = eAtx(O) d t Q + P dt ;U(PQ) =
Using this relationship, we o bserve t hat ~[eAtx] = (~eAt)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 yeAt t o yield x (t) = e A(tto)x(to) + t eA(tr)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 efthand side of Eq. (13.55b) is
Hence ~[eAt] 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 closedform
: 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.65 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 eroinput c omponent + t eA(tr)Bf(r) dr io
" ',." T his is t he desired s...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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