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Unformatted text preview: olution. T he first t erm o n t he r ighthand side represents x (t)
w hen t he i nput f (t) = O. H ence i t i s t he z eroinput component. T he s econd term,
by a similar argument, is seen t o b e t he z erostate component.
T he r esults of Eq. (13.57a) c an b e e xpressed more conveniently in t erms o f t he
m atrix convolution. We c an define t he c onvolution o f two matrices in a m anner
s imilar t o t he m ultiplication of two matrices, except t hat t he m ultiplication of two
elements is replaced by t heir convolution. F or e xample, e At = = 92]
9394 = [ (11*91+12*93) ( 11*92+12*94)] ( h*91+f4*93) ... A~1 e>'n t (13.59b)
(13.59c) E xample 1 3.7
F ind t he s olution t o t he p roblem i n E xample 13.5 using t he t imedomain m ethod.
F or t his c ase, t he c haracteristic r oots a re g iven by ( h*92+f4*94) U sing t his definition of m atrix convolution, we c an e xpress Eq. (13.57a) as .c 1 [ .(s)]
.c 1 [(sI  A)I] Thus, e At a nd . (s) a re a Laplace t ransform p air. To b e c onsistent with Laplace
transform n otation, e At is often d enoted b y " ,(t), t he s tate t ransition m atrix
( STM): • * [91 A; (13.57a) z erostate c omponent 11 12]
[ 13 f4 An a nd A I, A2, . .. , An a re t he n c haracteristic values (eigenvalues) of A .
We c an a lso d etermine e At by comparing Eqs. (13.57a) a nd (13.35b). I t is clear
t hat P remultiplying Eq. (13.56c) b y e At a nd using Eq. (13.53b), we have
x (t) = 1 ( 3nl lsI  AI = I S+12 I 36 s ~
2
=s +
+1 13s + 36 = (s + 4)(s+9) = 0 ;
8 08 13 S tateSpace A nalysis T he roots are A1 =  4 a nd A2 =  9, so PO] = [ 1
[Ih
1  4]
9 1 4
[ e ']
e  9t 13.3 S olution o f S tate E quations 809 S ubstitution for t he above convolution integrals from t he convolution table (Table
yields 4
9t
=.!. [ ge ' _ 4 e ]
5
e  4t _ e  9t 2.1) and _(9 4t  e 9t) [1 0] + (1e 4'  e 9t) [12
4
1
e  5 5 0 1 (;3 e4t + ~e9t)
[ ¥( _e 4' + e9') i s(e 5 4t 5 9t
 e )  36 = ~] [(is + ioe4t ~(e4t 9t
lse )u(t)] _ e  9t )u(t) (13.61b) 1 The sum of t he two components [Eq. (13.61a) a nd Eq. (13.61b)J now gives t he desired
solution for x(t): 1
(13.60) (~e4t _ ~e9t) (13.61c) T he zeroinput component is given by [see Eq. (13.57a)J
This result is consistent with t he solution obtained by using t he frequencydomain method
[see E q. (13.36b)J. Once t he s tate variables X l a nd X 2 are found for t 2: 0, all t he remaining
variables are determined from t he o utput equation. • T he O utput
T he o utput e quation is given b y
(13.61a) y (t) = C x(t) + D f(t) T he s ubstitution o f t he s olution for x [Eq. (13.57)J i n t his e quation y ields
N ote t he presence of u (t) in Eq. (13.61a), indicating t hat t he response begins a t t
T he zerostate component is eAt * B f [see Eq. (13.57b)J, where
B f= [.!]
3 1 = O. y (t) = C [eAtx(O) + e At * B f(t)J + D f(t) ( 13.62a) S ince t he e lements o f B a re c onstants,
u (t) = [ .!U(t)]
3
u (t) e At * B f(t) = e AtB * f(t) W ith t his r esult, Eq. ( 13.62a) b ecomes a nd y (t) N ote again t...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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