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

30b figure 137 shows t he b lock d iagram o f t he r

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Unformatted text preview: olution. T he first t erm o n t he r ight-hand side represents x (t) w hen t he i nput f (t) = O. H ence i t i s t he z ero-input component. T he s econd term, by a similar argument, is seen t o b e t he z ero-state 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 ime-domain 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 ero-state 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 ( 3n-l lsI - AI = I S+12 I 36 s -~ 2 =s + +1 13s + 36 = (s + 4)(s+9) = 0 ; 8 08 13 S tate-Space 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] + (1-e -4' - -e -9t) [-12 4 1 -e - 5 5 0 1 (-;3 e-4t + ~e-9t) [ ¥( _e- 4' + e-9') i s(e- 5 4t 5 9t - e- ) - 36 = ~] [(is + ioe-4t ~(e-4t 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) (~e-4t _ ~e-9t) (13.61c) T he zero-input component is given by [see Eq. (13.57a)J This result is consistent with t he solution obtained by using t he frequency-domain 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 zero-state 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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