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

2 a s ystematic p rocedure f or d etermining s tate e

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Unformatted text preview: aknXn + b klh + bk2/2 + ... + b kj!j (13.31a) We shall t ake t he L aplace t ransform o f t his e quation. L et so t hat Also, let ! i(t) = Fi(S) T he L aplace t ransform o f Eq. (13.31a) yields F ig. 1 3.7 R ealization of a second-order system. I t is clear from t he a bove discussion t hat a s tate-space d escription is n ot unique. For any realization of H (s) u sing integrators, scalar multipliers, a nd a dders, a corresponding state-space description exists. Since t here a re m any possible realizations of H (s), t here a re m any possible state-space descriptions. S Xk(S) - Xk(O) = a kIXI(s) + a k2X2(s) + ... + a knXn(S) + bkIFI(S) + bk2 F2(S) + ... + b kjFj(s) (13.31b) T aking t he L aplace t ransforms o f all n s tate e quations, we o btain X I(S) Consider a second-order system with a single i nput ! , a single o utput y , a nd two s tate variables, X l a nd X2. T he s ystem e quations a re XI(O) a ll a l2 aln X I(S) X 2(S) Realization X2(O) a 21 a22 a2n X 2(S) S .... . ..... . . ....... X n(s) anI Xn(O) '-v---' X (s) (13.30a) a n2 ' --v---' ann x(O) X n(S) ,'-v---' X (s) v A a nd (13.30b) Figure 13.7 shows t he b lock d iagram o f t he r ealized system. T he i nitial conditions XI(O) a nd X2(O) s hould b e a pplied a t NI a nd N2. T his p rocedure c an b e easily e xtended t o general multiple-input, m ultiple-output s ystems w ith n s tate variables. + (13.32a) bnl 1 3.3 Solution o f S tate Equations T he s tate e quations o f a linear system a re n s imultaneous linear differential equations of t he first order. We s tudied t he t echniques of solving l inear differential equations in C hapters 2 a nd 6. T he s ame techniques c an b e applied t o s tate equations without any modification. However, i t is more convenient t o c arry o ut t he s olution in t he framework o f m atrix n otation. . bn2 . .. bnj F j(s) ''-.---' B F (s) Defining t he v ectors, a s i ndicated above, we have or a nd s X(S) - x(O) = A X(s) + B F(s) s X(s) - AX(s) = x(O) + B F(s) . ...................--.... ~------------------------- 13 S tate-Space Analysis 800 (sI - A)X(s) = x(O) + B F(s) (13.32b) where I is t he n x n identity matrix. From Eq. 13.32b, we have X (s) = ( sI - A)-l[X(O) = ~(s)[x(O) + BF(s)] x(O) + BF(s) = X(s) = ~(s)[x(O) [ (13.34) (sI - A)-l + ~(s)BF(s) [ = + Cl[~(s)BF(s)] Cl[~(s)]x(O) ' ----v---'" z ero-input c omponent ~ ~] ( 8+4)(8+9) [¥.] . til - 36 8 +12 ( '+4)(S+9) 8 8 -59 ( .+4)(.+9) 1 (13.35b) z ero-state c omponent 21 8 8 +4 H ._...2iL = ~ [ Equation (13.35b) gives t he desired solution. Observe t he two components of t he solution. T he first component yields x (t) when t he i nput f (t) = O. Hence t he first component is t he zero-input component. In a similar manner, we see t hat t he second component is t he zero-state component. [~] 38 2 82+38+1 ] 8 (8+4)(8+9) (13.35a) and x (t) = ( .+4)(8+9) Thus, from Eq. (13.33b), X (s) = ~(s)x(O) 3. + BF(s)] 8 +1 ( 8+4)(.+9) ~(s) = 1-] l+~ (13.33b) where -~ 1 3. ~ + 8 +9] !L -s:f4 + 8+9 T he...
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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.

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