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

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Unformatted text preview: -Domain Analysis of Discrete-Time Systems y[k + 2] = y[k + 1] - + J[k + 2 ]- 2f[k + 1] (9.5) S etting k = - 2 a nd t hen s ubstituting y [-I] = 2, y [-2] = 1, frO] = f [-I] = 0 (recall t hat 0.24y[k] f [k] = k s tarting a t k = 0), we o btain vIol =2- 0.24(1) +0 - 0 S etting k = - 1 i n Eq. (9.5) a nd t hen s ubstituting vIOl w e o btain y[l] = 1.76 - 0.24(2) = 1.76 = 1. 76, Y [-I] = 2, f [l] = 1, frO] = 0, +I - 0 = 2.28 = 1.76, Y[I] 2(1) = 1.8576 S etting k = 0 i n Eq. (9.5) a nd t hen s ubstituting vIOl = I yields = 2.28, J[2] =2 a nd I P] y[2] a nd so on. = 2.28 - 0.24(1.76) +2- • 9.1 577 Discrete-Time System equations We shall see in t he f uture t hat t he s olution o f a difference equation o btained in this direct (iterative) way is useful in many situations. Despite t he m any uses of this method, a closed-form solution o f a difference equation is far more useful in s tudy o f t he s ystem behavior a nd i ts dependence on t he i nput a nd t he various system parameters. For t his r eason we s hall develop a systematic procedure t o a nalyze discrete-time systems along lines similar t o t hose used for continuous-time systems. Operational Notation I n difference equations i t is convenient t o use operational n otation s imilar t o t hat used in differential equations for t he sake o f c ompactness. I n c ontinuoustime systems we used t he o perator D t o denote t he o peration of differentiation. For discrete-time systems we s hall use the o perator E t o denote t he o peration for advancing t he sequence by one t ime u nit. T hus Note carefully t he i terative or recursive n ature o f t he c omputations. From t he E f[k] == f [k n i nitial conditions (and input) we o btained y[O] first. T hen, u sing this value of y[O] a nd t he previous n - 1 initial conditions (along with t he i nput), we found y[l]. E2 f [k] == f [k N ext, using y[O], Y[l] along with the initial conditions a nd i nput, we o btained y[2], a nd so on. This m ethod is general a nd c an b e a pplied t o a difference e quation o f a ny order. T he i terative procedure cannot be applied t o c ontinuous-time systems. I t is i nteresting t hat t he h ardware realization o f Eq. (9.3a) depicted in Fig. 8.17 ( with'Y = 0.5) g enerates t he solution precisely in this (iterative) fashion. + 1] + 2] (9.6) T he first-order difference e quation o f t he s avings account problem was found t o b e [see Eq. (8.25b)] + 1 ]- ay[k] + 1] 6 . E xercise E 9.l Using t he i terative m ethod, find t he first t hree t erms o f y[k] i f y[k + 1] - 2y[k] = i [k] T he i nitial c ondition is y [-I] = 10 a nd t he i nput I[k] = 2 s tarting a t k = o. A nswer: vIOl = 20,Y[I] = 42, andy[2] = 86. 'V Using t he o perational notation, we c an e xpress t his e quation as o or C omputer E xample C 9.1 Solve E xample 9.2 using MATLAB. i .(c91.m) Y =[l 2]; Y =Y'; k =-2:2; k =k'; f=[O 0 0]'; f or m =1:length(k}-2 y =Y(m+l}-0.24*Y(m)+f(m+2}-2*f(m+l}; Y =[Y;y]; F =m;f=[f;F]; e nd s tem(k,Y} [ kY] y[k ( E - a)y[k] = E f[k] (9.8)...
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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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