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

# O therwise t he p rocedure is t he s ame a s will b

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Unformatted text preview: t will prevent ki Y4>[k] from having a characteristic mode term. For example, when the input is r k, t he forced response in the right-hand column is of the form c r k . But if r k happens t o be a natural mode of the system, the correct form of the forced response is c krk(see Pair 2). • E xample 9 .10 Solve (E2 - 5E + 6)y[k] = ( E - 5)f[k] (9.63) (9.65) if the input f [k] = (3k + 5)u[k] and the auxiliary conditions are y[O] = 4, y[1] = 13. T he characteristic equation is S ubstitution of t his e quation in Eq. (9.62) yields Q[E]y.p[k] = P [E]f[k] r o;f ' Y;(i rk (9.62) B ut s ince Yn[k] is made u p o f c haracteristic modes, Q[E]Yn[k] = 0 rk 2. A s in t he case of LTIC s ystems, we c an a nalyze LTID systems by using t he F inding N atural and Forced Response A s e xplained earlier, t he n atural response of a s ystem c onsists o f all t he c haracteristic m ode t erms in t he response. T he r emaining noncharacteristic m ode t erms form t he forced response. I f Yn[k] a nd y.p[k] d enote t he n atural a nd t he forced r esponse respectively, t hen t he t otal r esponse is given by Forced Response y.p[k] (9.64) T he n atural response is a linear combination o f c haracteristic modes. T he a rbitrary c onstants (multipliers) are determined from suitable auxiliary conditions usually given as y[O], y [I], . .. , y[n - 1]. We now t urn o ur a ttention t o t he forced response. Therefore, the natural response is Forced Response To find the form of forced response Y4>[k], we use Table 9.2, Pair 4 with r = 1, m = 1. This yields Yn[k] = B l(2)k + B2(3)k W e have shown t hat t he forced response y.p[k] satisfies t he s ystem Equation (9.64) Therefore Y4>[k + 1] Q[E]y.p[k] = P [ElJ[k] B y d efinition, t he forced response contains only nonmode terms. To determine t he f orced response, we shall use a m ethod o f undetermined coefficients t he s ame m ethod u sed for t he c ontinuous-time system. However, r ather t han re~racing all t he s teps o f t he c ontinuous-time system, we shall present a t able ( Table 9.2) listing Y4>[k = c l(k+ 1) +CO = c lk+Cl + co + 2] = cr(k + 2) + Co = c lk + 2 Cl + Co Also J[k] = 3k + 5 and (9.66) 600 9 T ime-Domain Analysis o f D iscrete-Time Systems 9.5 601 C lassical Solution o f L inear Difference E quations A C omment on Initial Conditions I [k + 1] = 3 (k + 1) + 5 = 3 k + 8 T his m ethod r equires auxiliary conditions y[O], y[I], . .. , y[n -IJ for t he reason'Sc e xplained o n p. 142. I f we are given t he i nitial conditions y [-I], y [-2J, . .. , y [-n]" we c an derive t he c onditions y[O], y[I], . .. , y[n - IJ using iterative procedure. Substitution of the above results in Eq. (9.64) yields c lk + 2Cl + Co - 5(Clk + Cl + co) + 6(Clk + co) = 3k + 8 - 5(3k + 5) An Exponential Input or As in t he case o f c ontinuous-time systems, we can show t hat for a s ystem specified by t he e quation 2Clk - 3Cl + 2co = - 12k - 17 Comparison of similar terms on the two sides yields - 3Cl+2co=-17 Cl = - 6 = t he forced...
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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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