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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 righthand 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 ontinuoustime system. However, r ather t han re~racing all
t he s teps o f t he c ontinuoustime 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 imeDomain Analysis o f D iscreteTime 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 ontinuoustime 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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