Appeiidix A Solutioiis t o thr Exercises 530 - x[ k] + 'l'hc init,ial condiiion is fulfilled by [O] ~ ~ O.r)z2[0! ~[O] ~ [0] 1I = ;y/int[O] + +4 O.r).x[O] = 3, and S o t e : Thc internal part alone is suficierit to det,rrmirie t,he initial states DF 11 R.S shown in Fig 14.7 ancl Fig 14.9 Initial condition: DF ITI: * z[O]= scc also Fig 2.5 'yiI,t[0; = 0.52[0]] b) z[O]= 3-[0] yint[O] = (I) =+ ,401 = + z [ k + 11 = z [ k ] y[k] + 0.54k.l (11) ' yik] = 2X[[k:] (I) in (11) : yjk] = 2:c[k]+ O.S(.r[k - I] Differerice equation: [ k ] 0; 5j/[k - + lj[k - I ] ) I] = 2x[k)+ 0,5s\k I] - MultipIying the difference rqua,tiori by yields the coefficients from a)