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

928 can be expressed as lyilk 1 lyi k o e qehk

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Unformatted text preview: int: You m ay n eed t o u se t he s hift p roperty (9.46) o f t he c onvolution. + 25y[k - 2] = 2 f[k]- 4f[k - 1] For t he g eneral n th-order difference Eq. ( 9.9), if h[k] = bn6[k] + 1] + 2y[k] if y [-l] = 10, a nd t he i nput f [k] = e-ku[k]. R epeat P rob. 9.3-1 if y [k]- 6y[k - 1] 9 .:J-5 = (3)k cos ( ik - 0.5)u[k] + 1] + 2y[k] = f [k + 1] (E2 - 6E + 9)y[k] 9 .3-4 9 .4-6 f [k] R epeat P rob. 9.3-1 if y[k 9 .3-3 h[k] F ind t he u nit i mpulse response h[k] o f a s ystem specified b y t he e quation + 1] + bo6[k - n] H int: Express t he s ystem e quation i n delay-operator form. ( b) F ind t he i mpulse response o f a n onrecursive LTID s ystem d escribed by t he e quation y[k] = 3 f[k]- 5f[k - 1]- 2f[k - 3] w here r is t he i nterest r ate p er d ollar p er m onth. H int: T his p roblem c an b e m odeled by Eq. ( 8.24) w ith t he p ayments o f P d ollars s tarting a t k = 1. T he p roblem c an b e a pproached i n two ways: (1) C onsider t he l oan a s t he i nitial c ondition yolO] = - M, a nd t he i nput f [k] = PU[k - 1]. T he l oan b alance is t he s um o f t he z ero-input c omponent ( due t o t he i nitial c ondition) a nd t he z ero-state c omponent h[k] * f [k]. (2) I n t he s econd a pproach t he l oan is considered a s a n i nput - M a t k = 0; t hat is, - M6[k]. T he t otal i nput is, therefore, f [k] = - M6[k] + Pu[k - 1], a nd t he l oan b alance is h[k] * f [k]. B ecause t he l oan is p aid off i n N p ayments, s et y[N] = O. O bserve t hat t he i mpulse response h as o nly a finite ( n) n umber o f n onzero elements. For t his r eason, such s ystems a re c alled f inite i mpulse r esponse ( FIR) s ystems. For a general recursive case [Eq. (9.9)], t he i mpulse response has a n i nfinite n umber o f n onzero elements, a nd s uch s ystems a re c alled i nfinite i mpulse r esponse ( UR) 9 .4-1 s ystems. F ind t he ( zero-state) response y[k] o f a n L TID s ystem whose u nit i mpulse response is h[k] a nd t he i nput is f [k] 9 .4-2 = (_2)ku[k] = e-ku[k]. 9 .4-10 A p erson receives a n a utomobile loan o f $10,000 f rom a b ank a t t he i nterest r ate o f 1.5% p er m onth. His m onthly p ayment is $500, w ith t he first p ayment d ue o ne m onth a fter he receives t he loan. C ompute t he n umber o f p ayments r equired t o p ayoff t he l oan. N ote t hat t he l ast p ayment m ay n ot b e e xactly $500. H int: Follow t he p rocedure in P rob. 9.4-9 t o d etermine t he b alance y[k]. T o d etermine N , t he n umber o f p ayments, s et y[N] = O. I n g eneral, N will n ot b e a n i nteger. T he n umber o f p ayments K is t he l argest integer :S N. T he r esidual p ayment is ly[K]I· 9 .4-11 Using t he s liding-tape a lgorithm, show t hat R epeat P rob. 9.4-1 i f h[k] = ~[6[k]- (_2)k]u[k] 2 9 .4-3 R epeat P rob. 9.4-1 i f t he i nput f [k] = (3)k+ u[k], a nd 9 .4-4 R epeat P rob. 9.4-1 i f t he i nput J[k] = (3)-ku[k], a nd h[k] = 3k(2)k u [k] 9 .4-5 R epeat P rob. 9.4-1 i f t he i nput f [k] = (2)kU[k], a nd ( a) u[k] * u[k] + l )u[k] m]) * u[k] = = (k (...
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