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

T hen u sing this value of yo a nd t he previous n 1

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Unformatted text preview: ave a r epeated c haracteristic r oot a t 'Y = - 3. T he c haracteristic m odes a re ( _3)k a nd k ( _ 3)k. H ence, t he z ero-input response is j =chl k c os (f3k + 0) ( 9.22) We c an d etermine t he a rbitrary c onstants Cl a nd C2 f rom t he i nitial c onditions following t he p rocedure i n p art ( a). I t is left a s a n e xercise for t he r eader t o s how t hat Cl = 4 a nd C2 = 3 s o t hat w here c a nd 0 a re a rbitrary c onstants d etermined f rom t he a uxiliary c onditions. T his i s t he s olution i n r eal f orm, w hich a voids d ealing w ith c omplex n umbers . • york] E xample 9 .3 ( a) F or a n L TID s ystem d escribed by t he difference e quation y[k + 2J - 0.6y[k + IJ - 0.16y[kJ = 5J[k + 2J ¥, (E2 - 0.6E - 0.16)y[kJ = 5E2 J[kJ 'Y2 - 0.6'Y - 0.16 = (')' + 0.2)(,,( - 0.8) = ( E + 3)f[k] york] = c(0.9)k cos (2Jk C 2 =-cos 0.9 (9.24) + 0) (7r ) =0- [v'23 1.] - -+0 -cosO+-smO 6 .9 2 T he c haracteristic r oots a re 'Yl = - 0.2 a nd 'Y2 = 0.8. T he z ero-input response is e 1 =--cos (0.9)2 (9.25) T o d etermine a rbitrary c onstants Cl a nd C2, we s et k = - 1 a nd - 2 i n Eq. (9.25), t hen s ubstitute y o[-IJ = 0 a nd y o[-2J = t o o btaint ¥ T herefore find t he z ero-input r esponse o f a n L TID T o d etermine t he a rbitrary c onstants C a nd 0, we s et k = - 1 a nd - 2 i n t his e quation a nd s ubstituting t he i nitial conditions y o[-I] = 2, yo[-2] = 1, we o btain T he c haracteristic e quation is ('Y + 0.2)('Y - 0.8) = 0 US w hen t he i nitial conditions a re y o[-I] = 2 a nd yo[-2] = 1. T he c haracteristic p olynomial is (')'2 -1.56'Y+0.81) = ( ')'-0.78- j0.45)('Y-0. 7 8+j0.45). T he c haracteristic r oots a re 0.78 ± jO.45; t hat is, 0 .ge±jii. T hus, [')'[ = 0 .9 a nd f3 = 7r/6, a nd t he z ero-input response, according t o E q. (9.22), is given by (9.23b) T he c haracteristic polynomial is 0=-5Cl+~C2} (4 + 3 k)( - 3l (E2 - 1.56E + 0.81)y[k] (9.23a) f ind t he t otal r esponse i f t he i nitial conditions a re y [-IJ = 0 a nd Y[-2J = a nd i f t he i nput f[kJ = 4 - ku[kJ. I n t his e xample we shall d etermine t he z ero-input c omponent yo[kJ o nly. T he z ero-state c omponent is d etermined l ater i n E xample 9.7. T he s ystem e quation i n o perational n otation is 25 = ( c) F or t he case o f c omplex r oots, let s ystem d escribed by t he e quation C (-7r) =0- - (12 v'3.] -+0 -cosO+-smO 3 .81 2 C or ~ e cos 0 + /8 c sin 0 = 2 1.~2 c cos 0 + ; 12 e sin 0 = 1 = ;. 25 " 4 = 25cr + TIi C2 T hese a re t wo simultaneous e quations in two unknowns c cos 0 a nd e s in t hese e quations yields (9.26) c cos 0 =2.308 t The initial conditions Y[-I] a nd y[-2] are the conditions given on the total response. B ut because t he input does not s tart until k = 0, the zero-state response is zero for k < O. Hence, a t k = - 1 a nd - 2 t he total response consists of only the zero-input component, so t hat y[ - 1] = YO [ -1] and y[-2] = Yo[-2]. c sin 0 = - 0.397 Dividing c sin 0 b y e cos 0 yields o. S olution o f 5 82 9 T ime-Domain A nalysis o f D iscrete-Time S ystems t an () = - 0.397 = - 0.172 2.308 1 9 .3 5 83 9.3 The Unit Impuls...
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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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