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

580 9 t ime domain a nalysis o f d iscrete time s

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n d oes n ot y ield a c losed-form e xpression for h [k]. N evertheless, d etermining a few v alues o f h [k] c an b e u seful i n d etermining t he c losed-form s olution, as t he f ollowing d evelopment s hows. T he Closed-Form Solution o f h[k) R ecall t hat h [k] i s t he s ystem r esponse t o i nput 6[k], w hich is z ero f or k > o. W e k now t hat w hen t he i nput i s zero, o nly t he c haracteristic m odes c an b e s ustained b y t he s ystem. T herefore, h [k] m ust b e m ade u p o f c haracteristic m odes for k > O. A t k = 0, i t m ay h ave s ome n onzero v alue Ao, s o t hat h [k] c an b e e xpressed as h [k] = Aob[k] + Yn[k] ( 9.32) 5 84 9 T ime-Domain A nalysis o f D iscrete-Time S ystems 9.4 S ystem r esponse t o E xternal I nput: T he Z ero-State R esponse w here, Yn[k] is a linear c ombination o f t he c haracteristic m odes. I n A ppendix 9.1 we show t hat Ao = bo (9.33) ao Therefore M oreover, b ecause h[k] is causal, we m ust m ultiply Yn[k] b y u [k]. Therefore,:j: 585 o h [k] = bo a[k] ao + Yn[k]u[k] (9.34) T he n u nknown coefficients i n Yn[k] ( on t he r ight-hand s ide) c an b e d etermined f rom a k nowledge o f n v alues o f h[k]. F ortunately, i t is a s traightforward t ask t o d etermine v alues o f h[k] i teratively, a s d emonstrated i n E xample 9.4. We c ompute n v alues h[O], h[1], h [2]'· . . , h [n - 1] i teratively. Now, s etting k = 0, 1 ,2,··· , n - 1 in E q. (9.34), we c an d etermine t he n u nknowns in Yn[k]. T his p oint will become c lear in t he following example. h[kJ N =20; b =[5 0 0 ]; a =[1 - 0.6 - 0.16]; f =[1,zeros(1,N-1)]; y =filter(b,a,f) ; k =0:1:N-1; s tem(k,y) x label('k');ylabel('h[k]'); 0.6y[k + 1J - 0.16y[kJ = 5 f[k + 2J (9.35) or (E2 - 0.6E - 0.16)y[kJ = 5E2 f[kJ A lthough d etermination o f t he i mpulse response h[k] u sing t he p rocedure i n t his s ection is relatively simple, i n C hapter 11 we s hall d iscuss a nother, m uch s impler, m ethod o f z -transform. I n t he p resent m ethod, w hen ao = 0, Ao c annot b e d etermined f rom Eq. (9.33). I t is shown i n A ppendix 9.1 t hat in s uch a c ase t he i mpulse r esponse c ontains a n a dditional i mpulse a t k = 1, a nd h[k] c an b e e xpressed a s ( 9.36) h[k] = Aoa[k] T he characteristic polynomial is "'12 - 0.6"'1 - 0.16 = h 0 C omment = 5f[kJ This equation can be expressed in the advance operator form a s + 2J - + 0 .2)h - 6 1] + Yn[k]u[k] E xercise E9.5 Find h[kJ, the unit impulse response of the LTID systems specified by the following equations: ( a) y[k + 2) - 5y[k + 1) ( b) y[k + 2) - 4y[k + 1) (9.37) o. + A I8[k - Now h[k] c ontains n + 2 u nknowns (Ao, A I, a nd n coefficients o f Yn[kJ), w hich c an b e d etermined f rom n + 2 v alues o f h[k] o btained i teratively. 0.8) T he characteristic modes are ( -0.2)k and (0.8)k. Therefore Also, from Eq. ( 9.36), we have ao = - 0.16 a nd bo = ( 9.39) C omputer E xample C 9.3 Solve Example 9.5 using Matlab. T here are several ways of finding impulse response by Matlab. In...
View Full Document

This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online