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

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

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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 &gt; 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 &gt; 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 &quot;'12 - 0.6&quot;'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...
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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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