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

# 5 2fzj z 05 z 05 z3 fzj z3 foj z2 j ij

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Unformatted text preview: 0 .5)(z _ 1) = 3 -(k+ 1 )u[k] ( a) F ind t he s ystem r esponse t o i nput j [k] i f all initial conditions a re zero. ( b) W rite t he d ifference e quation r elating t he o utput y[k] t o i nput j [k] for t his s ystem. Answers: ( a) y[k] = i [~ - 0.8( -0.5)k + 0.3 (i t] u[k] ( b) y[k + 2J - 0.5y[k + 1J - 0.5y[kJ = f [k + 1 ]- 0.5j[kJ V 1 1.4 System Realization or (E2 + E + 0.16)y[k] = (E + 0.32)f[k] for t he input f [k] = ( -2)-ku[k] and with all the initial conditions zero (system in zerO s tate initially). From the difference equation we find P[z] z + 0.32 H [z]- - -=---'--- Q[z] - Z2 + z + 0.16 z z +0.5 a nd Y[z] - F[z]H[z] _ - (z2 z(z + 0.32) + z + 0.16)(z + 0.5) Therefore z (z + 0.32) (Z2 + z + 0.16)(z 2 (z + 0.32) + 0.5) (z 2 =+/3.2 -z-/3.8 + -0.5 - - 8 - z +2 z0 +0 (11.39) T his t ransfer f unction is identical t o t he g eneral n th-order c ontinuous-time t ransfer f unction H (s) i n Eq. (6.70) w ith s r eplaced b y z. I t is r easonable t o believe t hat t he r ealization o f H[z] i n (11.39) would b e i dentical t o t hat o f H(s) w ith s r eplaced b y z. F ortunately t his h appens t o b e t he case. I n r ealizations o f H (s) t he b asic element used was a n i ntegrator w ith t ransfer f unction l /s. I n r ealizations o f H[z] t he b asic element is u nit d elay w ith t ransfer f unction 1/ z. T herefore, all t he r ealizations o f H(s) s tudied in Sec. 6.6 are also t he r ealizations o f H[z] i f we r eplace i ntegrators b y u nit delays. To d emonstrate t his p oint, c onsider a realization o f a t hird-order t ransfer f unction. For the input f [k] = ( -2)-kU[k] = [ (_2)-1]ku(k) = (-0.5)ku[k] F [z]- W e now discuss ways t o realize a n n th-order d iscrete-time s ystem d escribed b y a t ransfer f unction H [z] + 0.2)(z + 0.8)(z + 0.5) (11.37) b b2Z + = ~3Z3++ 2z2 ++abIZ+ abo ~--~-------­ z3 a Iz o (11.40) F igure 11.7 s hows Fig. 6.21 w ith all t he i ntegrators ( with t ransfer f unction l is) r eplaced w ith u nit d elays ( with t ransfer f unction 1 / z). We shall now show t hat t his 6 94 11 Discrete-Time Systems Analysis Using t he Z - Transform 11.5 Connection between T he Laplace and t he Z -Transforms 695 • E xample 1 1.8 R ealize t he following transfe~ f unctions, u sing o nly t he c ascade f orm for p art a a nd u sing o nly t he p arallel f orm for p art b . ( a) H[z] = 6.28. /':, 4z + 28 ( b) H[z] = 7Z2 + 37z + 51 z2 + 6z + 5 ( z+2)(z +3)2 I dentical t ransfer f unctions for c ontinuous-time s ystems a re r ealized i n Figs. 6.27 a nd • E xercise E l1.12 Give the canonical realization of the following transfer functions. ( a) () c Z z +5 ( d) z2 2z + 3 + 6z+25 2 z +5 ( b) z + 8 z+5 Answer: See Example 6.18. Replace l is b y l iz a nd make appropriate changes in coefficients. 1 1.5 F ig. 1 1.7 A c anonical r ealization o f H[z]. r ealization indeed represents H[z] in Eq. (11.40). Let t he signal a t t he o utput o f t he t hird delay be X[z]. Consequently, signals a t t he i nputs of t he second a nd t he f irst delay are z X [z] a nd z2 X [z]. T he first summer o utput z3 X [z] is equal t o t he...
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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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