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

T his goal is readily accomplished by dlvldmg fzj by

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Unformatted text preview: operty has set t he s tage for solving linear difference equations w ith c onstant coefficients. As in t he case o f t he Laplace transform with differential equations, t he z-transform converts difference equations into algebraic equations which are readily solved t o find t he s olution in t he zdomain. Taking t he inverse z-transform of t he z-domain solution yields t he desired time-domain solution. T he following examples demonstrate t he p rocedure. • Proof: E xample 1 1.5 S olve Z{-ykf[k]u[k]} = f >kf[k]z-k = f f[k] (~)-k = F [~] k=O k=O l' l' y[k + 2 ]- 5y[k + I] + 6y[k] if t he i nitial conditions are y [-I] = 3J[k + I] + 5f[k] (11.24) = ¥, y [-2] = ¥S, a nd t he i nput f[k] = (2)-kU[k]. 6 86 11 D iscrete-Time S ystems A nalysis U sing t he Z -Transform As we s hall see, difference e quations c an b e solved by using t he r ight-shift o r t he leftshift property. Because t he difference e quation (11.24) is in a dvance-operator form, t he u se o f t he l eft-shift p roperty i n Eqs. ( 1l.17a) a nd (11.17b) m ay s eem a ppropriate for i ts s olution. U nfortunately, a s s een from Eqs. (11.17a) a nd ( 1l.17b), t hese p roperties r equire a k nowledge o f a uxiliary c onditions y[O], y [I]' . .. ,y[n - IJ r ather t han o f t he i nitial conditions y [-IJ,y[-2],· .. ,y[-nJ, which a re g enerally given. T his difficulty c an b e overcome by expressing t he difference e quation (11.24) in delay o perator form ( obtained b y replacing k w ith k - 2) a nd t hen u sing t he r ight-shift p roperty.t E quation (11.24) in delay o perator form is (11.25) T able 1 1.2 Z - Transform O perations O peration F[zJ f[kJ + h[kJ + F2[ZJ A ddition J,[kJ S calar m ultiplication af[kJ aF[zJ R ight-shift f [k - mJu[k - mJ zmF[zJ f [k - mJu[kJ 2m F [zJ + 2.. f J[-kJz k .. z zm F'[zJ 1 k=l 1 687 11.3 Z -Transform S olution o f L inear D ifference E quations We now use t he r ight-shift p roperty t o t ake t he z -transform o f t his e quation. B ut b efore proceeding, we m ust b e c lear a bout t he m eaning o f a t erm like y[k - IJ. Does i t m ean y[k - IJu[k - IJ o r y[k - IJu[kJ? T he a nswer becomes clear when we recognize t hat t he u se o f t he u nilateral t ransform i mplies t hat we a re c onsidering t he s ituation for k ~ 0, a nd t hat e very signal in Eq. (11.25) m ust b e c ounted f rom k = O. T herefore, t he t erm y[k - jJ m eans y[k - jJu[kJ. R emember also t hat a lthough we a re c onsidering t he s ituation for k ~ 0, y[kJ is p resent e ven before k = 0 (in t he form o f i nitial conditions). Now + f [-IJ f [k - IJu[kJ ;F[zJ f [k - 2Ju[kJ 2"F[zJ z 1 + - f[-IJ + f[-2J z f [k - 3Ju[kJ - F[zJ z3 1 + -2 f[-lJ + - f[-2J + f[-3J z z -=- Y[zJ 1 1 II IJu[kJ -=- ;Y[zJ + y [-IJ = ;Y[zJ + "6 1 1 1 II 37 2Ju[kJ -=- ;2Y[zJ + ; y[-IJ + y [-2J = ;2Y[zJ + 6z + 3 6 y[kJu[kJ 1 1 y[k y[k - 1 Also m -l Left-shift f[k+mJu[kJ zmF[zJ - zm L f[kJz-k k=Q J[k + IJu[kJ zF[zJ - zf[OJ J[k + 2Ju[kJ Z2 F[zJ - z2 f[OJ - zf[IJ liz 1 + 3Ju[kJ -=- ;F[zJ + f [-IJ = f [k - 2Ju[kJ...
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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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