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

# 5 u sing o nly t he f act t hat ukj z 1 a nd t he r

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Unformatted text preview: ons are linear w ith c onstant coefficients. z) Y z - 5 ( -z ) -2 (- - ] [ ]- [ z- 2 z- 3 C omment a nd .. E xercise E Il.S S olve t he e quation below if t he i nitial conditions are y [-l] = 2, y [-2] = 0, [2( +26 ( -z- ) - 232 ( --z- ) +8- -z- ) ] -z2 15 z - 0.5 5 z- 3 ' .. y y[k] = [?(2)k , z ero-state z ero-input S ometimes auxiliary conditions y[O], y [I]'···, y[n - 1] ( instead of initial conditions y [-l]' y [-2]' . .. , y [-n]) a re given to solve a difference equation. In this case, t he e quation can be solved by expressing it in the advance operator form a nd t hen using t he left-shift property (see Exercise E l1.9 below). I:;; 'V ~ 2(3)k, - ,¥(2)k z ero-mput + ~{:)k + ~{0.5)k,] u[k] z ero-state a nd t he i nput j [k] = u[k]: y[k + 2] - ~y[k + 1] + h [k] = 5 j[k + 1 ]- j[k] A nswer:y[k] = I:;; [12 - 15(!)k + ¥(W] u[k] \J I:;; E xercise E ll.9 S olve t he following equation if t he a uxiliary conditions a re y[O] = 1, y[1] = 2, a nd t he i nput j [k] = u[k]: y[k + 2] + 3y[k + 1] + 2y[k] = a conclusion, which agrees with t he r esult in Eq. (11.28). j [k + 1] + 3 j[k] E xercise E ll.IO Solve y[k + 2] - h[k + 1] i f t he i nitial conditions a re y [-1] = 2, y [-2] into zero-input and zero-state components. + ~y[k] = 5 f[k + 1 ]- j[k] = 0, a nd t he i nput j [k] = u[k]. S eparate t he r esponse 6 90 11 Discrete- Time S ystems A nalysis Using t he Z - Transform 1 11.3 Z -Transform S olution o f L inear Difference E quations -1[ o"Hld"!f- Answer: -'F:....>::[Z..!.l...._ _ _ > y[kJ = {[:(!)k ~ ~(~l! +!12 - 18(!~k + 6(~)k!} u[kJ z ero-lDput F ig. 1 1.5 691 Y _--3I __[Z_l_=_F[ZlH[zl _ T he t ransformed r epresentation o f a n LTID s ystem. z ero-state T herefore 1 1.3-1 Zero-State Response o f l T ID S ystems: T he T ransfer Function (11.32) C onsider a n n th-order LTID system specified by t he difference e quation ( I1.31a) Q[EJy[kJ = P[EJf[kJ - P[zJ [J - Q[z{ z (11.33) or We have shown in Eq. (11.20) t hat Y[zJ = F[zJH[zJ. Hence, i t follows t hat ( En + a n_lEn-I + ... + a lE + ao)y[kJ = (bnEn + b n_IE n - 1 + ... + b lE + bo)f[kJ ( I1.3Ib) n n1 H[zJ = P[zJ = bnz + b n_Iz - + ... + biZ + bo n 1 + .. , + a lz + ao Q[zJ z n + an_IZ (11.34) or y[k + nJ + an_Iy[k + n = + ... + a ly[k + IJ + aoy[kJ bnJ[k + nJ + ... + b d[k + IJ + bof[kJ IJ ( I1.3Ic) W e now derive t he general expression for t he z ero-state response; t hat is, t he s ystem response t o i nput f[kJ w hen all t he i nitial conditions y [-IJ = y [-2J = . .. = y [-nJ = o ( zero s tate). T he i nput f[kJ is assumed t o b e c ausal so t hat f [-IJ = J [-2J = '" = f [-nJ = 0. E quation (11.3Ic) can b e e xpressed in the delay o perator form as y[kJ B ecause y [-rJ + an_Iy[k - IJ + ... + aoy[k - nJ = bnf[kJ + b n-I![k - IJ + ... + bof[k - nJ ( I1.3Id) = f [-rJ = 0 for r = 1, 2, . .. , n y[k - mJu[kJ {=} 1 - Y[zJ zm f[k - mJu[kJ {=} - F[zJ zm 1 m = 1, 2, . .. , n N ow t he z -transform of Eq. (11.31d) is given by ( an-I 1 + -- z an-2 ao + -2 + ... + -n ) Y [zJ z z = bn ( bn + - - I + - -2 + ... + - o ) bn b z z2 M ultiplication of b oth sides by z n yie...
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