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

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: 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...
View Full Document

Ask a homework question - tutors are online