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

5 is 05k uk o r 2 k uk v relationship between 681 112

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Unformatted text preview: f [k f [k - IJu[kJ 1 1 -=- ;2F[zJ + ;1f[-1J + f[-2J = ;1 + 0 + 0 = z(z _ 0.5) 2F[zJ ; z _ 0.5 z - 0.5 z3 F[zJ - z3 f[OJ - z2 J [IJ - zf[2J N ote t hat for c ausal i nput f[kJ, [~] M ultiplication b y-/ ·If[kJu[kJ F M ultiplication by k kf[kJu[kJ d - z dzF[zJ f [-IJ = f [-2J = . .. = f [-nJ = 0 H ence f [k - rJu[kJ * h [kJ +0 = F'[zJF2[ZJ T ime C onvolution J,[kJ F requency Convolution h [kJh[kJ 2;j I nitial value f[OJ Iimz~ooF[zJ F inal value IimN~oof[NJ limz~l(z 1 -=- ;;:F[zJ T aking t he z -transform o f E q. (11.25) a nd s ubstituting t he a bove results, we o btain f F'[uJF2 [~l u - 1 3 7] 2Y[zJ [Ill] + 6 [;1 + 6IzI + 36 = z - 30.5 + z(z -5 0.5) du or - I)F[zJ poles o f (z - I)F[zJ i nside t he u nit circle. (11.26a) (1_~ + ~2) Y[zJ _ (3 _.!2) = _-3_ + -z,-_5_= zz z Z 0.5 (z - 0.5) (11.26b) Y[zJ - 5 ;Y[zJ + "6 t Another approach is t o find y[O], y[I], y[2],"" y[n - 1] from y [-I], y[-2],' " , y[-n] iteratively, as in Sec. 9.1-1, and then apply the left-shift property to E q. (11.24) 688 11 Discrete-Time Systems Analysis Using t he Z - Transform i -1 Answer:y[k] = [~+2(-lJk - ~(_2)k] u[k] a nd (1_ ~ z + ~) Y[z] z2 = (3 _ 3z 2 g) + z(zz-+0.5) 3 5 z 2 I n Example 11.5 we found t he t otal s olution of t he difference equation. I t is relatively easy t o s eparate t he s olution into zero-input a nd z ero-state components. All we have t o do is t o s eparate t he response into terms arising from t he i nput a nd t erms arising from initial conditions. We c an s eparate t he response in Eq. (11.26b) as follows: 9.5z + 10.5 z (z - 0.5) - 2 ) - 5z \J Zero-Input and Zero-State Components M ultiplication o f b oth s ides b y z2 y ields (z + 6 Y[z] = z (3z - 9.5z + lO.5) (z _ 0.5) 3 ,z - 0.5 = -- 2 Y[z] = z(3z - 9.5z + lO.5) (z - 0.5)(z2 - 5z + 6) (11.27) Therefore a nd (1- ~ z + ~) Y[z] z2 5 z{z - 0.5), + -;----,~ (11.29) . t erms a rising f rom i nput i nitial c ondition t erms s o t hat 689 11.3 Z -Transform S olution o f L inear Difference Equations (3z + 5) + -:,---::-,,:-:- = z{z - 0.5) '---v--' i nitial c ondition t erms 3z 2 - 9.5z + lO.5 (z - 0.5)(z - 2)(z - 3) Y[z] z = ( 26/15) _ (7/3) z - 0.5 z- 2 Multiplying b oth sides by z2 yields + ( 18/5) z {3z - 11) ( z2-5z+6)Y[z]= z- 3 '---v--' i nitial c ondition t erms T herefore Y[z] = a nd 26 ( 15 z z - 0.5 ) 7( - :3 Z) 18 ( z ) z - 2 + "5 z - 3 a nd 3 • + 5) -----z - 0.5 input t erms (11.30) +-;---::-::7-;-;;---::-----::-:- ~ (11.28) 5 z{3z + z{3z + 5) ,(z - 0.5)(z2 - 5z z{3z - 11) z2 - 5z + 6 Y[z] = y[k] = [ 26 (0.5)k _ ~(2)k + 18 (3)k] u[k] 15 i nput t erms z ero-input r esponse + 6), z ero-state r esponse We e xpand b oth t erms o n t he r ight-hand side into modified partial fractions t o yield T his example demonstrates t he ease with which linear difference equations with c onstant coefficients can be solved b y z-transform. This method is general; i t c an be u sed t o solve a single difference equation or a set of simultaneous difference equations of any order as long as t he e quati...
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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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