The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j

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1 The -Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discrete-time LTI systems. 7. The -trasform Derivatio of the -trasform: x[] = LTI system, h[] = re j y[] = x[] h[]. Iput x[] = : x r cos jr si. (7.) Fig. 7.. The output of the LTI system with impulse respose h[]: y H x h x k We use x[] = to obtai h k x k.

2 The -Trasform Damped cosie Damped sie Figure 7. (p. 554) Real ad imagiary parts of the sigal.

3 The -Trasform y h k h k k k k k.. We defie trasfer fuctio: k k k H h (7.) H y H. H e j. The form of a eigefuctio = eigefuctio; H() = eigevalue where H() is expressed i polar form, i.e., H() = H()e j (). Usig = r e j ad applyig Euler s formula, we obtai j j j j y H re r cos re j H re r si re. The system modifies the amplitude of the iput by H(re j ) ad shifts the phase of the siusoidal compoets by (re j ). 4. Represetatio of arbitrary sigals as a weighted superpositio of the eigefuctio :

4 The -Trasform Substitutig = re j ito Eq. (7.) yields j j j. H re h re h r e h j j r H re e d H(re j ) correspods to DTFT of h[]r. Hece, the iverse DTFT of H(re j ) must be h[]r, so we may write Multiplyig this result by r gives r j j j j h H re e d H re re d. where h The symbol d (7.) j H, re j = ; d = jre j d ; d = (/j) d deotes the itegratio aroud a circle of radius = r i a couterclockwise directio. 4

5 5. The -trasform pair: The -Trasform Iverse -trasform x, (7.4) x. Notatio: x d j I Eq. (7.5), the sigal x[] is expressed as a weighted superpositio of complex potetial. The weights are (/j)x() d. (7.5) 7.. Covergece. The -trasform exists whe the ifiite sum i Eq. (7.4) coverges.. A ecessary coditio for covergece is absolute summability of x[]. Sice x[] = x[]r, we must have x r. The -trasform exists for sigals that do ot have a DTFT. Illustratio: Fig

6 The -Trasform Figure 7. (p. 556) Illustratio of a sigal that has a -trasform, but does ot have a DTFT. (a) A icreasig expoetial sigal for which the DTFT does ot exist. (b) The atteuatig factor r associated with the -trasform. (c) The modified sigal x[]r is absolutely summable, provided that r >, ad thus the -trasform of x[] exists. a 6

7 The -Trasform 7.. The -Plae. The -plae: Fig If x[] is absolutely summable, the the DTFT obtaied from the -trasform by settig r =, or substitutig = e j ito Eq. (7.4). That is, j j e. e (7.6). The equatio = e j describes a circle of uit radius cetered o the origi i the -plae. Fig Uit circle i the -plae. Figure 7. (p. 557) The -plae. A poit = re j is located at a distace r from the origi ad a agle relative to the real axis. The frequecy i the DTFT correspods to the poit o the uit circle at a agle with respect to the positive real axis. 7

8 The -Trasform Example 7. The -Trasform ad The DTFT Determie the -trasform of the sigal x,,,, 0, 0 otherwise Use the -trasform to determie the DTFT of x[]. <Sol.>. We substitute the prescribed x[] ito Eq. (7.4) to obtai Figure 7.4 (p. 557) The uit circle, = e j, i the -plae... We obtai the DTFT from X() by substitutig = e j : j j j j e e e e. 8

9 The -Trasform 7.. Poles ad Zeros. The -trasform i terms of two polyomials i : b 0 M. a 0 b a ~ b b a N M k k N M N c k. d k bwhere b 0 / a 0 gai factor. The c k are the roots of the umerator polyomial the eros of X(). The d k are the roots of the deomiator polyomial the poles of X().. Symbols i the -plae: poles; eros Example 7. -TRANSFORM OF A CAUSAL EXPONENTIOAL SIGNAL Determie the -trasform of the sigal x u. Depict the ROC (regio of covergece) ad the locatio of poles ad eros of X() i the -plae. 9

10 The -Trasform <Sol.>. Substitutig x[] = α u[] ito Eq. (7.4) yields u 0.. This is a geometric series of ifiite legth i the ratio α/; the sum coverges, provided that α/ <, or > α. Hece,,,. (7.7). There is thus a pole at = α ad a ero at = 0, as illustrated i Fig 7.5. the ROC is depicted as the shaded regio of the -plae. Figure 7.5 (p. 559) Locatios of poles ad eros of x[] = u[] i the -plae. The ROC is the shaded area. a 0

11 The -Trasform Example 7. -TRANSFORM OF AN ANTICAUSAL EXPONENTIOAL SIGNAL Determie the -trasform of the sigal y u. Depict the ROC ad the locatio of poles ad eros of X() i the -plae. <Sol.>. We substitutig y[] = -α u[- -] ito Eq.(7.4) ad write Y u k k 0 k k.. The sum coverges, provide that / <, or < α. Hece, Y,,, (7.8). The ROC ad the locatio of poles ad ero are depicted i Fig 7.6.

12 The -Trasform Figure 7.6 (p. 560) ROC ad locatios of poles ad eros of x[] = u[ ] i the - plae. a Example 7.4 -TRANSFORM OF A TWO SIDED SIGNAL Determie the -trasform of x u u. Depict the ROC ad the locatio of poles ad eros of X() i the -plae. <Sol.>. Substitutig for x[] i Eq. (7.4), we obtai

13 The -Trasform CHAPTER u u k k. Both sum must coverge i order for X() to coverge. This implies that we must have > / ad <. Hece,. The ROC ad the locatio of poles ad ero are depicted i Fig 7.7. i this case, the ROC is a rig i the -plae. Figure 7.7 (p. 560) ROC ad locatios of poles ad eros i the -plae for Example 7.4. /,, /,

14 The -Trasform 7. Properties of Regio of Covergece The ROC caot cotai ay poles. X() must be fiite for all i the ROC. If d is a pole, the X(d) =, ad the -trasform does ot coverge at the pole. The ROC for a fiite-duratio sigal icludes the etire -plae, except possible = 0 or = (or both).. Suppose x[] is oero oly o the iterval. We have x.. This sum will coverge, provided that each of its terms is fiite. ) If a sigal has ay oero causal compoets ( > 0), the the expressio for X() will have a term ivolvig, ad thus the ROC caot iclude = 0 (because X()->). ) If a sigal is ocausal, a power of ( < 0), the the expressio for X() will have a term ivolvig, ad thus the ROC caot iclude =. 4

15 The -Trasform ) Coversely, If 0, the the ROC will iclude = 0. If a sigal has o oero ocausal compoets ( 0), the the ROC will iclude =. This lie of reasoig also idicates that x[] = c[] is the oly sigal whose ROC is the etire -plae. Ifiite-duratio sigals:. The coditio for covergece is X() <. We may write x x x.. Splittig the ifiite sum ito egative- ad positive-term portios, we defie Note that x X ( ) x. (7.9) ad (7.0) 0 If I () ad I + () are fiite, the X() is guarateed to be fiite, too. 5

16 The -Trasform. 4. If x A r for -, the The last sum coverges if ad oly if < r. If x A r for 0, the r A r A 0 0 r A r A A k r This sum coverges if ad oly if > r +. k k = Hece, if r + < < r, the both I + () ad I () coverge ad X() also coverges. 5. Coclusios: Fig ) The ROC of a right-sided sigal (x[] 0 for < 0) is of the form > r +. ) The ROC of a left-sided sigal (x[] 0 for 0) is of the form > r. ) The ROC of a two-sided sigal is of the form r + < < r. 6

17 The -Trasform Figure 7.8 (p. 564) The relatioship betwee the ROC ad the time extet of a sigal. (a) A right-sided sigal has a ROC of the form > r +. (b) A left-sided sigal has a ROC of the form < r. (c) A two-sided sigal has a ROC of the form r + < < r. 7

18 The -Trasform Example 7.5 ROCs OF A TWO-SIDED SIGNALS Idetify the ROC associated with -trasform for each of the followig sigal: / / 4 x u u ; / ( / 4) y u u ; w / u ( / 4) u. <Sol.>. Begiig with x[], we use Eq. (7.4) to write 0 k. 0 4 k ) The first series coverge for </, while the secod coverge for > /4. ) Both series must coverge for X() to coverge, so the ROC is /4 < < /. the ROC of this two-side sigal is depicted i Fig. 7.9(a).. Summig the two geometric series, we obtai Poles at = ½ ad = /4, 4 8

19 The -Trasform Figure 7.9 (p. 565) ROCs for Example 7.5. (a) Two-sided sigal x[] has ROC i betwee the poles. (b) Right-sided sigal y[] has ROC outside of the circle cotaiig the pole of largest magitude. (c) Left-sided sigal w[] has ROC iside the circle cotaiig the pole of smallest magitude. 9

20 The -Trasform 7.4 Properties of The -Trasform. Assume that Z, w ith RO C x x R Z, with RO C y y Y R The ROC is chaged by certai operatios.. Liearity: The ROC ca be larger tha the itersectio if oe or more terms i x[] or y[] cacel each other i the sum. Z, with RO C at least x y ax by a by R R Example 7.6 Pole-Zero Cacellatio Suppose (7.) Z x u u, ad Z y u u Y 4 4 4, with ROC: / < < / with ROC: > / 0

21 The -Trasform Evaluate the -trasform of ax[] + by[]. <Sol.>. The pole-ero plots ad ROCs for x[] ad y[] are depicted i Fig. 7.0 (a) ad (b), respectively. The liearity property give by Eq. (7.) idicates that Z ax by a b 4 I geeral, the ROC is the itersectio of idividual ROCs, or / < < / i this example, which correspods to the ROC depicted i Fig. 7.0(a). 4. Figure 7.0 (p. 567) ROCs for Example 7.6. (a) ROC ad pole-ero plot for X(). (b) ROC ad pole-ero plot for Y(). (c) ROC ad pole-ero plot for a(x() ad Y()).

22 The -Trasform. However, whe a = b: ax ay a u u, 4 ad we see that the term (/) u[] has be caceled i time-domai sigal. The ROC is ow easily verified to be /4 < < /, as show i Fig. 7.0(c). This ROC is larger tha the itersectio of the idividual ROCs, because the term (/) u[] is o loger preset.. Combiig -trasforms ad usig the liearity property gives a ay a a a a 4 4 Note: The ero at = / cacels the pole at = /. The ROC elarges whe the pole is removed. 4.

23 The -Trasform Time Reversal x Z, with RO C R x (7.) Time reversal, or reflectio, correspods to replacig by. Hece, if R x is of the form a < < b, the ROC of the reflected sigal is a < / < b, or /b < < /a. Time Shift Z 0 x, w ith R O C R, except possibly 0 or 0 x. Multiplicatio by o itroduces a pole of order o at = 0 if o > 0.. If o < 0, the multiplicatio by o itroduces o poles at ifiity. If these poles are ot caceled by eros at ifiity i X(), the the ROC of o X() caot iclude =. Multiplicatio by a Expoetial Sequece. Let be a complex umber. The (7.), with RO C x R x (7.4)

24 The -Trasform. The otatio, R x implies that the ROC boudaries are multiplied by.. If R x is a < < b, the the ew ROC is a < < b. 4. If X() cotais a factor d i the deomiator, so that d is pole, the X(/) has a factor d i the deomiator ad thus has a pole at d. 5. If c is a ero of X(), the X(/) has a ero at c. 6. This idicates that the poles ad eros of X() have their radius chaged by, ad their agles are chaged by arg{}. Fig. 7. Covolutio Z, with RO C at least x y x y Y R R (7.5). Covolutio of time-domai sigals correspods to multiplicatio of -trasforms.. The ROC may be larger tha the itersectio of R x ad R y if a pole-ero cacellatio occurs i the product X()Y(). 4

25 The -Trasform Differetiatio i the -Domai d x R d Z, with ROC x (7.6). Multiplicatio by i the time domai correspods to differetiatio with respect to ad multiplicatio of the result by i the -domai.. This operatio does ot chage the ROC. Example 7.7 Applyig Multiple Problems Fid the -trasform of the sigal x u u 4 <Sol.>,. First we fid the -trasform of w[] = (-/) u[]. We kow from Example 7. that u, w ith RO C 5

26 The -Trasform Thus, the -domai differetiatio property of Eq. (7.6) implies that d w u W, with RO C d W ( ), w ith R O C. Next, we fid the -trasform of y[] = (/4) u[-]. We do this by applyig the time-reversal property give i Eq. (7.) to the result of Example 7.. Notig that Z u[ ], with RO C 4 / 4 4 we see that Eq.(7.) implies that Z 4 y Y, with ROC 4 4, 4 6

27 The -Trasform. Last, we apply the covolutio property give i Eq. (7.5) to obtai X() ad thus write Z, w ith RO C w y x w y W Y R R X ( ), with RO C 4 4 7

28 The -Trasform 7.5 Iversio of The -Trasform Two method for determiig iverse -trasform:. The method of partial fractio basic -trasform pairs ad -trasform properties. A right-sided time sigal has a ROC that lies outside the pole radius, while a leftsided time sigal has a ROC that lies iside the pole radius.. Ispectio method Express X() as a power series i of the form Eq. (7.4) 7.5. Partial-Fractio Expasios. The -trasform expressed as a ratioal fuctio of : M B b b b 0 M N A a a a 0 N (7.7) where M < N.. If M N, the we may use log divisio to express X() i the form ~ M N k B f k k 0 A. First term: usig the pair Z [ ] Secod term: Partial fractio ad the time-shift property 8

29 9 The -Trasform CHAPTER Example:, We factor from the umerator ad from the deomiator 0. 0 Simple pole case:. Factorig the deomiator polyomial ito a product of the first-order terms:, 0 0 N k k M M d a b b b. N k k k d A. Rewrite X() as a sum of the first-order terms:

30 The -Trasform. Depedig o the ROC, the iverse -trasform associated with each term is the determied by usig the appropriate trasform pair. Ak A d u, with RO C d d or k k k k Ak A d u, with RO C d d k k k Repeated pole case: k. If a pole d i is repeated r times, the there are r terms i the partial-fractio expasio associated with that pole: A d i, A i,, d d r i A i i r. If the ROC is of the form > d i, the the right-sided iverse -trasform is chose: i m m Z A A d u, w ith RO C d m! d i i 0

31 The -Trasform. If the ROC is of the form < d i, the the left-sided iverse -trasform is chose: i m m A A d u, w ith RO C d m! d i i Example 7.9 Iversio by Partial-Fractio Expasio Fid the iverse -trasform of, with RO C <Sol.>. We use a partial fractio expasio to write. A A A. Solvig for A, A, ad A gives.

32 The -Trasform Figure 7. (p. 574) Locatios of poles ad ROC for Example 7.9. u. 4. The ROC also has a radius less tha the pole at =, so this term has the left-sided iverse trasform u. 5. Fially, the ROC has a radius greater tha the pole at =, so this term has the right-sided iverse -trasform u. 6. Combiig the idividual term gives x u u u.

33 The -Trasform Example 7.0 Iversio of a Improper ratioal Fuctio Fid the iverse -trasform of <Sol.> 0 4 4, w ith RO C 4. The pole at = - ad = are foud by determiig the roots of the deomiator polyomial. The ROC ad pole locatios i the -plae are depicted i Fig. 7.. We covert X() ito a ratio of polyomials i i accordace with Eq. (7.7). We do this by factorig from the umerator ad from the deomiator, yieldig 0 4, 4 Figure 7. (p. 575) Locatios of poles ad ROC for Example The factor (/) is easily icorporated later by usig the time-shift property, so we focus o the ratio of polyomials i paretheses.

34 4 The -Trasform CHAPTER Usig log divisio to reduce the order of the umerator polyomial, we have Thus, we may write

35 The -Trasform. Next, usig a partial-fractio expasio, we have 5 ad thus defie W, (7.8) where W, with ROC 4. The ROC has a smaller radius tha either pole, as show i Fig. 7., so the iverse -trasform of W() is w u u. 5

36 The -Trasform 5. Fially, we apply the time-shift property (Eq.(7.)) to obtai x w x 7.5. Power Series Expasios u u.. Express X() as a power series i of the form Eq. (7.4).. The values of x[] are the give by the coefficiets associated with. Oly oe-sided sigal is applicable! DT sigals with ROC of the form < a or > a.. If the ROC is > a, the we express X() as power series i, so that we obtai a right-sided sigal. 4. If the ROC is < a, the we express X() as power series i, so that we obtai a left-sided sigal. 6

37 7 The -Trasform CHAPTER Example 7. Iversio by Meas of Log Divisio <Sol.> Fid the iverse -trasform of, with RO C usig a power series expasio.. We use log divisio to write X() as a power series i, sice ROC idicates that x[] is right sided. We have X

38 The -Trasform Thus, comparig X() with Eq.(7.4), we obtai x.... If the ROC is chaged to < /, the we expad X() as a power series i ; X I this case, we therefore have x A advatage of the power series approach is the ability to fid iverse - trasforms for sigals that are ot a ratio of polyomials i. 8