3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

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1 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [ ] < k = As show i Chapter, the FT does ot always exist. If this occurs, it may be icoveiet to aalye the correspodig sigal/system. I some cases, we ca get aroud the problem by acceptig the impulse fuctio i the frequecy domai as a valid fuctio to represet the sigal i the frequecy domai. A more geeral, ad very ofte more coveiet, solutio is to adopt the so-called Z-trasform (ZT). As you will see i a momet, it is possible to obtai the ZT of sigals whose FT do ot exist. 3-

2 The ZT ca be viewed as a more geeral versio of the FT. Its relatioship with the FT is aalogous to that betwee the Laplace trasform ad the Fourier trasform of a cotiuous time sigal. 3. Properties of Z-Trasform The ZT of a discrete time sigal x [ ] is X() = xk [ ] k = where is a complex umber. Whe we evaluate the ZT o the uit circle, i.e. we replace i the above equatio by k j = e ω, the the FT of the sigal is obtaied (if it exists i the first place). Example : Determie the ZT of the right-sided expoetial fuctio where u [ ] is the uit-step fuctio. Solutio: x [ ] = au [ ], k k k ( ) X() = xk [ ] = a = a k= k= k= k 3-

3 If the above summatio is to exist i the fiite magitude sese, the a < or equivaletly > a. I this case k X() = ( a ) = = ; > a a a k = I the above example, the regio where the ZT exists, i.e. > a, is called the regio of covergece (ROC), the poit = a where the ZT goes to ifiity is called a pole, ad the poit = where the ZT goes to ero is called a ero, The ROC ad the pole-ero plot of Example are show below (assumig a is real ad positive). Uit circle Im{} ROC pole ero Re{} Circle of radius a Example : Determie the ZT of the left-sided sigal x [ ] =au [ ] 3-3

4 Solutio: m k k k ( ) ( ) X() = xk [ ] = a = a = a k= k= m= m= m The sum exists oly if a < or < a. I this case m X() = ( a ) = = ; < a a a m = The aswer is similar to that i Example, except for the ROC. Uit circle Im{} ROC pole ero Re{} Circle of radius a Through these two examples, it becomes evidiet that it is very importat to specify the ROC of the ZT. Example 3: Determie the ZT of x [ ] = au [ ] + bu [ ] 3-

5 Solutio: Usig the result from Example ad the fact that the ZT is a liear operatio, we have X( ) = + a b The first term exists if > a ad the secod term exists if > b. So the ROC of X() is the overlap of the idividual ROCs, i.e. the exterior of a circle of radius R = max { a, b} As for poles ad eros, we first rewrite X() as X() a+ b ( ) ( ) ( ) = + = a b a b ; > R So the poles are located at ad = ( a+ b)/. = a ad = b; the eros are located at = The ROC ad the pole-ero plot for a = 3/, b = / is show below. Im{} Uit circle ROC pole Circle of radius a 3 ero Re{} 3-5

6 Example : Determie the ZT of the -sided sigal Solutio x [ ] = au [ ] bu[ ] Usig results from both Examples &, ad the fact that ZT is a liear operatio, we have X( ) = + a b The first term exists if > a while the secod term exists if < b. So the ROC of X() is the overlap of the idividual ROCs. Case I: If a < b, the the ROC is the rig a < < b I this case, the poles are located at = a ad = b, ad the eros are located at = ad = ( a+ b)/; see Example 3. Case II: If a > b, the ROC ad hece X() does ot exist. The ROC ad pole-ero plot for a = /, b = 3/ is 3-6

7 Uit circle 3 Im{} ROC pole ero Re{} I summary, what the ZT does is that it describe a DT sigal i terms of a p, a set of eros { }, ad a ROC: set of poles { j } M ( ) i a i i= i ( ) = ; ROC b j ( p ) j = j X I the above equatio, a i ad b j represet respectively the multiplicity of the i th ero ad the j th pole. The FT, if exists, is obtaied by evaluatig the ZT o the uit circle. Specifically, the magitude of the FT depeds o the separatios betwee j the poit e ω ad the poles ad eros accordig to M jω i ( ) = j= X e a jω i e i jω = ; e ROC jω b j e p j 3-7

8 ROC of the ZT of a fiite-duratio sigal: - The sigal x [ ] is of fiite duratio if x [ ] = for < ad >, where both ad are fiite, with. - I this case, the ZT is - It ca be easily show that ( ) = X x [ ] = ( ) [ ] [ ] max X = x x x = = = where xmax = max x [ ]. So the ZT is fiite for ay fiite but o-ero value of. If, the the ZT is also fiite at =. - The ROC of a fiite duratio sigal is the etire -plae, excludig ifiity ad perhaps. ROC of the ZT of a right-sided sigal: - By a right-sided sigal, we mea x [ ] = for < for some fiite. - I this case, the ZT is X ( ) = x [ ] = 3-8

9 - Let q j = re θ represets a poit o a circle of radius r i the -plae. If the ZT exists o this circle, the for ay θ ( ) [ ]( ) o = o X q = x q < j - The ZT o the circle q = re φ, if exists, is q ρ = = q = ( ) = [ ]( ) = [ ] = [ ]( ) X q x q x q x q where ρ = q q Defie ( ) g [ ] = x [ ] q u [ ], f[ ] = ρ u[ ], ad h [ ] = g [ ] f[ ] = gk [ ] f[ k]. k = The it ca be easily show that 3-9

10 π jω jω h[] = X ( q ) = G( e ) F( e ) dω π π F e ω are the FTs of g [ ] ad f[ ]. It is j where G( e ω j ) ad ( ) jω iterestig to poit out that G( e ) X ( q ) ω= =, ad i geeral jω jω ( ) X ( qe ) G e =. j The importat thig is that G( e ω j ) exists. The FT F( e ω ) will also exist if ρ <, i.e. whe r > r. If this coditio holds, the the ZT o the circle of radius r exists. I summary, the ROC of a right-sided sigal is to the exterior of a circle. The radius of this circle equals the largest pole (i magitude sese). ROC of the ZT of a left-sided sigal: - By a left-sided sigal, we mea x [ ] = for > for some fiite. - The ROC is to the iterior of a circle. The radius of the circle equals the the smallest pole (i the magitude sese). ROC of the ZT of a two-sided sigal: - By a -sided sigal, we mea a [ ] x that stretches from = to =. 3-

11 - Sice a -sided sigal ca be treated as the sum of a left sided sigal with a ROC of < R ad a right-sided sigal with a ROC of > R +, its ROC is the overlap of the two. - If R > R+, the ROC is Otherwise, the ZT does ot exist. R+ < < R+. Exercise: Show that the ROC must be a cotiuous regio. Exercise: I this exercise, we use the otatio x [ ] Z X( ); ROC: R x to represet the sigal x [ ] ad its ZT X( ) whose ROC is followig R x. Show the (a) ax[ ] + bx[ ] Z ax() + bx( ); ROC: Rx R x where represets itersectio of two ROCs. (b) x X R Z [ ] ( ); ROC: (except o x for the possible addito or delectio of =, or = ) x [ ] X( / ); ROC: R x (c) Z 3-

12 (d) (e) (f) Z dx( ) x[ ] ; ROC: R d * * * x [ ] Z X ( ); ROC: R x Z * * x[ ] X (/ ); ROC: / R x x x[ ] x [ ] Z X ( X ) ( ); ROC: R R (g) x x 3. Iverse Fourier Trasform There are a umber of approaches to determie a sigal from its ZT. 3..: By Ispectio A ZT (or a compoet of a ZT) that you will ecouter from time to time is X( ) = ; ROC: > a a This is because may ZTs express the sigal i terms of poles (ad eros). Accordig to a earlier example, the time domai sigal is x [ ] au [ ] =. 3-

13 Exercise: Deteremie x [ ] if d X( ) = ; ROC: > a a Aswer: This is simply the sigal x[ ] = au [ ] delayed by d samples (see Part (b) of the last exercise i Sectio 3.). Cosequetly ( d) x [ ] = a u [ d] 3.. Power Series Expasio I its most primitive form, the ZT is a power series i beig x [ ]. coefficiet associated with ( ), with the Cosequetly, the coefficiets i the power series represetatio of the ZT give you the sigal itself. Example: Determie x [ ] whe X( ) = e ; < Solutio: The power series represetatio of X () is m X( ) = e = = ; <!! = m= ( m) 3-3

14 where! deotes factorial. From this expressio, we ca deduce that x [ ] = u[ ]! ( ) Partial Fractio Expasio We assume that the ZT is give i the form X() = b a M ( c k ) k= ( d i ) i= where ck, k =,,..., M are the eros, di, i =,,..., are the poles, ad a, b are costats. The idea behid the Partial Fractio Expasio techique is to express the above ZT as a weighted sum of terms of the form ( d i ) where m is a iteger. The iverse ZT of the above ca be determied aalytically. Specifically, let m Z f [ ] F ( ) = ( d ) m m m 3-

15 where d represets i geeral a pole of X( ). Sice d F F dm ( ) d m( ) = m( ) d, = Fm ( ) dm ( ) d this meas fm[ ] = ( + ) fm [ + ] ; dm ( ) refer to (b) ad (d) i the last exercise i Sectio 3.. Sice f [ ] [ ] du =, so f [ ] = ( + ) du [ + ], f3[ ] = ( + ) du [ + ], f[ ] = ( + 3) du [ + 3] 3, or i geeral fm[ ] = ( + m ) du [ + m] ( m )! 3-5

16 To describe the partial fractio expasio techique, we cosider three cases, (a) (b) (c) the umber of eros is less tha the umber of poles ad there are o multiple-order poles, the umber of eros is greater tha the umber of poles but still there are o multiple-order poles, ad there are more eros tha poles ad there is oe multiple-order pole. Commets: - A pole is of multiple-order if it is repeated more tha time. - The sceario of more tha oe multiple-order poles ca be cosidered as a straightforward extesio of case (c) above. Case M < ad o multiple-order pole All the poles are differet. I this case, the origial ZT ca be expressed as X() b M ( c k ) k= = = a i= ( d i ) i= d A A A = d d d A i i 3-6

17 Cosequetly x [ ] = Ai ( di) u [ ] i= The questio that remais to be aswered is: what are the Ai s? If we multiply both sides of X( ) by ( ) d, the result is ( ) ( ) A d A d d X() = A d d ow, if we evalue the above at = d, we simply have ( ) () = d d X = A I geeral, ( i ) d X() = Ai = d i Example: A left-sided sigal has a ZT of X( ) = ( )( ) 3-7

18 Determie the ROC ad the iverse ZT. Solutio: The ZT ca be expressed as X( ) A = = + ( )( ) ( ) ( ) A where ( ) A = X() = = / ad ( ) A = X() = = / Cosequetly X( ) = + ( ) ( ) Give that x [ ] is a left-sided sigal, the first term must have a regio of covergece of < / ad the secod term must have a regio of covergece of < /. The overall ROC is the overlap of the two, which is simply < /. The iverse ZTs of the first ad secod terms are ( ) / ( ) u[ ] respectively. So / u[ ] ad 3-8

19 ( ) ( ) {( ) ( ) } u x [ ] = / u[ ] / u[ ] = / / [ ]. Exercise: Repeat the last example for a -sided sigal. Case M greater tha or equal to ad o multiple-order pole As i Case, all the poles are differet. I this case, the origial ZT ca be expressed as X M ( c k ) b = = + a k = () Q () ( d i ) ( d i ) i= i= R () where M Q ( ) = B r r= r ad R ( ) are respectively the quotiet ad remaider polyomials (i ) whe / M b a ( c k ) is divided by ( d i ). k = i= 3-9

20 Sice the degree of the remaider polyomial (i ) is less tha, it meas the secod term i X( ) ca be expressed i the same form as that i Case. Cosequetly, we ca rewrite X( ) as X() = Q () + ( d i ) i= R () A M r B r r= i= i = + i ( d ), where agai ( i ) Ai = d X() = d i Example: Determie the IZT of X( ) = ; < ( )( ) Solutio: - From the ROC, we kow that this is a left-sided sigal. - The deomiator ca be writte as D () = Dividig the umerator C ( ) = by the deomiator, usig log divisio, yields Q () = + 3-

21 ad R ( ) = +. 3 This meas + X() = + + ( )( ) 3 A A = + + +, ( ) ( ) where A + + = ( ) = 3 6 ( )( ) = / A + + = ( ) = 3 6 ( )( ) = / - The iverse ZT is thus ( ) ( ) x [ ] = δ[ ] + δ[ ] + u[ ] + u[ ] Case 3 M greater tha or equal to ad multiple-order pole The pole d is repeated s times, i.e. 3-

22 d = d = = d. s Each of the remaiig s poles has a order of. This meas the ZT trasform, i product form, is X() = b a k = M ( c k ) s ( p ) ( d i ) i= s+ The ZT, i summatio form, is X() A = + + M s r i m B r r i ( ) d = = i m= ( d ) C m The determiatio of the Ai s ad the ( i ) Br s are the same as before, i.e Ai = d X( ) ; i= s+, s+,...,, = d i ad B r is the r-th coefficiets of the quotiet polyomial whe the umerator ( b / a) M ( c ) k= k of X( ) is divided by its deomiator s ( p ) ( d i= s+ i ). The Cm s ca be obtaied by first replacig the by w ad the compute 3-

23 ( )!( ) s {( ) ( )} sm d Cm = dw X w sm sm sm d dw ote that ( dw) X ( w ) = ( dw) Bw + A + s r ( dw) Bw ( dw) ( dw) w= d M s s s r i m r r= i= m= ( dw i ) ( dw) M s i r r= i= i = + s + C r= r s r A. ( dw) Commets: - Evaluatig the ( s m) -th order derivatives of the first two terms at w= d will always yield ero. s r - Evaluatig the ( s m) Cr dw at w= d will be ero except whe r = m. ote that whe r > m, we will at some poit ed up with differetiatig a costat. O the other had whe r < m, the result after differetiatio will iclude power of the term ( dw ), which whe evaluated at w= d equal to ero. -th order derivative of ( ) C m 3-3