The -Trasform 7. Itroductio The trasform is the discrete-time couterpart of the Laplace trasform. Other descriptio: see page 553, textbook. 7. The -trasform Derivatio of the -trasform: x[] re jω LTI system, h[] y[] x[] h[]. Iput x[] : x [] r cos( Ω) + jr si( Ω). (7.) Fig. 7.. The output of the LTI system with impulse respose h[]: y [ ] H{ x[ ] } h[ ] x[ ] h k We use x[] to obtai [][ k x k].
The -Trasform Damped cosie Damped sie Figure 7. (p. 554) Real ad imagiary parts of the sigal.
The -Trasform y hk hk k k k k [ ] [ ] [ ]. 3. We defie trasfer fuctio: ( ) h[] k k H (7.) k { } H ( ). H y jφ ( ) [] H ( ) e. The form of a eigefuctio eigefuctio; H() eigevalue where H() is expressed i polar form, i.e., H() H() e jφ (). Usig re jω ad applyig Euler s formula, we obtai y jω jω jφ jω [ ] H ( re ) r cos( Ω + φ( re ) + j H ( re ) r Ω + φ( re ) si( ). The system modifies the amplitude of the iput by H(re jω ) ad shifts the phase of the siusoidal compoets by φ (re jω ). 4. Represetatio of arbitrary sigals as a weighted superpositio of the eigefuctio : 3
The -Trasform Substitutig re j Ω ito Eq. (7.) yields jω jω jω ( ) [ ]( ) ( [ ] ). H re h re h r e h π π jω jω [] r H ( re ) e dω π H(re j Ω ) correspods to h[]r. Hece, the iverse DTFT of H(re j Ω ) must be h[]r, so we may write Multiplyig this result by r gives r π π jω jω jω jω h[ ] H ( re ) e d H ( re )( re ) d. π Ω π π Ω π where h The symbol j d (7.3) π [] H ( ), re j Ω ; d jre j Ω d Ω; dω (/j) d deotes the itegratio aroud a circle of radius r i a couterclockwise directio. 4
5. The -trasform pair: The -Trasform Iverse -trasform Χ ( ) x[], (7.4) x[] ( ). Notatio: x [] Χ( ) πj Χ d I Eq. (7.5), the sigal x[] is expressed as a weighted superpositio of complex potetial. The weights are (/πj)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. 7.. 5
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 > α, ad thus the -trasform of x[] exists. a α 6
The -Trasform 7.. The -Plae. The -plae: Fig. 7.3.. If x[] is absolutely summable, the the DTFT obtaied from the -trasform by settig r, or substitutig e jω ito Eq. (7.4). That is, Χ jω ( ) Χ( ) jω e. e (7.6) 3. The equatio e jω describes a circle of uit radius cetered o the origi i the -plae. Fig. 7.4. Uit circle i the -plae. Figure 7.3 (p. 557) The -plae. A poit re jω is located at a distace r from the origi ad a agle Ω relative to the real axis. The frequecy Ω i the DTFT correspods to the poit o the uit circle at a agle Ω with respect to the positive real axis. 7
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 Χ ( ) + +.. We obtai the DTFT from X() by substitutig e jω : Χ jω jω jω jω ( e ) e + e + e. Figure 7.4 (p. 557) The uit circle, e j Ω, i the -plae. 8
The -Trasform 7..3 Poles ad Zeros. The -trasform i terms of two polyomials i : Χ b 0 M ( ). a + b 0 Χ + a ( ) + K + b ~ + Ka N b M k k N M N ( c ) k ( ). d k where 0 0 b% b / a 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(). 3. Symbols i the -plae: poles; eros Example 7. -TRANSFORM OF A CAUSAL EXPONENTIOAL SIGNAL Determie the -trasform of the sigal x [] α u[]. Depict the ROC ad the locatio of poles ad eros of X() i the -plae. 9
The -Trasform <Sol.>. Substitutig x[] α u[] ito Eq. (7.4) yields α Χ ( ) α u[ ] 0.. This is a geometric series of ifiite legth i the ratio α/; the sum coverges, provided that α/ <, or > α. Hece, Χ ( ), α > α, α > α. (7.7) 3. 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[] α u[] i the -plae. The ROC is the shaded area. a α 0
The -Trasform Example 7.3 -TRANSFORM OF A ANTICAUSAL EXPONENTIOAL SIGNAL Determie the -trasform of the sigal y [] α 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 ( ) α u[ ] α k α k 0 k α k.. The sum coverges, provide that /α <, or < α. Hece,, < α, α, < α α ( ) Y (7.8) 3. The ROC ad the locatio of poles ad ero are depicted i Fig 7.6.
The -Trasform Figure 7.6 (p. 560) ROC ad locatios of poles ad eros of x[] α u[ ] i the - plae. a α 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
3 The The -Trasform Trasform CHAPTER ( ) [] [ ] u u Χ 0. 0 0 + k k. Both sum must coverge i order for X() to coverge. This implies that we must have > / ad <. Hece, 3. The ROC ad the locatio of poles ad ero are depicted i Fig 7.7 Fig 7.7. i this case, the ROC is a rig i the -plae. Figure 7.7 (p. 560) Figure 7.7 (p. 560) ROC ad locatios of poles ad eros i the -plae for Example 7.4. ( ) /, < < + Χ ( ) ( )( ), /, 3 < <
The -Trasform 7.3 Properties of Regio of Covergece The ROC (regio of covergece) 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 caot lie i the ROC. 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 Χ ( ) 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. ) 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
The -Trasform 3) 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[] cδ[] 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[ ]. Χ. Splittig the ifiite sum ito egative- ad positive-term portios, we defie Ι ( ) x[] Note that ( ) ( ) X ( ) Ι +Ι + + 0 ( ) x[]. (7.9) ad Ι (7.0) If I () ad I + () are fiite, the X() is guarateed to be fiite, too. A sigal that satisfies these two bouds grows o faster tha (r + ) for positive ad (r ) for egative. 5
The -Trasform 3. If the boud give i Eq. (7.9) is satisfied, the r ( ) A ( r ) A A k r Ι The last sum coverges if ad oly if r. 4. If the boud give i Eq. (7.0) is satisfied, the r + Ι+ ( ) A+ ( r+ ) A+ 0 0 This sum coverges if ad oly if r +. k k Hece, if r + < < r, the both I + () ad I () coverge ad X() also coverges. If r + > r, the there are o values of for which covergece is guarateed. 5. Left-sided sigal: x[] 0 for 0, right-sided sigal: x[] 0 for < 0, ad a two-sided sigal is defied as a sigal that has ifiite duratio i both the positive ad egative directios. The, for sigals x[] satisfy the expoetial bouds of Eqs. (7.9) ad (7.0), we 6
The -Trasform have the followig coclusios: ) The ROC of a right-sided sigal is of the form > r +. ) The ROC of a left-sided sigal is of the form > r. 3) The ROC of a two-sided sigal is of the form r + < < r. Fig. 7.8. Detailed discussio: see page 563, textbook. Figure 7.8 (p. 564) 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 +. 7
The -Trasform Figure 7.8 (p. 564) Figure 7.8 (p. 564) The relatioship betwee the ROC ad the time extet of a sigal. (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. 8
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 Χ ( ) + ( ) +. 0 4 k 0 0 4 ) 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 Χ + ( ) +, Poles at ½ ad /4 4 k 9
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. 0
The -Trasform Note that the ROC is a rig-shaped regio located betwee the poles. 3. y[] is a right-sided sigal, ad agai usig the defiitio of the -trasform give by Eq. (7.4) we write Y 4 ( ) +. 0 9 ) The first series coverge for >/, while the secod coverge for > /4. ) Hece, the combie ROC is > /, as depicted i Fig. 7.9(b). I this case, we write Poles at ½ ad /4 Y ( ) +, + 4 The ROC is outside a circle cotaiig the pole of largest radius, - /. 4. The last sigal, w[], is left sided ad has -trasform give by 0 0 k k W ( ) + ( ) + ( 4 ), 4 k 0 k 0 The first series coverge for < /, while the secod coverge for < /4, givig a combied ROC of < /4, as depicted i Fig. 7.9(c).
The -Trasform I this case, we have W + 4 ( ) +, Poles at ½ ad /4 The ROC is iside a circle cotaiig the pole of smallest radius, /4. This example illustrates that the ROC of a two-side sigal is a rig, the ROC of a right sided sigal is the exterior of a circle, ad the ROC of a left-sided sigal is the iterior of a circle. I each case the poles defie the boudaries of the ROC. 7.4 Properties of The -Trasform. Assume that [ ] Χ Z ( ), with ROC x x R [ ] Z ( ), with ROC 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 Χ ( ) + ( ), with ROC at least x y ax by a by R R (7.)
The -Trasform Example 7.6 Pole-Zero Cacellatio Suppose 3 Z x [ ] u [ ] u[ ] Χ ( ), ad Z y [ ] u [ ] u [ ] Y( ) 4 3 ( )( ) 4 ( )( ) 4, with ROC: / < < 3/ with ROC: > / 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 [ ] [ ] 3 ( )( ) ( )( ) 4 I geeral, the ROC is the itersectio of idividual ROCs, or / < < 3/ i this example, which correspods to the ROC depicted i Fig. 7.0(a). 4. 3
The -Trasform Figure 7.0 (p. 567) 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()). 4
5 The The -Trasform Trasform CHAPTER. Note, however, what happes whe a b: We have [] [] [ ] [], 4 3 + + u u a ay ax ad we see that the term (/) u[] has be caceled i time-domai sigal. The ROC is ow easily verified to be /4 < < 3/, 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. 3. Combiig -trasforms ad usig the liearity property gives ( ) ( ) ( )( ) ( )( ) + + Χ 4 4 3 a ay a ( ) ( ) ( )( )( ) 3 4 4 3 4 a ( ) ( )( )( ). 3 4 4 5 a The ero at / cacels the pole at /, so we have
The -Trasform aχ ( ) + ay ( ) a ( )( ). 3 4 5 4 Hece, cacellatio of the (/) u[] term i the time domai correspods to cacellatio of the pole at / by a ero i -domai. This pole defied the ROC boudary, so the ROC elarges whe the pole is removed. Time Reversal x Z Χ, with ROC 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 [ ] ( ) x0 Χ, with ROC Rx, except possibly 0 or. 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 <. (7.3) 6
The -Trasform Multiplicatio by a Expoetial Sequece. Let α be a complex umber. The α x Χ α [ ],with ROC α R x (7.4). The otatio, α R x implies that the ROC boudaries are multiplied by α. 3. If R x is a < < b, the the ew ROC is α a < < α b. 4. If X() cotais a factor d i the deomiator, so that d is pole, the X(/α) has a factor αd i the deomiator ad thus has a pole at αd. 5. If c is a ero of X(), the X(/α) has a ero at αc. 6. This idicates that the poles ad eros of X() have their radii chaged by α, ad their agles are chaged by arg{α}. Fig. 7. Covolutio [ ] [ ] Χ Z ( ) ( ), with ROC 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(). 7
The -Trasform Figure 7. (p. 569) The effect of multiplicatio by α o the poles ad eros of a trasfer fuctio. (a) Locatios of poles ad eros of X(). (b) Locatios of poles ad eros of X(/α). Differetiatio i the -Domai d x R d Z [ ] Χ( ), 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. 8
The -Trasform 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,with ROC > [ ] + Thus, the -domai differetiatio property of Eq. (7.6) implies that d w [ ] u [ ] W( ), with ROC > d + + ( ) + 9
( + ) The -Trasform W( ), with ROC >. 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 ROC > 4 /4 4 we see that Eq.(7.) implies that Z [ ] Y( ) y, with ROC 4 > 4, with ROC < 4 4 3. Last, we apply the covolutio property give i Eq. (7.5) to obtai X() ad thus write [ ] [ ] [ ] Χ Z ( ) ( ) ( ), with ROC w y x w y W Y R R 4, 30
( 4)( + ) The -Trasform X( ), with ROC < < 4 Example 7.8 -Trasform of A Expoetially Damped Cosie Use the properties of liearity ad multiplicatio by a complex expoetial to fid the -trasform of x [] a cos( Ω0) u[ ], where a is real ad positive. <Sol.>. First we ote form Example 7. that y[] α u[] has the -trasform Y ( ), with ROC > a. a. Now we rewrite x[] as the sum x Ω jω0 j 0 [] e y[] + e y[] ad apply the property of multiplicatio by a complex expoetial give i Eq. (7.4) to each term, obtaiig 3
The -Trasform jω 0 jω0 Χ ( ) Y ( e ) + Y ( e ), with ROC > a + jω0 jω0 ae ae jω 0 jω0 ae + ae jω Ω ( )( ) 0 j 0 ae ae acos( Ω0 ), with ROC > a. acos Ω + a ( ) 7.5 Iversio of The -Trasform Two method for determiig iverse -trasform: 0. 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 left-sided 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) 3
The -Trasform 7.5. Partial-Fractio Expasios. The -trasform expressed as a ratioal fuctio of : ( ) ( ) ( ) Χ B M b0 + b + L+ bm A N a0 + a + L+ an (7.7) where M > N.. If M N, the we may use log divisio to express X() i the form ~ M N k B Χ( ) f k + k 0 A ( ). δ Z [ ] Secod term: Partial fractio First term: usig the pair ad the time-shift property Example: Χ ( ), 3 3 6 + 0 + 9 We factor from the umerator ad 3 3 from the deomiator Χ ( ) + 3 3 3 3 + 0 3 + 0 3 + 3. 33
The -Trasform Simple pole case:. Factorig the deomiator polyomial ito a product of the first-order terms: Χ ( ) b 0 + b a N 0 k + L + b ( d ) k M M. Rewrite X() as a sum of the first-order terms: Χ N k ( ). k A d k, 3. 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 ROC > d d or k k k k Ak A ( d ) u[ ], with ROC < d d k k k k 34
The -Trasform Repeated pole case:. If a pole d i is repeated r times, the there are r terms i the partial-fractio expasio associated with that pole: A i d, A i, L, r ( d ) ( d ) r i A i i. If the ROC is of the form > d i, the the right-sided iverse -trasform is chose: ( + ) L( + ) ( i ) [ ] ( m ) m Z A A d u, with ROC > d m! ( d i ) 3. If the ROC is of the form < d i, the the left-sided iverse -trasform is chose: ( + ) L( + ) ( i ) [ ] ( m ) m A A d u, with ROC < d m! ( d i ) Discussio about ROC: see page 573, textbook. i i 35
The -Trasform Example 7.9 Iversio by Partial-Fractio Expasio Fid the iverse -trasform of + Χ, with ROC < < ( ) ( )( )( ) <Sol.>. We use a partial fractio expasio to write Χ 3 ( ) + +. A A A. Solvig for A, A, ad A 3 gives Χ ( ) + +. 3. Now we fid the iverse -trasform of each term, usig the relatioship betwee the locatios of the poles ad the ROC of X(), each of which is depicted i Fig. 7.. The figure shows that the ROC has a radius greater tha the pole at /, so this term has the right-sided iverse trasform 36
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 37
The -Trasform x [] u[] ( ) u[ ] u[ ]. Example 7.0 Iversio of a Improper ratioal Fuctio Fid the iverse -trasform of 3 0 4+ 4 Χ ( ), with ROC < 4 <Sol.>. 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. 7.3. We covert X() ito a ratio of polyomials i i accordace with Eq. (7.7). We do this by factorig 3 from the umerator ad from the deomiator, yieldig Χ 0 4 + 4 3 ( ),. The factor (/) is easily icorporated later by usig the time-shift property, so we focus o the ratio of polyomials i paretheses. 38
39 The The -Trasform Trasform CHAPTER Figure 7.3 (p. 575) Figure 7.3 (p. 575) Locatios of poles ad ROC for Example 7.0. Usig log divisio to reduce the order of the umerator polyomial, we have 5 3 3 6 8 6 4 3 0 4 4 3 3 + + + + + + Thus, we may write
The -Trasform 0 4 + 4 3 5 + 3 + 5 + 3 + ( )( ). 3. Next, usig a partial-fractio expasio, we have 5 ( + )( ) + ad thus defie Χ ( ) W ( ), (7.8) where ( 3 ) + 3 +, + with ROC < 3 4. The ROC has a smaller radius tha either pole, as show i Fig. 7.3 Fig. 7.3, so the iverse -trasform of W() is 40
The -Trasform ( ) δ [ ] + 3δ [ ] ( ) u[ ] + 3( ) u[ ]. w 5. Fially, we apply the time-shift property (Eq.(7.3)) to obtai x [] w[ + ] x 3 + [] δ [] + δ [ + ] ( ) u[ ] + 3( ) u[ ]. Causality, stability, or the existece of the DTFT is sufficiet to determie the iverse trasform. ) If a sigal is kow to be causal, the the right-sided iverse trasforms are chose. ) If a sigal is stable, the it is absolutely summable ad has a DTFT. Hece, stability ad the existece of the DTFT are equivalet coditios. I both cases, the ROC icludes the uit circle i the -plae,. The iverse - trasform is determied by comparig the locatios of the poles with the uit circle. If a pole is iside the uit circle, the the right-sided iverse -trasform is chose; if a pole is outside the uit circle, the the left-sided iverse - trasform is chose. 4
The -Trasform 7.5. Power Series Expasios. 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. 3. 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. Example 7. Iversio by Meas of Log Divisio Fid the iverse -trasform of ( ) +, with ROC Χ > usig a power series expasio. <Sol.>. We use log divisio to write X() as a power series i, sice ROC idicates that x[] is right sided. We have 4
43 The The -Trasform Trasform CHAPTER... 3 3 3 + + + + + ( )... 3 + + + + X Thus, comparig X() with Eq.(7.4), we obtai [] [] [ ] [ ] [ ]... 3 + + + + x δ δ δ δ. If the ROC is chaged to < /, the we expad X() as a power series i ;
The -Trasform 8 6 + + 4 4 8 8 8 3 6 6 I this case, we therefore have x 3 +... [] δ [] 8δ [ + ] 6δ [ + ] 3δ [ + 3]... X 3 ( ) 8 6 3... A advatage of the power series approach is the ability to fid iverse - trasforms for sigals that are ot a ratio of polyomials i. Example 7. Iversio via a Power Series Expasio Fid the iverse -trasform of 44
The -Trasform X( ) e, with ROC all except <Sol.>. Usig the power series represetatio for e u, vi., e a Thus k a 0 k! k [ ] X X ( ) K 0 0, > 0 or odd, otherwise! 7.6 The Trasfer Fuctio x[] re jω LTI system, h[] ( ) k! k K 0 k! k y[] x[] h[] 45
The -Trasform. Output of LTI system: [ ] h[ ] x[ ] Takig -trasform [ ] H[ ] X [ ] y *. Trasfer fuctio: [] [] Y H [] (7.0) X Y (7.9) This defiitio applies at all i the ROC of X() ad Y() for which X() is oero. Example 7.3 System Idetificatio The problem of fidig the system descriptio from kowledge of iput ad output is kow as system idetificatio. Fid the trasfer fuctio ad impulse respose of a causal LTI system if the iput to the system is x [] (/ 3) u[] ad the output is y [] 3 ( ) u[] + (/ 3) u[ ] <Sol.> Sol.>. The -trasforms of the iput ad output are respectively give by 46
The -Trasform X( ) with ROC / 3 (/3) > + ad 3 Y( ) + + (/3) 4, with ROC > + ( )( (/ 3) ). We apply Eq.(7.0) the obtai the trasfer fuctio: 4( (/ 3) ) ( )( (/ 3) ) H( ), with ROC > + 3. The impulse respose of the system is obtai by fidig the iverse - trasform of H(). Applyig a partial fractio expasio to H() yields H( ) +, with ROC > + (/3) 4. The impulse respose is thus give by 47
The -Trasform h [] ( ) u[] + (/ 3) u[ ] 7.6. Relatig the Trasfer Fuctio ad the Differece equatio. Nth-order differece equatio: K 0 a k y M [ k] b x[ k] K 0 k. The trasfer fuctio H() is a eigevalue of the system associated with the eigefuctio. If x[], the the output of a LTI system is y[] H(). Substitutig x[ k] k H() ito the differece equatio gives the relatioship N K 0 a k k H ( ) 3. Trasfer fuctio: M K 0 b N k K 0 a k k k M k bk K 0 H ( ) N (7.) k a K 0 k Ratioal trasfer fuctio See page 58, textbook. 48
The -Trasform Example 7.4 Fidig the Trasfer Fuctio ad Impulse Respose Determie the trasfer fuctio ad the impulse respose for the causal LTI system described by the differece equatio y [] ( / 4) y[ ] (3/ 8) y[ ] x[ ] + x[ ] <Sol.>. We obtai the trasfer fuctio by applyig Eq.(7.): + H( ) (/4) (3/8). The impulse respose is foud by idetifyig the iverse -trasform of H(). Applyig a partial-fractio expasio to H() give H + (/ ) (3/ 4) ( ) + 3. The system is causal, so we choose the right-side iverse -trasform for each term to obtai the followig impulse respose: h [] ( / ) u[ ] + (3/ 4) u[ ] 49
The -Trasform Example 7.5 Fidig a Differece-Equatio Descriptio Fid the differece-equatio descriptio of a LTI system with trasfer fuctio 5 + H ( ) ( + 3 + ) <Sol.>. We rewrite H() as a ratio of polyomials i. Dividig both the umerator ad deomiator, we obtai 5 + H ( ) ( + 3 + ). Comparig trasfer fuctio with Eq.(7.), we coclude that M, N, b 0 0, b 5, b, a 0, a 3, ad a. Hece, this system is described by the differece equatio y [] + 3y[ ] + y[ ] 5x[ ] + x[ ] Trasfer fuctio i pole-ero form: where c k eros; d k poles; ad b% b / a gai factor 0 0 ~ M b k ( ck ) H ( ) N ( d ) k k (7.) 50
The -Trasform A pth-order pole at 0 occurs whe b 0 b b p 0, while a lth-order ero at 0 occurs whe a 0 a a l 0. I this case, we write ~ p M p b k ( ck ) H ( ) (7.3) N b ( d ) where k b% b / a gai factor p l 7.7 Causality ad Stability k. The impulse respose of a causal LTI system is ero for < 0. Right-sided Trasfer fuctio Impulse respose iverse trasform Pole d k iside the uit circle, i.e., d k < Pole d k outside the uit circle, i.e., d k > Fig. 7.4 Expoetially decayig term Expoetially icreasig term 5
The -Trasform Figure 7.4 (p. 583) Figure 7.4 (p. 583) The relatioship betwee the locatio of a pole ad the impulse respose characteristics for a causal system. (a) A pole iside the uit circle cotributes a expoetially decayig term to the impulse respose. (b) A pole outside the uit circle cotributes a expoetially icreasig term to the impulse respose. 5
The -Trasform. The impulse respose of a stable LTI system is absolutely summable ad the DTFT of the impulse respose exists. The ROC must icludes the uit circle i the -plae. Pole d k iside the uit circle, i.e., d k < Left-sided iverse trasform Expoetially decayig term Pole d k outside the uit circle, i.e., d k > Right-sided iverse trasform Expoetially icreasig term Fig. 7.5 Note that a stable impulse respose caot cotai ay icreasig expoetial or siusoidal terms, sice the the impulse respose is ot absolutely summable. LTI system are both stable ad causal must have all their poles iside the uit circle. Fig. 7.6 53
The -Trasform Figure 7.5 (p. 583) Figure 7.5 (p. 583) The relatioship betwee the locatio of a pole ad the impulse respose characteristics for a stable system. (a) A pole iside the uit circle cotributes a right-sided term to the impulse respose. (b) A pole outside the uit circle cotributes a left-sided term to the impulse respose. 54
The -Trasform Figure 7.6 (p. 584) A system that is both stable ad causal must have all its poles iside the uit circle i the -plae, as illustrated here. Example 7.6 Causally ad Stability A LTI system has the trasfer fuctio 3 H ( ) + + π π j j 4 4 + 0.9e 0.9e Fid the impulse respose, assumig that the system is (a) stable or (b) causal. Ca this system both stable ad causal? 55
The -Trasform <Sol.> Figure 7.7 (p. 584) Figure 7.7 (p. 584) Locatios of poles i the -plae for the system i Example 7.6.. The give system has poles at 0.9 e jπ/4, 0.9 e -jπ/4, ad -, depicted i Fig. 7.7. If the system is stable, the the ROC icludes the uit circle. The two poles iside the uit circle cotribute the right-sided term to the impulse respose, while the pole outside the uit circle cotributes a leftsided term.. For case (a): π j π j 4 4 h( ) (0.9e ) u[ ] + (0.9e ) π 4(0.9) cos( ) u[ ] 3( ) 4 u[ ] 3( ) u[ ] u[ ] 56
The -Trasform 3. If the system is assumed causal, the all poles cotribute right-sided terms to the impulse respose. 4. For case (b), we have 4 4 h( ) (0.9e ) u[ ] + (0.9e 4(0.9) π j π j π cos[ ] u[ ] + 3( ) 4 ) u[ ] + 3( ) u[ ] u[ ] 5. Note that LTI system caot be both stable ad causal, sice there is a pole outside the uit circle. Example 7.7 First-Order Recursive System: Ivestmet Computatio I Example.5, we showed that first-order recursive equatio y [ ] ρy[ ] x[ ] may be used to describe the value y[] of a ivestmet by settig ρ + r/00, where r us the iterest rate per period, expressed i percet. Fid the trasfer fuctio of this system ad determie whether it ca be both stable ad causal. <Sol.>. The trasfer fuctio is determied by substitutig ito Eq.(7.) to obtai 57
The -Trasform H ρ ( ). This LTI system caot be both stable ad causal, because the pole at ρis outside the uit circle. 7.7. Iverse System. The impulse respose of a iverse system, h iv [], satisfies h iv [ ] * h[ ] δ [ ] where h[] is the impulse respose of the system to be iverted.. Takig -trasform: H iv ( ) H ( ) H iv ( ) H ( ) 3. If H() is writte i pole-ero form show i Eq. (7.3), the H iv ( ) ~ b p N l k M p k ( d k ( c k ) ) (7.4) Ay system described by a ratioal trasfer fuctio has a iverse system of this form. 58
The -Trasform Figure 7.8 (p. 586) A system that has a causal ad stable iverse must have all its poles ad eros iside the uit circle, as illustrated here. 4. H iv () is both stable ad causal if all of its poles are iside the uit circle. Sice the poles of H iv () are the eros of H (), we coclude that a stable ad causal iverse of a LTI system H() exists if ad oly if all the eros of H() are iside the uit circle. 5. A system with all its poles ad eros iside the uit circle, as illustrated i Fig. 7.8, is termed a miimum-phase system. The phase respose of a miimum-phase system is uiquely determied by the magitude respose, ad vice versa. Example 7.8 Stable ad Causal Iverse System A LTI system is described by the differece equatio 59
The -Trasform y 4 [] y[ ] + y[ ] x[] + x[ ] + x[ ] 4 Fid the trasfer fuctio of the iverse system. Does a stable ad causal LTI iverse system exist? <Sol.>. We fid the trasfer fuctio of the give system by applyig Eq.(7.) to obtai H ( ) + ( 4 ( + 4 8 )( + ). The iverse system the has the trasfer fuctio ( ) ( + ) H iv ( 4 )( + ) ) 8 60
The -Trasform 3. The poles of the iverse system are at /4 ad -/. Both of these poles are iside the uit circle, ad therefore the iverse system ca be both stable ad causal. Note that this system is also miimum phase, sice the double ero at / is locate iside the uit circle. Example 7.9 Multipath Commuicatio Chael: Iverse System Recall from Sectio.0 that a discrete-time LTI model for a two-path commuicatio chael is y [] x[ ] + ax[ ] Fid the trasfer fuctio ad differece-equatio descriptio of the iverse system. What must the parameter a satisfy for the iverse system to be stable ad causal? <Sol.>. We use Eq.(7.) to obtai the trasfer fuctio of multipath system: H ( ) + a. The iverse system has trasfer fuctio H ( ) + a H iv ( ) 6
The -Trasform which satisfies the differece-equatio descriptio [] + ay[ ] x[ ] y 3. The iverse system is both stable ad causal whe a <. 7.8 Determiig the Frequecy Respose from Poles ad Zeros Relatioship betwee the locatios of poles ad eros i the -plae ad the frequecy respose of the system:. Trasfer fuctio: Substitutig e jω for i H(). Assume that the ROC icludes the uit circle. Substitutig e jω ito Eq. (7.3) gives ~ jpω M p jω jω ( ) ( ) b e k cke H e jlω N l Ω e ( j k d ke ) ~ j( N M ) Ω M p jω jω b e k ( e ck ) H ( e ) N l jω (7.5) ( e d ) k k 6
The -Trasform 3. The magitude of H(e jω ) at some fixed value of Ω, say, Ω 0, is defied by ~ o b ( e ck ) jω H ( e ) d ) M p jω k N l jω0 k ( e ) e jω 0 a vector from the origi to e jω 0 ; g a vector from the origi to g. ) We assess the cotributio of each pole ad ero to the overall frequecy respose by examiig e jω 0 g as Ω 0 chages. k e jω 0 g product term, where g represets either a pole or a ero. 4. Vector represetatio of e jω g: Fig. 7. 9. Figure 7.9 (p. 589) Figure 7.9 (p. 589) Vector iterpretatio of e jω 0 g i the -plae. 63
The -Trasform Figure 7.0 (p. 589) The quatity e jω g is the legth of a vector from g to e jω i the -plae. (a) Vectors from g to e jω at several frequecies. (b) The fuctio e jω g. 5. Fig. 7.0(a) depicts the vector e jω 0 g for several differet values of Ω; while Fig. 7.0(b) depicts e jω 0 g as a cotiuous fuctio of frequecy. 64
The -Trasform If Ω arg{g}, the e jω g attais its miimum value of g whe g is iside the uit circle ad takes o the value g whe g is outside the uit circle. Hece, if g is close to the uit circle ( g ), the e jω g becomes very small whe Ω arg{g}. 6. If g represets a ero, the e jω g cotributes to the umerator of H(e jω ). Thus, at frequecies ear arg{g}, H(e jω ) teds to have a miimum. 7. If g represets a pole, the e jω g cotributes to the deomiator of H(e jω ). Whe e jω g decreases, H(e jω ) icreases, with the sie of the icrease depedet o how far the pole is from the uit circle. Example 7.0 Multipath Commuicatio Chael:Magitude Respose I Example 7.9 the trasfer fuctio of the discrete-time model for a two path commuicatio system is foud to be H ( ) + a Sketch the magitude respose of this system ad the correspodig iverse system for a 0.5e jπ/4, a 0.8e jπ/4, ad a 0.95e jπ/4. <Sol.>. The multipath chael has a sigal ero at a, while the iverse system has a sigle pole at a, as show i Fig 7.(a) ad (b). 65
The -Trasform Figure 7. (p. 590) (a) Locatio of ero for multipath chael. (b) Locatio of pole for iverse of the multipath chael.. The magitude respose are sketched i Figs 7.(a)-(c (c). Both the miimum of H(e jω) ad maximum of H iv (e jω ) occur at the frequecy correspodig to the agle of the ero of H(), amely, Ω π/4. The miimum of H(e jω) is - a. Hece, as a approaches uity, the chael magitude respose at Ω π /4 approaches ero ad the two path chael suppresses ay compoets of the iput havig frequecy Ω π /4. 66
The -Trasform Figure 7. (p. 59) Magitude respose of multipath chael (left pael) ad iverse system (right pael). (a) a 0.5e jπ/4. (b) a 0.8e jπ/4. (c) a 0.95e jπ/4. p π; v Ω 67
The -Trasform 3. The iverse system maximum occurs at Ω π/4 ad is give by /( - a ). Thus, as a approaches uity the magitude respose of the iverse system approaches ifiity. If the multipath chael elimiate the compoet of the iput frequecy Ω π/4, the iverse system caot restore this compoet to its origial value. Large values of gai i the iverse system are geerally udesirable, sice oise i the received sigal would the be amplified. Furthermore, iverse system is highly sesitive to small chage i a as a approaches uity. Example 7. Magitude Respose from Poles ad Zeros Sketch the magitude respose for a LTI system havig the trasfer fuctio H ( ) <Sol.> ( 0.9e π j 4 + )( 0.9e π j 4 ) The system has a ero at - ad poles at 0.9e jπ/4 ad 0.9e jπ/4 as depicted i Fig.7.3(a). Hece, the magitude respose will be ero at Ω π ad will be large at Ω ± π/4, because the poles are close to the uit circle. Figures 7.3 (b)-(d) (d) depict the compoet of the magitude respose associated with the ero ad each poles. Multiplicatio of these cotributios gives the overall magitude respose sketched i Fig. 7.3(e). 68
The -Trasform Figure 7.3a (p. 59) Solutio for Example 7.. (a) Locatios of poles ad eros i the -plae. (b) The compoet of the magitude respose associated with a ero is give by the legth of a vector from the ero to e jω. p π; v Ω 69
The -Trasform p π; v Ω Figure 7.3a (p. 59) Solutio for Example 7.. (c) The compoet of the magitude respose associated with the pole at e jπ/4 is the iverse of the legth of a vector from the pole to e jω. 70
The -Trasform Figure 7.3b (p. 593, cotiued) (d) The compoet of the magitude respose associated with the pole at is the iverse of the legth of a vector from the pole to e jω. (e) The system magitude respose is the product of the respose i parts (b) (d). p π; v Ω 7
The -Trasform The phase of H(e jω) may also be evaluated i terms of the phase associated with each pole ad ero.. Usig Eq.(7.5), we obtai arg{ H ( e jω )} arg{ b ~ } + ( N M ) Ω + M P k arg{ e. The phase of H(e jω) ivolves the sum of the phase agles due to each ero mius the phase agle due to each pole. Discussio: see pp. 59-594, textbook. Fig. 7.5. jω c k } N l k arg{ e jω d k } Figure 7.5 (p. 593) Figure 7.5 (p. 593) The quatity arg{e jω g} is the agle of the vector from g to e jω with respect to a horiotal lie through g, as show here. 7
The -Trasform 7.9 Computatioal Structures for Implemetig Discrete- Time LTI Systems. Block diagram represetatio: Fig..33. Cascade Form (Direct Form I) Figure.33 (p. 6) Block diagram represetatio of a discrete-time LTI system described by a secod-order differece equatio.. Differece equatio: 73
The -Trasform ( + a + a ) Y( ) ( b0 + b + b ) X ( ) (7.6) 3. Takig the -trasform of this differece equatio gives ( + a + a ) Y( ) ( b0 + b + b ) X ( ) 4. Block diagram: Replacig the shift operators i Fig..33 with. Fig. 7.6. Figure 7.6 (p. 595) Block diagram of the trasfer fuctio correspodig to Fig..33. 74
The -Trasform 5. Trasfer fuctio Fig. 7.6: Y ( ) H ( ) X ( ) ( b0 + b ( + a + b + a Suppose we write H() H ()H (), where H ( ) b 0 + b + b ) ) ad ) ( + a 6. The direct form II implemetatio for H() is obtaied by writig Y ( ) H ( ) F( ) (7.8) ad F ( ) H ( ) X ( ) (7.9) H ( + a The block diagram depicted i Fig. 7.7 (a) implemets Eqs. (7.8) ad (7.9). The blocks i H () ad H () geerate idetical quatities ad thus may be combied to obtai the direct form II block diagram depicted i Fig. 7.7 (b). 7. The pole-ero form of the trasfer fuctio leads to two alterative system implemetatios: cascade ad parallel forms. ) 75
The -Trasform Figure 7.7 (p. 596) Figure 7.7 (p. 596) Developmet of the direct form II represetatio of a LTI system. (a) Represetatio of the trasfer fuctio H() as H ()H (). (b) Direct form II implemetatio of the trasfer fuctio H() obtaied from (a) by collapsig the two sets of blocks. 76
The -Trasform ) Cascade form: H ( ) p H i ) Parallel form: i i ( ) H( ) Hi( ) p The H i () cotai distict subsets of the poles ad eros H(). The H i () cotai distict sets of the poles of H(). Example 7. Cascade Implemetatio Cosider the system represeted by the trasfer fuctio ( + j )( j )( + ) H ( ) π π π π j j j j 4 4 8 8 ( e )( e )( e )( e ) 3 4 Depict the cascade from for this system, usig real-valued secod-order sectios. Assume that each secod-order sectio is implemeted as a direct from II represetatio. <Sol.>. We combie complex-cojugate poles ad ero ito the sectios, obtaiig 3 4 77
The -Trasform H + ) π ( cos( ) ( 4 + 4 ) ad + ( ) 3 π 9 ( cos( 8 ) + 6 The block diagram correspodig to H () H () is depicted i Fig. 7.8. Note that this solutio is ot uique, sice we could have iterchaged the order of H () ad H () or iterchaged the pairig of poles ad eros. H ) Figure 7.8 (p. 597) Cascade form of implemeta tio for Example 7.. 78
The -Trasform 7.0 The Uilateral -Trasform The uilateral, or oe-sided, -trasform is evaluated by usig the portio of a sigal associated with oegative values of the time idex ( 0). The uilateral -trasform is appropriate for problems ivolvig causal sigals ad LTI systems. Two major advatages to usig uilateral -trasform: ) We do ot eed to use ROCs. ) The uilateral trasform allow us to study LTI systems described by differece equatios with iitial coditios. 7.0. Defiitio ad Properties. The uilateral -trasform of a sigal x[] is defied as X ( ) x[ ] (7.30) 0 Iverse -trasform: x [ ] Χ( ) d. π j (7.5) x[] for 0 x[] for 0 79
The -Trasform u. Notatio of uilateral -trasform: x ( ) Z X( ) The uilateral ad bilateral -trasform are equivalet for causal sigals. Example: Z u au [ ] α ad a cos( Ω 0 a cos( Ω ) u[ ] u a cos( Ω ) Time-shift property. Let w[] x[ ].. The -trasforms of x[] ad w[]: 3. Substitutig w[] x[ ], we obtai X ( ) x[ ] ad W ( ) 0 0 ( m+ ) W( ) x [ ] x[ ] + x [ ] x[ ] + xm [ ] 0 m [ ] [ ] [ ] ( ) x + x m x + X 0 0 0 0 ) W[ ] + a 80
The -Trasform 4. A oe-uit time-shift results i multiplicatio by ad additio of the costat x[]. The time-shift property for delays greater tha uity:. If u x( ) X ( ) the x [ k] u x[ k] + x k + + L + x + X k > k+ k+ [ ] [ ] ( )for 0. I the case of a time advace, the time-shift property chages somewhat. Here, we obtai [ ] u [0] k [] k x+ k x x L xk [ ] + k X( ) for k> 0 (7.3) (7.3) Both time-shift properties correspod to the bilateral time-shift property, with additioal terms that accout for values of the sequece that are shifted ito or out of the oegative time portio of the sigal. 7.0. Solvig Differece Equatios with Iitial Coditios. Cosider takig the uilateral -trasform of both sides of the differece equatio N M a y[ k] b x[ k] k 0 k 0 k 8
The -Trasform where A ( ) Y ( ) + C( ) B( ) X ( ) M M N A k ( ) a B k k ( ) b k N C( ) ak y[ k + m] k 0 k 0 m 0k m+. We have assumed that x[] is causal, so that u k [ ] Z ( ) x k X 3. The term C() depeds o the N iitial coditios y[ ], y[ ],, y[ N] ad the a k. C() is ero if all the iitial coditios are ero. Solvig for Y() yields Y ( ) B( ) X ( ) A( ) Forced respose: C( ) A( ) B ( ) X ( ) A ( ) Natural respose: C ( ) A( ) m Due to iitial coditios The poles of the atural respose are the roots of A(), which are also the poles of the trasfer fuctio. If the system is stable, the the poles must lie iside the uit circle. 8
The -Trasform Example 7.3 First-Order Recursive System: Ivestmet Computatio Recall from Example.5 that the growth i a asset due to compoud iterest is described by the first-order differece equatio [] py[ ] x[ ] y where ρ + r/00, r is the iterest rate per period i percet, ad y[] represets the balace after the deposit or withdrawal represet by x[]. Assume that a bak accout has a iitial balace of $0,000 ad ear 6% aual iterest compouded mothly. Startig i the first moth of the secod year, the ower withdraws $00 per moth from the accout at the begiig of each moth. Determie the balace at the star of each moth (followig ay withdrawals) ad how may moths it will take for the accout balace to reach ero. <Sol.>. We take the uilateral -trasform of both side of the differece equatio ad use the time-shift property of Eq. (7.3) to obtai Y ( ) p( y[ ] + Y ( )) X ( ). Now we rearrage this equatio to determie Y(). We have 83
The -Trasform ( p ) Y ( ) Y X ( ) + X ( ) p py[ ] py[ ] + p ( ) Note that Y() is give as the sum of two terms: oe that depeds o the iput ad aother that depeds o the iitial coditio. The iput-depedet term represets the forced respose of the system; the iitial coditio term represets the atural respose of the system. 3. The iitial balace of $0,000 at the start of the first moth is the iitial coditio y[ -], ad there is an offset of two betwee the time idex ad the moth idex. That is, y[] represets the balace i the accout at the start of the + d moth. We have ρ + (6/)/00.005. Sice the ower withdraws $00 per moth at the start of moth 3 ( ), we may express the iput to the system as x[] -00u[ -]. Thus, 00 X ( ) ad we have ( 00.005(0,000) Y ) + ( )(.005 ) (.005 ) 84
The -Trasform 4. Now we perform a partial-fractio expasio o the first term of Y(), obtaiig 0,000 0,000 0,050 Y ( ) + ( ) (.005 ) (.005 ) 5. The mothly accout balace is obtaied by iverse -trasformig Y(), resultig i y[ ] 0,000u[ ] 0,000(.005) u[ ] + 0,050(.005) u[ ] The last term, 0,050(.005) u[] is the atural respose associated with the iitial balace, while the first two terms represet the forced respose associated with the withdrawals. The accout balace, atural respose, ad forced respose for the first 60 moth are illustrated i Fig.7.30 as a fuctio of the moth, ot. the accout balace reaches ero durig the withdrawal at the start of moth 63. 85
The -Trasform Figure 7.30a (p. 60) Figure 7.30a (p. 60) Solutio to Example 7.3, depicted as a fuctio of the moth. (a) Accout balace at the start of each moth followig possible withdrawal. 86
The -Trasform Figure 7.30b (p. 60) Figure 7.30b (p. 60) (b) Natural respose. 87
The -Trasform Figure 7.30c (p. 60) Figure 7.30c (p. 60) Forced respose. 88
The -Trasform 7. Explorig Cocepts with MATLAB 7.. Poles ad Zeros MATLAB commad: roots Fid the roots of polyomial Example Fid the roots of + 4 + 3. <Sol.>. MATLAB commad: roots([, 4, 3]) MATLAB commad: plae(b, a) Example <Sol.> Zplae fids the roots of the umerator ad deomiator polyomials represeted by b ad a, respectively, ad the displays the poles ad eros i the -plae. Fid pole-ero plot of y( ) 0. MATLAB commads: 4 4 3 89
The -Trasform >> b[, -0, -4, 4]; >> a[, -, -4]; >> plae(b, a). Pole-ero plot: 7.. Iverse -trasform. The residue commad computes partial-fractio expasios for -trasform expressed as a ratio of two polyomials i.. Sytax: [r, p, k] residue(b, a) Imagiary Part where b ad a are vectors represetig the umerator ad deomiator polyomial coefficiets, ordered i descedig powers of. 4 3 0 - - -3-4 0 4 6 8 0 Re al Part 90
The -Trasform Example Fid partial-fractio expasio for -trasform give i Example 7.0: 3 0 4 + 4 X ( ) ( 4) <Sol.>. We first write X() Y(), where y( ) 0 4 4 3. Now we use residue to fid the partial-fractio expasio for Y() as follows: >> [r,p,k]residue([,-0,-4,4],[,-,-4]) r -.5000 0.5000 p - k.5000 -.0000 3. This implies a partial-fractio expasio of the form.5 0.5 Y ( ) + +. 5 + which, as expected, correspods to ½W() i Example 7.0. 9
The -Trasform 7..3 Trasform Aalysis of LTI Systems. The MATLAB Sigal Processig Toolbox cotais several routies for covertig betwee LTI system descriptios.. If b ad a cotai the coefficiets of the trasfer fuctio umerator ad deomiator polyomials, respectively, ordered i descedig powers of, the tfss(b, a) determie a state-variable descriptio of the system ad tfp(b, a) determie the pole-ero-gai descriptio of the system. 3. pss ad pstf covert from pole-ero-gai descriptios to state-variable ad trasfer-fuctio descriptios, while sstf ad ssp, respectively, covert from state-variable to trasfer-fuctio ad pole-ero-gai forms. 4. The frequecy respose of a system described by a differece equatio is evaluated from the trasfer fuctio with the use of freq. Example:. Cosider a LTI system with trasfer fuctio 3 4 0.094( + 4 + 6 + 4 + ) H ( ) 4 ( + 0.4860 + 0.077 ). We may depict the poles ad eros of H() i the -plae ad plot the system s magitude respose with the followig commads: 9
>>b.094*[,4,6,4,]; >>a[, 0, 0.486, 0, 0.077]; >>plae(b, a) >>[H,w] freq(b, a, 50); >>plot(w,abs(h)) The -Trasform Fig. 7.3 ad Fig. 7.3 Figure 7.3 (p. 604) Figure 7.3 (p. 604) Locatio of poles ad eros i the -plae obtaied by usig MATLAB. 93
The -Trasform Figure 7.3 (p. 605) Magitude respose evaluated by usig MATLAB. 94
The -Trasform 7..4 Computatioal Structures for Implemetig Discrete-Time LTI Systems. The MATLAB Sigal Processig Toolbox cotais several routies for covertig a state-variable or pole-ero-gai descriptio of a system to a cascade coectio of secod-order sectios. MATLAB commads: sssos ad psos.. Sytax: sos psos (, p, k) where ad p are vectors cotaiig eros ad poles, respectively, ad k is the gai. The matrix sos is L by 6, where each row cotais the coefficiets of the trasfer fuctio for that sectio. The first three elemets of the row cotai the umerator coefficiets, while the last three elemets cotai the deomiator coefficiets. 3. The commads sosp, sosss, ad sostf covert from a cascade of the secod-order sectios to pole-ero-gai, state-variable, ad trasfer-fuctio descriptios. Example 7. (repeated) Usig MATLAB, obtai a represetatio of the system as a cascade of secodorder sectios. 95
The -Trasform <Sol.>. The trasfer fuctio is give i pole-ero-gai forms: H( ) ( + j )( j )( + ) π π π π j 4 j 4 3 j 8 3 j 8 ( e )( e )( e )( e ) 4 4. The system has eros at ± j ad, while the poles are at π ± j 4 e ad 3 4 e π ± j 8 3. We employ psos to covert from pole-ero-gai form to secod-order sectios as follows: >> [ -, -j, j]; >> p [ 0.5*exp(j*pi/4), 0.5*exp(-j*pi/4), 0.75*exp(j*pi/8), 0.75exp(-j*pi/8) ]; >> k ; >> sos psos(, p, k) sos 0.706 0.706 0.0000-0.707 0.500 3.6955 0 3.6955.0000 -.3858 0.565 96
The -Trasform 4. Hece, the system is described as a cascade of secod-order sectios give by 0. 706 + 0. 706 F ( ) ad F ( ) 0. 707 + 0. 5 3. 6955 + 3. 6955. 3858 + 0. 565 97
The -Trasform 98