X (z) = n= 1. Ã! X (z) x [n] X (z) = Z fx [n]g x [n] = Z 1 fx (z)g. r n x [n] ª e jnω

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1 3 The z-transform ² Two advantages with the z-transform:. The z-transform is a generalization of the Fourier transform for discrete-time signals; which encompasses a broader class of sequences. The z-transform exists for some sequences for which the Fourier transform does not converge.. The advantage and power of complex variable theory allow one to obtain z-transforms easily. 3. -Transform ² Defn: The (two-sided or bilateral) z-transform of a sequence x [n] is de ned as X x [n] z n where z is a complex variable. We shall adopt the notation or throughout. x [n] n Ã! X (z) fx [n]g x [n] fx (z)g ² Note that the Fourier transform X (e jω ) of x [n] is only a special case of X (z) since X e jω X (z)j ze jω. If we express z by a polar form, i.e., z re jω, then X n r n x [n] ª e jnω whose region of convergence (ROC) may or may not contain the circle r (unit circle on z-plane). Fig. 3.

2 The su cient and necessary condition for a z-transform to exist is jx (z)j < 8z ROC. A su cient condition is that the sequence fx [n] z n g is absolutely summable for z ROC, i.e., jx (z)j X n x [n] z n X n jx [n]j jzj n X n jx [n]j r n < 8z ROC. This su cient condition implies that if z z ROC, then all z on the circle de ned by jzj jz j will also be in ROC. Therefore, the ROC consists of a ring in the z-plane. Fig. 3. Note: if ROC does not contain the unit circle, then the Fourier transform of x [n] does not exist. ² It is mentioned in Section.7 that if the sequence fx [n] z n g is absolutely summable for z ROC, i.e. then X M (z) with X n X M (z) jx [n]j r n < MX n M x [n] z n converges uniformly to a continuous function of z for z ROC. ² It is shown by theory that the z-transform and all its derivatives are continuous functions of z within ROC. Thus if ROC contains unit circle, then the Fourier transform X (e jω ) and all its derivatives w.r.t. ω are continuous function of ω. ² Some useful z-transform pairs are summarized in Table 3.. Table 3.

3 ² Ex: Right-Sided Exponential Sequence x [n] a n u [n], a real ) X n Convergence of X (z) requires su ciently that X n az < n0 a n u [n] z n X az n. n0 which implies the ROC fz : jzj > jajg. Thus, for jzj > jaj, Notes: X az n n0 az z z a.. if jaj, x [n] a n u [n] does not have a Fourier transform.. X (z) simpli es to a rational function. Any rational z-transform can be described by its poles and zeros. 3. if a > 0, z a is the pole of X (z), while z 0 is the zero. ² Ex: Left-Sided Exponential Sequence Fig. 3.3 x [n] a n u [ n ], a real X X ) a n z n a n z n n For jzj < jaj, X (z) converges to Notes: n a z z z a. X a z n.. The z-transform here equals to that in the previous example for x [n] a n u [n] with the same pole-zero plot, while ROC is di erent. Fig n0

4 . It is necessary to specify the algebraic expression for X (z) and the ROC for the bilateral z-transform of a given sequence. ² Ex: Sum of Two Exponential Sequences µ n µ x [n] u [n] + n u [n] 3 X ½µ ) n X µ n X z n + n0 n µ u [n] + n ¾ u [n] z 3 n n0 µ 3 n z n + for z + 3 z z z for jzj > z z +. 3 Fig. 3.5 z ² Ex: Sum of Two Exponential Sequences (Revisited) µ n µ x [n] u [n] + n u [n] 3 µ n ) u [n] Ã! jzj > z µ n u [n] Ã! z jzj > 3 < and 3 z < µ n µ u [n] + n u [n] Ã! 3 + z + jzj > 3 z This indicates that the z-transform is linear and that the resultant ROC is the intersection of two individual ROCs. 4

5 ² Ex: Two-Sided Exponential Sequence µ x [n] n µ n u [n] u [ n ] 3 ) Notes: µ 3 n u [n] µ n u [ n ] µ n u [n] 3 Ã! jzj > + 3 z 3 Ã! µ jzj < z n u [ n ] Ã! z z 3 < jzj < z(z ) (z + 3 )(z ).. The rational function here is identical to that in the previous example, but the ROC is di erent. Fig This sequence does not have a Fourier transform. ² Ex: Finite-Length Truncated Exponential Sequence ½ a x [n] n, 0 n N 0, otherwise N X ) a n z n (az ) N z N a N az z N z a. n0 The ROC 0 is determined by N X n0 az n <. As long as jaz j is nite, i.e., jaj < and z 6 0, X (z) exists. Assuming a nite jaj, X (z) has a pole at z 0 and N zeros at z k a expfjπk/ng for k,,..., N. ² See Table 3. for useful pairs. Fig

6 3. Properties of the ROC for the z-transform ² Assuming that X (z) is rational and x [n] has nite amplitude for < n <.. The ROC is generally of the form 0 r R < jzj < r L (an annulus).. For a given sequence x[n], X (e jω ) converges absolutely i the ROC of X (z) includes the unit circle. 3. The ROC can not contain any poles. 4. If x [n] is a nite-duration sequence, i.e., it is nonzero only for nite n s, then the ROC is the entire z-plane except possibly z 0 or jzj. 5. If x [n] is a right-sided sequence, i.e., it is zero for n < N <, then the ROC is of the form either r max < jzj or r max < jzj <, with r max the largest magnitude of a nite pole. 6. If x [n] is a left-sided sequence, i.e., it is zero for n > N >, then the ROC is of the form either jzj < r min or 0 < jzj < r min, with r min the smallest magnitude of a nite pole. 7. If x [n] is a two-sided sequence, i.e., it is zero < N < n < N < for nite N and N, then the ROC is of the form 0 r R < jzj < r L with r R and r L the magnitudes of two distinct poles, and not containing any poles inside. 8. The ROC must be a connected region (since X (z) is continuous). ² See section 3. for proof and Fig. 3.8 ² Ex: Non-Overlapping Regions of Convergence µ n µ x [n] u [n] n u [ n ]. 3 ) + z + for jzj > 3 z and jzj < 3 Thus, x [n] has no z-transform. 6

7 3.3 The Inverse -Transform ² Defn: The inverse z-transform of X (z) is de ned by the contour integral x [n] I X (z) z n dz πj C where C is a counterclockwise closed contour in the ROC of X (z) and encircling the origin z 0. ² Derivation of this formal de nition: From Cauchy integral theorem, I ½, k z k dz πj 0, k 6 for such C. Since C X n x [n] z n δ [k ] ) I X (z) z k dz I X x [n] z n+k dz πj C πj C n X I x [n] z n+k dz πj n C {z } x [k]. δ[n k+ ]δ[n k] ² If the ROC includes unit circle and we let C be unit circle, I x [n] X e jω e jω(n ) de jω ze jω πj unit circle π X e jω e jωn dω. π π ² We shall discuss below various useful procedures of evaluating inverse z-transform, i.e., nding x [n] from X (z) Inspection Method ² The method is to inspect Table 3. to nd the desired x [n]. For example, with pair 5 of Table 3. a n u [n] Ã!, jzj > jaj az 7

8 we can nd the x [n] of as 0 n u [n] + z +, jzj > 3 z Ã! 3.3. Partial Fraction Expression, jzj z n A 3 u [n] Ã!, jzj > + 3 z 3 µ n µ ) x [n] u [n] + n u [n]. 3 ² The method is to applying partial fraction expression to a rational X (z) rst, and use the inspection method to obtain x [n]. Let P M k0 b kz k P N k0 a kz for a k and b Then, zn P M k0 b kz M k z M P N k0 a kz N k ² In general, µ implying M zeros and N poles at nonzero locations at most P M k0 b kz k P N k0 a kz k for a and b can be expressed as b Q M 0 k ( c kz ) Q a N 0 k ( d kz ) where c k s are nonzero zeros of X (z) and d k s are nonzero poles of X (z). Suppose that c k s are di erent from d k s and that d, d,..., d N s+ are di erent and d N s+,..., d N are the same. Then, we can expand X (z) into M N X r0 N s + + X k sx m B r z r (exists if M N) A k (for poles of order one) d k z C m ( d N s+ z ) m (for pole of order s). 8

9 (For the cases with other di erent-order poles, above expression can be modi ed accordingly.), where. B r s can be derived by long division of the numerator by the denominator,. A k ( d k z ) X (z) j zdk, 3. C m ² Ex: Find x [n] if X (z) is given by Now, (s m)! ( d N s+ ) s m ½ d s m ( dn dw s m s+ w) s X w ¾. w/d N s+ + z + z 3 z + for jzj >. z + z + z z ( z ). Fig. 3. We have M N and two poles of order one, thus where. B 0 is derived B 0 + A z + A z z 3 z +, z + z + z 3z + 5z ) B 0 9

10 . A µ z X (z) + z + z z 9 z z A z X (z) z + z + z z z 8. That is, By inspection with Table 3., jzj >. z z So, δ [n] µ n u [n] u [n] x [n] δ [n] 9 Ã! all z Ã! z jzj > Ã! z jzj >. µ n u [n] + 8u [n] Power Series Expansion ² If we can expand X (z) in terms of power series (Laurent series) X X(z) x [n] z n n then x [n] can be obtained as the coe cient of z n. 0

11 ² Ex: z µ z + z z jzj > 0 z z + z 8 >< ) x [n] >:, n, n, n 0, n 0, otherwise ) x [n] δ [n + ] δ [n + ] δ [n] + δ [n ]. ² Ex: Since we have ² Ex: Consider Since we have ² Ex: Consider log + az log ( + x) X n jzj > jaj n ( )n+ x n for jxj < X ( ) n+ a n z n n ) x [n] ( ) n+ a n u [n ]. n n, jzj > jaj. az x + x + x +... for jxj < X az n ) x [n] a n u [n]. n0, jzj < jaj. az

12 Now Since we have µ a z a z a z a z + x X x n for jxj < n0. a z 0X n X n X a n z n a z n0 a n z (n ) ( a n ) z n ) x [n] a n u [ n ] Cauchy Residue Theorem 0X n a n z n ² From Cauchy Residue Theorem, x [n] I X (z) z n dz πj C X residues of X (z) z n at the poles inside C ² If X (z) is a rational function of z, then X (z) z n has the form of X (z) z n ª (z) (z d 0 ) s where X (z) z n has a pole of order s at z d 0 and ª (z) has no poles nor zeros at z d 0. Then, Res X (z) z n ds ª (z) at z d 0 (s )! dz. s zd 0 If s, Res X (z) z n at z d 0 ª (d0 ).

13 ² Ex: Consider, jzj > jaj az ) X (z) z n zn z a.. For n 0, there is only one pole of order one at z a and Res X (z) z n at z a a n, n 0.. For n < 0, there is jnj-th order pole at z 0 and rst order pole at z a, and Res X (z) z n at z a a n Res X (z) z n at z 0 ) x [n] a n a n 0 for n < Therefore, from cases and, 3.4 -Transform Properties ² We shall adopt the notations and x [n] x [n] x [n] (jnj )! d jnj (z a) dz jnj ( ) jnj (jnj )! (jnj )! ( ) ( ) jnj ( a) jnj a jnj a n. x [n] a n u [n]. Ã! X (z), ROC R x Ã! X (z), ROC R x Ã! X (z), ROC R x fx [n]g x [n] fx (z)g. 3 z0 (z a) jnj z0

14 ² Properties:. Linearity: ax [n] + bx [n]. Time-Shifting: x [n n 0 ] Pf: X n Ã! ax (z) + bx (z), ROC R x \ R x. Ã! z n0 X (z), ROC R x (may exclude z 0 or jzj ). x [n n 0 ] z n z n 0 X m z n 0 X (z). x [m] z m (m n n 0 ) (ROC should be reconsidered at z 0 and jzj.) QED Ex: Let z, jzj > 4. 4 First, from partial fraction expansion, Since ) x [n] 4δ [n] + 4 z, jzj > 4 z z δ [n] Ã!, 8z a n u [n] Ã! µ 4, jzj > jaj az n u [n] Alternatively, since µ n u [n] Ã! 4, jzj > 4 z 4 µ n u [n ]. 4 we have µ n u [n ] Ã! z 4 4 z from the time-shifting property. 4

15 3. Multiplication by z n 0 : Pf: z n 0 x [n] X new (z) Ã! X X n µ z µ z X z 0 z 0, ROC jz 0 j R x. z n 0 x [n] z n X n z z 0 R x. µ n z x [n] z 0 QED Ex: Find X (z) of x [n] r n cos (ω 0 n) u [n]. Now, x [n] re jω 0 n u [n] + re jω 0 n u [n]. Since, u [n] Ã! z re jω 0 n u [n] Ã! re jω 0 n u [n], jzj > Ã! z re jω 0 z re jω 0, jzj > re jω 0 r, jzj > r ) re jω0 z +, jzj > r re jω0 z r cos ω 0 z, jzj > r. r cos ω 0 z + r z 4. Di erentiation of X (z): nx [n] Pf: dx (z) Ã! z, ROC R x (may exclude z 0 or jzj ). dz X n nx [n] z n z 5 X n z d dz z nx [n] z n X n dx (z) dz x [n] z n QED

16 Ex: Find X (z) of na n u [n]. Since a n u [n] Ã!, jzj > jaj az ) na n u [n] Ã! z d µ dz az z (az ) ( az, jzj > jaj ) az, jzj > jaj ( az, jzj > jaj. ) ( jzj can be included in the ROC) 5. Conjugation of a Complex Sequence: x [n] Ã! X (z ), ROC R x. Pf: X x [n] z n n Ã X x [n] z n n! Ã! X x [n] (z ) n n X (z ), jz j R x ) jzj R x. QED 6. Time Reversal: x [ n] Ã! X µ, z R x. z Pf: X x [ n] z n n 0 n n X n 0 h x n 0i z n 0 X z, z R x. QED Ex: Find z-transform of a n u [ n]. Since a n u [n] Ã!, jzj > jaj az ) a n u [ n] Ã! az, jzj < a. 6

17 7. Convolution of Sequences: Pf: Ex: x [n] x [n] X n X n k Ã X m Ã! X (z) X (z), ROC R x \ R x. (x [n] x [n]) z n X x [k] x [n k] z (n k) z k x [m] z m! Ã X k x [k] z k! X (z) X (z), z R x and z R x. QED x [n] a n u [n], jaj < x [n] u [n] x [n] x [n] Ã! X (z) Ã! By partial fraction expansion, Y (z) a Ã! X (z), jzj > z, jzj > z {z az } Y (z) µ z a az (m n k), jzj > jaj az, jzj > ) y [n] a (u [n] a an u [n]) an+ a u [n]. 8. Initial Value Theorem: If x [n] 0 for n < 0, then Pf: x [0] lim z! X (z). X x [n] z n n0 ) lim z! X n ² See Table 3. for a list of properties. x [n] lim z! z n + x [0] x [0]. QED 7

18 3.5 Complex Convolution Theorem (Supplemental) ² Question: What is the z-transform of x [n] x [n]? Answer: Let Then W (z) w [n] x [n] x [n]. X n x [n] x [n] z n. From contour-integral de nition of inverse z-transform, x [n] I X (v) v n dv πj C ) W (z) πj πj X I ³ z n x [n] X (v) v dv n C v I " X # ³ z n x [n] X (v) v dv. v C n {z } X ( z v) If we choose C as a closed contour in the overlap of ROCs of X z v and X (v), then W (z) I ³ z X X (v) v dv. πj C v Alternatively, if we choose C as a closed contour in the overlap of ROCs of X (v) and X z v, then W (z) I ³ z X (v) X v dv. πj C v ² Question: What is the ROC for W (z)? Answer: Let R x : r R < jzj < r L R x : r R < jzj < r L ) 8

19 . C is determined by. C is determined by ¾ r R < jvj < r L r R < z ) r < R r R < jzj < r L r L. rl v ¾ r R < jvj < r L r R < z ) r < R r R < jzj < r L r L. rl v ) The ROC of W (z) should contain r R r R < jzj < r L r L, denoted by R w which may be di erent from R x \ R x (since X (v) X z v v may have di erent poles for z). ² Note: As v e jθ, z e jω, C C unit circle, we have W e jω π π π X e jθ X e j(ω θ) dθ which is exactly the windowing theorem in the Fourier transform. ² Ex: Given x [n] a n u [n] and x [n] b n u [n] where jaj < and jbj <, nd w [n] x [n] x [n]. Now, where C is chosen as with jzj jaj X (z), jzj > jaj az X (z), jzj > jbj bz ) W (z) I z/a dv πj C v z/a {z } v {z b } X ( z X v) (v)v Fig. 44-F > jbj, i.e., jzj > jabj. Now, from Cauchy Residue Theorem, z/a Res v z/a v b at v b z/a b z/a ) W (z), jzj > jabj abz ) w [n] (ab) n u [n] as expected! Note: In this example R w is di erent from R x \ R x. 9

20 3.6 Parseval s Relation (Supplemental) ² For two complex sequences x [n] and x [n], Parseval s Relation states that X x [n] x [n] I µ X (v) X v dv πj C v n where P n x [n] x [n] is assumed to exist and C is in the overlap of ROCs of X (v) and X v. Pf: Let y [n], x [n] x [n]. From the Complex Convolution Theorem, Y (z) I ³ z X (v) U v dv πj v C where U (z) fx [n]g and C is in the overlap of ROC s of X (v) and U z v. Now, from conjugation of a complex sequence, (x [n] Ã! X (z ), ROC R x ), Furthermore, U (z) X (z ), ROC R x ) Y (z) I µ z X (v) X v dv. πj v C Y (z)j z X n y [n] yields the relation X x [n] x [n] I µ X (v) X v dv. πj C v ² Notes: n QED. When v e jω, we have the Parseval s Relation in terms of the Fourier Transform, X n x [n] x [n] π π π which is di cult, in general, to evaluate. X e jω X e jω dω. Parseval s relation in terms of z-transform is easy to evaluate by use of the Cauchy Residue Theorem. 0

21 3. When x [n] x [n] x [n] is real, X n jx [n] j energy of x [n] I µ X (v) X v dv πj C v where C ROC x \ (ROC x ), with ROC x fzjr R < jzj < r L g and (ROC x ) fzjr L < jzj < r R g. Fig. 45-F ² Ex: Find the energy of real sequence x [n] with Now, X n x [n] I πj C ( az ) ( bz ), jaj <, jbj <, a 6 b jzj > max fjaj, jbjg ( av ) ( bv ) ( av) ( bv) v dv with C being unit circle. This can be simpli ed to X n x [n] I πj C v (v a) (v b) ( av) ( bv) dv. Fig. 45-B... Res v (v a) (v b) ( av) ( bv) at v a a (a b) ( a ) ( ab) v Res (v a) (v b) ( av) ( bv) at v b b (b a) ( ab) ( b )

22 Thus, X x [n] n a (a b) ( ab) a b b (a b) + ab (a b) (a b) ( ab) ( a ) ( b ) + ab ( ab) ( a ) ( b ). 3.7 z-transforms and LTI Systems ² For an LTI system with input x [n], output y [n], and impulse response h [n], y[n] x[n] h[n] () Y (z) X (z) H (z), ROC R y R x \ R h where fx [n]g for z R x, Y (z) fy [n]g for z R y, and H (z) fh [n]g for z R n, by convolution property. Here, H (z) is called the system function of the LTI system with impulse response h[n]. ² Ex: Let h[n] a n u[n] with jaj < and x[n] Au[n]. Now, the system function is H(z), jzj > jaj az and the z-transform of the input is X(z) Thus, the z-transform of the output is Y (z) A, jzj >. z A ( z )( az ), jzj > Az (z )(z a) A a ( z a ). az Fig. 3. Taking the inverse z-transform of Y (z) yields y[n] A a ( an+ )u[n].

23 Alternatively, we can evaluate y[n] by convolution sum as y[n] Au[n] a n u[n] X Au[k]a n k u[n k] A k nx a n k u[n] k0 A a ( an+ )u[n]. Note that the z-transform is useful for describing the input-output relation of an LTI system. ² Consider the causal LTI system described by a linear constant-coe cient di erence equation y[n] NX ( a MX k )y[n k] + ( b k )x[n k] a 0 a 0 k with a Let input x[n] and output y[n] be both causal. Taking z- transform, the equation becomes after applying linearity and time-shift properties Y (z) k k0 NX ( a MX k )z k Y (z) + ( b k )z k X(z) a 0 a 0 ) Y (z) H(z)X(z) k0 with the system function of the causal LTI system given by H(z) for jzj > r R (since h[n] is causal). Notes: P M k0 ( b k a0 )z k P N k0 ( a k a 0 )z k. r R is the magnitude of the pole of H(z) farthest from the origin.. If r R <, all poles of H(z) locate inside the unit circle. Thus, the system is stable and the frequency response H(e jω ) of the LTI system exists. 3

24 Recall that an LTI system is stable i its impulse response is absolutely summable, i.e., P jh[n]j <. Now, if the ROC of H(z) contains the unit circle, then h[n] is absolutely summable and this guarantees the system stability. ² Ex: First-Order Causal LTI System Consider y[n] ay[n ] + x[n]. By inspection, the system function is given by H(z), jzj > jaj az, h[n] a n u[n]. 3.8 The Unilateral z-transform ² Defn: The unilateral z-transform of x [n] is de ned and denoted by X x [n] z n. n0 ² Note:. Conventional z-transform is bilateral!. If x [n] 0 for n < 0, i.e., the sequence is causal, unilateral and bilateral z-transforms will be identical; and share the same properties. 3. X (z) has a ROC with the form jzj > r R. Thus, if X (z) is a rational function, r R is de ned by the pole with the largest magnitude. ² Ex: x [n] δ [n] ) For unilateral z-transform, ) For bilateral z-transform, X δ [n] z n n0 X n δ [n] z n. 4

25 ² Ex: x [n] δ [n + ] ) For unilateral z-transform, X δ [n + ] z n 0 ) For bilateral z-transform, n0 X n δ [n] z n z. ² Note: For x [n] 6 0, n < 0, i.e., any noncausal sequence, X (z) and X (z) are di erent. ² Further Notes:. The timing properties for unilateral and bilateral z-transforms are di erent. (a) For bilateral z-transform, x [n] Ã! X (z) x [n m] (b) For unilateral z-transform, let y [n] x [n m], m > 0. X Y (z) y [n] z n n0 X x [n m] z n n0 X n 0 m (n 0 n m) h x n 0i z n0 m Ã! z m X (z) x [ m] + x [ m + ] z + x [ m + ] z + 0 X h... + x [ ] z m x n 0i z n0 A z m n 0 0 mx x [ m + (k )] z k+ k {z } di erent from that of bilateral z-transform 5 + z m X (z)

26 . The linearities for both are, however, the same. ² Ex: Consider a system with () I/O relationship () x [n], n 0, and (3) y [ ]. y [n] y [n ] x [n] ~ Now, applying unilateral transform to both sides of ~ yields 0 Since we have Now, Y (z) Y y [ ] {z } initial output X x [n] z n n0 z + z Y (z) A X (z). X z n n0 z µ y [ ] + z + z z ( z ) z + z z {z } partial fraction expansion z. z! z inverse unilateral z-transform u [n]! z inverse unilateral z-transform Ã ) y [n] µ! n+ u [n]. µ n u [n] ² Note: Unilateral z-transform is good for analyzing systems described by LCCD equations with nonzero initial conditions (i.e., noncausal output sequence). 6

The Z-Transform. Fall 2012, EE123 Digital Signal Processing. Eigen Functions of LTI System. Eigen Functions of LTI System

The Z-Transform. Fall 2012, EE123 Digital Signal Processing. Eigen Functions of LTI System. Eigen Functions of LTI System The Z-Transform Fall 202, EE2 Digital Signal Processing Lecture 4 September 4, 202 Used for: Analysis of LTI systems Solving di erence equations Determining system stability Finding frequency response