Z-Transform Z-Transform Z-Transform Z-Transform Z-Transform Z-Transform

z-Transform z-Transform • A generalization of the DTFT defined by • The DTFT provides a frequency-domain ∞ representation of discrete-time signals and X (e jω) = ∑ x[n]e− jωn LTI discrete-time systems n=−∞ • Because of the convergence condition, in leads to the z-transform many cases, the DTFT of a sequence may • z-transform may exist for many sequences not exist for which the DTFT does not exist • As a result, it is not possible to make use of • Moreover, use of z-transform techniques such frequency-domain characterization in permits simple algebraic manipulations these cases 1 2 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

z-Transform z-Transform • Consequently, z-transform has become an important tool in the analysis and design of • If we let z = r e j ω , then the z-transform digital filters reduces to ∞ • For a given sequence g[n], its z-transform jω −n − jωn G(z) is defined as G(r e ) = ∑ g[n]r e n=−∞ ∞ G(z) = g[n]z−n • The above can be interpreted as the DTFT ∑ −n n=−∞ of the modified sequence{g[n]r } where z = Re(z) + jIm(z) is a complex •Forr = 1 (i.e., |z| = 1), z-transform reduces variable to its DTFT, provided the latter exists 3 4 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

z-Transform z-Transform • The contour |z| = 1 is a circle in the z-plane of unity radius and is called the unit circle • From our earlier discussion on the uniform convergence of the DTFT, it follows that the • Like the DTFT, there are conditions on the series convergence of the infinite series ∞ jω −n − jωn ∞ G(r e ) = ∑ g[n]r e ∑ g[n]z−n n=−∞ n=−∞ converges if { g [ n ] r − n } is absolutely • For a given sequence, the set of values of R summable, i.e., if z for which its z-transform converges is ∞ called the region of convergence (ROC) ∑ g[n]r−n < ∞ 5 6 n=−∞ Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

1 z-Transform z-Transform •Example- Determine the z-transform X(z) n • In general, the ROC R of a z-transform of a of the causal sequence x [ n ] = α µ [ n ] and its sequence g[n] is an annular region of the z- ROC ∞ ∞ plane: •Now X (z) = ∑αnµ[n]z−n = ∑αnz−n Rg − < z < Rg + n=−∞ n=0

where 0 ≤ Rg − < Rg + ≤ ∞ • The above power series converges to •Note:The z-transform is a form of a Laurent 1 X (z) = , for α z−1 <1 series and is an analytic function at every 1− α z−1 point in the ROC • ROC is the annular region |z| > |α| 7 8 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

z-Transform z-Transform •Example- The z-transform µ(z) of the unit step sequence µ[n] can be obtained from •Note: The unit step sequence µ[n] is not 1 absolutely summable, and hence its DTFT X (z) = , for α z−1 <1 1− α z−1 does not converge uniformly by setting α = 1: 1 •Example- Consider the anti-causal µ(z) = , for z−1 <1 1− z−1 sequence n • ROC is the annular region 1< z ≤ ∞ y[n] = −α µ[−n −1]

z-Transform z-Transform •Its z-transform is given by −1 ∞ n −n −m m •Note: The z-transforms of the two Y (z) = ∑− α z = − ∑α z sequences α n µ [ n ] and − α n µ [ − n − 1 ] are n=−∞ m=1 identical even though the two parent ∞ −1 −1 −m m α z sequences are different = −α z ∑α z = − −1 m=0 1− α z • Only way a unique sequence can be 1 −1 associated with a z-transform is by = −1 , for α z <1 1− α z specifying its ROC • ROC is the annular region z < α

2 z-Transform z-Transform z-Transform •Example- The finite energy sequence jω • The DTFT G ( e ) of a sequence g[n] sin ωcn converges uniformly if and only if the ROC hLP[n] = πn , − ∞ < n < ∞ of the z-transform G(z) of g[n] includes the has a DTFT given by unit circle jω ⎧1, 0 ≤ ω ≤ ωc • The existence of the DTFT does not always H LP (e ) = ⎨ imply the existence of the z-transform ⎩0, ωc < ω ≤ π which converges in the mean-square sense

Table 6.1: Commonly Used z- z-Transform Transform Pairs

• However, h L P [ n ] does not have a z-transform as it is not absolutely summable for any value of r

• Some commonly used z-transform pairs are listed on the next slide

Rational z-Transforms Rational z-Transforms

• In the case of LTI discrete-time systems we •The degree of the numerator polynomial are concerned with in this course, all P(z) is M and the degree of the denominator pertinent z-transforms are rational functions polynomial D(z) is N −1 of z • An alternate representation of a rational z- • That is, they are ratios of two polynomials transform is as a ratio of two polynomials in in z − 1 : z: P(z) p + p z−1 + .... + p z−(M −1) + p z−M p zM + p zM −1 + .... + p z + p G(z) = = 0 1 M −1 M G(z) = z(N −M ) 0 1 M −1 M −1 .... −(N −1) −N N N −1 .... D(z) d0 + d1z + + dN −1z + dN z d0z + d1z + + dN −1z + dN 17 18 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

3 Rational z-Transforms Rational z-Transforms

• A rational z-transform can be alternately • At a root z = ξ of the numerator polynomial l written in factored form as G(ξ ) = 0, and as a result, these values of z l M −1 are known as the zeros of G(z) p0∏ (1−ξ z ) G(z) = l=1 l • At a root z = λ of the denominator N −1 l d0∏ (1− λ z ) polynomial G ( λ ) → ∞ , and as a result, l=1 l l M these values of z are known as the poles of p0∏ (z −ξ ) = z(N −M ) l=1 l G(z) d N (z − λ ) 0∏l=1 l 19 20 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Rational z-Transforms Rational z-Transforms •Example- The z-transform • Consider M 1 (N −M ) p0∏ =1(z −ξ ) µ (z) = , for z >1 G(z) = z l l −1 N 1− z d0∏ (z − λ ) l=1 l has a zero at z = 0 and a pole at z = 1 •NoteG(z) has M finite zeros and N finite poles •IfN > M there are additional N − M zeros at z = 0 (the origin in the z-plane) •IfN < M there are additional M − N poles at z = 0 21 22 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Rational z-Transforms Rational z-Transforms

• A physical interpretation of the concepts of poles and zeros can be given by plotting the log-magnitude 20 log 10 G ( z ) as shown on next slide for 1− 2.4 z−1 + 2.88 z−2 G(z) = 1− 0.8 z−1 + 0.64 z−2

4 ROC of a Rational Rational z-Transforms z-Transform • Observe that the magnitude plot exhibits •ROC of a z-transform is an important very large peaks around the points concept z = 0.4 ± j 0.6928 which are the poles of • Without the knowledge of the ROC, there is G(z) no unique relationship between a sequence • It also exhibits very narrow and deep wells and its z-transform around the location of the zeros at • Hence, the z-transform must always be z =1.2 ± j1.2 specified with its ROC

ROC of a Rational ROC of a Rational z-Transform z-Transform • Moreover, if the ROC of a z-transform • The ROC of a rational z-transform is includes the unit circle, the DTFT of the bounded by the locations of its poles sequence is obtained by simply evaluating • To understand the relationship between the the z-transform on the unit circle poles and the ROC, it is instructive to • There is a relationship between the ROC of examine the pole-zero plot of a z-transform the z-transform of the impulse response of a • Consider again the pole-zero plot of the z- causal LTI discrete-time system and its transform µ(z) BIBO stability

ROC of a Rational ROC of a Rational z-Transform z-Transform •Example- The z-transform H(z) of the sequence h [ n ] = ( − 0 . 6 ) n µ [ n ] is given by 1 H(z) = , 1+ 0.6 z−1 z > 0.6 • In this plot, the ROC, shown as the shaded area, is the region of the z-plane just outside the circle centered at the origin and going • Here the ROC is just outside the circle through the pole at z = 1 29 30 going through the point z = −0.6 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

5 ROC of a Rational ROC of a Rational z-Transform z-Transform • A sequence can be one of the following •Example- Consider a finite-length sequence types: finite-length, right-sided, left-sided g[n] defined for − M ≤ n ≤ N , where M and and two-sided N are non-negative integers and g[n] < ∞ • In general, the ROC depends on the type of •Its z-transform is given by the sequence of interest N ∑N +M g[n − M ]z N +M −n G(z) = g[n]z−n = 0 ∑ N n=−M z

ROC of a Rational ROC of a Rational z-Transform z-Transform •Note:G(z) has M poles at z = ∞ and N poles •Example- A right-sided sequence with at z = 0 nonzero sample values for n ≥ 0 is • As can be seen from the expression for sometimes called a causal sequence G(z), the z-transform of a finite-length • Consider a causal sequence u1[n] bounded sequence converges everywhere in •Its z-transform is given by the z-plane except possibly at z = 0 and/or at ∞ z = ∞ −n U1(z) = ∑u1[n]z n=0

ROC of a Rational ROC of a Rational z-Transform z-Transform • It can be shown that U ( z ) converges 1 •Example- A left-sided sequence with exterior to a circle z = R 1 , including the point z = ∞ nonzero sample values for n ≤ 0 is sometimes called a anticausal sequence • On the other hand, a right-sided sequence u 2[n] with nonzero sample values only for n ≥ − M • Consider an anticausal sequence v1[n] •Its z-transform is given by with M nonnegative has a z-transform U 2 ( z) with M poles at z = ∞ 0 −n V1(z) = ∑v1[n]z • The ROC of U 2 ( z ) is exterior to a circle n=−∞ z = R2 , excluding the point z = ∞ 35 36 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

6 ROC of a Rational ROC of a Rational z-Transform z-Transform • It can be shown that V ( z ) converges 1 •Example- The z-transform of a two-sided interior to a circle z = R , including the 3 sequence w[n] can be expressed as point z = 0 ∞ ∞ −1 −n −n −n • On the other hand, a left-sided sequence W (z) = ∑ w[n]z = ∑ w[n]z + ∑ w[n]z with nonzero sample values only for n ≤ N n=−∞ n=0 n=−∞ • The first term on the RHS, ∞ w [ n ] z − n , with N nonnegative has a z-transform V 2 ( z) ∑n=0 with N poles at z = 0 can be interpreted as the z-transform of a • The ROC of V ( z ) is interior to a circle right-sided sequence and it thus converges 2 exterior to the circle z = R z = R4 , excluding the point z = 0 5 37 38 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

ROC of a Rational ROC of a Rational z-Transform z-Transform •Example- Consider the two-sided sequence −1 −n n • The second term on the RHS, ∑ n = −∞ w [ n ] z , u[n] = α can be interpreted as the z-transform of a left- where α can be either real or complex sided sequence and it thus converges interior •Its z-transform is given by to the circle z = R6 ∞ ∞ −1 U (z) = αnz−n = αnz−n + αnz−n • If R < R , there is an overlapping ROC ∑ ∑ ∑ 5 6 n=−∞ n=0 n=−∞ given by R5 < z < R6 • The first term on the RHS converges for • If R 5 > R 6 , there is no overlap and the z > α , whereas the second term converges z-transform does not exist for z < α 39 40 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

ROC of a Rational ROC of a Rational z-Transform z-Transform • There is no overlap between these two • The ROC of a rational z-transform cannot regions contain any poles and is bounded by the • Hence, the z-transform of u [ n ] = α n does poles not exist • To show that the z-transform is bounded by the poles, assume that the z-transform X(z) has simple poles at z = α and z = β • Assume that the corresponding sequence x[n] is a right-sided sequence 41 42 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

7 ROC of a Rational ROC of a Rational z-Transform z-Transform • Then x[n] has the form • The condition x[n] = (r αn + r βn )µ[n − N ], α < β ∞ 1 2 o ∑ γnz−n < ∞ where N is a positive or negative integer o n=No •Now, the z-transform of the right-sided holds for z > γ but not for z ≤ γ n sequence γ µ [ n − N o ] exists if • Therefore, the z-transform of ∞ n n ∑ γnz−n < ∞ x[n] = (r1α + r2β )µ[n − No ], α < β n=No has an ROC defined by β < z ≤ ∞

ROC of a Rational ROC of a Rational z-Transform z-Transform • Likewise, the z-transform of a left-sided • The ROC is thus bounded on the outside by sequence the pole with the smallest magnitude that n n x[n] = (r1α + r2β )µ[−n − No ], α < β contributes for n < 0 and on the inside by has an ROC defined by 0 ≤ z < α the pole with the largest magnitude that contributes for n ≥ 0 • Finally, for a two-sided sequence, some of the poles contribute to terms in the parent • There are three possible ROCs of a rational sequence for n < 0 and the other poles z-transform with poles at z = α and z = β α < β contribute to terms n ≥ 0 ( ) 45 46 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

ROC of a Rational ROC of a Rational z-Transform z-Transform • In general, if the rational z-transform has N poles with R distinct magnitudes, then it has R +1 ROCs • Thus, there are R + 1 distinct sequences with the same z-transform • Hence, a rational z-transform with a specified ROC has a unique sequence as its inverse z-transform 47 48 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

8 ROC of a Rational ROC of a Rational z-Transform z-Transform • The ROC of a rational z-transform can be • [num,den] = zp2tf(z,p,k) easily determined using MATLAB implements the reverse process [z,p,k] = tf2zp(num,den) • The factored form of the z-transform can be determines the zeros, poles, and the gain obtained using sos = zp2sos(z,p,k) constant of a rational z-transform with the • The above statement computes the numerator coefficients specified by the coefficients of each second-order factor vector num and the denominator coefficients given as an L × 6 matrix sos specified by the vector den 49 50 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

ROC of a Rational ROC of a Rational z-Transform z-Transform

⎡b01 b11 b21 a01 a11 a12 ⎤ • The pole-zero plot is determined using the ⎢ ⎥ function zplane sos = b02 b12 b22 a02 a12 a22 ⎢ ⎥ •The z-transform can be either described in ⎢ M M M M M M ⎥ ⎣b0L b1L b2L a0L a1L a2L ⎦ terms of its zeros and poles: zplane(zeros,poles) where • or, it can be described in terms of its L b + b z−1 + b z−2 G(z) = 0k 1k 2k numerator and denominator coefficients: ∏ −1 −2 k=1a0k + a1k z + a2k z zplane(num,den) 51 52 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

ROC of a Rational z-Transform Inverse z-Transform •Example- The pole-zero plot of • General Expression: Recall that, for z = r e j ω , 2 z4 +16 z3 + 44 z2 + 56 z + 32 the z-transform G(z) given by G(z) = 4 3 2 ∞ −n ∞ −n − jωn 3z + 3z −15z +18z −12 G(z) = ∑n=−∞ g[n]z = ∑n=−∞ g[n]r e obtained using MATLAB is shown below is merely the DTFT of the modified sequence −n 2 g[n]r 1 × −pole • Accordingly, the inverse DTFT is thus given 0 o −zero by

9 Inverse z-Transform Inverse z-Transform • But the integral remains unchanged when • By making a change of variable z = r e j ω , is replaced with any contour C encircling the previous equation can be converted into the point z = 0 in the ROC of G(z) a contour integral given by • The contour integral can be evaluated using 1 the Cauchy’s residue theorem resulting in g[n] = G(z) zn−1dz ∫ ⎡residuesof G(z)zn−1 ⎤ 2πj C′ g[n] = ∑ ⎢at the poles inside C⎥ where C ′ is a counterclockwise contour of ⎣ ⎦ integration defined by |z| = r • The above equation needs to be evaluated at all values of n and is not pursued here 55 56 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Inverse Transform by Inverse Transform by Partial-Fraction Expansion Partial-Fraction Expansion • A rational G(z) can be expressed as • A rational z-transform G(z) with a causal M −i inverse transform g[n] has an ROC that is P(z) ∑ pi z G(z) = = i=0 exterior to a circle D(z) N −i ∑i=0 di z • Here it is more convenient to express G(z) in a partial-fraction expansion form and • If M ≥ N then G(z) can be re-expressed as then determine g[n] by summing the inverse M −N G(z) = η z−l + P1(z) transform of the individual simpler terms in ∑ l D(z) the expansion l=0 where the degree of P 1 ( z ) is less than N 57 58 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Inverse Transform by Inverse Transform by Partial-Fraction Expansion Partial-Fraction Expansion

• The rational function P 1 ( z ) / D ( z ) is called a • Simple Poles: In most practical cases, the proper fraction rational z-transform of interest G(z) is a •Example- Consider proper fraction with simple poles 2 + 0.8 z−1 + 0.5 z−2 + 0.3z−3 • Let the poles of G(z) be at z = λ k ,1≤ k ≤ N G(z) = 1+ 0.8 z−1 + 0.2 z−2 • A partial-fraction expansion of G(z) is then • By long division we arrive at of the form −1 N ⎛ ρ ⎞ 5.5 + 2.1z G(z) = ⎜ l ⎟ −1 ∑⎜ −1 ⎟ G(z) = −3.5 +1.5 z + =1 1− λ z 1+ 0.8 z−1 + 0.2 z−2 l ⎝ l ⎠ 59 60 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

10 Inverse Transform by Inverse Transform by Partial-Fraction Expansion Partial-Fraction Expansion • The constants ρ in the partial-fraction • Therefore, the inverse transform g[n] of l expansion are called the residues and are G(z) is given by N given by n −1 g[n] = ρ (λ ) µ[n] ∑ l l ρ = (1− λ z )G(z) z=λ l l l l=1 • Each term of the sum in partial-fraction •Note:The above approach with a slight expansion has an ROC given by z > λ l modification can also be used to determine and, thus has an inverse transform of the the inverse of a rational z-transform of a form ρ (λ )nµ[n] l l noncausal sequence 61 62 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Inverse Transform by Inverse Transform by Partial-Fraction Expansion Partial-Fraction Expansion •Example- Let the z-transform H(z) of a •Now causal sequence h[n] be given by 1+ 2 z−1 ρ = (1− 0.2 z−1)H (z) = = 2.75 z(z + 2) 1+ 2 z−1 1 z=0.2 −1 H(z) = = 1+ 0.6 z z=0.2 −1 −1 (z − 0.2)(z + 0.6) (1− 0.2 z )(1+ 0.6 z ) and • A partial-fraction expansion of H(z) is then 1+ 2 z−1 of the form −1 ρ2 = (1+ 0.6 z )H (z) z=−0.6 = −1 = −1.75 ρ ρ 1− 0.2 z H (z) = 1 + 2 z=−0.6 1− 0.2 z−1 1+ 0.6 z−1 63 64 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Inverse Transform by Inverse Transform by Partial-Fraction Expansion Partial-Fraction Expansion • Hence • Multiple Poles: If G(z) has multiple poles, 2.75 1.75 the partial-fraction expansion is of slightly H(z) = − 1− 0.2 z−1 1+ 0.6 z−1 different form • The inverse transform of the above is • Let the pole at z = ν be of multiplicity L and therefore given by the remaining N − L poles be simple and at z = λ 1≤ ≤ N − L l , l h[n] = 2.75(0.2)nµ[n]−1.75(−0.6)nµ[n]

11 Inverse Transform by Partial-Fraction Expansion Partial-Fraction Expansion Using MATLAB • Then the partial-fraction expansion of G(z) is of the form • [r,p,k]= residuez(num,den) M −N N −L ρ L γ develops the partial-fraction expansion of G(z) = η z−l + l + i ∑ l ∑ −1 ∑ −1 i a rational z-transform with numerator and =0 =1 1− λ z i=1(1− ν z ) l l l denominator coefficients given by vectors where the constants γ i are computed using num and den L−i 1 d −1 L r γi = [](1− ν z ) G(z) , • Vector contains the residues (L − i)!(−ν)L−i d(z−1)L−i z=ν 1≤ i ≤ L • Vector p contains the poles • The residues ρ are calculated as before • Vector k contains the constantsη l l 67 68 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Partial-Fraction Expansion Inverse z-Transform via Long Using MATLAB Division

• [num,den]=residuez(r,p,k) •The z-transform G(z) of a causal sequence {g[n]} can be expanded in a power series in z−1 converts a z-transform expressed in a • In the series expansion, the coefficient partial-fraction expansion form to its −n rational form multiplying the term z is then the n-th sample g[n] • For a rational z-transform expressed as a ratio of polynomials in z − 1 , the power series expansion can be obtained by long division 69 70 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Inverse z-Transform via Long Inverse z-Transform Using Division MATLAB •Example- Consider • The function impz can be used to find the 1+ 2 z−1 inverse of a rational z-transform G(z) H (z) = 1+ 0.4 z−1 − 0.12 z−2 • The function computes the coefficients of • Long division of the numerator by the the power series expansion of G(z) denominator yields • The number of coefficients can either be user specified or determined automatically H(z) =1+1.6z−1 − 0.52 z−2 + 0.4 z−3 − 0.2224 z−4 + .... • As a result {h[n]} = {1 1.6 − 0.52 0.4 − 0.2224 ....}, n ≥ 0 71 72 ↑ Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

12 Table 6.2: z-Transform Properties z-Transform Properties •Example- Consider the two-sided sequence v[n] = αnµ[n]−βnµ[−n −1] • Let x [ n ] = α n µ [ n ] and y [ n ] = − β n µ [ − n − 1] with X(z) and Y(z) denoting, respectively, their z-transforms 1 •Now X (z) = , z > α 1− α z−1 1 and Y (z) = , z < β 1−β z−1 73 74 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

z-Transform Properties z-Transform Properties • Using the linearity property we arrive at •Example- Determine the z-transform and 1 1 its ROC of the causal sequence V (z) = X (z) +Y(z) = + −1 −1 n 1−α z 1− β z x[n] = r (cosωon)µ[n] • The ROC of V(z) is given by the overlap • We can express x[n] = v[n] + v*[n] where regions of z > α and z < β v[n] = 1 rne jωonµ[n] = 1 αnµ[n] 2 2 • If α < β , then there is an overlap and the •The z-transform of v[n] is given by ROC is an annular region α < z < β 1 1 • If α > β , then there is no overlap and V(z) V (z) = 1 ⋅ = 1 ⋅ , z > α = r 2 1− α z−1 2 1− r e jωo z−1 does not exist 75 76 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

z-Transform Properties z-Transform Properties • Using the conjugation property we obtain •or, 1− (r cosω )z−1 the z-transform of v*[n] as X (z) = o , z > r −1 2 −2 1 1 1− (2r cosωo )z + r z V *(z*) = 1 ⋅ = 1 ⋅ , 2 1− α* z−1 2 1− r e− jωo z−1 •Example- Determine the z-transform Y(z) z > α and the ROC of the sequence n • Finally, using the linearity property we get y[n] = (n +1)α µ[n] X (z) = V (z) +V *(z*) • We can write y [ n ] = n x [ n ] + x [ n ] where ⎛ 1 1 ⎞ n = 1 ⎜ + ⎟ x[n] = α µ[n] 2 ⎜ jωo −1 − jωo −1 ⎟ 77 ⎝1− r e z 1− r e z ⎠ 78 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

13 z-Transform Properties z-Transform Properties

•Now, the z-transform X(z) of x [ n ] = α n µ [ n ] • Using the linearity property we finally is given by obtain 1 −1 X (z) = , z > α 1 α z −1 Y (z) = + 1− α z 1− α z−1 (1− α z−1)2 • Using the differentiation property, we arrive 1 n x[n] = , z > α at the z-transform of as (1− α z−1)2 dX (z) α z−1 − z = , z > α dz (1− α z−1) 79 80 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

LTI Discrete-Time Systems in LTI Discrete-Time Systems in the Transform Domain the Transform Domain • An LTI discrete-time system is completely • Such transform-domain representations characterized in the time-domain by its provide additional insight into the behavior impulse response sequence {h[n]} of such systems • Thus, the transform-domain representation • It is easier to design and implement these systems in the transform-domain for certain of a discrete-time signal can also be equally applications applied to the transform-domain • We consider now the use of the DTFT and representation of an LTI discrete-time the z-transform in developing the transform- system domain representations of an LTI system 81 82 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Finite-Dimensional LTI Finite-Dimensional LTI Discrete-Time Systems Discrete-Time Systems • Applying the z-transform to both sides of • In this course we shall be concerned with the difference equation and making use of LTI discrete-time systems characterized by the linearity and the time-invariance linear constant coefficient difference properties of Table 6.2 we arrive at equations of the form: N M −k −k N M ∑dk z Y(z) = ∑ pk z X (z) ∑dk y[n − k] = ∑ pk x[n − k] k=0 k=0 k=0 k=0 where Y(z) and X(z) denote the z-transforms of y[n] and x[n] with associated ROCs, respectively 83 84 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

14 Finite-Dimensional LTI Discrete-Time Systems • A more convenient form of the z-domain representation of the difference equation is given by N M ⎛ −k ⎞ ⎛ −k ⎞ ⎜ ∑dk z ⎟Y(z) = ⎜ ∑ pk z ⎟X (z) ⎝ k=0 ⎠ ⎝ k=0 ⎠