Inverse z-Transform Chapter 6B Part B ◆Inverse z.Transform ●●● ●●● ●●● z-Transform Properties Inverse z-Transform z-Transform 09 and ZT Properties 3 1.Inverse z-Transform 1.1 General Expression 1.1 General Expression ●Recall that,,for=red“,thez-transform G(2) ◆By making a change of variable z=re“,the 1.1 General Expression given by previous equation can be converted into a 1.2 Inverse z-Transform by Partial-Fraction Expansion ()=G(re)=g(n)r"e contour integral given by 1.3 Partial-Fraction Using MATLAB gm=2[G2]=2 G(=)2"d is merely the DTFT of the modified sequence 1.4 Inverse z-Transform via Long Division g(n)rn where c is a counterclockwise contour of 1.5 Inverse z-Transform Using MATLAB integration defined by=r Accordingly,the inverse DTFT is thus given But the integral remains unchanged when c is replaced with any contour c'encircling the point z=0 in the ROC of G(z) 4 5
Chapter 6B z-Transform Part B Inverse z-Transform and ZT Properties 3 Inverse z-Transform Inverse z Inverse z-Transform z-Transform Properties Transform Properties 4 1. Inverse z-Transform 1.1 General Expression 1.2 Inverse z-Transform by Partial Transform by Partial-Fraction Expansion 1.3 Partial 1.3 Partial-Fraction Using MATLAB 1.4 Inverse z-Transform via Long Division 1.5 Inverse z-Transform Using MATLAB 5 1.1 General Expression Recall that, for z= rej¹, the z-transform G(z) given by is merely the DTFT of the modified sequence g(n)r-n Accordingly, the inverse DTFT is thus given by - () ( ) () j n jn n G z G re g n r e 1 () ( ) 2 n j jn g n r G re e d 6 1.1 General Expression By making a change of variable z= rej¹ , the previous equation can be converted into a contour integral given by where c is a counterclockwise contour of integration defined by |z|= r But the integral remains unchanged when c is replaced with any contour c’ encircling the point z=0 in the ROC of G(z) 1 1 1 ( ) [ ( )] ( ) 2 n c g n G z G z z dz j
1.1 General Expression 1.1 General Expression 1.1 General Expression Example: (2-0.5z The contour integral can be evaluated using 1-0.52 Zer0s:2=0z=0.5 Ge=7+075z+02四 the Cauchy's residue theorem resulting in G8)=1+0.752+0.1252 Poles: 2=-0.5z=-025 Case 1:0.5 7 1.1 General Expression 1.1 General Expression 1.1 General Expression If n<0,there are one first-order pole and one nlth-order pole (2-0.5)z g=-(e+0.52+0.752+0.12可 Case2:0.25<H<0.5 at 2=-0.25 and :=0 inside c,respectively.Thus,we can If n0,there is only one pole at ==-0.25 inside c compute the summation of the residues outside e instead of that inside (2-0.5)z -e+020z+0.75z+0.12 (2-0.5z”1 (z-05)z -a 8(m)=(2+0.25) 8(m=-(z+0.5) 02+0.75z+0.125 z2+0.75z+0.125 =-4(-0.5)°+3(-0.25) nS-1 =-4-0.5”n≤-1 =-3-0.25H n20 Hence.we can rewrite g(n)as follows g(m)=-3-0.25yu(m-4(-0.5yu(-n-1) 10 12
7 1.1 General Expression The contour integral can be evaluated using the Cauchy’s residue theorem resulting in The above equation needs to be evaluated at all values of n and is not pursued here 1 1 ( ) residues of ( ) at the poles inside residues of ( ) at the poles outside only if there are any higher-order poles inside n n gn Gzz c Gzz c c 8 1.1 General Expression Example: Example: 1 1 2 1 0.5 ( ) 1 0.75 0.125 z G z z z Zeros: z 0 z 0.5 Poles: z 0.5 z 0.25 -1 -0.5 0 0.5 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Real Part Imaginary Part Three ROCs: z 0.25 0.25 0.5 z z 0.5 9 1.1 General Expression Case 1: Case 1: z 0.25 1 2 ( 0.5) ( ) 0.75 0.125 n n z z Gzz z z If nı0, there is no poles inside c. Thus, g(n)=0 when nı0 If n <0, there is an |n|-order pole at z=0 which is inside c. In this case, we can compute the summation of the residues outside c instead of that inside gn z z ( ) Res { 0.5}+Res{ 0.25} 10 1.1 General Expression 2 0.5 2 0.25 ( 0.5) ( ) ( 0.5) 0.75 0.125 ( 0.5) ( 0.25) 0.75 0.125 4( 0.5) 3( 0.25) 1 n z n z n n z z gn z z z z z z z z n 11 1.1 General Expression Case 2: Case 2: If nı0, there is only one pole at z=ˉ0.25 inside c 0.25 0.5 z 2 0.25 ( 0.5) ( ) ( 0.25) 0.75 0.125 3( 0.25) 0 n z n z z gn z z z n 12 1.1 General Expression If n <0, there are one first-order pole and one |n|th-order pole at z=ˉ0.25 and z=0 inside c, respectively. Thus, we can compute the summation of the residues outside c instead of that inside Hence, we can rewrite g(n) as follows 2 0.5 ( 0.5) ( ) ( 0.5) 0.75 0.125 4( 0.5) 1 n z n z z gn z z z n ( ) 3( 0.25) ( ) 4( 0.5) ( 1) n n gn un u n
1.2 Inverse z-Transform by Partial- 1.1 General Expression 1.1 General Expression Fraction Expansion Ifthere are two first-order poles and one th-order pole Case3:>0.5 at z=-0.25,2=-0.25 and z=0 inside c,respectively.Thus,we .A rational z-transform G(z)with a causal If20,there are two first-order poles at ==-0.25 and ==-0.5 can compute the summation of the residues outside c instead of that inside.Because there is no poles outside c.Thus. inverse transform g(n)has an ROC that is insidec g(n)=0 in this case exterior to a circle (2-0.5z gm=e+0.5)g+075z+0.12西 Here it is more convenient to express G(z)in a Summary: partial-fraction expansion form and then (z-05)z 「-4-0.5y(-n-1)+3(-0.25y°u(-n-1.0.5 the expansion =4H-0.5)°-3-025 n20 13 14 1.2 Inverse z-Transform by Partial- 1.2 Inverse z-Transform by Partial- 1.2 Inverse z-Transform by Partial- Fraction Expansion Fraction Expansion Fraction Expansion .A rational G(z)can be expressed as Solutions: Example: ) 品-/2r Step 1-Converting G(z)into the form of G2=2+0.82+0522+0.32 proper fractions by long division 1+0.8z1+0.22 If then G(z)can be re-expressed as By long division we arrive at Step 2--Summing the inverse transform of ce-艺a-号 Proper Fraction the individual simpler terms in the expansion G(z)--3.5+1.52 5.5+2.1z Assume that g(n)is causal 1+0.82+0.22 where the degree of P(z)is less than N -3.56n)↓ 1 分 1.56(n-1) 18
13 1.1 General Expression Case 3: Case 3: If nı0, there are two first-order poles at z=ˉ0.25 and z=ˉ0.5 inside c z 0.5 2 0.5 2 0.25 ( 0.5) ( ) ( 0.5) 0.75 0.125 ( 0.5) ( 0.25) 0.75 0.125 4( 0.5) 3( 0.25) 0 n z n z n n z z gn z z z z z z z z n 14 1.1 General Expression If n<0, there are two first-order poles and one |n|th-order pole at z=ˉ0.25, z=ˉ0.25 and z=0 inside c, respectively. Thus, we can compute the summation of the residues outside c instead of that inside. Because there is no poles outside c. Thus, g(n)=0 in this case Summary: Summary: 4( 0.5) ( 1) 3( 0.25) ( 1), 0.25 ( ) 3( 0.25) ( ) 4( 0.5) ( 1), 0.25 0.5 4( 0.5) ( ) 3( 0.25) ( ), 0.5 n n n n n n un un z gn un u n z un un z 15 1.2 Inverse z-Transform by PartialFraction Expansion A rational z-transform G(z) with a causal inverse transform g(n) has an ROC that is exterior to a circle Here it is more convenient to express G(z) in a partial-fraction expansion form and then determine g(n) by summing the inverse transform of the individual simpler terms in the expansion 16 1.2 Inverse z-Transform by PartialFraction Expansion A rational G(z) can be expressed as If then G(z) can be re-expressed as 1 0 ( ) ( ) ( ) M N l l l P z Gz z D z where the degree of is less than N Proper Fraction Proper Fraction (ⵏ࠶ᮠ) 0 0 ( ) ( ) ( ) M N i i i i i i P z G z pz dz D z 1P( )z 17 1.2 Inverse z-Transform by PartialFraction Expansion Solutions: Solutions: Step 1-- Converting G(z) into the form of proper fractions by long division Step 2-- Summing the inverse transform of the individual simpler terms in the expansion Assume that g(n) is causal 18 1.2 Inverse z-Transform by PartialFraction Expansion Example: Example: 123 1 2 2 0.8 0.5 0.3 ( ) 1 0.8 0.2 z z z G z z z 1 1 1 2 5.5 2.1 ( ) 3.5 1.5 1 0.8 0.2 z Gz z z z By long division we arrive at 3.5 ( ) n 1.5 ( 1) n
1.2 Inverse z-Transform by Partial- 1.3 Partial-Fraction Expansion 1.3 Partial-Fraction Expansion Fraction Expansion Using MATLAB Using MATLAB Let H(z)= 5.5+2.121 .[r.p.c]=residuez(num,den)develops the .[num,den]=residuez(r.p.c)converts a z- partial-fraction expansion of a rational z- 1+0.82+0.22 transform expressed in a partial-fraction transform with numerator and denominator expansion form to its rational form 2.75+0.25i 2.75-0.257i coefficients given by vectors num and den 1-(-0.4+02021-(-0.4-0.202 Vector r contains the residues ↑ Vector p contains the poles (2.75+0.250-0.4+0.20m Vector c contains the constants (2.75-0.250-0.4-0.2i)(m 19 20 21 1.4 Inverse z-Transform via 1.4 Inverse z-Transform via Long Division Long Division 1.5 Inverse z-Transform Using MATLAB The z-transform G(z)of a causal sequence Example (g(n))can be expanded in a power series in ●Consider X(z)= 1+221 .The function impz can be used to find the 1+0.42-0.122 inverse of a rational z-transform G(z) In the series expansion,the coefficient The function computes the coefficients of the Long division of the numerator by the multiplying the term z-"is then the n-th denominator yields power series expansion of G(z) sample g(n) X(z)=1+1.6z-0.52z2+0.4z3-0.224z4+… The number of coefficients can either be user .For a rational z-transform expressed as a specified or determined automatically ●Hence ratio of polynomials in z-,the power series (m)}={1,1.6-0.52,0.4,-0.2224,.}n≥0 expansion can be obtained by long division. 中 -0 23 24
19 1.2 Inverse z-Transform by PartialFraction Expansion 1 1 2 1 1 5.5 2.1 ( ) 1 0.8 0.2 2.75 0.25 2.75 0.25 1 ( 0.4 0.2 ) 1 ( 0.4 0.2 ) z H z z z i i iz iz Let (2.75 0.25 )( 0.4 0.2 ) ( ) n i i un (2.75 0.25 )( 0.4 0.2 ) ( ) n i i un 20 1.3 Partial-Fraction Expansion Using MATLAB [r,p,c]= residuez(num,den) develops the partial-fraction expansion of a rational ztransform with numerator and denominator coefficients given by vectors num and den Vector r contains the residues Vector p contains the poles Vector c contains the constants l 21 1.3 Partial-Fraction Expansion Using MATLAB [num,den]=residuez(r,p,c) converts a ztransform expressed in a partial-fraction expansion form to its rational form 22 1.4 Inverse z-Transform via Long Division The z-transform G(z) of a causal sequence {g(n)} can be expanded in a power series in zˉ1 In the series expansion, the coefficient multiplying the term zˉn is then the n-th sample g(n) For a rational z-transform expressed as a ratio of polynomials in zˉ1, the power series expansion can be obtained by long division. 23 1.4 Inverse z-Transform via Long Division Example Example Consider Long division of the numerator by the denominator yields Hence {x(n)}={ 1, 1.6, ˉ0.52, 0.4, ˉ0.2224,…} nı0 1 1 2 1 2 ( ) 1 0.4 0.12 z X z z z 1 23 4 Xz z z z z ( ) 1 1.6 0.52 0.4 0.224 n=0 24 1.5 Inverse z-Transform Using MATLAB The function impz can be used to find the inverse of a rational z-transform G(z) The function computes the coefficients of the power series expansion of G(z) The number of coefficients can either be user specified or determined automatically
2.z-Transform Properties 2.z-Transform Properties 2.z-Transform Properties Some useful properties of z-Transform are Let x(n)=a"u(m)y(n)=B"u(-n-1) 1 1 listed in Table 6.2 with X(z)and Y(=)denoting,respectively,their z- V()=X()+Y()=1-a+1-B This section is devoted to the computation of transforms .The ROC of V(=)is given by the overlap regions z-Transform by means of these properties ●Now H>bl of >a andthen there is no overlap and V(=) n)=a"u()-B"u(-n-1) does not exist .Using the linearity property we arrive at 26 27 2.z-Transform Properties 2.z-Transform Properties Example 2 Using the differentiation property,we arrive at the z-transform of n(n)as y(n)=(n+I)a"u(n) -rae yn)can be rewritten as y(n)=n(n)+x(n) where x(n)=a"u(n) da-alal Using the linearity property we finally obtain .The z-transform of x(n)is given by Y(2) -a小网 (I-az H>l回 X(z)= 28
25 2. z-Transform Properties Some useful properties of z-Transform are listed in Table 6.2 This section is devoted to the computation of z-Transform by means of these properties Example 1 Consider the two-sided sequences ( ) ( ) ( 1) n n vn un u n 26 2. z-Transform Properties Let with X(z) and Y(z) denoting, respectively, their ztransforms Now Using the linearity property we arrive at 1 1 ( ) 1 Xz z z () () n xn un ( ) ( 1) n yn u n 1 1 ( ) 1 Yz z z 27 2. z-Transform Properties The ROC of V(z) is given by the overlap regions of and If , then there is an overlap and the ROC is an annular region If , then there is no overlap and V(z) does not exist 1 1 1 1 () () () 1 1 Vz Xz Yz z z z z z 28 2. z-Transform Properties Example 2 ( ) ( 1) ( ) n yn n un y(n) can be rewritten as where The z-transform of x(n) is given by y n nx n x n () () () () () n xn un 1 1 ( ) 1 Xz z z 29 2. z-Transform Properties Using the differentiation property, we arrive at the z-transform of nx(n) as 1 1 2 ( ) (1 ) dX z z z z dz z Using the linearity property we finally obtain 2 1 1 ( ) 1 Yz z z
