Topic 4: The Z Transform
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1 ELEN E480: Digital Signal Processing Topic 4: The Z Transform. The Z Transform 2. Inverse Z Transform
2 . The Z Transform Powerful tool for analyzing & designing DT systems Generalization of the DTFT: G(z) = Z gn [ ] z is complex... z = e jω DTFT z = r e jω n= { }= g[n]z n n gn [ ]r n e jωn Z Transform DTFT of r -n g[n] 2
3 Region of Convergence (ROC) Critical question: Does summation G(z) = converge (to a finite value)? n= x[n]z n In general, depends on the value of z Region of Convergence: Portion of complex z-plane for which a particular G(z) will converge ROC z > λ Im{z} λ Re{z} z-plane 3
4 ROC Example e.g. x[n] = λ n µ[n] X(z) = Σ converges for λz - < i.e. ROC is z > λ λ n z n = n= λz (see previous slide) λ < (e.g. 0.8) - finite energy sequence λ > (e.g..2) - divergent sequence, infinite energy, DTFT does not exist but still has ZT when z >.2 (ROC) n 4
5 About ROCs ROCs always defined in terms of z circular regions on z-plane (inside circles/outside circles/rings) If ROC includes unit circle ( z = ), g[n] has a DTFT (finite energy sequence) Unit circle lies in ROC DTFT OK Im{z} Re{z} z-plane 5
6 Another ROC example Anticausal (left-sided) sequence: xn [ ]= λ n µ n Xz n ( )z n ()= λ n µ [ n ] [ ] 2 3 4n = λ n z n = λ m z n = λ z n= m= λ z = λz ROC: λ > z Same ZT as λ n µ[n], different sequence? 6
7 -4 ROC is necessary! To completely define a ZT, you must specify the ROC: x[n] = λ n µ[n] X(z) = Im λz n ROC z > λ x[n] = -λ n µ[-n-] X(z) = λz n ROC z < λ -4 z-plane A single G(z) can describe several sequences with different ROCs 7 λ Re DTFTs?
8 Gz Rational Z-transforms G(z) can be any function; rational polynomials are important class: ()= Pz () Dz () = p 0 + p z +K+ p M z (M ) + p M z M d 0 + d z +K+ d N z (N ) + d N z N By convention, expressed in terms of z - matches ZT definition (Reminiscent of LCCDE expression...) 8
9 Gz Factored rational ZTs Numerator, denominator can be factored: ( ζ l z ) d 0 N ( l= λ l z ) = z M p 0 M l= z N N d 0 l= ()= p 0 M l= z ζ l ( ) z λ l ( ) {ζ l } are roots of numerator G(z) = 0 {ζ l } are the zeros of G(z) {λ l } are roots of denominator G(z) = {λ l } are the poles of G(z) 9
10 Pole-zero diagram Can plot poles and zeros on complex z-plane: o zeros ζ l o o Im{z} o poles λ l (cpx conj for real g[n]) Re{z} z-plane 0
11 Z-plane surface G(z): cpx function of a cpx variable Can calculate value over entire z-plane Slice between surface and unit cylinder ( z = z = e jω ) is G(e jω ), the DTFT
12 Z-plane and DTFT Unwrapping the cylindrical slice gives the DTFT: 2
13 ZT is Linear G(z) = Z{ gn [ ]}= g[n]z n n y[n] = αg[n] + βh[n] Y(z) = Σ(αg[n]+βh[n])z -n Z Transform Linear =Σαg[n]z -n + Σβh[n]z -n = αg(z) + βh(z) Thus, if then y[n] = α λ n µ[n]+ α 2 λ 2n µ[n] Y (z) = α λ z + α 2 λ 2 z ROC: z > λ, λ 2 3
14 ZT of LCCDEs LCCDEs have solutions of form: Hence ZT y c [n] = α i λ i n µ n [ ]+... Y c [z] = λ i z Each term λ n i in g[n] corresponds to a pole λ i of G(z)... and vice versa α i +L LCCDE sol ns are right-sided ROCs are z > λ i outside circles (same λs) 4
15 z-plane ROCs and sidedness Two sequences have: Im λ Re G(z) = ROC z > λ g[n] = λ n µ[n] RIGHT-SIDED ROC z < λ g[n] = -λ n µ[-n-] LEFT-SIDED λz Each ZT pole region in ROC outside or inside λ for R/L sided term in g[n] Overall ROC is intersection of each term s n ( λ < ) n 5
16 -λ n µ[-n-] -λ 2n µ[-n-] ROC intersections G(z) = λ z + Consider λ 2 z with λ <, λ 2 >... Two possible sequences for λ term... Similarly for λ 2... n n λ n µ[n] no ROC specified 4 possible g[n] seq s and ROCs... 6 or or λ 2n µ[n] n n
17 ROC intersections: Case G(z) = λ z + λ 2 z ROC: z > λ and z > λ 2 Im g[n] = λ n µ[n] + λ 2n µ[n] both right-sided: n Im Re λ λ 2 7
18 G(z) = ROC intersections: Case 2 λ z + g[n] = -λ n µ[-n-] - λ 2n µ[-n-] λ 2 z both left-sided: n ROC: z < λ and z < λ 2 Im Im Re λ λ 2 8
19 ROC intersections: Case 3 G(z) = λ z + λ 2 z ROC: z > λ and z < λ 2 Im g[n] = λ n µ[n] - λ 2n µ[-n-] two-sided: n Im Re λ λ 2 9
20 G(z) = ROC intersections: Case 4 λ z + λ 2 z ROC: z < λ and z > λ 2? Im g[n] = -λ n µ[-n-] + λ 2n µ[n] two-sided: n Re λ λ 2 no ROC... 20
21 ROC intersections Note: Two-sided exponential gn [ ]= α n = α n µ n [ ]+ α n µ n [ ] ROC z > α < n < ROC z < α No overlap in ROCs ZT does not exist (does not converge for any z) Im α Re n 2
22 Some common Z transforms g[n] G(z) ROC δ[n] z µ[n] z z > α n µ[n] z > α r n cos(ω 0 n)µ[n] z > r r n sin(ω 0 n)µ[n] z > r αz r cos( ω 0 )z 2r cos( ω 0 )z +r 2 z 2 r sin( ω 0 )z 2r cos( ω 0 )z +r 2 z 2 poles at z = re ±jω 0 sum of re jω 0 n + re -jω 0 n conjugate pole pair 22
23 Z Transform properties g[n] G(z) w/roc Rg Conjugation g * [n] G * (z * ) Rg Time reversal g[-n] G(/z) /Rg Time shift g[n-n 0 ] z -n 0G(z) Rg (0/?) Exp. scaling α n g[n] G(z/α) α Rg Diff. wrt z ng[n] z dg(z) dz Rg (0/?) 23
24 Z Transform properties g[n] G(z) ROC Convolution g[n] h[n] G(z)H(z) Modulation g[n]h[n] Parseval: 2πj C gn [ ]h * [ n] = n= 2πj Gv ()H z v C ( )v dv Gv ()H * at least Rg Rh at least RgRh ( v )v dv 24
25 ZT Example xn [ ]= r n cos( ω 0 n)µ [ n] ; can express as µ n 2 [ ] re jω 0 Hence, ( ) n + ( re jω ) n 0 Xz ()= Vz ()+ V * z * = vn [ ]+ v * n 25 [ ] v[n] = / 2 µ[n]α n ; α = r ejω 0 V(z) = /(2(- re jω 0z - )) ROC: z > r ( ) ( ) = 2 re jω 0 + z re jω 0 z r cos( ω = 0 )z 2r cos( ω 0 )z +r 2 z 2
26 Another ZT example Xz ()= Y z ()= yn [ ]= n + ( )α n µ n [ ] = x[n]+ nx[n] where x[n] = α n µ[n] αz z dx z () dz = z d dz = αz 26 αz ( αz ) 2 αz + αz ( αz ) 2 = ( αz ) 2 repeated root - IZT
27 2. Inverse Z Transform (IZT) Forward z transform was defined as: G(z) = Z gn [ ] n= { }= g[n]z n 3 approaches to inverting G(z) to g[n]: Generalization of inverse DTFT Power series in z (long division) Manipulate into recognizable pieces (partial fractions) the useful one 27
28 IZT #: Generalize IDTFT If z = re jω then Gz ()= Gre ( jω )= gn [ ]r n e jωn so z = re jω dω = dz/jz gn [ ]r n = 2π = 2πj π Gre ( jω )e jωn dω π C Gz ()z n dz Counterclockwise closed contour at z = = DTFT{ gn [ ]r n } IDTFT Any closed contour around origin will do Cauchy: g[n] = Σ[residues of G(z)z n- ] Im Re 28
29 IZT #2: Long division G(z) = g[n]z n n= if we could express G(z) as a simple power series G(z) = a + bz - + cz then can just read off g[n] = {a, b, c,...} Since Typically G(z) is right-sided (causal) and a rational polynomial Gz ()= P(z) D(z) Can expand as power series through long division of polynomials 29
30 IZT #2: Long division Procedure: Express numerator, denominator in descending powers of z (for a causal fn) Find constant to cancel highest term first term in result Subtract & repeat lower terms in result Just like long division for base-0 numbers 30
31 IZT #2: Long division e.g. Hz ()= + 2z + 0.4z 0.2z 2 Result +.6z 0.52z z z 0.2z 2 ) + 2z + 0.4z 0.2z 2.6z + 0.2z 2.6z z z z z
32 IZT#3: Partial Fractions Basic idea: Rearrange G(z) as sum of terms recognized as simple ZTs especially α n µ n αz [ ] or sin/cos forms i.e. given products rearrange to sums P(z) ( αz ) ( βz )L A αz + B +L βz 32
33 Partial Fractions Note that: A αz + B βz + C γz = A( βz ) ( γz )+ B( αz ) ( γz )+ C( αz ) ( βz ) ( αz ) ( βz ) ( γz ) order 3 polynomial Can do the reverse i.e. P(z) go from N to ( λ l z ) Π l= if order of P(z) is less than D(z) order 2 polynomial u + vz - + wz -2 ρ N l l= λ l z else cancel w/ long div. 33
34 Partial Fractions Procedure: F(z) = no repeated poles! P(z) N ( λ l z ) = ρ N l l= λ l z Π l= order N- N f[ n]= ρ l λ l where ρ l = ( λ l z )Fz () z=λl i.e. evaluate F(z) at the pole but multiplied by the pole term dominates = residue of pole 34 l= ( ) n µ [ n] (cancels term in denominator)
35 Partial Fractions Example + 2z Given Hz ()= (again) + 0.4z 0.2z 2 factor: + 2z = ( + 0.6z ) ( 0.2z ) = ρ + 0.6z + ρ 2 0.2z where: ρ = ( + 0.6z )Hz () = z= 0.6 ρ 2 = + 2z + 0.6z z=0.2 = z 0.2z z= =.75
36 Partial Fractions Example Hence Hz ()= z z If we know ROC z > α i.e. h[n] causal: hn [ ]=.75 ( ) 0.6 ( ) n µ n [ ] ( ) 0.2 ( ) n µ n [ ] =.75{ } +2.75{ } = { } same as long division! 36
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