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ELEN E4810: Digital Signal Processing Topic 4: The Z Transform

ELEN E4810: Digital Signal Processing Topic 4: The Z Transform. The Z Transform Inverse Z Transform. 1. The Z Transform. Powerful tool for analyzing & designing DT systems Generalization of the DTFT: z is complex ... z = e j w  DTFT z = r · e j w . Z Transform.

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ELEN E4810: Digital Signal Processing Topic 4: The Z Transform

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  1. ELEN E4810: Digital Signal ProcessingTopic 4: The Z Transform • The Z Transform • Inverse Z Transform Dan Ellis

  2. 1. The Z Transform • Powerful tool for analyzing & designing DT systems • Generalization of the DTFT: • z is complex... • z = ejw DTFT • z = r·ejw Z Transform DTFT of r-n·g[n] Dan Ellis

  3. Im{z} Re{z} l z-plane ROC|z| > l Region of Convergence (ROC) • Critical question:Does summationconverge (to a finite value)? • In general, depends on the value of zRegion of Convergence:Portion of complex z-planefor which a particularG(z)will converge Dan Ellis

  4. n -2 -1 1 2 3 4 ROC Example • e.g. x[n] = lnm[n] • S converges for |lz-1| < 1i.e. ROC is |z| > |l| • |l| < 1 (e.g. 0.8) - finite energy sequence • |l| > 1 (e.g. 1.2) - divergent sequence, infinite energy, DTFT does not existbut still has ZT when |z| > 1.2 (ROC) (see previous slide) Dan Ellis

  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| = 1), g[n] has a DTFT (finite energy sequence) Im{z} Re{z} 1 Unit circlelies in ROC DTFT OK z-plane Dan Ellis

  6. -5 -4 -3 -2 -1 n 1 2 3 4 Another ROC example • Anticausal (left-sided) sequence: • Same ZT as lnm[n], different sequence? ROC:|l| > |z| Dan Ellis

  7. Im Re n |l| -4 -3 -2 -1 1 2 3 4 z-plane -4 -3 -2 -1 n 1 2 3 4 ROC is necessary! • To completely define a ZT, you must specify the ROC: x[n] = lnm[n] ROC|z| > |l| x[n] = -lnm[-n-1] ROC|z| < |l| • A single G(z) can describe several sequences with different ROCs DTFTs? Dan Ellis

  8. Rational Z-transforms • G(z) can be any function;rational polynomials are important class: • By convention, expressed in terms of z-1– matches ZT definition • (Reminiscent of LCCDE expression...) Dan Ellis

  9. Factored rational ZTs • Numerator, denominator can be factored: • {zl} are roots of numeratorG(z) = 0{zl} are thezerosofG(z) • {ll} are roots of denominatorG(z) = {ll} are the poles ofG(z) Dan Ellis

  10. Im{z} poles ll(cpx conj for real g[n]) o  Re{z}  o o 1  o zeros zl z-plane Pole-zero diagram • Can plot poles and zeros on complex z-plane: Dan Ellis

  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| = 1 z = ejw) is G(ejw), the DTFT Dan Ellis

  12. Z-plane and DTFT • Unwrapping the cylindrical slice gives the DTFT: Dan Ellis

  13. ZT is Linear • Thus, ifthen Z Transform • y[n] = ag[n] + bh[n] • Y(z) = S(ag[n]+bh[n])z-n =Sag[n]z-n + Sbh[n]z-n = aG(z) + bH(z) Linear ROC: |z|>|l1|,|l2| Dan Ellis

  14. ZT of LCCDEs • LCCDEs have solutions of form: • Hence ZT • Each termlin in g[n] corresponds to a poleli of G(z) ... and vice versa • LCCDE sol’ns are right-sided ROCs are |z| > |li| (samels) outsidecircles Dan Ellis

  15. Im n Re n |l| z-plane ROCs and sidedness • Two sequences have: • Each ZT poleregion in ROCoutside or inside|l| for R/L sided term in g[n] • Overall ROC is intersection of each term’s ROC |z| > |l| g[n] = lnm[n] RIGHT-SIDED ( |l| < 1 ) ROC |z| < |l| g[n] = -lnm[-n-1] LEFT-SIDED Dan Ellis

  16. n n ROC intersections • Consider with |l1| < 1 , |l2| > 1 ... • Two possible sequences for l1 term... • Similarly for l2 ...  4 possible g[n] seq’s and ROCs ... no ROC specified -l1nm[-n-1] l1nm[n] or -l2nm[-n-1] n l2nm[n] or n Dan Ellis

  17. Im Im Re |l1| |l2| n ROC intersections: Case 1 ROC:|z| > |l1| and |z| > |l2| g[n] = l1nm[n] + l2nm[n] both right-sided: Dan Ellis

  18. Im Im Re |l1| |l2| n ROC intersections: Case 2 g[n] = -l1nm[-n-1] - l2nm[-n-1] both left-sided: ROC:|z| < |l1| and |z| < |l2| Dan Ellis

  19. Im Im Re |l1| |l2| n ROC intersections: Case 3 ROC:|z| > |l1| and |z| < |l2| g[n] = l1nm[n] - l2nm[-n-1] two-sided: Dan Ellis

  20. n Im Re |l1| |l2| ROC intersections: Case 4 ROC:|z| < |l1| and |z| > |l2| ? g[n] = -l1nm[-n-1] + l2nm[n] two-sided: no ROC ... Dan Ellis

  21. n Im Re |a| ROC intersections • Note: Two-sided exponential • No overlap in ROCs ZT does not exist(does not converge for any z) ROC|z| > |a| ROC|z| < |a| Dan Ellis

  22. Some common Z transforms g[n]G(z)ROC d[n] 1 z m[n] |z| > 1 anm[n] |z| > |a| rncos(w0n)m[n] |z| > r rnsin(w0n)m[n] |z| > r sum of rejw0n + re-jw0n “conjugate polepair”  poles at z = re±jw0  Dan Ellis

  23. Z Transform properties g[n] G(z)w/ROCRg Conjugation g*[n]G*(z*)Rg Time reversal g[-n] G(1/z) 1/Rg Time shift g[n-n0] z-n0G(z)Rg(0/?) Exp. scaling ang[n] G(z/a)|a|Rg Diff. wrt zng[n] Rg(0/?) Dan Ellis

  24. Z Transform properties g[n] G(z)ROC Convolution g[n] h[n]G(z)H(z) Modulation g[n]h[n] Parseval: at leastRgRh * at leastRgRh Dan Ellis

  25. ZT Example • ; can express as • Hence, v[n] = 1/2m[n]an ; a = rejw0V(z) = 1/(2(1- rejw0z-1))ROC: |z| > r Dan Ellis

  26. Another ZT example where x[n] = anm[n] repeatedroot - IZT Dan Ellis

  27. 2. Inverse Z Transform (IZT) • Forward z transform was defined as: • 3 approaches to invertingG(z) to g[n]: • Generalization of inverse DTFT • Power series in z (long division) • Manipulate into recognizable pieces (partial fractions) the usefulone Dan Ellis

  28. Im Re IZT #1: Generalize IDTFT • If then • so • Any closed contour around origin will do • Cauchy: g[n] = S[residues of G(z)zn-1] IDTFT z = rejw dw = dz/jz Counterclockwiseclosed contour at |z| = 1 Dan Ellis

  29. IZT #2: Long division • Since if we could express G(z) as a simple power series G(z) = a + bz-1 + cz-2... then can just read off g[n] = {a, b, c, ...} • Typically G(z) is right-sided (causal)and a rational polynomial • Can expand as power series through long division of polynomials Dan Ellis

  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-10 numbers Dan Ellis

  31. IZT #2: Long division • e.g. Result ) ... Dan Ellis

  32. IZT#3: Partial Fractions • Basic idea: Rearrange G(z) as sum of terms recognized as simple ZTs • especiallyor sin/cos forms • i.e. given products rearrange to sums Dan Ellis

  33. Partial Fractions • Note that: • Can do the reverse i.e.go from to • if order of P(z) is less than D(z) order 2 polynomialu + vz-1 + wz-2 order 3 polynomial else cancelw/ long div. Dan Ellis

  34. Partial Fractions order N-1 • Procedure: • where i.e. evaluateF(z)at the pole but multiplied by the pole term  dominates = residue of pole no repeatedpoles! (cancels term indenominator) Dan Ellis

  35. Partial Fractions Example • Given (again) factor: • where: Dan Ellis

  36. Partial Fractions Example • Hence • If we know ROC |z| > |a| i.e. h[n] causal: = –1.75{ 1 -0.6 0.36 -0.216 ...} +2.75{ 1 0.2 0.04 0.008 ...} = {1 1.6 -0.52 0.4 ...} same aslong division! Dan Ellis

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