1 / 36

Understanding Z-Transform in Digital Signal Processing

Explore the power of Z-Transform for analyzing and designing discrete-time systems, the concept of Region of Convergence (ROC), rational Z-Transforms, pole-zero diagrams, and properties of Z-Transform. Learn how to determine ROC, solve Linear Constant Coefficient Difference Equations (LCCDEs), and apply Z-Transform in various sequences and transformations.

nlebouef
Download Presentation

Understanding Z-Transform in Digital Signal Processing

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related