Dan Ellis 1 ELEN E4810: Digital Signal Processing Topic 4: The Z Transform 1.The Z Transform 2.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 = e j   DTFT z = r·e j   Z Transform DTFT of r -n ·g[n]

Dan Ellis 3 Region of Convergence (ROC) Critical question: Does summation converge (to a finite value)? In general, depends on the value of z  Region of Convergence: Portion of complex z -plane for which a particular G(z) will converge z-plane Re{z} Im{z} ROC |z| >

Dan Ellis 4 ROC Example e.g. x[n] = n  [n]  converges for | z -1 | | | | | < 1 (e.g. 0.8) - finite energy sequence | | > 1 (e.g. 1.2) - divergent sequence, infinite energy, DTFT does not exist but still has ZT when |z| > 1.2 (ROC) (see previous slide) n

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) z-plane Re{z} Im{z} Unit circle lies in ROC  DTFT OK 

Dan Ellis 6 Another ROC example Anticausal (left-sided) sequence: Same ZT as n  [n], different sequence? n ROC: | | > |z|

Dan Ellis 7 ROC is necessary! To completely define a ZT, you must specify the ROC: x[n] = n  [n] ROC |z| > | | x[n] = - n  [-n-1] ROC |z| < | | A single G(z) can describe several sequences with different ROCs n z-plane Re Im  n 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: {  } are roots of numerator  G(z) = 0  {  } are the zeros of G(z) { } are roots of denominator  G(z) =   { } are the poles of G(z)

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

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 = e j  ) is G(e j  ), the DTFT

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

Dan Ellis 13 ZT is Linear Thus, if then Z Transform y[n] =  g[n] +  h[n]  Y(z) =  (  g[n]+  h[n])z -n =   g[n]z -n +   h[n]z -n =  G(z) +  H(z) ROC: |z|>| 1 |,| 2 | Linear

Dan Ellis 14 ZT of LCCDEs LCCDEs have solutions of form: Hence ZT Each term i n in g[n] corresponds to a pole i of G(z)... and vice versa LCCDE sol’ns are right-sided  ROCs are |z| > | i | (same s) outside circles

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

Dan Ellis 16 ROC intersections Consider with | 1 | 1... Two possible sequences for 1 term... Similarly for 2...  4 possible g[n] seq’s and ROCs... no ROC specified  n  [n] n -  n  [-n-1] n or n -  n  [-n-1] n  n  [n] or

Dan Ellis 17 ROC intersections: Case 1 ROC: |z| > | 1 | and |z| > | 2 | Im Re  Im n g[n] = 1 n  [n] + 2 n  [n] both right-sided:

Dan Ellis 18 ROC intersections: Case 2 Im Re  Im n g[n] = - 1 n  [-n-1] - 2 n  [-n-1] both left-sided: ROC: |z| < | 1 | and |z| < | 2 |

Dan Ellis 19 ROC intersections: Case 3 ROC: |z| > | 1 | and |z| < | 2 | Im Re  Im n g[n] = 1 n  [n] - 2 n  [-n-1] two-sided:

Dan Ellis 20 ROC intersections: Case 4 ROC: |z| | 2 | ? Re  Im n g[n] = - 1 n  [-n-1] + 2 n  [n] two-sided: no ROC ...

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

Dan Ellis 22 Some common Z transforms g[n]G(z) ROC  [n] 1  z  [n] | z | > 1  n  [n] | z | > |  | r n cos(  0 n)  [n] | z | > r r n sin(  0 n)  [n] | z | > r sum of re j  0 n + re -j  0 n poles at z = re ±j  0   “conjugate pole pair”

Dan Ellis 23 Z Transform properties g[n]  G(z) w/ROC R g Conjugation g * [n]G * (z * ) R g Time reversal g[-n]G(1/z)1/ R g Time shift g[n-n 0 ]z -n 0 G(z) R g (0/  ?) Exp. scaling  n g[n]G(z/  )  R g Diff. wrt z ng[n] R g (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 least R g  R h at least R g R h

Dan Ellis 25 ZT Example ; can express as Hence, v[n] = 1 / 2  [n]  n ;  = r ej  0  V(z) = 1/(2(1- re j  0 z -1 )) ROC: |z| > r

Dan Ellis 26 Another ZT example where x[n] =  n  [n] repeated root - IZT

Dan Ellis Inverse Z Transform (IZT) Forward z transform was defined as: 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

Dan Ellis 28 IZT #1: Generalize IDTFT If then so Any closed contour around origin will do Cauchy: g[n] =  [residues of G(z)z n-1 ] IDTFT z = re j   d  = dz/jz Counterclockwise closed contour at |z| = 1 Re Im

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 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 especially or 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 3 polynomial order 2 polynomial u + vz -1 + wz -2 else cancel w/ long div.

Dan Ellis 34 Partial Fractions Procedure: where i.e. evaluate F(z) at the pole but multiplied by the pole term  dominates = residue of pole order N-1 (cancels term in denominator) no repeated poles!

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

Dan Ellis 36 Partial Fractions Example Hence If we know ROC |z| > |  | i.e. h[n] causal: = –1.75{ } +2.75{ } = { } same as long division!