This slide explains the concept of a transform in the better way or beginners

This is a brief course of Applied physics d Digital signal processing and book followed by prokais

1. App. Phys. 602: Digital Signal Processing 2. Inverse Z Transform Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 1 The Z Transform

2. 10/06/2022 Digital Signal Processing – 602, Semester – VIII, 2021 1. The Z Transform 2  Powerful tool for analyzing & designing DT systems Generalization of the DTFT:  z is complex ... z = e j  → DTFT   G ( z )  Z  g  n     g [ n ] z  n n  Z Transform   g n r e  n  j  n n z = r · e j  →  DTFT of r -n · g [ n ]

3. Region of Convergence (ROC) Critical question: 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 x [ n ] z  n n   Does summa t ion G ( z )    z -plane R e { z } I m { z } ROC | z | > λ λ Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 4

4. R O C Example e.g. x [ n ] = ∏ n µ [ n ] ∑ converges only for |  z -1 | < 1 i.e. ROC is | z | > |  |  |  | < 1 (e . g. . 8) - f ini t e energy sequence |  | > 1 (e . g. 1 . 2) - divergent sequence, infinite energy, DTFT does not exist but still has ZT when | z | > 1.2 (in ROC)   n z  n  n    X ( z )   1 1   z  1 (previous slide) n 1 2 3 4 -2 -1 “closed form” when |  z -1 | < 1 R e { z } Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 5 I m { z } λ

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 R e { z } I m { z } Unit circle lies in ROC → DTFT OK  Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 7

6. Ano t her R O C example Anticausal (left-sided) sequence:  Same ZT as  n µ [ n ] , different sequence? x  n     n    n  1  n 1 2 3 4 -5 -4 -3 -2 -1 X  z   n         n  1 z  n n    n    1      m z m m  1       1 z   n z  n 1 1 1    1 z 1   z  1 ROC: | λ | > | z | Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 8

7. ROC is necessary! A closed-form expression for ZT must specify the ROC : x [ n ] =  n µ [ n ] ROC | z | > |  | ROC | z | < |  |  several sequences with different ROCs  X ( z )  1 1   z  1 x [ n ] = -  n µ [- n- 1]  X ( z )  1 1   z  1 n 1 2 3 4 -4 -3 -2 -1 z -plane Re Im  λ  n -4 -3 -2 -1 1 2 3 4 Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 9 A single G ( z ) expression can ma t ch D TF T s?

8. Rational Z-transforms G ( z ) expression can be any function; rational polynomials are important class:  By convention, expressed in terms of z -1 – matches ZT definition (Reminiscent of LCCDE expression...)  G  z   P  z  D  z    1 p  p 1 z    p M  1 M z  ( M  1 )  p z  M  d  d z 1    d 0 1 N  1 N z  ( N  1 )  d z  N Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 10

9. F ac t ored ra t ional Z T s Numerator, denominator can be factored :  {   } are roots of numerator → G ( z ) = → {   } are the zeros of G ( z ) {   } are roots of denominator → G ( z ) = ∞ → {   } are the poles of G ( z )  G  z     1 p  M  1    z  1    1 d  N  1   z  1      1  z M p  M  z      1 z N d  N  z     Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 11

10. Pole-zero diagram Can plot poles and zeros on complex z -plane:  (Value of) expression determined by roots z -plane R e { z } I m { z }  × o o o o × × poles   (cpx conj for real g [ n ] ) zeros   Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 12

11. Z-plane surface Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 13 G ( z ) : cpx function of a cpx variable Can calculate value over entire z-plane M ROC not shown ! !

12. − 1 − 0.5 0.5 1 − 1 − 0.5 0.5 1 10 8 6 4 2 R O Cs and sidedness  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 1 1   z  1 z -p la ne  λ  Two sequences have: G ( z )  RO C | z | > |  | → g [ n ] =  n µ [ n ] n RO C | z | < |  | → n g [ n ] = -  µ [- n- 1] n L E F T -S I DED Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 14 R IG H T -S I DED ( |  | < 1 )

13. ZT is Linear   Thus, if then  n G ( z )  Z  g  n     g [ n ] z  n Z Transform y [ n ]    n  [ n ]    n  [ n ] 1 1 2 2 Y ( z )  1 1 1   z  1    2 2 1   z  1 y [ n ] =  g [ n ] +  h [ n ] Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 15 → Y ( z ) = ∑ (  g [ n ] +  h [ n ] ) z - n = ∑  g [ n ] z - n + ∑  h [ n ] z - n =  G ( z ) +  H ( z ) ROC:  z  >  λ 1  ,  λ 2  Linear 

14. G ( z )  1 1 1   z  1  1 2 1   z  1 ROC intersections Consider wit h |  1 | < 1 , |  2 | > 1 ...  Two possible sequences for  1 term...  Similarly for  2 ... no ROC specified n or   n µ [ n ] -  1 n µ [- n -1] n n n → 4 possible g [ n ] seq’s and ROCs ... -   n µ [- n -1]  Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 16  n µ [ n ] or

15. ROC intersections: Case 1 ROC: | z | > |  1 | and | z | > |  2 | Im Im Re  λ    λ   G ( z )   1 1 1 2 1   z  1 1   z  1 n g [ n ] =  n µ [ n ] +  n µ [ n ] 1 2 both right -sided: Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 17

16. ROC intersections: Case 2 Im Im Re  λ    λ   G ( z )   1 1 1 2 1   z  1 1   z  1 n Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 18 both left -sided: ROC: | z | < |  1 | and | z | < |  2 |

17. ROC intersections: Case 3 ROC: | z | > |  1 | and | z | < |  2 | Im Im Re  λ    λ   G ( z )   1 1 1 2 1   z  1 1   z  1 n g [ n ] =  n µ [ n ] -  n µ [- n -1] 1 2 two -sided: Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 19

18. ROC intersections: Case 4 ROC: | z | < |  1 | and | z | > |  2 | ? Re  λ    λ   Im G ( z )   1 1 1 2 1   z  1 1   z  1 n g [ n ] = -  1 n µ [- n -1] +  2 n µ [ n ] Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 20 two -sided: no ROC → ...

19. ROC intersections Note: Two-sided exponential  No overlap in ROCs → ZT does not exist (does not converge for any z ) g  n    n   n   n    n    n  1  -  < n <  ROC | z | > |  | ROC | z | < |  | Re  α  Im n Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 21

20. ZT of LCCDEs LCCDEs have solutions of form:    LCCDE sol’ns are right-sided → R O Cs are | z | > |  i | c i i n   y [ n ]     n  ... c   Hence Z T Y z   i i Each term  i n in g [ n ] corresponds to a pole  i of G ( z ) ... and vice versa 1   z  1   (same  s) ou ts id e circles Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 22

21. Z -plane and D TF T Slice between surface and unit cylinder ( | z | = 1 → z = e j  ) is G ( e j  ) , the DTFT | G ( e j  )| z = e j  º  / r a d / s am p Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 23

22. Some common Z transforms g [ n ] G ( z ) ROC Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 24

23. Z Transform properties Conjugation g [ n ] — G ( z ) w/ROC R g g * [ n ] G * ( z * ) R g Time reversal Time shift Exp. scaling g [- n ] G (1/ z ) 1/ R g g [ n - n ] z - n G ( z ) R g (0/ ∞ ?) α n g [ n ] G ( z/ α )  α  R g Diff. wrt z ng [ n ] R g (0/ ∞ ?)  z d G ( z ) dz Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 25

24. Z Transform properties Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 26

25. ZT Example can express as  1 2  n   j  n re  re  j  0 0     n x  n   r n cos   n    n  ;        v n  v n *     v [ n ] = 1 / 2 µ [ n ] α n ; α = r e j  → V ( z ) = 1 / (2(1- r e j  z -1 )) ROC: | z | > r Hence, X  z   V  z   V *  z *   1 2 1 j   1  1  j  1  re z 1  re z   1   1  r c o s    z  1   1  2 r cos  z  r z  1 2  2 Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 27

26. Another ZT example y  n    n  1   n   n   x [ n ]  n x [ n ] where x [ n ] =  n µ [ n ] X  z   ( | z | > |  | ) 1 1   z  1   z dX  z  dz   z 1 d  dz  1   z     1    z  1 ( 1   z )  1 2  Y  z   ROC | z | > |  | 1 1   z  1   z  1 ( 1   z  1 2  ) ( 1   z ) 1  1 2 repeated root - IZT Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 28

27. 2. Inverse Z Transform (IZT) Forward z transform was defined as:    Generalization of inverse DTFT Power series in z (long division) Manipulate into recognizable pieces (partial fractions)  G ( z )  Z  g  n     g [ n ] z  n n  3 approaches to inverting G ( z ) to g [ n ] : the useful one Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 29

28. IZT #1: Generalize IDTFT Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 30

29. IZ T #2: Long division 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 ra t ional polynomial G  z   P ( z )  D ( z ) Can expand as power series through long division of polynomials g [ n ] z  n n    Since G ( z )   Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 31

30. IZ T #2: Long division Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 32 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

31. IZ T #2: Long division e.g. H  z   1  2 z  1 1  0. 4 z  0.1 2 z  1  2 1  1. 6 z  1  0.5 2 z  2  0. 4 z  3 ... 1  0. 4 z  1  0.1 2 z  2 ) 1  2 z  1 1  0. 4 z  1  0.1 2 z  2 1. 6 z  1  0.1 2 z  2 1. 6 z  1  0.6 4 z  2  0.19 2 z  3  0.5 2 z  2  0.19 2 z  3 Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 33 ... Result

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 1 1   z  1   n   n  P ( z )  1   z  1   1   z  1   A  B 1   z  1 1   z  1   Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 34

33. Partial Fractions  Can do the reverse i.e. go from  order 3 polynomial P ( z )  N   1  ( 1   z  1 )     1 1    z  1 Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 35 t o  N else cancel if order of P ( z ) is less than D ( z ) w/ long div.

34. Partial Fractions Procedure:  i.e. evaluate F ( z ) at the pole but multiplied by the pole term denominator) → dominates = residue of pole F ( z )  P ( z )   1  N  ( 1   z  1 ) where     1    z  1  F  z    N      1 1   z  1 z      n  n       1  f  n    N   order N -1 (cancels term in no repeated poles! Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 36

35. Partial Fractions Example G iven H  z   (again) factor:  where: 1  2 z  1 1  0. 4 z  0.1 2 z  1  2  1  2 z  1  1  0. 6 z  1   1  0. 2 z  1    1 1  0. 6 z  1   2 1  0. 2 z  1 1   1  0. 6 z  1   H z   z   0.6  1  2 z  1 1  0. 2 z  1 z   0.6   1.75 2   1  2 z  1 1  0. 6 z  1 z  0.2 Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 37  2.75

36. Partial Fractions Example Hence H  z   If we know ROC | z | > |  | i.e. h [ n ] causal:  h  n     1.7 5    0. 6  n   n    2.7 5   0. 2  n   n  = –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 ...}  1   1.75 2.75 1  0. 6 z 1  0. 2 z  1 same as long division! Digital Signal Processing – 602, Semester – VIII, 2021 10/06/2022 38