z-Transform Chapter 6 Chapter 6 D I Dr. Iyad Jafar d J f - PowerPoint PPT Presentation

z-Transform Chapter 6 Chapter 6 D I Dr. Iyad Jafar d J f

Outline Outline � Definition � Relation Between z Transform and DTFT � Relation Between z-Transform and DTFT � Region of Convergence g g � Common z-Transform Pairs � The Rational z-Transform h l f � The Inverse z-Transform The Inverse z Transform � z-Transform Properties � LTI Systems & z-Transform � Filt � Filter Design using Pole-zero Placement D i i P l Pl t 2

Def Definition nition � The DTFT does not exist for all sequences! Can we still analyze such sequences in the frequency domain? Additionally, the DTFT is a complex function of ω , which Addi i ll h DTFT i l f i f hi h makes it hard to manipulate. � One way around these issues is to use the z-transform ! O d h h f ! � The z-transform � A generalization of the DTFT and it may exist for sequences for which the DTFT does not exist. � For a real sequence, the z-transform is a real function of the F l h f i l f ti f h complex variable z.This makes provides easier manipulation. � For a sequence g[n] the z transform is defined by � For a sequence g[n], the z-transform is defined by � ∑ � ∑ -n G(z) g[n] z n ��� where z = Re(z) + j Im(z) 3

Relation be lation betw tween een z- z-Transf ransform and D orm and DTFT FT � The z-transform is defined in terms of the complex variable j Im(z) = r e � � j z = Re(z) ( ) j ( ) Im(z) Im(z) � So, the z-transform can written as Z=re j ω j � ∑ r =1 � ∑ � � � � -n � � j j G( ) G(z) g[n] re [ ] � j � z re ω ��� n Re(z) � � -1 1 � ∑ � � � n j n g[n] r e ��� n -j j � for r = 1, � ∑ ∑ � � � � � j n j G(z) G(z) g[n] e g[n] e G(e ) G(e ) � � j j � z re ��� n � So, the DTFT is the z-transform evaluated at the unit , circle (r=1) and varying ω 4

Relation be lation betw tween een z- z-Transf ransform and D orm and DTFT FT � Thus, G(e j ω ) can be computed from th e clockwise or counter clockwise on the unit circle in the complex p plane for 0 ≤ ω ≤ 2 π on the circle (r=1) Im(z) ( ) j j Z=re j ω r =1 ω Re(z) Re(z) -1 1 -j j ω = 0 � z = 1 � G(z=1) = G(e j0 ) ( ) ( ) ω = π /2 � z = j � G(z=1) = G(e j π /2 ) � z = 1 � G(z=1) = G(e j π ) � z = -1 � G(z=1) = G(e j π ) ω = π ω = π 5

Relation be lation betw tween een z- z-Transf ransform and D orm and DTFT FT 6 4 (z)| |G( 2 DTFT DTFT 0 2 2 2 0 0 -2 -2 -2 2 I Im(z) ( ) Re(z) 6

Region of Con gion of Convergence ergence � The z-transform exists if the summation � ∑ � ∑ -n G(z) G(z) g[n] z g[n] z ��� n converges to a finite value! � In other words, the summation should be absolutely summable � � � � ∑ ∑ � � -n � � ⇒ � ⇒ � G(z) < -n j g[n] z < g[n] re < ��� ��� n n � � � ∑ ∑ ∑ � � -n � � � ⇒ � � j -n j -n g[n] re g[n] r e = g[n] r < ��� � � ��� � � � � ��� n n n n n n � This implies that the z-transform may exist even if g[n] is not not absolutely absolutely summable summable given given that that we we specify specify appropriate values for r 7

Region of Con gion of Convergence ergence � The set of values of r for which the z-transform exists is called the region of convergence (ROC) � In general, the ROC for a sequence g[n] is represented as � R z < R � � g g g g � � � where 0 R < R � � g g � The ROC is always defined in terms of |z| as circular region The ROC is always defined in terms of |z| as circular region in the complex plane (inside a circle/outside a circle / rings) α < |z| < β |z| > α |z| < α - β β - α α - α α - α α 8

Region of Con gion of Convergence ergence Example 6.1 : Let x[n] = α n u[n], the determine X(z). |z| > α - α α ROC for causal sequences is the region outside |z| = α 9

Region of Con gion of Convergence ergence Example 6.2 : Let x[n] = - α n u[-n-1], the determine X(z). - α α ROC for anti causal sequences is the region inside |z| = α ROC for anti-causal sequences is the region inside |z| = α To completely define the z-transform, we must define the ROC 10

Common z- Common z-Transf ansform P orm Pair irs 11

The Rational z- The Rational z-Transf ransform orm � Most LTI systems have rational z-transforms of the form where P(z) and D(z) are two polynomials in z -1 of degrees M where P(z) and D(z) are two polynomials in z of degrees M and N, respectively. � The rational z-transform can be written as The rational z transform can be written as � � M M � � � � � 1 p (1 z ) p (z ) 0 k 0 k � � � � (N M) � G(z) G(z) k 1 z z k 1 � � N N N N � � � � � 1 d (1 z ) d (z ) 0 k 0 k � � k 1 k 1 � For z = ζ G(z) = 0 � { ζ } are called the zeros of G(z) ζ l , G(z) 0 � { ζ l } are called the zeros of G(z) � For z � For z = λ l , G(z) = ∞ � { λ l } are called the poles of G(z) � If N > M � dditi � If N > M � additional (N – M) finite zeros at z = 0 l (N M) fi it t = 0 � If N < M � additional (M – N) finite poles t z = 0 12

The Rational z- The Rational z-Transf ansform orm Example 6.3: � 2 z z � H(z) ( ) � � 2 2 z 3z 2 H(z) has zeros at z = 0 and z = 1 (circles in the z-plane‘o’) H(z) has zeros at z 0 and z 1 (circles in the z plane o ) H(z) has poles at z = -1 and z = -2 (crosses in the z-plane‘x’) H(z) has no additional zeros or poles at z = ∞ since M = N H(z) has no additional zeros or poles at z = ∞ since M = N Im(z) R ( ) Re(z) -2 1 z-plane l 13

The Rational z- The Rational z-Transf ansform orm � The z-transform is not defined at the poles. � Thus, the ROC for any z-transform should not include , y the poles. � The rational z-transform can be completely specified by � The rational z transform can be completely specified by the locations of poles and zeros and the gain p 0 /d 0 1 z Example 6.4: � � ⇒ � H(z) H(z) 1 z � � 1 z 1 H(z) = 0 at z = 0 H( ) = ∞ t H(z) = ∞ at z = 1 = 1 ROC: |z| > 1 (causal) ROC : |z| < 1 (anti-causal) 14

The Rational z- The Rational z-Transf ansform orm Example 6.5: x[n] = (0.5) n u[n] + (-0.3) n u[n] 1 1 � � � X(z) X(z) � � � � 1 1 1 0.5z 1 0.3z the z transform exists if |z| > 0 5 and |z| > 0 3 the z-transform exists if |z| > 0.5 and |z| > 0.3 � ROC: ( Z > 0.5) ( Z > 0.3) ⇒ ROC: Z > 0.5 15

The Rational z- The Rational z-Transf ansform orm Example 6.6: x[n] = α n 16

The Rational z- The Rational z-Transf ansform orm � � 2 2 z 2z 3 � Example 6.7: X(z) � � � 3 2 z 3z z 5 All possible regions of convergence are All possible regions of convergence are 17

The In The Inverse z- e z-Transf ansform orm � The general form for computing the inverse z-transform is through evaluating the complex integral g g p g 1 � ∫ � ∫ � � � n 1 g[n] G(z)z dz 2 j 2 j � However, the integral has to be evaluated for all values of n � Alternatively, we consider two approaches to compute the inverse z-transform � Power series in z (Long division) g � Manipulate G(z) into recognizable pieces (partial fraction expansion) and use lookup tables p ) p 18

The In The Inverse z- e z-Transf ansform orm Example 6.8: h[n] = {1, 1.6, -0.52, 0.4, …} h[n] {1, 1.6, 0.52, 0.4, …} 19

The Inverse z- The In e z-Transf ansform orm � Partial fraction expansion � rearrange G(z) as a sum of recognizable terms of simple and known z-transforms P(z) P( ) P(z) � � G(z) � � � � � � � � � 1 1 1 (1 z )(1 z )(1 z )... D(z) 1 2 3 � � � � � � � � 1 2 3 ... � � � � � � � � � 1 1 1 (1 z ) (1 z ) (1 z ) 1 2 3 � Thus, a rational z-transform of the form G(z) = h l f f h f P(z)/D(z) can be expressed as N N � ∑ P(z) � � G(z) k z � � � 1 D(z) 1 k � k 1 where N is the order of the denominator, λ k are poles λ and ρ k are called the residues and are computed by � � � � � 1 (1 z )G(z) k k �� z 20 k

The Inverse z- The In e z-Transf ansform orm Example 6.9: Determine the causal sequence h[n] which has the following z-transform g � z(z 2) � H(z) � � (z 0.2)(z 0.6) 21

The In The Inverse z- e z-Transf ansform orm Example 6.9 - Continued. 22

The In The Inverse z- e z-Transf ansform orm � Partial fraction expansion � What if G(z) has poles with multiplicity L? l ith lti li it L? � G(z) is expressed by � N L L ∑ ∑ � ∑ ∑ � P(z) � � � G(z) k m � � � � � � � � 1 m D(z) D(z) 1 1 z z � � � � 1 1 1 1 z k � � k 1 m 1 where σ is a pole with multiplicity L and the residues μ m are computed d � L m L m 1 1 d ⎡ ⎤ � � � � � 1 L (1 z ) G(z) ⎣ ⎦ m � �� � � � L m 1 L m (L m)!( ) d(z ) �� z m � � 1 m L 23