z-transform Chapter 6 Dr. Iyad djafar
Outline 2 Definition Relation Between z-transform and DTFT Region of Convergence Common z-transform Pairs The Rational z-transform The Inverse z-transform z-transform Properties LTI Systems & z-transform Filter Design using Pole-zero Placement
3 Definition The DTFT does not exist for all sequences! Can we still analyze such sequences in the frequency domain? Additionally, i the DTFT is acomplex function of ω, which h makes it hard to manipulate. One way around these issues is to use the z-transform!! 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 areal function of the complex variable z.this makes provides easier manipulation. For a sequence g[n], the z-transform is defined by where z = Re(z) + j Im(z) n G(z) g[n] z -n
Relation between z-transform and DTFT 4 The z-transform is defined in terms of the complex j variable z = Re(z) j Im(z) = r e Im(z) So, the z-transform can written as j G(z) j g[n] re zre for r = 1, n g[n] r n n e j jn -n r =1 j n j G(z) j g[n] e G(e ) zre n ω -1 1 So, the DTFT is the z-transform evaluated at the unit circle (r=1) and varying ω -j Z=re jω Re(z)
Relation between z-transform and DTFT Thus, G(e jω ) can be computed from the clockwiseor counter clockwise on the unit circle in the complex plane for 0 ω 2π on the circle (r=1) r =1 j Im(z) Z=re jω ω -1 1 Re(z) -j ω = 0 z = 1 G(z=1) = G(e j0 ) ω = π/2 z = j G(z=1) = G(e jπ/2 ) ω = π z= -1 G(z=1) = G(e jπ ) 5
Relation between z-transform and DTFT 6 4 G( (z) 2 DTFT 0 2 0 0 2 6 Im(z) -2-2 Re(z)
Region of Convergence 7 The z-transform exists if the summation converges to a finite value! n G(z) g[n] z In other words, the summation should be absolutely summable G(z) < -n j g[n] z < g[n] re < -n n -n n j -n j -n g[n] re g[n] r e = g[n] r < -n n n n This implies that the z-transform may exist even if g[n] is not absolutely summable given that we specify appropriate values for r
Region of Convergence 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 where 0 R g < R g The ROC is always defined in terms of z as circular region in the complex plane (inside a circle/outside a circle / rings) g z > α z < α α < z < β -α α -α α -β -α α β 8
Region of Convergence Example 6.1: Let x[n] = α n u[n], the determine X(z). z > α -α α 9 ROC for causal sequences is the region outside z = α
Region of Convergence Example 6.2: Let x[n] = -α n u[-n-1], the determine X(z). -α α 10 ROC for anti-causal sequences is the region inside z = α To completely define the z-transform, we must define the ROC
11 Common z-transform Pairs
The Rational z-transform Most LTI systems have rational z-transforms of the form 12 where P(z) and D(z) are two polynomials in z -1 of degrees M and N, respectively. The rational z-transform can be written as G(z) M M 1 0 k 0 k k1 (NM) k1 z N N 1 0 k 0 k k1 k1 p (1 z ) p (z ) d (1 z ) d (z ) For z = ζ l, G(z) = 0 {ζ l } are called the zeros of G(z) For z = λ l,g(z)= {λ l } are called the poles of G(z) If N > M additional (N M) finiteit zerosat z = 0 If N < M additional (M N) finite poles t z = 0
The Rational z-transform Example 6.3: H(z) () z 2 z 2 z 3z2 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 Im(z) -2 1 Re(z) 13 z-plane
The Rational z-transform The z-transform is not defined at the poles. Thus, the ROC for any z-transform should not include the poles. The rational z-transform can be completely specified by the locations of poles and zeros and the gain p 0 /d 0 Example 6.4: H(z) 1 1 z H(z) 1 z z 1 14 H(z) = 0 at z = 0 H(z) = at z = 1 ROC: z > 1 (causal) ROC : z < 1 (anti-causal)
The Rational z-transform Example 6.5: x[n] = (0.5) n u[n] + (-0.3) n u[n] 15 X(z) 1 1 10.5z 10.3z 1 1 the z-transform exists if z > 0.5 and z > 0.3 ROC : ( Z > 0.5) ( Z > 0.3) ROC : Z > 0.5
The Rational z-transform Example 6.6: x[n] = α n 16
The Rational z-transform Example 6.7: X(z) 2 z 2z 3 3 2 z 3z z5 All possible regions of convergence are 17
The Inverse z-transform The general form for computing the inverse z-transform is throughevaluating g the complex integral 1 2 j n1 g[n] G(z)z dz 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) Manipulate G(z) into recognizable pieces (partial fraction expansion) and use lookup tables 18
The Inverse z-transform Example 6.8: h[n] = {1, 1.6, -0.52, 0.4, } 19
The Inverse z-transform Partial fraction expansion rearrangeg(z)asasumof recognizable terms of simple and known z-transforms P(z) P(z) G(z) 1 1 1 D(z) (11z )(12z )(13z )... 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) = P(z)/D(z) can be expressed as G(z) where N is the order of the denominator, λ k are poles and ρ k are called the residues and are computed by N P(z) k D(z) 1 z k1 (1 z ) G(z) 1 20 k k zk k 1
The Inverse z-transform Example 6.9: Determine the causal sequence h[n] which has the following z-transform H(z) z(z 2) (z0.2)(z0.6) 21
The Inverse z-transform Example 6.9 -Continued. 22
The Inverse z-transform Partial fraction expansion What if G(z) has poles with multiplicity li it L? G(z) is expressed by G(z) NL L P(z) k 1 k1 k m1 D(z) 1 z 1 z where σ is a pole with multiplicity L and the residues μ m are computed m 1 m L m 1 d 1 L m (1z ) G(z) Lm 1 Lm (L m)!( ) d(z ) 1 m L 23 z m
The Inverse z-transform Example 6.10: Determine the causal sequence g[n] which has the following z-transform G(z) z (z0.5)(z1) 2 24
The Inverse z-transform Example 6.10 -Continued. 25
The Inverse z-transform Example 6.11: Determine all possible sequences whose z- transform is 1 11 G(z) 3 2 1 z z 3z 1 3 6 1 5 6 6 2 1 z z 1 26
The Inverse z-transform Example 6.11 -Continued. 27
28 z-transform Properties
z-transform Properties Example 6.12: Determine the z-transform of x[n] = r n cos(ω o n) u[n] 29
z-transform Properties Example 6.13: Determine the z-transform of A x[n] + B x[n-1] = C δ[n] + D δ[n-1] 30
z-transform Properties Example 6.14: Determine the z-transform of x[n] = (n+1) a n u[n] + a n u[n] 31
z-transform Properties Example 6.15: Using z-transform, compute the convolution between h[n] = {0, 3, 4, 5} and g[n] = {2, 7, 1} 32
z-transform Properties Example 6.16: Using z-transform, compute the convolution between h[n] = 3 (-0.6) n u[n] g[n] = 2(0.2) 2) n u[n] + 1.2(0.2) 2) n-1 u[n-1] 33
LTI Systems and z-transform The impulse response h[n] of a LTI system is completely ltl characterized in frequency domain using the frequency response H(e jω ) However, H(e jω ) is a complex function of ω, which makes it difficult to manipulate for realization How about the characterization of LTI systems using the z-transform? 34
35 LTI Systems and z-transform For a LTI system, the input-output relation is given by linear convolution y[n] x[n k] h[k] Computing the z-transform of both sides k Y(z) x[n k] h[k] z n n k Interchanging the order of summation Y(z) h[k] x[n k] z k n n h[k] X(z) z k Y(z) X(z) h[k] z X(z) H(z) k H(z) is called the Transfer Function of the impulse response k k
36 LTI Systems and z-transform Consider a FIR LTI system h[n], N 1 n N 2.The transfer function H(z) is H(z) N 2 h[n] z N 1 If h[n] is causal, 0 N 1 <N 2, then all the poles of H(z) will be at z = 0. Thus, the ROC is the entire z-plane except at z =0 If h[n] is anticausal, N 1 <N 2 < 0, then te all the tepoles of H(z) will be at z =. Thus,theROCistheentirezplane except atz= If h[n] is two-sided, then the poles of H(z) will be at z = 0 and z =. Thus, the ROC is the entire z-plane except atz=0andz= n
LTI Systems and z-transform Example 6.17: consider the M-point moving average filter M1 1 y[n] x[n k] M k 0 37
LTI Systems and z-transform 38 Consider an IIR LTI system described by the difference q equation N M k k0 k0 The transfer function H(z) () is d y[n k] px[nk] N M k k dy(z)z Y(z) k0 k () p k X(z)z () H(z) () N X(z) k0 k0 k M p z -1 k 0 Thus, H(z) is a rational polynomial in z. In case h[n] is causal IIR, the ROC is the exterior of the circlepassing i through hthe furthest pole from theorigin ii In case h[n] is an anti-causal IIR, the ROC is the interior of the circle passing through the closet pole from the origin k dz k k k
LTI Systems and z-transform Example 6.18: consider the transfer function for a causal LTI system. Draw the pole-zero diagram and determine all possible regions of convergence. X(z) 2 z 0.2z 0.15 3 2 z 0.4z 0.36z0.144 39
LTI Systems and z-transform 40 We showed earlier that the DTFT is basically the z- j transform evaluated at the unit circle, i.e. For a rational transfer function, the frequency response is computed by H(e ) M M 0 k pz (z ) j k 1 N N 0 k dz (z ) p e d k1 M 0 j (NM) k1 N 0 j k1 j ze j (e ) k (e ) H(e ) H(z) j k ze
LTI Systems and z-transform Accordingly, the magnitude response is M H(e ) And the phase response is e j j p0 k1 N d 0 j e k1 k k M N p0 j j ( ) ( ) + (N-M) + (e k )- (e k) d0 k1 k1 Effectively, the computation tti of H(e jω ) can be done by geometric vector manipulation in the z-plane 41
LTI Systems and z-transform Example 6.19: Compute the frequency response at ω = π for z 1 X(z) z(z j)(z j) 42
LTI Systems and z-transform A LTI is said to be BIBO stable if its impulse response is absolutely summable, i.e. S h n h[n] Can we determine the stability of a LTI system based on its transfer function? 43 H(z) h[n]z h[n]z h[n] z n n n n n n on the unit circle, r = 1 j H(z) j H(e ) h[n] ze This implies that a BIBO stable LTI system has H(e jω ) <. In other words, the DTFT should exist. So, the ROC of A BIBO stable LTI system should include the unit circle. n
LTI Systems and z-transform Inverse System Consider the impulse response of two systems h 1 [n] and h 2 [n] that are cascaded x[n] h 1 [n] h 2 [n] y[n] 44 If y[n] equals x[n], then h 2[n]issaidtobetheinverse system of h 1 [n] and vice versa. It can be easily verified that the relation between the transfer functions of a system and its inverse is given by H(z) 1 1 H(z) 2
LTI Systems and z-transform Example 6.20: determine the transfer function for the causal inverse system for z 0.2z 0.6 1) H(z), z 0.5 (z0.3)(z0.5) 2) z4z5 H(z), z 0.5 (z0.5)(z0.3) 45
LTI Systems and z-transform Example 6.21: determine the transfer function of the causal inverse system for h[n] = 1.9δ[n] + 0.5 (-0.2) n u[n] -0.6(0.7) n u[n] 46
LTI Systems and z-transform Deconvolution Given that the relation between the input and output of a LTI system is convolution, y[n] = x[n] * h[n], can we determine x[n] if we know h[n] and y[n]? Y(z) Y(z) = X(z) H(z) X(z) = H(z) Example 6.22: y[n] = {6, 10, 3, -2, 5, -6} h[n] = {2, 4, 1, -3} 47
Filter Design Using Pole-Zero Placement Recall the rational z-transform 48 The frequency response (magnitude and phase) at specific frequency can be computed geometrically based on the location of the poles and zeros of the z-transform. Specifically the magnitude response is M M e j k j p0 k1 k1 N N d 0 j e k k1 k1 H(e ) K Dis tan ce to zeros Dis tan ce to poles Note how the magnitude is large nearthepoles and small near the zeros Given some filter type andspecifications, can we design the filter by proper placement of the poles and zeros?
Filter Design Using Pole-Zero Placement 49 Example 6.23: Design a single-pole lowpass filter with unity magnitude at ω =0and0.5atω = π/2 1) place the pole at ω =0 2) place on zero at ω = π 3) the transfer function will be z 1 H(z) K z r 4) at ω = 0, the magnitude response is 1 j0 e 1 j0 H(z) j0 K 1 2K = 1 r ze e r 5) at ω = π/2 j /2 e 1 H(z) z e j K 0.5 /2 j /2 e r Solving K = 3/8 and r = 1/4 r
Filter Design Using Pole-Zero Placement Example 6.23: continued 1 H(z) 3 z 1 3 1 z 8 z0.25 8 10.25z 1 0.8 1 3 1 Y(z) 1 0.25z X(z) 1 z 0.4 8 0.2 3 y[n] 0.25y[n 1] x[n] x[n 1] 0 8 5 0 1 H(e j ) 0.6-4 -2 0 2 4 x[n] y[n] -5 50 0 5 10 15 20-10 n
Related functions ztrans iztrans zplane tf residuez conv freqz Matlab 51