Chapter 7: the Z-Transform

Chapter 7: The z-Transform

Chih-Wei Liu Outline

 Introduction  The z-Transform  Properties of the Region of Convergence  Properties of the z-Transform  Inversion of the z-Transform  The Transfer Function  Causality and Stability  Determining Frequency Response from Poles & Zeros  Computational Structures for DT-LTI Systems  The Unilateral z-Transform

2 Introduction

 The z-transform provides a broader characterization of discrete-time LTI systems and their interaction with signals than is possible with DTFT  Signal that is not absolutely summable z-transform DTFT  Two varieties of z-transform:  Unilateral or one-sided  Bilateral or two-sided  The unilateral z-transform is for solving difference equations with initial conditions.  The bilateral z-transform offers insight into the nature of system characteristics such as stability, causality, and frequency response.

3 A General Complex Exponential zn

 Complex exponential z= rej with magnitude r and angle  n zn  rn cos(n)  jrn sin(n) Re{z }: exponential damped cosine Im{zn}: exponential damped sine r: damping factor : sinusoidal frequency  < 0

exponentially damped cosine exponentially damped sine

 zn is an eigenfunction of the LTI system

4 Eigenfunction Property of zn

x[n] = zn y[n]= x[n]  h[n] LTI system, h[n]  y[n]  h[n] x[n]  h[k]x[n  k]   k k  Hz  hk z Transfer function   k  h[k]znk k n  H(z) is the eigenvalue of the eigenfunction z  n  k  j (z)  z  h[k]z   Polar form of H(z): H(z) = H(z)e    k  H(z)amplitude of H(z); (z)  phase of H(z)  zn H (z) Then yn  H ze j z z n . Let z= rej yn  H re j r n cosn  re j  j H re j r n sinn  re j .

The LTI system changes the amplitude of the input by H(rej) and shifts the phase of the sinusoidal components by (rej).

5 The z-Transform

xnz z  z   xn z n , n  Definition: The z-transform of x[n]: 1  Definition: The inverse z-transform of X(z): xn  z z n1dz. 2j 

 A representation of arbitrary signals as a weighted superposition of eigenfunctions zn with z= rej. We obtain  H (re j )  h[n](re j )n  n DTFT j n h[n]r  H (re )  n  h[n]rn e jn z-transform is the DTFT of h[n]r Hence n 1  1  j j n hn r n  H re j e jn d h[n]  H (re )(re ) d    2   2 z= rej 1 n1 dz = jrej d  H (z)z dz 6 2j  d = (1/j)z  1dz Convergence of Laplace Transform

 z-transform is the DTFT of x[n]rn  A necessary condition for convergence of the z-transform is the absolute summability of x[n]rn :   xn r n  . n

 The range of r for which the z-transform converges is termed the region of convergence (ROC).  Convergence example: 1. DTFT of x[n]=a n u[n], a>1, does not exist, since x[n] is not absolutely summable. 2. But x[n]r n is absolutely summable, if r>a, i.e. ROC, so the z-transform of x[n], which is the DTFT of x[n]r n, does exist.

7 The z-Plane, Poles, and Zeros

 To represent z= rej graphically in terms of complex plane  Horizontal axis of z-plane = real part of z; pole; zero  vertical axis of z-plane = imaginary part of z.

 Relation between DTFT and z-transform: j e  z| j    ze . the DTFT is given by the z-transform evaluated on the unit circle The frequency  in the DTFT corresponds to the point on the unit circle at an angle  with respect to the positive real axis  z-transform X(z): ~ M b  b z   b z M b 1 c z 1  z  0 1 1  M . z  k 1 k . 1 N N 1 a0  a1 z aN z 1 d z k 1 k

ck = zeros of X(z); dk = poles of X(z) 8 bba 00/ gain factor Example 7.2 Right-Sided Signal Determine the z-transform of the signal xn  nu n . Depict the ROC and the location of poles and zeros of X(z) in the z-plane. n nn  By definition, we have zunz . nn 0 z This is a geometric series of infinite length in the ratio α/z; X(z) converges if |α/z| < 1, or the ROC is |z| > |α|. And, 1 zz,  1z1 z ,.z z    There is a pole at z = α and a zero at z = 0 

Right-sided signal  the ROC is |z| > |α|. 9 Example 7.3 Left-Sided Signal Determine the z-transform of the signal yn  nu n 1. Depict the ROC and the location of poles and zeros of Y(z) in the z-plane.  1 n n n   By definition, we have Yz   u n 1 z      n n  z  k k   z    z      1   . k 1   k 0   Y(z) converges if |z/α| < 1, or the ROC is |z| < |α|. And, 1 Yz1,, z  1 z 1 z , z  z    There is a pole at z = α and a zero at z = 0 Left-sided signal  the ROC is |z| < |α|. Examples 7.2 & 7.3 reveal that the same z-transform but different ROC. 10 This ambiguity occurs in general with signals that are one sided Properties of the ROC

 1. The ROC cannot contain any poles If d is a pole, then |X(d)| = , and the z-transform does not converge at the pole  2. The ROC for a finite-duration x[n] includes the entire z-plane, except possibly z=0 or |z|= n2 For finite-duration x[n], we might suppose that z   xn z n . X(z) will converge, if each term of x[n] is finite. nn1 1) If a signal has any nonzero causal components, then the expression for X(z) will  1 have a term involving z for n2 > 0, and thus the ROC cannot include z = 0. 2) If a signal has any nonzero noncausal components, then the expression for X(z)

will have a term involving z for n1 < 0, and thus the ROC cannot include |z| = .

 3. x[n]=c[n] is the only signal whose ROC is the entire z-plane

n2 Consider z   xn z n . nn1 If n2  0, then the ROC will include z = 0.

If a signal has no nonzero noncausal components (n1  0), then the ROC will 11 include |z| =  . Properties of the ROC

 4. For the infinite-duration signals, then   n zxnzxnzxnznn    . nn  n 

The condition for convergence is |X(z)| < . We may write X ()zzz     That is, we split the infinite sum into negative- and positive-term portions: 1  n n and   z   xn z   z   xn z . n n0

Note that If I(z) and I+(z) are finite, then |X(z)| is guaranteed to be finite, too. n A signal that satisfies these two bounds grows no faster than (r+) for positive n n and (r) for negative n. That is, n x[n]  A (r ) , n  0 (7.9)

n x[n]  A (r ) , n  0 (7.10)

12 If the bound given in Eq. (7.9) is satisfied, then k =  n n k 11n  n r z zA rz    A   A  nnk zr 1 

I(z) converges if and only if |z| < r. If the bound given in Eq. (7.10) is satisfied, then n n n r zA rz    A   I+(z) converges if and only if |z| > r+. nn00z

Hence, if r+ <|z|  |

Note that if r+ > r , then the ROC = 

For signals x[n] satisfy the exponential bounds of Eqs. (7.9) and (7.10) , we have

(1). The ROC of a right-sided signal is of the form |z| > r+.

(2). The ROC of a left-sided signal is of the form |z| < r.

(3). The ROC of a two-sided signal is of the form r+ < |z|  |

13 A right-sided signal has an ROC of the form |z| > r+.

A left-sided signal has an ROC of the form |z| < r–.

A two-sided signal has an ROC of the form r < |z| < r . 14 + – Example 7.5 Identify the ROC associated with z-transform for each of the following signal: nn n x nunun1/2   2 1/4  ； yn  1/2 un   2(1/4)n un ； wn   1/2n u n 2(1/4)n u n .

nnn k 0 11    1 1. zz 222.      nnkn24zz 00    0  4 z The first series converge for |z|<1/2, while the second converge for |z|>1/4. So, the ROC is 1/4 < |z| < 1/2. Hence, 1 2z z   , 1 Poles at z = 1/2 and z = 1/4 1 2z z  4 n n 2.   1   1  Yz     2  . n0  2z  n9  4z  The first series converge for |z| >1/2, while the second converge for |z| > 1/4. Hence, the ROC is |z| > 1/2, and z 2z Yz   , Poles at z = 1/2 and z = 1/4 15 1 1 z  2 z  4 00nn 11  kk 3. Wz2224,   z z  nnkk24zz   00  The first series converge for |z| < 1/2, while the second converge for |z| < ¼. So, the ROC is |z| < 1/4, and 1 2 W z   , Poles at z = 1/2 and z = 1/4 1 2z 1 4z

. This example illustrates that the ROC of a two-side signal (a) is a ring, i.e. in between the poles, the ROC of a right sided signal (b) is the exterior of a circle, and the ROC of a left-sided signal (c) is the interior of a circle. In each case the poles define the boundaries of the ROC. 16 Outline

17 Properties of the z-Transform

 Most properties of the z-transform are analogous to those of the DTFT.  Z Assume that xn  z, with ROC Rx Z yn  Y z, with ROC Ry  Linearity: Z ax n bynazbYz    , with ROC at least RRx  y The ROC can be larger than the intersection if one or more terms in x[n] or y[n] cancel each other in the sum. z  Time Reversal: x[n] X (1/ z), with ROC 1/ Rx

1 Time reversal, or reflection, corresponds to replacing z by z . Hence, if Rx is of the form a < |z| < b, the ROC of the reflected signal is a < 1/|z| < b, or 1/b < |z| < 1/a.  Time Shift:

Z n0 xn  n0  z z, with ROC Rx ,except possibly z  0 or z  n 1. Multiplication by z o introduces a pole of order no at z = 0 if no > 0. n 2. If no < 0, then multiplication by z o introduces no poles at . If they are not canceled 18 n by zeros at infinity in X(z), then the ROC of z oX(z) cannot include |z| < . Properties of the z-Transform

 Multiplication by an Exponential Sequence:

n z z xn, with ROC Rx  Is a complex number  1.If X(z) contains a factor (1dz 1) in the denominator, so that d is pole, then X(z/) has a factor (1dz 1) in the denominator and thus has a pole at d. 2. If c is a zero of X(z), then X(z/) has a zero at c. The poles and zeros of X(z) have their radii changed by || in X(z/), as well as their angles are changed by arg{} in X(z/)

X(z) X(z/)

19 Properties of the z-Transform

Z  Convolution: xn  yn   zY  z, with ROC at least Rx  Ry Convolution of time-domain signals corresponds to multiplication of z-transforms.

The ROC may be larger than the intersection of Rx and Ry if a pole-zero cancellation occurs in the product X(z)Y(z).

 Differentiation in the z-Domain: d nx nZ  z  z,withROC R dz x Multiplication by n in the time domain corresponds to differentiation with respect to z and multiplication of the result by  z in the z-domain. This operation does not change the ROC.

20 Example 7.6 Suppose nn 13  z xn un u n1,Z  z  and   1 3 22  zz22 nn 11   1 z yn un unZ  Y z 4 ,   11 42  zz42 Evaluate the z-transform of ax[n] + by[n]. zz 1 ax n by n Z  a b 4 . 1113 zz22 zz  42 In general, the ROC is the intersection of individual ROCs However, when a = b: We see that the term (1/2)n u[n] has be canceled in ax[n] + by[n].  5 z az  aY z  a 4 . 1 3 z  4 z  2  The pole at z=1/2 is canceled !!  the ROC enlarges 21 Example 7.7 nn 11  Find the z-transform of the signal xn n un u n,    24  n 11z z First, we know that un, with ROC z  1 22z  2 Apply the z-domain differentiation property, we have n 11z dz wn n un  W z z, with ROC z   1 22dz z  2 n  11Z z Next, we know that un[] , withROC z 41/44z  Apply the time-reversal property, we have 1 11 ynZ  Y z z , with ROC  11 z  4 z 4, Last, we apply the convolution property to obtain X(z), i.e.

Z xn  wn  ynzWzYzRR      ,withROC w  y 21z Xz() 2 , withROC z 4 zz4 1 2 22 2 Example 7.8 n Find the z-transform of xn a cos0 nun, where a is real and positive. 1 Let y[n] = αnu[n]. Then we have the z-transform Yz ,withROC z a . 1 az1 Now we rewrite x[n] as the sum 1 1 xn  e j0n yn  e  j0n yn 2 2 Then, apply the property of multiplication by a complex exponential, we have 11 zYezYezjj00   ,withROC za  22 1 1 1 1   2 1 ae j0 z 1 2 1 ae j0 z 1 1 1 ae  j0 z 1 1 ae j0 z 1      j0 1  j0 1  2  1 ae z 1 ae z  1cosaz  1  0 ,withROCza . 122 12cosazaz0 23 Inversion of the z-Transform

 Direct evaluation of the inversion integral for inversion of the z-transform requires the complex variable theory  We apply the method of partial-fraction expression, based on z-transform pairs and z-transform properties, to inverse transform.  The inverse transform can be obtained by expressing X(z) as a sum of terms for which we already known the time function, which relies on the property of the ROC  A right-/left- sided time signal has an ROC that lies outside/inside the pole radius Bz 1 M a rational function of z1   bbz01 bzM  Suppose that z  Azaaz1 azN M < N 01  N ~ M N B(z) If M  N, we may use long division to express X(z) as k X (z)   fk z  k0 A(z) 1. Using the time-shift property and the pair 1[]Z  n We obtain M N M N f [n  k]z f zk 24  k  k k0 k0 Inversion by Partial-Fraction Expansion 2. Factor the denominator polynomial as a product of pole factors to obtain 1 M b0  b1 z    bM z z  N , a 1 d z 1 M < N 0 k 1 k

 Case I, If all poles dk are distinct: N A n A z  k . z k  1 Akkdun1 ,withROC zd k k 1 1 d k z 1 dzk

n z Ak or Akkdun1 1 , with ROC zdk 1 dzk

 Case II: If a pole di is repeated r times: A A A i1 i2 ir 1 , 2 ,, r 1 d z 1 1 1 1 di z 1 di z

1. If the ROC is of the form |z| > di, then the right-sided inverse z-transform is chosen:

nnm11  n Z A A duni ,withROC zdi 1 m 25 m 1! 1 dz i Appendix B !! Inversion by Partial-Fraction Expansion

 Case II: If a pole di is repeated r times: 2. If the ROC is of the form |z| < di, then the left-sided inverse z-transform is chosen:

nnm11  n z A A duni 1,withROC  zdi m 1! 1 m 1 dzi

 Example 7.9 Find the inverse z-transform of 1zz12 zz, with ROC 1  2 1 111 11212 zzz By partial fraction expansion, we obtain 1 2 2 z    . 1 1 1 1 1 2 z 1 2z 1 z right-sided left-sided n  1  n 26 xn    un  22 u n 1  2u n .  2  Example 7.10 zzz3210 4 4 Find the inverse z-transform of  zz, with ROC 1 224zz2  First, convert X(z) into a ratio of polynomials in z 1 1 110z 1  4z 2  4z 3  1 z  z , zW (z)  1 2  2  1 z  2z  2 Since 3>2, 110z 1  4z 2  4z 3  5z 1  2  5z 1  2  2z 1  3   2z 1  3  . 1 z 1  2z 2 1 z 1  2z 2 1 z 1 1 2z 1 13 Wz2 z1  3   , with ROC z  1 112zz11  left-sided ww[nz]  2 n 1 3 n 1n u n 1 32n u n 1. 1 Finally, apply the time-shift property to obtain xn  wn 1  2 3 1 27 xn   n   n 1  1 n1 u n  2  32 n u n  2 . 2 2 Remarks

 Causality, stability, or the existence of the DTFT is sufficient to determine the inverse transform

 Causality:  If a signal is known to be causal, then the right-sided inverse transforms are chosen  Stability:  If a signal is stable, then it is absolutely summable and has a DTFT. Hence, stability and the existence of the DTFT are equivalent conditions. In both cases, the ROC includes the unit circle in the z-plane, |z| = 1. The inverse z- transform is determined by comparing the locations of the poles with the unit circle.  If a pole is inside the unit circle, then the right-sided inverse z-transform is chosen; if a pole is outside the unit circle, then the left-sided inverse z- transform is chosen

28 Inversion by Power Series Expansion

 Only one-sided signal is applicable!  Express X(z) as a power series in z1 or in z. 1. If the ROC is |z| < a, then we express X(z) as power series in z 1, so that we obtain a right-sided signal. 2. If the ROC is |z| > a, then we express X(z) as power series in z, so that we obtain a left-sided signal. 21 z1  Ex 7.11 Find the inverse z-transform of  zz,withROC 12 1 z1 1 2 1 3 2 2  2z  z  2 z  ... 1 1 1 1 2 z 2  z 1 2 1 3 1 X z  2  2z  z  z  ... 2  z 2 1 1 2z xn 2 n  2 n 1  n  2 2  n  3 ... 1 2 2z  z This may not lead to a closed-form expression !! z 2

2 1 3 z  2 z 29 1 3 2 z Example 7.12 An advantage of the power series approach is the ability to find inverse z-transforms for signals that are not a ratio of polynomials in z.

2 Find the inverse z-transform of Xz( ) ez , with ROC all z except z

 k a a Using the power series representation for ea: e   k 0 k!  (z2 )k z2 (z2 )2 X (z)    1   k0 k! 1! 2! [n  2] [n  4] Thus x[n]  [n]    1! 2!   0, n  0 or n odd  1 x[n]   , otherwise  n ( 2 ) !

30 Outline

31 The Transfer Function

x[n] = zn y[n] = x[n]  h[n] LTI system, h[n] z = rej  k  Recall that the transfer function H(z) of an LTI system is Hz  hk z k  If we take the z-transform of both sides of y[n], then Y z Yz  HzX z Hz  This definition applies only at X z values of z for which X[z]  0

 M  From difference equation: ak yn  k  bk x n  k  K 0 K 0 After Substituting zn for x[n] and znH(z) for y[n], we obtain rational transfer function

M k ~ N M b z b  M (1 c z 1 ) z n a z k H (z) z n b z k H (z)  k0 kH (z)  k 1 k  k  k N k N 1 K 0 K 0 a z  k1 (1 d k z ) k0 k 32 ~ Knowledge of the poles dk, zeros ck, and factor b  b 0 a 0 completely determine the system Example 7.13

Find the transfer function and impulse response of a causal LTI system if the input to the n system is xn  (1/ 3) un and the output is yn  3(1) n un (1/ 3) n un The z-transforms of the input and output are respectively given by 1 3 1 X (z)  , ROC z  1 and Y (z)   , ROC z  1 1 (1 3)z1 3 1 z1 1 (1 3)z1 Hence, the transfer function is

4(1 (1 3)z1) H (z)  , ROC z  1 (1 z1)(1 (1 3)z1) The impulse response of the system is obtain by finding the inverse z-transform of H(z). Applying a partial fraction expansion to H(z) yields 22 Hz()  , withROC z 1 11(1/3)zz11

hn 2(1) n un 2(1/ 3) n u n 33 Examples 7.14 & 7.15 Determine the transfer function and the impulse response for the causal LTI system described by yn  (1/ 4)yn 1 (3/ 8)yn 1 xn 2xn 1 We first obtain the transfer function by taking the z-transform: 12z1  2 1 Hz() H (z)   1(1/4)zz12 (3/8) 1 (1/ 2)z 1 1 (3/ 4)z 1 The system is causal, so we choose the right-side inverse z-transform for each term to obtain the following impulse response: hn 2(1/ 2) n un (3/ 4) n un Find the difference-equation description of an LTI system with transfer function 5z  2 H (z)  2 z  3z  2 5z 1  2z 2 We rewrite H(z) as a ratio of polynomials in z 1: H (z)  (1 3z 1  2z 2 )

yn 3y[n 1]  2y[n  2]  5xn 1 2xn  2

34 Causality and Stability

 The impulse response h(t) is the inverse LT of the transfer function H(z)  Causality Pole dk inside the unit right-sided circle, i.e., |d | < 1 Exponentially decaying term k inverse z-transform

Exponentially increasing term  Stability Pole dk inside the unit circle, i.e., |d | < 1 the ROC includes k unit circle in z-plane Causal system

Pole dk outside the unit circle, i.e., |dk| > 1 35 Non-caucal system Causal and Stable LTI System

 To obtain a unique inverse transform of H(z), we must know the ROC or have other knowledge(s) of the impulse response  The relationships between the poles, zeros, and system characteristics can provide some additional knowledges  Systems that are stable and causal must have all their poles inside the unit circle of the z-plane:

36 Example 7.16  2 2 3 An LTI system has the transfer function H (z)     1 j  j 1 2z 1 0.9e 4 z 1 1 0.9e 4 z 1 Find the impulse response, assuming that the system is (a) stable, or (b) causal. (c) Can this system both stable and causal? a. If the system is stable, then the ROC includes the unit circle. The two conjugate poles inside the unit circle contribute the right-sided term  to the impulse response, while the pole outside the unit circle contributes a left-sided term.  j  j h(n)  2(0.9e 4 ) n u[n]  2(0.9e 4 ) n u[n]  3(2) n u[n 1]

n  n  4(0.9) cos( n)u[n]  3(2) u[n 1] 4 b. If the system is causal, then all poles contribute right-sided terms to the impulse response.  j  j h(n)  2(0.9e 4 ) n u[n]  2(0.9e 4 ) n u[n]  3(2) n u[n]   4(0.9) n cos[ n]u[n]  3(2) n u[n] 4 c. the37 LTI system cannot be both stable and causal, since there is a pole outside the unit circle. Example 7.17 The first-order recursive equation y[n]  y[n 1]  x[n] may be used to describe the value y[n] of an investment by setting ρ=1+r/100, where r is the interest rate per period, expressed in percent. Find the transfer function of this system and determine whether it can be both stable and causal. 1 The transfer function is determined by using z-transform: H (z)  1 z 1

This LTI system cannot be both stable and causal, because the pole at z= ρ >1

38 Inverse System

 Given an LTI system with impulse response h[n], the impulse response of the inverse system, h inv[n], satisfies the condition hinv n* hn  n inv 1 H (z)H (z)  1 or H inv (z)  H (z) the poles of the inverse system Hinv(z) are the zeros of H(z), and vice versa a stable and causal inverse system exists only if all of the zeros of H(z) are inside the unit circle in the z-plane. A (stable and causal) H(z) has all of its poles and zeros inside the unit circle H(z) is minimum phase.

A nonminimum-phase system cannot have a stable and causal inverse system. 39 Example 7.18 An LTI system is described by the difference equation 1 1 1 yn  y[n 1]  y[n  2]  xn  x[n 1]  x[n  2] 4 4 8 Find the transfer function of the inverse system. Does a stable and causal LTI inverse system exist? 1 1 z 1  1 z 2 The transfer function of the given system is H (z)  2 8 1 1 2 1 z  4 z (1 1 z 1 )(1 1 z 1 )  4 2 1 1 2 (1 2 z ) Hence, the inverse system then has the transfer function Both of the poles of the inverse system, i.e. z=1/4 and (1 1 z1)2 H inv (z)  2 z=1/2, are inside the unit circle. The inverse system 1 1 1 1 (1 4 z )(1 2 z ) can be both stable and causal. Note that this system is also minimum phase, since all zeros and poles of the system are inside the unit circle. 40 Example 7.19

Recall a two-path communication channel is described by yn x[n]  ax[n 1] Find the transfer function and difference-equation description of the inverse system. What must the parameter a satisfy for the inverse system to be stable and causal? First find the transfer function of the two-path multiple channel system: H (z)  1 az 1 1 1 Then, the transfer function of the inverse system is H inv (z)   H (z) 1 az 1

 The corresponding difference-equation representation is yn ayn 1 xn

The inverse system is both stable and causal when |a| < 1.

41 Outline