Chapter 7: the Z-Transform

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]znk 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 yn H ze j z z n . Let z= rej yn H re j r n cosn re j j H re j r n sinn 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 xnz z z xn z n , n Definition: The z-transform of x[n]: 1 Definition: The inverse z-transform of X(z): xn z z n1dz. 2j 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 ) z-transform is the DTFT of h[n]r n h[n]rn e jn Hence n 1 j j n n 1 j jn h[n] H (re )(re ) d hn r H re e d 2 2 z= rej 1 n1 dz = jrej d H (z)z dz 6 2j d = (1/j)z 1dz Convergence of Laplace Transform z-transform is the DTFT of x[n]rn A necessary condition for convergence of the z-transform is the absolute summability of x[n]rn : xn 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 ze . 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 M b 1 c z 1 b0 b1 z1 bM z k 1 k z . z . 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 xn nu n . Depict the ROC and the location of poles and zeros of X(z) in the z-plane. <Sol.> 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, 1z1 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 yn nu n 1. Depict the ROC and the location of poles and zeros of Y(z) in the z-plane. <Sol.> 1 n n n By definition, we have Yz 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 Yz1,, 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 xn z n . X(z) will converge, if each term of x[n] is finite. nn1 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 xn z n . nn1 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 zxnzxnzxnznn . 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 xn z z xn z . n n0 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+. nn00z Hence, if r+ <|z| |<r, then both I+(z) and I(z) converge and |X(z)| also converges. 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| |<r. 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 nunun1/2 2 1/4 ; yn 1/2 un 2(1/4)n un ; n wn 1/2 u n 2(1/4)n u n . <Sol.> nnn k 0 11 1 1. zz 222. nnkn24zz 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 Yz 2 . n0 2z n9 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 Yz , Poles at z = 1/2 and z = 1/4 15 1 1 z 2 z 4 00nn 11 kk 3. Wz2224, z z nnkk24zz 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 .

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    42 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us