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Chapter 7: the Z-Transform

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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 .
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