5.8 THE BILATERAL z[-TRANSFORM](#page-11-0)

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5.8 THE BILATERAL z**-TRANSFORM**

Situations involving noncausal signals or systems cannot be handled by the (unilateral) z-transform discussed so far. Such cases can be analyzed by the bilateral (or two-sided) z-transform defined in Eq. (5.1) as

$$X[z] = \sum_{n=-\infty}^{\infty} x[n]z^{-n}$$

As in Eq. (5.2), the inverse z-transform is given by

$$x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz$$

These equations define the bilateral z-transform. Earlier, we showed that

$$\gamma^n u[n] \Longleftrightarrow \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.51}$$

In contrast, the z-transform of the signal −γ nu[−(n+1)], illustrated in Fig. 5.27a, is

$$\mathcal{Z}\{-\gamma^n u[-(n+1)]\} = \sum_{-\infty}^{-1} -\gamma^n z^{-n} = \sum_{-\infty}^{-1} -\left(\frac{\gamma}{z}\right)^n$$ $$= -\left[\frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right]$$ $$= 1 - \left[1 + \frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right]$$ $$= 1 - \frac{1}{1 - \frac{z}{\gamma}} \qquad \left|\frac{z}{\gamma}\right| < 1$$ $$= \frac{z}{z - \gamma} \qquad |z| < |\gamma|$$

Figure 5.27 (a) ^−γ nu[−(n+1)] and (b) the region of convergence (ROC) of its ^z-transform.

Therefore,

$$\mathcal{Z}\{-\gamma^{n}u[-(n+1)]\} = \frac{z}{z-\gamma} \qquad |z| < |\gamma| \tag{5.52}$$

A comparison of Eqs. (5.51) and (5.52) shows that the z-transform of γ nu[n] is identical to that of −γ nu[−(n + 1)]. The regions of convergence, however, are different. In the former case, X[z] converges for |z| > |γ |; in the latter, X[z] converges for |z| < |γ | (see Fig. 5.27b). Clearly, the inverse transform of X[z] is not unique unless the region of convergence is specified. If we add the restriction that all our signals be causal, however, this ambiguity does not arise. The inverse transform of z/(z − γ ) is γ nu[n] even without specifying the ROC. Thus, in the unilateral transform, we can ignore the ROC in determining the inverse z-transform of X[z].

As in the case of the bilateral Laplace transform, if ^x[n] = %^{k i}=¹ xi[n], then the ROC for X[z] is the intersection of the ROCs (region common to all ROCs) for the transforms X1[z],X2[z],…,Xk[z].

The preceding results lead to the conclusion (similar to that for the Laplace transform) that if z = β is the largest magnitude pole for a causal sequence, its ROC is |z| > |β|. If z = α is the smallest magnitude nonzero pole for an anticausal sequence, its ROC is |z| < |α|.

REGION OF CONVERGENCE FOR LEFT-SIDED AND RIGHT-SIDED SEQUENCES

Let us first consider a finite duration sequence xf[n], defined as a sequence that is nonzero for N¹ ≤ n ≤ N2, where both N¹ and N² are finite numbers and N² > N1. Also,

$$X_f[z] = \sum_{n=N_1}^{N_2} x_f[n]z^{-n}$$

For example, if N¹ = −2 and N² = 1, then

$$X_f[z] = x_f[-2]z^2 + x_f[-1]z + x_f[0] + \frac{x_f[1]}{z}$$

Assuming all the elements in xf[n] are finite, we observe that Xf[z] has two poles at z = ∞ because of terms xf[−2]z² +xf[−1]z and one pole at z = 0 because of term xf[1]/z. Thus, a finite-duration sequence could have poles at z = 0 and z = ∞. Observe that Xf[z] converges for all values of z except possibly z = 0 and z = ∞.

This means that the ROC of a general signal x[n] + xf[n] is the same as the ROC of x[n] with the possible exception of z = 0 and z = ∞.

A right-sided sequence is zero for n < N² < ∞ and a left-sided sequence is zero for n > N¹ > −∞. A causal sequence is always a right-sided sequence, but the converse is not necessarily true. An anticausal sequence is always a left-sided sequence, but the converse is not necessarily true. A two-sided sequence is of infinite duration and is neither right-sided nor left-sided.

A right-sided sequence xr[n] can be expressed as xr[n] = xc[n] +xf[n], where xc[n] is a causal signal and xf[n] is a finite-duration signal. Therefore, the ROC for xr[n] is the same as the ROC for xc[n] except possibly z = ∞. If z = β is the largest magnitude pole for a right-sided sequence xr[n], its ROC is |β| < |z|≤∞. Similarly, a left-sided sequence can be expressed as xl[n] = xa[n]+xf[n], where xa[n] is an anticausal sequence and xf[n] is a finite-duration signal. Therefore, the ROC for xl[n] is the same as the ROC for xa[n] except possibly z=0. Thus, ifz=α is the smallest magnitude nonzero pole for a left-sided sequence, its ROC is 0 ≤ |z| < |α|.

EXAMPLE 5.17 Bilateral z**-Transform**

Determine the bilateral z-transform of

$$x[n] = \underbrace{(0.9)^n u[n]}_{x_1[n]} + \underbrace{(1.2)^n u[-(n+1)]}_{x_2[n]}$$

From the results in Eqs. (5.51) and (5.52), we have

$$X_1[z] = \frac{z}{z - 0.9} \qquad |z| > 0.9$$ $$X_2[z] = \frac{-z}{z - 1.2} \qquad |z| < 1.2$$

The common region where both X1[z] and X2[z] converge is 0.9 < |z| < 1.2 (Fig. 5.28b). Hence,

$$X[z] = X_1[z] + X_2[z]$$

= $\frac{z}{z - 0.9} - \frac{z}{z - 1.2}$
= $\frac{-0.3z}{(z - 0.9)(z - 1.2)}$ 0.9 < |z| < 1.2

The sequence x[n] and the ROC of X[z] are depicted in Fig. 5.28.

EXAMPLE 5.18 Inverse Bilateral z**-Transform**

Find the inverse bilateral z-transform of

$$X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)}$$

if the ROC is (a) |z| > 2, (b) |z| < 0.8, and (c) 0.8 < |z| < 2.

(a)

$$\frac{X[z]}{z} = \frac{-(z+0.4)}{(z-0.8)(z-2)} = \frac{1}{z-0.8} - \frac{2}{z-2}$$

and

$$X[z] = \frac{z}{z - 0.8} - 2\frac{z}{z - 2}$$

Since the ROC is |z| > 2, both terms correspond to causal sequences and

$$x[n] = [(0.8)^n - 2(2)^n]u[n]$$

This sequence appears in Fig. 5.29a.

(b) In this case, |z| < 0.8, which is less than the magnitudes of both poles. Hence, both terms correspond to anticausal sequences, and

$$x[n] = [-(0.8)^n + 2(2)^n]u[-(n+1)]$$

This sequence appears in Fig. 5.29b.

(c) In this case, 0.8 < |z| < 2; the part of X[z] corresponding to the pole at 0.8 is a causal sequence, and the part corresponding to the pole at 2 is an anticausal sequence:

$$x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)]$$

This sequence appears in Fig. 5.29c.

DR ILL 5.23 Inverse Bilateral z**-Transform**

Find the inverse bilateral z-transform of

$$X[z] = \frac{z}{z^2 + \frac{5}{6}z + \frac{1}{6}} \qquad \frac{1}{2} > |z| > \frac{1}{3}$$

ANSWER

− 1 3 n u[n] +6 − 1 2 n u[−(n+1)]

INVERSE TRANSFORM BY EXPANSION OF X[z] IN POWER SERIES OF z We have

$$X[z] = \sum_{n} x[n]z^{-n}$$

For an anticausal sequence, which exists only for n ≤ −1, this equation becomes

$$X[z] = x[-1]z + x[-2]z^{2} + x[-3]z^{3} + \cdots$$

We can find the inverse z-transform of X[z] by dividing the numerator polynomial by the denominator polynomial, both in ascending powers of z, to obtain a polynomial in ascending powers of z. Thus, to find the inverse transform of z/(z − 0.5) (when the ROC is |z| < 0.5), we divide z by −0.5+z to obtain −2z−4z2−8z3−· · ·. Hence, x[−1]=−2, x[−2]=−4, x[−3]=−8, and so on.

5.8-1 Properties of the Bilateral z**-Transform**

Properties of the bilateral z-transform are similar to those of the unilateral transform. We shall merely state the properties here, without proofs, for xi[n] ⇐⇒ Xi[z].

LINEARITY

$$a_1x_1[n]+a_2x_2[n] \Longleftrightarrow a_1X_1[z]+a_2X_2[z]$$

The ROC for a1X1[z] + a2X2[z] is the region common to (intersection of) the ROCs for X1[z] and X2[z].

SHIFT

x[n−m] ⇐⇒ 1 zm ^X[z] ^m is positive or negative integer

The ROC for X[z]/zm is the ROC for X[z] except for the addition or deletion of z = 0 or z = ∞ caused by the factor 1/z^m.

CONVOLUTION

$$x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z]$$

The ROC for X1[z]X2[z] is the region common to (intersection of) the ROCs for X1[z] and X2[z].

MULTIPLICATION BY γ ⁿ

$$\gamma^n x[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right]$$

If the ROC for X[z] is |γ1| < |z| < |γ2|, then the ROC for X[z/γ ] is |γ γ1| < |z| < |γ γ2|, indicating that the ROC is scaled by the factor |γ |.

MULTIPLICATION BY n

$$nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z]$$

The ROC for −z(dX/dz) is the same as the ROC for X[z].

TIME REVERSAL

$$x[-n] \Longleftrightarrow X[1/z]$$

If the ROC for X[z] is |γ1| < |z| < |γ2|, then the ROC for X[1/z] is 1/|γ1| > |z| > |1/γ2|.

COMPLEX CONJUGATION

x∗[n] ⇐⇒ X∗[z ∗]

The ROC for X∗[z∗] is the same as the ROC for X[z].

5.8-2 Using the Bilateral z**-Transform for Analysis of LTID Systems**

Because the bilateral z-transform can handle noncausal signals, we can use this transform to analyze noncausal linear systems. The zero-state response y[n] is given by

$$y[n] = \mathcal{Z}^{-1}\{X[z]H[z]\}$$

provided X[z]H[z] exists. The ROC of X[z]H[z] is the region in which both X[z] and H[z] exist, which means that the region is the common part of the ROC of both X[z] and H[z].

EXAMPLE 5.19 Zero-State Response by Bilateral z**-Transform**

For a causal system specified by the transfer function

$$H[z] = \frac{z}{z - 0.5}$$

find the zero-state response to input

$$x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)]$$ $$X[z] = \frac{z}{z - 0.8} - \frac{2z}{z - 2} = \frac{-z(z + 0.4)}{(z - 0.8)(z - 2)}$$

The ROC corresponding to the causal term is |z| > 0.8, and that corresponding to the anticausal term is |z| < 2. Hence, the ROC for X[z] is the common region, given by 0.8 < |z| < 2. Hence,

$$X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)} \qquad 0.8 < |z| < 2$$

Therefore,

$$Y[z] = X[z]H[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)}$$

Since the system is causal, the ROC of H[z] is |z| > 0.5. The ROC of X[z] is 0.8 < |z| < 2. The common region of convergence for X[z] and H[z] is 0.8 < |z| < 2. Therefore,

$$Y[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)} \qquad 0.8 < |z| < 2$$

Expanding Y[z] into modified partial fractions yields

$$Y[z] = -\frac{z}{z - 0.5} + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) - \frac{8}{3} \left( \frac{z}{z - 2} \right) \qquad 0.8 < |z| < 2$$

Since the ROC extends outward from the pole at 0.8, both poles at 0.5 and 0.8 correspond to causal sequence. The ROC extends inward from the pole at 2. Hence, the pole at 2 corresponds to anticausal sequence. Therefore,

$$y[n] = \left[ -(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] + \frac{8}{3}(2)^n u[-(n+1)]$$

EXAMPLE 5.20 Zero-State Response for an Input with No z**-Transform**

For the system in Ex. 5.19, find the zero-state response to input

$$x[n] = \underbrace{(0.8)^n u[n]}_{x_1[n]} + \underbrace{(0.6)^n u[-(n+1)]}_{x_2[n]}$$

The z-transforms of the causal and anticausal components x1[n] and x2[n] of the output are

$$X_1[z] = \frac{z}{z - 0.8} \qquad |z| > 0.8$$ $$X_2[z] = \frac{-z}{z - 0.6} \qquad |z| < 0.6$$

Observe that a common ROC for X1[z] and X2[z] does not exist. Therefore, X[z] does not exist. In such a case we take advantage of the superposition principle and find y1[n] and y2[n], the system responses to x1[n] and x2[n], separately. The desired response y[n] is the sum of y1[n] and y2[n]. Now

$$H[z] = \frac{z}{z - 0.5}$$

$|z| > 0.5$
\n

$$Y_1[z] = X_1[z]H[z] = \frac{z^2}{(z - 0.5)(z - 0.8)}$$

$|z| > 0.8$
\n

$$Y_2[z] = X_2[z]H[z] = \frac{-z^2}{(z - 0.5)(z - 0.6)}$$

$0.5 < |z| < 0.6$

Expanding Y1[z] and Y2[z] into modified partial fractions yields

$$Y_1[z] = -\frac{5}{3} \left( \frac{z}{z - 0.5} \right) + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) \qquad |z| > 0.8$$ $$Y_2[z] = 5 \left( \frac{z}{z - 0.5} \right) - 6 \left( \frac{z}{z - 0.6} \right) \qquad 0.5 < |z| < 0.6$$

Therefore,

$$y_1[n] = \left[ -\frac{5}{3}(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n]$$ $$y_2[n] = 5(0.5)^n u[n] + 6(0.6)^n u[-(n+1)]$$

and

$$y[n] = y_1[n] + y_2[n] = \left[\frac{10}{3}(0.5)^n + \frac{8}{3}(0.8)^n\right]u[n] + 6(0.6)^nu[-(n+1)]$$

DR ILL 5.24 Zero-State Response by Bilateral z**-Transform**

For the causal system in Ex. 5.19, find the zero-state response to input

$$x[n] = \left(\frac{1}{4}\right)^n u[n] + 5(3)^n u[-(n+1)]$$

ANSWER

^{− 1} 4 n +3 1 2 ⁿ u[n] +6(3)nu[−(n+1)]