Two-Sided Z-Transform 517

9.3 Two-Sided Z-Transform 517

Solution

To see the poles and the zeros more clearly let us express X 1 ( z) as a function of positive powers of z:

There are three poles at z = 0, the roots of D 1 ( z) = 0, and the zeros are the roots of N 1 (

Likewise, expressing X 2 ( z) as a function of positive powers of z,

1 − z)(1 + 2z) 2

N 2 ( z)

1+ 2z + z 2 D 2 ( z)

z) = 1 + 2 2z + z = 0, while the zeros of X 2 ( z) are the roots of N 2 (

The poles of X 2 ( z) are the roots of D 2 (

z) = (1 − z)(1 + 2z) 2 = 0. ■ The region of convergence depends on the support of the signal. If it is finite, the ROC is very much

the whole z-plane; if it is infinite, the ROC depends on whether the signal is causal, anti-causal, or noncausal. Something to remember is that in no case does the ROC include any poles of the Z-transform.

ROC of Finite-Support Signals

The ROC of the Z-transform of a signal x[n] of finite support [N 0 ,N 1 ] where −∞ < N 0 ≤n≤N 1 < ∞,

X(z) =

x[n]z −n

n=N 0 is the whole z-plane, excluding the origin z = 0 and/or z = ±∞ depending on N 0 and N 1 .

Given the finite support of x[n] its Z-transform has no convergence problem. Indeed, for any z 6= 0 (or z 6= ±∞ if positive powers of z occur in Equation (9.40)), we have

|X(z)| ≤

|x[n]||z −n | ≤ (N 1 −N 0 + 1) max |x[n]| max |z −n |<∞

n=N 0

The poles of X(z) are either at the origin of the z-plane (e.g., when N 0 ≥ 0) or there are no poles (e.g.,

when N 1 ≤ 0). Thus, only when z = 0 or z = ±∞ would X(z) go to infinity. The ROC is the whole z-plane excluding these values.

C H A P T E R 9: The Z-Transform

■ Example 9.3

Find the Z-transform of a discrete-time pulse

x[n] = 0 otherwise

Determine the region of convergence of X(z).

Solution

The Z-transform of x[n] is

X −10

X(z) =

That this sum equals the term on the right can be shown by multiplying the left term by the denominator 1 − z −1 and verifying the result is the same as the numerator in negative powers of z. In fact,

( 1−z −1 X ) 1z −n

= X 1z −n X − 1z −n−1

− (z −10 +···+z +z ) =1−z Since x[n] is a finite sequence there is no problem with the convergence of the sum, although X(z)

= (1 + z −1 +···+z )

in Equation (9.8) seems to indicate the need for z 6= 1 (z = 1 makes the numerator and denom- inator zero). From the sum, if we let z = 1, then X(1) = 10, so there is no need to restrict z to be different from 1. This is caused by the pole at z = 1 being canceled by a zero. Indeed, the zeros z k

of X(z) (see Eq. 9.8) are the roots of z 10 − 1 = 0, which are z j2π k/10 k =e for k = 0, . . . , 9. Therefore,

the zero when k = 0, or z 0 = 1, cancels the pole at 1 so that

k=1 ( z−e

Q 9 jπ k/5

X(z) =

That is, X(z) has nine poles at the origin and nine zeros around the unit circle except at z = 1. Thus, the whole z-plane excluding the origin is the region of convergence of X(z) .