Signals and Systems - Electrical Engineering

9.3 Two-Sided Z-Transform 517

Solution

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

ofz:

X 1 (z)=

z^3 ( 1 + 2 z−^1 + 3 z−^2 + 4 z−^3 )

z^3

=

z^3 + 2 z^2 + 3 z+ 4

z^3

=

N 1 (z)

D 1 (z)

There are three poles atz=0, the roots ofD 1 (z)=0, and the zeros are the roots ofN 1 (z)=z^3 +

2 z^2 + 3 z+ 4 =0.

Likewise, expressingX 2 (z)as a function of positive powers ofz,

X 2 (z)=

z^3 (z−^1 − 1 )(z−^1 + 2 )^2

z^3 (z−^1 (z−^2 +

√

2 z−^1 + 1 ))

=

( 1 −z)( 1 + 2 z)^2

1 +

√

2 z+z^2

=

N 2 (z)

D 2 (z)

The poles ofX 2 (z)are the roots ofD 2 (z)= 1 +

√

2 z+z^2 =0, while the zeros ofX 2 (z)are the roots

ofN 2 (z)=( 1 −z)( 1 + 2 z)^2 =0. n

The region of convergence depends on the support of the signal. If it is finite, the ROC is very much
the wholez-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 signalx[n]of finite support[N 0 ,N 1 ]where−∞<N 0 ≤n≤N 1 <∞,

X(z)=

∑N^1

n=N 0

x[n]z−n (9.7)

is the wholez-plane, excluding the originz= 0 and/orz=±∞depending onN 0 andN 1.

Given the finite support ofx[n] its Z-transform has no convergence problem. Indeed, for anyz6= 0
(orz6=±∞if positive powers ofzoccur in Equation (9.40)), we have

|X(z)|≤

∑N^1

n=N 0

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

The poles ofX(z)are either at the origin of thez-plane (e.g., whenN 0 ≥0) or there are no poles (e.g.,
whenN 1 ≤0). Thus, only whenz=0 orz=±∞wouldX(z)go to infinity. The ROC is the whole
z-plane excluding these values.