Signals and Systems - Electrical Engineering

9.3 Two-Sided Z-Transform 515

9.3 Two-Sided Z-Transform

Given a discrete-time signalx[n],−∞<n<∞, its two-sided Z-transform is

X(z)=

∑∞

n=−∞

x[n]z−n (9.4)

defined in a region of convergence (ROC) in thez-plane.

Considering the sampled signalx(nTs)a function ofnin Equation (9.3), we obtain the two-sided
Z-transform.

Remarks

n The Z-transform can be thought of as the transformation of the sequence{x[n]}into a polynomial X(z)
(possibly of infinite degree in positive and negative powers of z)where to each x[n 0 ]we attach a monomial
z−n^0. Thus, given a sequence of samples{x[n]}its Z-transform simply consists in creating a polynomial
with coefficients x[n]corresponding to z−n. Given a Z-transform as in Equation (9.4), its inverse is easily
obtained by looking at the coefficients attached to the z−nmonomials for positive as well as negative values
of the sample value n. Clearly, this inverse is not in a closed form. We will see ways to compute these later
in this chapter.
n The two-sided Z-transform is not useful in solving difference equations with initial conditions, just as the
two-sided Laplace transform was not useful either in solving differential equations with initial conditions.
To include initial conditions in the transformation it is necessary to define the one-sided Z-transform.

The one-sided Z-transform is defined for a causal signal,x[n]= 0 forn< 0 , or for signals that are made causal

by multiplying them with the unit-step signalu[n]:

X 1 (z)=Z(x[n]u[n])=

∑∞

n= 0

x[n]u[n]z−n (9.5)

in a region of convergenceR 1.

The two-sided Z-transform can be expressed in terms of the one-sided Z-transform as follows:

X(z)=Z

(

x[n]u[n]

)

+Z

(

x[−n]u[n]

)

|z−x[0] (9.6)

The region of convergence ofX(z)is

R=R 1 ∩R 2

whereR 1 is the region of convergence ofZ

(

x[n]u[n]

)

andR 2 is the region of convergence ofZ

(

x[−n]u[n]

)∣∣

z.

The one-sided Z-transform coincides with the two-sided Z-transform whenever the discrete-time sig-
nalx[n] is causal (i.e.,x[n]=0 forn<0). If the signal is noncausal, multiplying it byu[n] makes it