Signals and Systems - Electrical Engineering

9.4 One-Sided Z-Transform 521

RemarksThe uniqueness of the Z-transform requires that the Z-transform of a signal be accompanied by
a region of convergence. It is possible to have identical Z-transforms with different regions of convergence,
corresponding to different signals.

nExample 9.6

Letc[n]=α|n|, 0< α <1, be a discrete-time signal (it is actually an autocorrelation function

related to the power spectrum of a random signal). Determine its Z-transform.

Solution

To find its two-sided Z-transformC(z)we consider its causal and anti-causal components. First,

Z(c[n]u[n])=

∑∞

n= 0

αnz−n=

1

1 −αz−^1

with the region of convergence of|αz−^1 |<1 or|z|> α. For the anti-causal component,

Z(c[−n]u[n])z=

∑∞

n= 0

αnzn=

1

1 −αz

with a region of convergence of|αz|<1 or|z|<| 1 /α|.

Thus, the two-sided Z-transform ofc[n] is (notice that the term forn=0 was used twice in the

above calculations, so we need to subtract it)

C(z)=

1

1 −αz−^1

+

1

1 −αz

− 1 =

z

z−α

−

z

(z− 1 /α)

=

(α− 1 /α)z

(z−α)(z− 1 /α)

with a region of convergence of

|α|<|z|<

∣

∣

∣

∣

1

α

∣

∣

∣

∣

For instance, forα=0.5, we get

C(z)=

−1.5z

(z−0.5)(z− 2 )

0.5<|z|< (^2) n

9.4 One-Sided Z-Transform........................................................................

In most situations where the Z-transform is used the system is causal (its impulse response ish[n]= 0
forn<0) and the input signal is also causal(x[n]=0 forn< 0 ). In such cases the one-sided Z-
transform is very appropriate. Moreover, as we saw before, the two-sided Z-transform can be expressed