Signals and Systems - Electrical Engineering

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

n {R 2 :|z|<0.5}—the inside of a circle of radius 0.5, an anti-causal signalx 2 [n] can be associated

withX(z).

n {R 3 : 0.5<|z|< 2 }—a torus of radii 0.5 and 2, a noncausal signalx 3 [n] can be associated with

X(z).

Three different signals can be connected withX(z)by considering three different regions of

convergence. n

nExample 9.5

Find the regions of convergence of the Z-transforms of the following signals:

(a)x 1 [n]=

(

1

2

)n

u[n]

(b)x 2 [n]=−

(

1

2

)n

u[−n−1]

Determine then the Z-transform ofx 1 [n]+x 2 [n].

Solution

The signalx 1 [n] is causal, whilex 2 [n] is anti-causal. The Z-transform ofx 1 [n] is

X 1 (z)=

∑∞

n= 0

(

1

2

)n

z−n=

1

1 −0.5z−^1

=

z

z−0.5

provided that|0.5z−^1 |<1 or that its region of convergence isR 1 :|z|>0.5. The regionR 1 is the

outside of a circle of radius 0.5.

The signalx 2 [n] grows asndecreases from−1 to−∞, and the rest of its values are zero. Its Z-

transform is found as

X 2 (z)=−

∑−^1

n=−∞

(

1

2

)n

z−n=−

∑∞

m= 0

(

1

2

)−m

zm+ 1

=−

∑∞

m= 0

2 mzm+ 1 =

− 1

1 − 2 z

+ 1 =

z

z−0.5

with a region of convergence ofR 2 :|z|<0.5.

Although the signals are clearly different, their Z-transforms are identical. It is the corresponding

regions of convergence that differentiate them. The Z-transform ofx 1 [n]+x 2 [n] does not exist

given that the intersection ofR 1 andR 2 is empty. n