Signals and Systems - Electrical Engineering

9.5 One-Sided Z-Transform Inverse 543

nExample 9.13

Find the inverse Z-transform of

X(z)=

1

1 + 2 z−^2

|z|>

√

2

Solution

We can perform the long division to find thex[n] values, or equivalently let

X(z)=x[0]+x[1]z−^1 +x[2]z−^2 +···

and find the{x[n]}samples so that the productX(z)( 1 + 2 z−^2 )=1. Thus,

1 =( 1 + 2 z−^2 )(x[0]+x[1]z−^1 +x[2]z−^2 +···)

=x[0]+x[1]z−^1 +x[2]z−^2 +x[3]z−^3 +···

+ 2 x[0]z−^2 + 2 x[1]z−^3 +···

and comparing the terms on the two sides of the equality gives

x[0]= 1

x[1]= 0

x[2]+ 2 x[0]= 0 ⇒x[2]=− 2

x[3]+ 2 x[1]= 0 ⇒x[3]= 0

x[4]+ 2 x[2]= 0 ⇒x[4]=(− 2 )^2

..

.

So the inverse Z-transform isx[0]=1 and x[n]=(− 2 )log^2 (n)for n>0 and even, and zero

otherwise. Notice that this sequence grows asn→∞.

Another possible way to find the inverse is to use the geometric series equation

∑∞

k= 0

αn=

1

1 −α

|α|< 1

with−α= 2 z−^2 (notice that|α|= 2 /|z|^2 <1 or|z|>

√

2, the given ROC). Therefore,

X(z)=

1

1 + 2 z−^2

= 1 +(− 2 z−^2 )^1 +(− 2 z−^2 )^2 +(− 2 z−^2 )^3 +···

but this method is not as general as the long division. n