The z-transform - Continuous and Discrete Time Signals and Systems - page 591

The inverse of Y ( z ) takes two different forms depending on the values of a and
b :

1
(1 − az − 1 ) 2

a = b

Y ( z ) =

1
(1 − az − 1 )(1 − bz − 1 )

a = b .

We consider the two cases separately while calculating the inverse z-transform
of Y ( z ).

Case 1 ( a = b ) From Table 13.1, we know that

− 1
(1 − az

az

←→

ka k u [ k ]

− 1 ) 2 .
Applying the time-shifting property, we obtain

a
(1 − az − 1 ) 2 .

←→

( k + 1) a k + 1 u [ k + 1]

The output response is therefore given by

1
(1 − az − 1 ) 2

− 1

= ( k + 1) a k u [ k + 1] , y [ k ] = Z

which is the same as

y [ k ] = ( k + 1) a k u [ k ] .

Case 2 ( a = b ) Using partial fraction expansion, the function Y ( z ) is expressed
as follows:

1
(1 − az − 1 )(1 − bz − 1 )

A
1 − az − 1

B
1 − bz − 1 ,

Y ( z ) = ≡ + (13.25)

where the partial fraction coefficients are given by

1
1 − bz − 1

a
a − b

A = =

az − 1 = 1

and

1
1 − az

=− b

a − b .
Substituting the values of A and B into Eq. (13.25) and taking the inverse DTFT
yields

B =

− 1

bz − 1 = 1

a k + 1 − b k + 1

a
a − b

b
a − b

1
a − b

a k u [ k ] − b k u [ k ] = y [ k ] = u [ k ] .

Combining case 1 with case 2, we obtain

( k + 1) a k u [ k ]

a = b

y [ k ] = 1

a − b

(13.26)

a k + 1 − b k + 1 u [ k ] a = b ,

which is identical to the result of Example 11.15.