The z-transform - Continuous and Discrete Time Signals and Systems - page 577

can readily be applied to calculate the inverse transform. Otherwise, we would
be missing the factor of z in the numerator of Eq. (13.14a), and application of
Eq. (13.14b) would have been more complicated.

To illustrate the aforementioned procedure (steps (1)-(3)) for evaluating the
inverse z-transform using the partial fraction expansion, we consider the
following example.

Example 13.4
The z-transform of three right-sided functions is given below. Calculate the
inverse z-transform in each case.

z
z 2 − 3 z + 2 ;

(i) X 1 ( z ) =

1
( z − 0 . 1)( z − 0 . 5)( z + 0 . 2) ;

(ii) X 2 ( z ) =

2 z (3 z + 17)
( z − 1)( z 2 − 6 z + 25) .

X 3 ( z ) = (iii)

Solution
(i) The characteristic equation of X 1 ( z ) is given by z 2 − 3 z + 2 = 0, which has
two roots, at z

= 1 and 2. The z-transform X 1 ( z ) can therefore be expressed as

follows:

X 1 ( z )
z

1
z 2 − 3 z + 2

k 1
z − 1

k 2
z − 2 .

= ≡ +

Using Heaviside's partial fraction expansion formula, the coefficients of the
partial fractions k 1 and k 2 are given by

1
( z − 1)( z − 2)

1
z − 2

k 1 = ( z − 1) = =− 1

z = 1 z = 1

and

1
( z − 1)( z − 2)

1
z − 1

k 2 = ( z − 2) = = 1 .

z = 2 z = 2

The partial fraction expansion of X 1 ( z ) is therefore given by

− z
( z − 1 )
ROC: z > 1

z
( z − 2 )
ROC: z > 2

− 1
( 1 − z

1
( 1 − 2 z

X 1 ( z ) = + = + ,

− 1 ) − 1 )

ROC: z > 1 ROC: z > 2

where the ROC is obtained by noting that each term in X 1 ( z ) corresponds
to a right-hand-sided sequence. This follows directly from knowing that
x [ n ] is right-hand-sided; hence, each term in X 1 ( z ) should also correspond
to a right-hand sequence. Calculating the inverse z-transform of X 1 ( z ), we