The z-transform - Continuous and Discrete Time Signals and Systems - page 579

To determine the remaining partial fraction coefficients k 2 and k 3 , we expand
2(3 z + 17)
( z − 1)( z 2 − 6 z + 25)

2
z − 1

k 2 z + k 3
z 2 − 6 z + 25

≡ +

by cross-multiplying and equating the numerators, we obtain

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

Comparing coefficients of z 2 and z yields

coefficients of z 2 0 ≡ 2 + k 2 ⇒ k 2 =− 2;

coefficients of z 6 ≡− 12 − k 2 + k 3 ⇒ k 3 = 16 .

The partial fraction expansion of X 3 ( z ) can therefore be expressed as follows:

− 2 z + 16
z 2 − 6 z + 25

X 3 ( z )
z

2
z − 1

= +

or

z
z − 1

z ( z − 5 0 . 6)
z 2 − 2 5 z 0 . 6 + 5 2

X 3 ( z ) = 2 − 2

+ 5
2

z (5 0 . 8)
z 2 − 2 5 z 0 . 6 + 5 2 ,

where the final rearrangement makes the three terms in the above expres-
sion consistent with entries (4), (10), and (11) of Table 13.1, with α = 5,
and cos( Ω 0 ) = 0 . 6 and sin( Ω 0 ) = 0 . 8. Assuming that the three terms repre-
sent right-hand-sided sequences, the inverse z-transform for each term is given
by

z
z − 1

− 1
←−−→

z

term 1 2 2 u [ k ];

z ( z − 5 0 . 6)
z 2 − 2 5 z 0 . 6 + 5 2

− 1
←−−→ − 2 cos(cos

z

− 1 (0 . 6) k ) 5 k u [ k ];

term 2 − 2

5
2

z (5 0 . 8)
z 2 − 2 5 z 0 . 6 + 5 2

←−−→ 5
2

− 1

z

− 1 (0 . 6) k ) 5 k u [ k ] .

term 3 sin(cos

− 1 (0 . 6) = 0 . 9273, the three terms are combined as follows:

Substituting cos

x 3 [ k ] = 2 u [ k ] − 2 5 k cos(0 . 9273 k ) u [ k ] + 5
2

5 k sin(0 . 9273 k ) u [ k ] ,

which can be simplified to

2 + 3 . 2016 5 k cos(0 . 9273 k − 128 . 7

◦

x 3 [ k ] = ) u [ k ] .

The DT sequences x 1 [ k ], x 2 [ k ], and x 3 [ k ] are plotted in Fig. 13.3 for duration
0 ≤ k ≤ 6.