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Inversion of the z-Transform 113
where p k is the kth pole of X(z) and R k is the residue at p k .Itis
assumed that the poles are distinct for which the residues are given by
R k = ˜b 0 + ˜b 1 z −1 + ···+ ˜b N−1 z −(N−1)
1+a 1 z −1 + ···+ a N z −N (1 − p k z −1 ) ∣
∣
z=pk
For repeated poles the expansion (4.13) has a more general form. If a
pole p k has multiplicity r, then its expansion is given by
r∑
l=1
R k,l z −(l−1)
(1 − p k z −1 ) l = R k,1
1 − p k z −1 + R k,2z −1
(1 − p k z −1 ) 2 + ···+ R k,rz −(r−1)
(1 − p k z −1 ) r
(4.14)
where the residues R k,l are computed using a more general formula,
which is available in reference [23].
• assuming distinct poles as in (4.13), write x(n) as
N∑
[ ]
x(n) = R k Z −1 1
1 − p k z −1 +
k=1
M−N
∑
k=0
C k δ(n − k)
} {{ }
M≥N
• finally, use the relation from Table 4.1
[ ] {
Z −1 z
p n k
=
u(n) |z k|≤R x−
z − p k −p n k u(−n − 1) |z (4.15)
k|≥R x+
to complete x(n).
A similar procedure is used for repeated poles.
□ EXAMPLE 4.7 Find the inverse z-transform of x(z) =
z
3z 2 − 4z +1 .
Solution
Write
X(z) =
1
z
3(z 2 − 4 z + 1 ) = 3 z−1
1 − 4 3 3 3 z−1 + 1 3 z−2
or
=
1
3 z−1
1
(1 − z −1 )(1 − 1 3 z−1 ) = 2
1
1 − z − 2
−1 1 − 1 3 z−1
X(z) = 1 ( ) ( )
1
− 1 1
2 1 − z −1 2 1 − 1 3 z−1
Now, X(z) has two poles: z 1 =1and z 2 = 1 ; and since the ROC is not specified,
3
there are three possible ROCs as shown in Figure 4.5.
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