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116 Chapter 4 THE z-TRANSFORM
a =
1.0000
-1.3333
0.3333
so that
X(z) =
0+ 1 3 z−1
1 − 4 3 z−1 + 1 3 z−2 = z −1
3 − 4z −1 + z −2 = z
3z 2 − 4z +1
as before.
□
□ EXAMPLE 4.9 Compute the inverse z-transform of
1
X(z) =
, |z| > 0.9
(1 − 0.9z −1 ) 2 (1 + 0.9z −1 )
Solution
We will evaluate the denominator polynomial as well as the residues using the
MATLAB script:
>> b = 1; a = poly([0.9,0.9,-0.9])
a =
1.0000 -0.9000 -0.8100 0.7290
>> [R,p,C]=residuez(b,a)
R =
0.2500
0.5000
0.2500
p =
0.9000
0.9000
-0.9000
c =
[]
Note that the denominator polynomial is computed using MATLAB’s polynomial
function poly, which computes the polynomial coefficients, given its roots.
We could have used the conv function, but the use of the poly function is more
convenient for this purpose. From the residue calculations and using the order
of residues given in (4.16), we have
0.25
X(z) =
1 − 0.9z + 0.5
−1 (1 − 0.9z −1 ) + 0.25
, 2 1+0.9z−1 |z| > 0.9
( )
0.25 0.5 0.9z
−1
=
+
1 − 0.9z−1 0.9 z (1 − 0.9z −1 ) + 0.25
, 2 1+0.9z−1 |z| > 0.9
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