When using the
z-transform
Xz=∑n=-∞∞xnz-n
X
z
n
x
n
z
n
(1)
it is often useful to be able to find
xn
x
n
given
Xz
X
z
.
There are at least 4 different methods to do this:
This "method" is to basically become familiar with the z-transform pair tables
and then "reverse engineer".
When given
Xz=zz-α
X
z
z
z
α
with an ROC of
|z|>α
z
α
we could determine "by inspection" that
xn=αnun
x
n
α
n
u
n
When dealing with linear time-invariant systems
the z-transform often in the form
Xz=BzAz=∑k=0M
b
k
z-k∑k=0N
a
k
z-k
X
z
B
z
A
z
k
0
M
b
k
z
k
k
0
N
a
k
z
k
(2)
This can also expressed as
Xz=
a
0
b
0
∏k=1M1-
c
k
z-1∏k=1N1-
d
k
z-1
X
z
a
0
b
0
k
1
M
1
c
k
z
k
1
N
1
d
k
z
(3)
where
c
k
c
k
represents the nonzero zeros of
Xz
X
z
and
d
k
d
k
represents the nonzero poles.
If
M<N
M
N
then
Xz
X
z
can be represented as
Xz=∑k=1N
A
k
1-
d
k
z-1
X
z
k
1
N
A
k
1
d
k
z
(4)
This form allows for easy inversions of each term of the sum
using the
inspection
method and the
transform table. Thus if the numerator is
a polynomial then it is necessary to use
partial-fraction
expansion to put
Xz
X
z
in the above form. If
M≥N
M
N
then
Xz
X
z
can be expressed as
Xz=∑r=0M-N
B
r
z-r+∑k=0N-1
b
k
'
z-k∑k=0N
a
k
z-k
X
z
r
0
M
N
B
r
z
r
k
0
N
1
b
k
'
z
k
k
0
N
a
k
z
k
(5)
Find the inverse z-transform of
Xz=1+2z-1+z-21+-3z-1+2z-2
X
z
1
2
z
z
-2
1
-3
z
2
z
-2
where the ROC is
|z|>2
z
2
.
In this case
M=N=2
M
N
2
,
so we have to use long division to get
Xz=12+12+72z-11+-3z-1+2z-2
X
z
1
2
1
2
7
2
z
1
-3
z
2
z
-2
Next factor the denominator.
Xz=2+-1+5z-11-2z-11-z-1
X
z
2
-1
5
z
1
2
z
1
z
Now do partial-fraction expansion.
Xz=12+
A
1
1-2z-1+
A
2
1-z-1=12+921-2z-1+-41-z-1
X
z
1
2
A
1
1
2
z
A
2
1
z
1
2
9
2
1
2
z
-4
1
z
Now each term can be inverted using the inspection method
and the z-transform table. Thus, since the ROC is
|z|>2
z
2
,
xn=12δn+922nun+-4un
x
n
1
2
δ
n
9
2
2
n
u
n
-4
u
n
When the z-transform is defined as a power series in the form
Xz=∑n=-∞∞xnz-n
Xz
n
x
n
z
n
(6)
then each term of the sequence
xn
x
n
can be determined by looking at the coefficients of the
respective power of
z-n
z
n
.
Now look at the z-transform of a finite-length
sequence.
Xz=z21+2z-11-12z-11+z-1=z2+52z+12+-z-1
X
z
z
2
1
2
z
1
1
2
z
1
z
z
2
5
2
z
1
2
z
(7)
In this case, since there were no poles, we multiplied the
factors of
Xz
X
z
.
Now, by inspection, it is clear that
xn=δn+2+52δn+1+12δn+-δn-1
x
n
δ
n
2
5
2
δ
n
1
1
2
δ
n
δ
n
1
.
One of the advantages of the power series expansion method is
that many functions encountered in engineering problems have
their power series' tabulated. Thus functions such as log,
sin, exponent, sinh, etc, can be easily inverted.
Suppose
Xz=logn1+αz-1
X
z
n
1
α
z
Noting that
logn1+x=∑n=1∞-1n+1xnn
n
1
x
n
1
-1
n
1
x
n
n
Then
Xz=∑n=1∞-1n+1αnz-nn
X
z
n
1
-1
n
1
α
n
z
n
n
Therefore
Xz=
-1n+1αnnifn≥10ifn≤0
X
z
-1
n
1
α
n
n
n
1
0
n
0
Without going in to much detail
xn=12πⅈ∮rXzzn-1dz
x
n
1
2
z
r
X
z
z
n
1
(8)
where
r
r
is a counter-clockwise contour in the ROC of
Xz
X
z
encircling the origin of the z-plane. To further expand on
this method of finding the inverse requires the knowledge of
complex variable theory and thus will not be addressed in this
module.
"My introduction to signal processing course at Rice University."