Since the inversion integral must be taken in the ROC of the transform, it is
necessary to understand how this region is determined and what it means even
if the inversion is done by partial fraction expansion or long division. Since
all signals created by linear constant coefficient difference equations are
sums of geometric sequences (or samples of exponentials), an analysis of these
cases will cover most practical situations. Consider a geometric sequence
starting at zero.
f
(
n
)
=
u
(
n
)
a
n
f
(
n
)
=
u
(
n
)
a
n
(1)
with a z-transform
F
(
z
)
=
1
+
a
z
−
1
+
a
2
z
−
2
+
a
3
z
−
3
+
⋯
+
a
M
z
−
M
.
F
(
z
)
=
1
+
a
z
−
1
+
a
2
z
−
2
+
a
3
z
−
3
+
⋯
+
a
M
z
−
M
.
(2)
Multiplying by
a
z
−
1
a
z
−
1
gives
a
z
−
1
F
(
z
)
=
a
z
−
1
+
a
2
x
−
2
+
a
3
z
−
3
+
a
4
z
−
4
+
⋯
+
a
M
+
1
z
−
M
−
1
a
z
−
1
F
(
z
)
=
a
z
−
1
+
a
2
x
−
2
+
a
3
z
−
3
+
a
4
z
−
4
+
⋯
+
a
M
+
1
z
−
M
−
1
(3)
and subtracting from (2.32) gives
(
1
−
a
,
z
−
1
)
F
(
z
)
=
1
−
a
M
+
1
z
−
M
−
1
(
1
−
a
,
z
−
1
)
F
(
z
)
=
1
−
a
M
+
1
z
−
M
−
1
(4)
Solving for
F
(
z
)
F
(
z
)
results in
F
(
z
)
=
1
−
a
M
+
1
z
−
M
−
1
1
−
a
z
−
1
=
z
−
a
(
a
z
)
M
z
−
a
F
(
z
)
=
1
−
a
M
+
1
z
−
M
−
1
1
−
a
z
−
1
=
z
−
a
(
a
z
)
M
z
−
a
(5)
The limit of this sum as
M
→
∞
M
→
∞
is
F
(
z
)
=
z
z
−
a
F
(
z
)
=
z
z
−
a
(6)
for
|
z
|
>
|
a
|
|
z
|
>
|
a
|
.
This not only establishes the z-transform of
f
(
n
)
f
(
n
)
but gives the region in the
z
z
plane where the sum converges.
If a similar set of operation is performed on the sequence that exists for
negative
n
n
f
(
n
)
=
u
(
−
n
−
1
)
a
n
=
{
a
n
n
<
0
0
n
≥
0
f
(
n
)
=
u
(
−
n
−
1
)
a
n
=
{
a
n
n
<
0
0
n
≥
0
(7)
the result is
F
(
z
)
=
−
z
z
−
a
F
(
z
)
=
−
z
z
−
a
(8)
for
|
z
|
<
|
a
|
|
z
|
<
|
a
|
.
Here we have exactly the same z-transform for a different sequence
f
(
n
)
f
(
n
)
but with a different ROC. The pole in
F
(
z
)
F
(
z
)
divides the z-plane into two regions that give two different
f
(
n
)
f
(
n
)
.
This is a general result that can be applied to a general rational
F
(
z
)
F
(
z
)
with several poles and zeros. The z-plane will be divided into concentric
annular regions separated by the poles. The contour integral is evaluated in
one of these regions and the poles inside the contour give the part of the
solution existing for negative
n
n
with the poles outside the contour giving the part of the solution existing
for positive
n
n
.
Notice that any finite length signal has a z-transform that converges for all
z
z
.
The ROC is the entire z-plane except perhaps zero and/or infinity.
