The region of convergence, known as the ROC, is
important to understand because it defines the region where
the
z-transform
exists. The z-transform of a sequence is defined as
Xz=∑n=-∞∞xnz-n
Xz
n
x
n
z
n
(1)
The ROC for a given
xn
x
n
, is defined as the range of
z
z
for which the z-transform converges. Since the z-transform is
a
power series, it converges when
xnz-n
x
n
z
n
is absolutely summable. Stated differently,
∑n=-∞∞|xnz-n|<∞
n
x
n
z
n
(2)
must be satisfied for convergence.
The Region of Convergence has a number of properties that are
dependent on the characteristics of the signal,
xn
x
n
.
-
The ROC cannot contain any poles.
By definition a pole is a where
Xz
X
z
is infinite.
Since
Xz
X
z
must be finite for all zz for
convergence, there cannot be a pole in the ROC.
-
If
xn
x
n
is a finite-duration sequence, then the ROC is the
entire z-plane, except possibly
z=0
z
0
or
|z|=∞
z
.
A finite-duration sequence is a sequence that
is nonzero in a finite interval
n
1
≤n≤
n
2
n
1
n
n
2
.
As long as each value of
xn
x
n
is finite then the sequence will be absolutely summable.
When
n
2
>0
n
2
0
there will be a
z-1
z
term and thus the ROC will not include
z=0
z
0
.
When
n
1
<0
n
1
0
then the sum will be infinite and thus the ROC will not
include
|z|=∞
z
.
On the other hand, when
n
2
≤0
n
2
0
then the ROC will include
z=0
z
0
,
and when
n
1
≥0
n
1
0
the ROC will include
|z|=∞
z
.
With these constraints, the only signal, then, whose ROC
is the entire z-plane is
xn=cδn
x
n
c
δ
n
.
The next properties apply to infinite duration sequences. As
noted above, the z-transform converges when
|Xz|<∞
X
z
.
So we can write
|Xz|=|∑n=-∞∞xnz-n|≤∑n=-∞∞|xnz-n|=∑n=-∞∞|xn||z|-n
X
z
n
x
n
z
n
n
x
n
z
n
n
x
n
z
n
(3)
We can then split the infinite sum into positive-time and
negative-time portions. So
|Xz|≤Nz+Pz
X
z
N
z
P
z
(4)
where
Nz=∑n=-∞-1|xn||z|-n
N
z
n
-1
x
n
z
n
(5)
and
Pz=∑n=0∞|xn||z|-n
P
z
n
0
x
n
z
n
(6)
In order for
|Xz|
X
z
to be finite,
|xn|
x
n
must be bounded. Let us then set
|xn|≤
C
1
r
1
n
x
n
C
1
r
1
n
(7)
for
n<0
n
0
and
|xn|≤
C
2
r
2
n
x
n
C
2
r
2
n
(8)
for
n≥0
n
0
From this some further properties can be derived:
-
If
xn
x
n
is a two-sided sequence, the ROC will be a ring in the
z-plane that is bounded on the interior and exterior by
a pole.
A two-sided sequence is an sequence with
infinite duration in the positive and negative
directions. From the derivation of the above two
properties, it follows that if
r
2
<|z|<
r
2
r
2
z
r
2
converges, then both the positive-time and negative-time
portions converge and thus
Xz
X
z
converges as well. Therefore the ROC of a two-sided
sequence is of the form
r
2
<|z|<
r
2
r
2
z
r
2
.
To gain further insight it is good to look at a couple of
examples.
Lets take
x
1
n=12nun+14nun
x
1
n
1
2
n
u
n
1
4
n
u
n
(11)
The z-transform of
12nun
1
2
n
u
n
is
zz-12
z
z
1
2
with an ROC at
|z|>12
z
1
2
.
The z-transform of
-14nun
-1
4
n
u
n
is
zz+14
z
z
1
4
with an ROC at
|z|>-14
z
-1
4
.
Due to linearity,
X
1
z=zz-12+zz+14=2zz-18z-12z+14
X
1
z
z
z
1
2
z
z
1
4
2
z
z
1
8
z
1
2
z
1
4
(12)
By observation it is clear that there are two zeros, at
0
0
and
18
1
8
,
and two poles, at
12
1
2
,
and
-14
-1
4
.
Following the obove properties, the ROC is
|z|>12
z
1
2
.
Now take
x
2
n=-14nun-12nu-n-1
x
2
n
-1
4
n
u
n
1
2
n
u
n
1
(13)
The z-transform and ROC of
-14nun
-1
4
n
u
n
was shown in the
example above.
The z-transorm of
-12nu-n-1
1
2
n
u
n
1
is
zz-12
z
z
1
2
with an ROC at
|z|>12
z
1
2
.
Once again, by linearity,
X
2
z=zz+14+zz-12=z2z-18z+14z-12
X
2
z
z
z
1
4
z
z
1
2
z
2
z
1
8
z
1
4
z
1
2
(14)
By observation it is again clear that there are two zeros, at
0
0
and
116
1
16
,
and two poles, at
12
1
2
,
and
-14
-1
4
.
in ths case though, the ROC is
|z|<12
z
1
2
.
"My introduction to signal processing course at Rice University."