The region of convergence, known as the ROC, is
important to understand because it defines the region where
the
Laplace transform
exists. The Laplace transform of a sequence is defined as
The ROC for a given
ht
h
t
, is defined as the range of
t
t
for which the Laplace transform converges.
If we consider a
causal, complex exponential,
ht=e−(at)ut
h
t
a
t
u
t
,
we get the equation,
∫0∞e−(at)e−(st)d
t
=∫0∞e−((a+s)t)d
t
t
0
a
t
s
t
t
0
a
s
t
(2)
Evaluating this, we get
-1s+a(limit
t
→
∞
e−((s+a)t)−1)
-1
s
a
t
s
a
t
1
(3)
Notice that this equation will tend to infinity when
limit
t
→
∞
e−((s+a)t)
t
s
a
t
tends to infinity. To understand when this happens, we take one
more step by using
s=σ+iω
s
σ
ω
to realize this equation as
limit
t
→
∞
e−(iωt)e−((σ+a)t)
t
ω
t
σ
a
t
(4)
Recognizing that
e−(iωt)
ω
t
is sinusoidal, it becomes apparent that
e−(σat)
σ
a
t
is going to determine whether this blows up or not. What we
find is that if
σ+a
σ
a
is positive, the exponential will be to a negative power, which
will cause it to go to zero as
tt
tends to infinity. On the other hand, if
σ+a
σ
a
is negative or zero, the exponential will not be to a negative
power, which will prevent it from tending to zero and the system
will not converge. What all of this tells us is that for a
causal signal, we have convergence when
Alternatively, we can note that since the Laplace transform is
a
power series, it converges when
hte−(st)
h
t
s
t
is absolutely summable. Therefore,
∫−∞∞hte−(st)d
t
<∞
t
h
t
s
t
(6)
must be satisfied for convergence.
Although we will not go through the process again for anticausal
signals, we could. In doing so, we would find that the
necessary condition for convergence is when
The Region of Convergence has a number of properties that are
dependent on the characteristics of the signal,
ht
h
t
.
-
The ROC cannot contain any poles.
By definition a pole is a where
Hs
H
s
is infinite.
Since
Hs
H
s
must be finite for all ss for
convergence, there cannot be a pole in the ROC.
-
If
ht
h
t
is a finite-duration sequence, then the ROC is the
entire s-plane, except possibly
s=0
s
0
or
|s|=∞
s
.
A finite-duration sequence is a sequence that
is nonzero in a finite interval
t
1
≤t≤
t
2
t
1
t
t
2
.
As long as each value of
ht
h
t
is finite then the sequence will be absolutely summable.
When
t
2
>0
t
2
0
there will be a
s-1
s
term and thus the ROC will not include
s=0
s
0
.
When
t
1
<0
t
1
0
then the sum will be infinite and thus the ROC will not
include
|s|=∞
s
.
On the other hand, when
t
2
≤0
t
2
0
then the ROC will include
s=0
s
0
,
and when
t
1
≥0
t
1
0
the ROC will include
|s|=∞
s
.
With these constraints, the only signal, then, whose ROC
is the entire z-plane is
ht=cδt
h
t
c
δ
t
.
The next properties apply to infinite duration sequences. As
noted above, the z-transform converges when
|Hs|<∞
H
s
.
So we can write
|Hs|=|∫−∞∞hte−(st)d
t
|≤∫−∞∞|hte−(st)|d
t
=∫−∞∞|ht||e−(st)|d
t
H
s
t
h
t
s
t
t
h
t
s
t
t
h
t
s
t
(8)
We can then split the infinite sum into positive-time and
negative-time portions. So
|Hs|≤Ns+Ps
H
s
N
s
P
s
(9)
where
Ns=∫−∞-1|ht||e−(st)|d
t
N
s
t
-1
h
t
s
t
(10)
and
Ps=∫0∞|ht||e−(st)|d
t
P
s
t
0
h
t
s
t
(11)
In order for
|Hs|
H
s
to be finite,
|ht|
h
t
must be bounded. Let us then set
|ht|≤
C
1
r
1
t
h
t
C
1
r
1
t
(12)
for
t<0
t
0
and
|ht|≤
C
2
r
2
t
h
t
C
2
r
2
t
(13)
for
t≥0
t
0
From this some further properties can be derived:
-
If
ht
h
t
is a left-sided sequence, then the ROC extends inward
from the innermost pole in
Hs
H
s
.
A left-sided sequence is a sequence where
ht=0
h
t
0
for
t>
t
1
>−∞
t
t
1
.
Looking at the negative-time portion from the above
derivation, it follows that
Ps≤
C
1
∫−∞-1
r
1
te−(st)d
t
=
C
1
∫−∞-1
r
1
|e−s|td
t
=
C
1
∫1∞|e−s|
r
1
kd
k
P
s
C
1
t
-1
r
1
t
s
t
C
1
t
-1
r
1
s
t
C
1
k
1
s
r
1
k
(15)
-
If
ht
h
t
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
<|e−s|<
r
2
r
2
s
r
2
converges, then both the positive-time and negative-time
portions converge and thus
Hs
H
s
converges as well. Therefore the ROC of a two-sided
sequence is of the form
-r
2
<|e−s|<
r
2
-r
2
s
r
2
.
To gain further insight it is good to look at a couple of
examples.
Lets take
h
1
t=12tut+14tut
h
1
t
1
2
t
u
t
1
4
t
u
t
(16)
The Laplace-transform of
12tut
1
2
t
u
t
is
ss−12
s
s
1
2
with an ROC at
|s|>12
s
1
2
.
The z-transform of
-14tut
-1
4
t
u
t
is
ss+14
s
s
1
4
with an ROC at
|s|>-14
s
-1
4
.
Due to linearity,
H
1
s=ss−12+ss+14=2s(s−18)(s−12)(s+14)
H
1
s
s
s
1
2
s
s
1
4
2
s
s
1
8
s
1
2
s
1
4
(17)
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 above properties, the ROC is
|s|>12
s
1
2
.
Now take
h
2
t=-14tut−12tu(−t)−1
h
2
t
-1
4
t
u
t
1
2
t
u
t
1
(18)
The z-transform and ROC of
-14tut
-1
4
t
u
t
was shown in the
example above.
The Laplace-transorm of
(−12t)u(−t)−1
1
2
t
u
t
1
is
ss−12
s
s
1
2
with an ROC at
|s|>12
s
1
2
.
Once again, by linearity,
H
2
s=ss+14+ss−12=s(2s−18)(s+14)(s−12)
H
2
s
s
s
1
4
s
s
1
2
s
2
s
1
8
s
1
4
s
1
2
(19)
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
|s|<12
s
1
2
.
"My introduction to signal processing course at Rice University."