With the Laplace
transform, the s-plane represents a set of signals (complex exponentials). For
any given LTI system,
some of these signals may cause the output of the system to
converge, while others cause the output to diverge ("blow up").
The set of signals that cause the system's output to converge
lie in the region of convergence (ROC). This
module will discuss how to find this region of convergence for
any continuous-time, LTI system.

Recall the definition of the Laplace transform,

Hs=∫−∞∞hte−(st)d
t
H
s
t
h
t
s
t

(1)
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

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

Perhaps the best way to look at the region of convergence is
to view it in the s-plane. What we observe is that for a
single pole, the region of convergence lies to the right of it
for causal signals and to the left for anti-causal signals.

Once we have recognized this, the natural question becomes:
What do we do when we have multiple poles? The simple answer
is that we take the intersection of all of the regions of
convergence of the respective poles.

Find
Hs
H
s
and state the region of convergence for
ht=e−(at)ut+e−(bt)u−t
h
t
a
t
u
t
b
t
u
t

Breaking this up into its two terms, we get transfer
functions and respective regions of convergence of

∀s,ℜs>−a:
H
1
s=1s+a
s
s
a
H
1
s
1
s
a

(7)
and

∀s,ℜs<−b:
H
2
s=-1s+b
s
s
b
H
2
s
-1
s
b

(8)
Combining these, we get a region of convergence of

−b>ℜs>−a
b
s
a
.
If

a>b
a
b
, we can represent this graphically. Otherwise,
there will be no region of convergence.