|
What is the need to consider region of convergence
while determining the Laplace transform?
If we
consider the signals e-atu(t) and
-e-atu(-t) we note that although the
signals are differing, their Laplace Transforms are identical which is 1/(
s+a). Thus we conclude that to distinguish L.T's
uniquely
their ROC's must be specified. Further from the ROC we can
define many important conclusions which
help in the analysis of C.T. signals and CT-LTI systems.
A few important properties of the ROC are listed below: |
 |
The ROC of F(S) consists of strips parallel to
 in complex variable plane (S-plane).
We know that the ROC is dependent only on the real parts
of 'S' which is 'σ' , therefore the property.
|
|
|
|
The ROC
does not contain any poles. |
|
|
|
Since if this happens then the Laplace Transform becomes
infinity.
For example, if the F(S)= 1/(s-1) then the ROC cannot
contain s =1
because at this point the L.T becomes infinity.
|
|
|
|
If h(t) is a
time limited signal and is Laplace Transformable, then
its ROC will be the entire
S-plane. |
|
|
|
For
example the ROC for
 is the entire S-plane. |
|
|
|
The region of convergence is always between two vertical
lines in s-plane. These vertical lines need not be in
finite region. But note that the ROC is always
simply-connected but not multiply-connected in the
s-plane.
|
|
|
|
This fact can be
explained by the following illustration:
Let H1(s) and
H2(s) be the respective Laplace transforms of the first and second
terms.H1(s) and H2(s) converge in the region Re(s) > 1 and Re(s)
< 1 respectively. But, h(t) doesn't have any Laplace transform due to
no common ROC where both H1(s) and H2(s) converge.
|
|
|
|
