Region of Convergence (ROC)
The range variation of for which the Laplace transform converges is called region of convergence.
Properties of ROC of Laplace Transform
ROC contains strip lines parallel to jω axis in s-plane.
If x(t) is absolutely integral and it is of finite duration, then ROC is entire s-plane.
If x(t) is a right sided sequence then ROC : Re{s} > σo.
If x(t) is a left sided sequence then ROC : Re{s} < σo.
If x(t) is a two sided sequence then ROC is the combination of two regions.
ROC can be explained by making use of examples given below:
Example 1: Find the Laplace transform and ROC of $x(t) = e-^{at}u(t)$
$L.T[x(t)] = L.T[e-^{at}u(t)] = {1 \over S+a}$
$ Re{} \gt -a $
$ ROC:Re{s} \gt >-a$
Example 2: Find the Laplace transform and ROC of $x(t) = e^{at}u(-t)$
$ L.T[x(t)] = L.T[e^{at}u(t)] = {1 \over S-a} $
$ Re{s} < a $
$ ROC: Re{s} < a $
Example 3: Find the Laplace transform and ROC of $x(t) = e^{-at}u(t)+e^{at}u(-t)$
$L.T[x(t)] = L.T[e^{-at}u(t)+e^{at}u(-t)] = {1 \over S+a} + {1 \over S-a}$
For ${1 \over S+a} Re\{s\} \gt -a $
For ${1 \over S-a} Re\{s\} \lt a $
Referring to the above diagram, combination region lies from a to a. Hence,
$ ROC: -a < Re{s} < a $
Causality and Stability
For a system to be causal, all poles of its transfer function must be right half of s-plane.
A system is said to be stable when all poles of its transfer function lay on the left half of s-plane.
A system is said to be unstable when at least one pole of its transfer function is shifted to the right half of s-plane.
A system is said to be marginally stable when at least one pole of its transfer function lies on the jω axis of s-plane.
ROC of Basic Functions
| f(t) | F(s) | ROC |
|---|---|---|
| $u(t)$ | $${1\over s}$$ | ROC: Re{s} > 0 |
| $ t\, u(t) $ | $${1\over s^2} $$ | ROC:Re{s} > 0 |
| $ t^n\, u(t) $ | $$ {n! \over s^{n+1}} $$ | ROC:Re{s} > 0 |
| $ e^{at}\, u(t) $ | $$ {1\over s-a} $$ | ROC:Re{s} > a |
| $ e^{-at}\, u(t) $ | $$ {1\over s+a} $$ | ROC:Re{s} > -a |
| $ e^{at}\, u(t) $ | $$ - {1\over s-a} $$ | ROC:Re{s} < a |
| $ e^{-at}\, u(-t) $ | $$ - {1\over s+a} $$ | ROC:Re{s} < -a |
| $ t\, e^{at}\, u(t) $ | $$ {1 \over (s-a)^2} $$ | ROC:Re{s} > a |
| $ t^{n} e^{at}\, u(t) $ | $$ {n! \over (s-a)^{n+1}} $$ | ROC:Re{s} > a |
| $ t\, e^{-at}\, u(t) $ | $$ {1 \over (s+a)^2} $$ | ROC:Re{s} > -a |
| $ t^n\, e^{-at}\, u(t) $ | $${n! \over (s+a)^{n+1}} $$ | ROC:Re{s} > -a |
| $ t\, e^{at}\, u(-t) $ | $$ - {1 \over (s-a)^2} $$ | ROC:Re{s} < a |
| $ t^n\, e^{at}\, u(-t) $ | $$ - {n! \over (s-a)^{n+1}} $$ | ROC:Re{s} < a |
| $ t\, e^{-at}\,u(-t) $ | $$ - {1 \over (s+a)^2} $$ | ROC:Re{s} < -a |
| $ t^n\, e^{-at}\, u(-t) $ | $$ - {n! \over (s+a)^{n+1}} $$ | ROC:Re{s} < -a |
| $ e^{-at} \cos \, bt $ | $$ {s+a \over (s+a)^2 + b^2 } $$ | |
| $ e^{-at} \sin\, bt $ | $$ {b \over (s+a)^2 + b^2 } $$ |