- Signals and Systems Tutorial
- Signals & Systems Home
- Signals & Systems Overview
- Signals Basic Types
- Signals Classification
- Signals Basic Operations
- Systems Classification
- Signals Analysis
- Fourier Series
- Fourier Series Properties
- Fourier Series Types
- Fourier Transforms
- Fourier Transforms Properties
- Distortion Less Transmission
- Hilbert Transform
- Convolution and Correlation
- Signals Sampling Theorem
- Signals Sampling Techniques
- Laplace Transforms
- Laplace Transforms Properties
- Region of Convergence
- Z-Transforms (ZT)
- Z-Transforms Properties
- Signals and Systems Resources
- Signals and Systems - Resources
- Signals and Systems - Discussion
Region of Convergence (ROC)
The range variation of σ 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 $ | $$ {b \over (s+a)^2 + b^2 } $$ |