Region of Convergence (ROC)


Advertisements


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.

    strip lines
  • 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$

strip lines

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 $

strip lines

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 $

strip lines

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.

    Casual System
  • A system is said to be stable when all poles of its transfer function lay on the left half of s-plane.

    Stable System
  • 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.

    Unstable System
  • 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.

    Marginally Stable System

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 $$$ {b \over (s+a)^2 + b^2 } $$


Advertisements