Region of Convergence (ROC)

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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} > σ_o.

• If x(t) is a left sided sequence then ROC : Re{s} < σ_o.

• 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 $	$$ {b \over (s+a)^2 + b^2 } $$