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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.

    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
$u(t)$ $${1over s}$$ ROC: Re{s} > 0
$ t, u(t) $ $${1over s^2} $$ ROC:Re{s} > 0
$ t^n, u(t) $ $$ {n! over s^{n+1}} $$ ROC:Re{s} > 0
$ e^{at}, u(t) $ $$ {1over s-a} $$ ROC:Re{s} > a
$ e^{-at}, u(t) $ $$ {1over s+a} $$ ROC:Re{s} > -a
$ e^{at}, u(t) $ $$ – {1over s-a} $$ ROC:Re{s} < a
$ e^{-at}, u(-t) $ $$ – {1over s+a} $$ 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 $ $$ {b over (s+a)^2 + b^2 } $$

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