Region of Convergence (ROC)

The range of values of the complex variable s for which Laplace Transform $X(S)=\int_{-\infty }^{\infty }x(t)e^{-st}\ dt$ converges is called the Region of Convergence (ROC).

i.e, The region of Convergence (or) existence of signal’s Laplace transform X(S) is the set of values of s for which the integral defining the direct L T/F X(S) converges.

The ROC is required for evaluating the inverse L T/F of x(t) from X(S).

i.e, the operation of finding the inverse T/F requires an integration in the complex plane.

i.e, $x(t)= \frac{1}{2\pi j}\int_{\sigma -j\infty }^{\sigma +j\infty }X(S) e^{St} \ ds$ .

The path of integration is along S-plane $S = \sigma +j\omega$ that is along $\sigma +j\omega$ with $\omega$ varying from $-\infty \ to \ \infty$ and moreover , the path of integration must lie in the ROC for X(S).

for example the signal $e^{-at}u(t)$ , this is possible if $\sigma >-a$ so the path of integration is shown in the figure

Thus to obtain $x(t) = e^{-at}u(t)$ from $X(S) = \frac{1}{s+a}$ , the integration is performed through this path for the function $\frac{1}{s+a}$ . such integration in the complex plane requires a back ground in the theory of functions of complex variables.

so we can avoid this integration by compiling a Table of L T/F’s . so for inverse L T/F’s we use this table instead of performing complex integration.

specific constraints on the ROC are closely associated with time-domain properties of x(t).

Properties of ROC/ constraints (or) Limitations:-

1.The ROC of X(S) consists of strips parallel to the $j\omega$ axis in the S-plane.

i.e, The ROC of X(S) consists of the values of s for which Fourier T/F of $x(t)e^{-\sigma t}$ converges this is possible if $x(t)e^{-\sigma t}$ is fully integrable thus the condition depends only on $\sigma$ . Hence ROC is the strips (bands) which is only in terms of values of $\sigma$ .

3. For Rational Laplace T/F’s , the ROC does not contain any poles. This is because X(S) is finite at poles and the integral can not be converge at this point.

4. If x(t) is of finite duration and absolutely integrable, then the ROC is the entire S-plane.

5. If x(t) is right-sided and if the line $Re\left \{ s \right \} =\sigma _{o}$ is in the ROC, then all values of s for which $Re\left \{ s \right \} > \sigma _{o}$ will also be in the ROC.

i.e, if the signal is $x(t) = e^{-at}u(t)$ right-sided $[0 \ to \ \infty ]$ then $X(S) = \frac{1}{s+a}$ for ROC : $Re\left \{ s \right \} > -a$ .

6. If x(t) is left-sided and if the line $Re\left \{ s \right \} =\sigma _{o}$ is in the ROC, then all values of s for which $Re\left \{ s \right \} < \sigma _{o}$ will also be in the ROC.

7. If x(t) is two-sided and if the line $Re\left \{ s \right \} =\sigma _{o}$ is in the ROC, then the ROC consists of a strip in the s-plane that includes the line $Re\left \{ s \right \} = \sigma _{o}$ .

for the both sided signal , the ROC lies in the region $\sigma _{1} < Re\left \{ s \right \}<\sigma _{2}$ . This ROC is the strip parallel to $j\omega$ axis in the s-plane.

8. If the L T/F X(S) of x(t) is rational, then it’s ROC is bounded by poles (or) extends to infinity in addition no poles of X(s) are contained in the ROC.

If the function has two poles , then ROC will be area between these two poles for two sided signal, if for single sided signal the area extends from one pole to infinity.

But is does not include any pole.

9. If the L T/F X(S) of x(t) is rational, then if x(t) is right-sided. The ROC is the region in the s-plane to the right of the right most pole and if x(t) is left-sided, the ROC is the region in the s-plane to the left of the left most pole.

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems. View all posts by Lakshmi Prasanna Ponnala

Region of Convergence (ROC)

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