# ROC and its relation to the inverse Laplace transform

1. Nov 28, 2012

### elgen

This is a conceptual question on the region of convergence (ROC) and the inverse Laplace transform (ILT).

Here the bilateral laplace transform (LT) and the ILT are given by
$F(s)=L\{f(t)\}=\int_{-\infty}^{+\infty} f(t) e^{-st} dt$
and
$f(t)=L^{-1}\{F(s)\}=\frac{1}{i 2\pi}\int_{a-i\infty}^{a+i\infty} F(s)e^{st}ds$
where $a$ denotes the abssisa of absolute convergence.

If $f(t)=u(t)e^{-at}$ where $u(t)$ denotes the Heaviside step function, then $F(s)=\frac{1}{s+a}$ and the ROC is given by $\Re\{s\}+a > 0$.
When we take the inverse Laplace transform of $F(s)$, we use the Cauchy-Residue theorem.

Commonly, the textbook argues that, for ILT, when $t<0$, the integration contour in the form of the semi-circle is to the right of the vertical line passing $a$ and in parallel to the imaginary axis. Since there is no pole included, the contour integration is 0 and the ILT is 0.
When $t>0$, the integration contour is the semi-circle to the left. Since the contour encloses the pole at $s=-a$, the ILT is not 0.

My question is that since the ROC of $F(s)$is $\Re\{s\}+a > 0$, when taking the ILT for $t>0$, shouldn't the semi-circle still be in the ROC of the $F(s)$. Since the LT is not convergent to the left of the ROC, why is it sensible to talk about the ILT outside of the ROC?

My vague feeling is that the answer may be related to the concept of analytic continuation. A more lucid explaination is appreciated.

Thank you.

Last edited: Nov 28, 2012
2. Nov 28, 2012

### Ray Vickson

Please explain what are 'ROC' and 'ILT'.

3. Nov 28, 2012

### dydxforsn

I think he probably means "radius of convergence" and "inverse Laplace transform".

4. Nov 28, 2012

### Staff: Mentor

I didn't know what they were, either.

elgen, please confirm that ROC = radius of convergence and ILT = Inverse Laplace Transform.

5. Nov 28, 2012

### elgen

Yes. ROC = Region of convergence and ILT = Inverse Laplace Transform. Thank you.