What Does ROC Mean in Signals and Processing?

[jaus tail]

Homework Statement

Find laplace of e^-atu(t) where a > 0

Homework Equations

Laplace is integration from -inf to +inf f(t)e^-stdt
u(t) is 1 for t more than equal to 0.

The Attempt at a Solution

Well i got the answer as X(s) = 1/(s+a).

But the book said something like ROC like region of convergence and for that s must be more than -a
I understand that s more than minus a, gives positive 1/(s+a)
but why?

Why can't ROC be negative? What does ROC mean? I am studying Signals and Processing.

 

[jedishrfu]

I think ROC is actually radius of convergence:

https://en.wikipedia.org/wiki/Radius_of_convergence

 

[LCKurtz]

Homework Statement

Find laplace of e^-atu(t) where a > 0

Homework Equations

Laplace is integration from -inf to +inf f(t)e^-stdt
u(t) is 1 for t more than equal to 0.

The Attempt at a Solution

Well i got the answer as X(s) = 1/(s+a).

But the book said something like ROC like region of convergence and for that s must be more than -a
I understand that s more than minus a, gives positive 1/(s+a)
but why?

Why can't ROC be negative? What does ROC mean? I am studying Signals and Processing.

Your integral is $\int_1^\infty e^{-st}e^{-at}~dt = \int_1^\infty e^{-(s+a)t}~dt$. That integral will only converge if the exponent is negative, meaning $s+a>0$. ROC in this case means the region of convergence, the values of $s$ for which the transform is defined.