Whether the Laplace transform of a signal exists or not depends
on the complex variable as well as the signal itself. All complex values
of for which the integral in the definition converges form a
*region of convergence (ROC)* in the s-plane. exists if and only
if the argument is inside the ROC. As the imaginary part
of the complex variable
has no effect in terms of the
convergence, the ROC is determined solely by the real part .

**Example 1:** The Laplace transform of
is:

For this integral to converge, we need to have

and the Laplace transform is

As a special case where , and we have

**Example 2:** The Laplace transform of a signal
is:

Only when

will the integral converge, and Laplace transform is

Again as a special case when , we have

Comparing the two examples above we see that two different signals may have identical Laplace transform , but different ROC. In the first case above, the ROC is , and in the second case, the ROC is . To determine the time signal by the inverse Laplace transform, we need the ROC as well as .

**Example 3:**

The Laplace transform is linear, and is the sum of the transforms for the two terms:

If , i.e., decays when , the intersection of the two ROCs is , and we have:

However, if , i.e., grows without a bound when , the intersection of the two ROCs is a empty set, the Laplace transform does not exist.

**Example 4:**

The Laplace transform of this signal is

This exists only if the Laplace transforms of all three individual terms exist, i.e, the conditions for the three integrals to converge are simultaneously satisfied:

i.e., .

**Example 5:**

As the Laplace integration converges independent of , the ROC is the entire s-plane. In particular, when , we have