Region of convergence of a Laplace transform

[mjtsquared]

If a Laplace transform has a region of convergence starting at Re(s)=0, does the Laplace transform evaluated at the imaginary axis exist? I.e. say that the Laplace transform of 1 is 1/s. Does this Laplace transform exist at say s=i?

 

[jasonRF]

I know this thread is old, but in case you still interested here is an answer.

I am assuming you are using the single-sided Laplace transform, with the standard definition
$$F(s) = \int_0^\infty f(t) e^{-s t} dt$$
so the region of convergence a right half-plane $\Re(s)>s_0$ (where the notation $\Re(s)$ indicates the real part of $s$). This means that $F(s)$ is analytic in that half-plane, and in general is not defined outside of that half-plane. However, in many cases $F(s)$ can be extended so that it is analytic in a larger region of the complex plane (this is called analytic continuation). In your example you have $F(s) = 1/s$ with $\Re(s)>0$. Since $1/s$ is analytic everywhere in the complex plane except at $s=0$, $1/s$ is (trivially) the analytic continuation of $F(s)$ into the entire plane except the origin. So yes, in your example $1/s$ is defined at $s=i$.

If the term analytic is new to you, a function $F(s)$ is called analytic at a point if there is a converging Taylor series about that point. In your example, $1/s$ has a Taylor series about $i$: $F(s) = -i + (s-i) + i (s-i)^2 + \ldots$, which converges for $|s-i|<1$ .

Note that in some cases the process of analytic continuation is much more complicatd. However, in the types of problems you are likely to find in basic electronics, $F(s)$ is often a rational function, so is trivially its own analytic continuation into the entire complex plane except at the poles of $F(s)$.

Jason