Relation Fourier/Laplace Transform I have a question about the relation between Fourier and Laplace transforms.
I have seen in some places that the transfer functions in the Laplace space are represented as $G(s)$ where $s$ is the variable in the frequency domain (Laplace). In other places I have seen that the transfer functions are represented as $G(iw)$ where $w$ is the frequency in the Fourier space and $i$ the imaginary unit. 
I wonder how this connection applies because in some places it is supposed that the relation between both frequencies is $s=\sigma+iw$. But what is $\sigma$ then and why is it sometimes ignored?
If the relationship $s=iw$ would hold to be true, that means that we can Laplace transform by multiplying data times $i$ and doing the FFT?
 A: It is convenient to start with the complex Laplace transform.
Fourier transform
\begin{eqnarray*}
\tilde{f}(\omega ) &=&\int_{-\infty }^{+\infty }dt\exp [i\omega t]f(t) \\
f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i\omega t]%
\tilde{f}(\omega )
\end{eqnarray*}
Complex Laplace transform
$$
\hat{f}(z)=\int_{0}^{+\infty }dt\exp [izt]f(t),\;{Im}z>0
$$
With $\theta (t)$ the step function
$$
\hat{f}(z)=\int_{-\infty }^{+\infty }dt\theta (t)\exp [izt]f(t)
$$
Let $z=\omega +ia$, $a>0$. Then
$$
\hat{f}(\omega +ia)=\int_{-\infty }^{+\infty }dt\exp [i\omega t]\theta
(t)\exp [-at]f(t)
$$
so $\hat{f}(\omega +ia)$ is the Fourier transform of  $\theta (t)\exp
[-at]f(t)$ and
$$
\theta (t)\exp [-at]f(t)=\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega
\exp [-i\omega t]\hat{f}(\omega +ia)
$$
From this you see that the Laplace transform is essentially equivalent to
the Fourier transform of the product of the step function and $f(t)$.
A: If $f \in L^{2}[0,\infty)$, then $\mathscr{L}\{f\}$ is holomorphic in the right half plane where $\Re s > 0$. and the Laplace transform is square integrable on all vertical lines in the right half plane, with
$$
   \frac{1}{2\pi}\int_{-\infty}^{\infty}|\mathscr{L}\{f\}(v+iw)|^{2}dw \le \int_{0}^{\infty}|f(t)|^{2}dt,\;\;\; 0 < v < \infty.
$$
The inverse Laplace transform can be computed by an inverse Fourier transform of $\mathscr{L}\{f\}$ on any vertical line in the right half-plane. In fact, the section $l_{v}(w)=\mathscr{L}\{f\}(v+iw)$ has an $L^{2}$ limit as $v\downarrow 0$, and
$$
             f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{iwt}\mathscr{L}\{f\}(0^{+}+iw)dw.
$$
This is the Paley-Wiener Theorem. The stronger result is the following

Theorem [Paley-Wiener]: Let $F$ be a holomorphic function on the right half plane $\Re s > 0$. Then $F$ is the Laplace transform of a function $f \in L^{2}[0,\infty)$ iff $F$ is square integrable on every vertical line in the right half plane and there exists a constant $M$ such that
  $$
                   \int_{-\infty}^{\infty}|F(v+iw)|^{2}dw \le M.
$$
  For any such $F$, the limit $\lim_{v\downarrow 0}F(v+iw)=F_{0}(w)$ exists in $L^{2}(\mathbb{R})$ and
  $$
                    f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F_{0}(w)dw.
$$

The space of holomorphic functions as described above is known as the Hardy space $H^{2}(\Pi_{+})$ where $\Pi_{+}$ is the right half plane. $H^{2}(\Pi_{+})$ is a Hilbert space.
