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?

  • 1
    $\begingroup$ You have to consider the two sided laplace transform! if you do so, there is indeed a relation of the kind you describe $\endgroup$
    – tired
    Jul 12, 2015 at 20:00
  • $\begingroup$ @tired thanks for your comment. But in the textbook where i saw the $iw=s$ substitution, they don't mention the two sided version of the transform. $\endgroup$
    – Ambesh
    Jul 13, 2015 at 9:23

2 Answers 2


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 $f(t)$.


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.


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