This brief note is about Fourier transform and how related to Laplace Transform.

To begin, we first state the definition of the Fourier transform [$$:]\hat{f}(\omega) = \int_{-\infty}^\infty f(x) e^{-j\omega x} dx[/:$$] and its inverse [$$:]f(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(\omega) e^{j\omega x} d\omega[/:$$] There are two notable features. One, with Laplace transform, the lower integration limit is zero, not negative infinity. Second, Laplace transform uses a transform variable in the complex plane. The transform variable in Fourier transform is a pure imaginary number, restricted to [$:]s = j\omega[/:$].

Generally, Laplace transform is for functions that are semi-infinite or piecewise continuous, as in the step or rectangular pulse functions. We also impose the condition that the function is zero at negative times: [$$:]f(t)=0,\;t<0[/:$$] More formally, we say the function must be of exponential order as [$:]t[/:$] approaches infinity so that the transform integral converges.

A function is of exponential order if there exists(real) constants [$:]K[/:$],[$:]c[/:$] and [$:]T[/:$] such that [$$:]|f(t)| < Ke^{ct}\; \text{ for }\; t > T[/:$$] or in other words, the quantity [$:]e^{-ct} |f(t)|[/:$] is bounded. If [$:]c[/:$] is chosen sufficiently large, the so-called abscissa of convergence, then [$:]e^{-ct} |f(t)|[/:$] should approach zero as [$:]t[/:$] approaches infinity. In terms of the Laplace transform integral[$:]\int_0^{\infty}f(t) e^{-st} dt[/:$], it means that the real part of [$:]s[/:$] must be larger than the real part of all the poles of [$:]f(t)[/:$] in order the integral to converge. Otherwise, we can force a function to be transformable with [$:]e^{-\gamma t} f(t)[/:$] if we can choose [$:]\gamma > c[/:$] such that [$:]Ke^{-(\gamma -c)t}[/:$] approaches zero as [$:]t[/:$] goes to infinity.

We now do a quick two-step to see how the definition of Laplace transform may arise from that of Fourier transform. First, we write the inverse transform of the Fourier transform of the function [$:] e^{-\gamma t} f(t)[/:$], which of course, should recover the function itself: [$$:]e^{-\gamma t} f(t) = \frac{1}{2 \pi}\int_{-\infty}^\infty e^{j\omega t} d \omega \int_0^\infty e^{-\gamma \tau} f(\tau) e^{-j\omega \tau} d\tau[/:$$] where we have changed the lower integration limit of the Fourier transform from[$:]-\infty[/:$] to [$:]0[/:$] because [$:] f(t) = 0[/:$] when [$:]t < 0[/:$]. Next, we move the exponential function [$:]e^{-\gamma t}[/:$] to the RHS to go with the inverse integral and then combine the two exponential functions in the transform integral to give [$$:] f(t) = \frac{1}{2 \pi}\int_{-\infty}^\infty e^{(\gamma + j\omega)t} d \omega \int_0^\infty f(\tau) e^{-(\gamma + j\omega) \tau} d\tau[/:$$] Now we define [$$:] s = \gamma + j\omega[/:$$] So, [$$:] f(t) = \frac{1}{2 \pi j}\int_{\gamma-j\infty}^{\gamma+j\infty} e^{st} ds \int_0^\infty f(\tau) e^{-s \tau} d\tau [/:$$] From this form, we can extract the definitions of Laplace transform and its inverse: [$$:]F(s) = \int_0^\infty f(t) e^{-st} dt [/:$$] and [$$:]f(t) = \frac{1}{2 \pi j}\int_{\gamma-j\infty}^{\gamma+j\infty} F(s) e^{st} ds[/:$$] where again, [$:]\gamma[/:$] must be chosen to be larger than the real parts of all the poles of [$:]F(s)[/:$]. Thus the path of integration of the inverse is the imaginary axis shift by the quantity [$:]\gamma[/:$] to the right.