# Relation between Laplace and Fourier Transform

The Fourier transform  of a signal x(t) is given as

$X(j\omega&space;)&space;=&space;\int_{-\infty&space;}^{\infty&space;}&space;x(t)&space;e^{-j\omega&space;t}dt----EQN(I)$

Fourier Transform exists only if $\int_{-\infty&space;}^{\infty&space;}&space;\left&space;|&space;x(t)&space;\right&space;|dt<&space;\infty$

we know that $s=\sigma&space;+&space;j\omega$

$X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}&space;x(t)&space;e^{-s&space;t}dt$

$X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}&space;\left&space;|&space;x(t)e^{-\sigma&space;t}&space;\right&space;|&space;e^{-j\omega&space;t}dt----EQN(II)$

if we compare Equations (I) and (II) both are equal when  $\sigma&space;=0$.

i.e, $X(S)&space;=X(j\omega)|&space;\right&space;|_{s=j\omega&space;}$ .

This means that Laplace Transform is same as Fourier transform when $s=j\omega$.

Fourier Transform is nothing but the special case of Laplace transform where  $s=j\omega$ indicates the imaginary axis in complex-s-plane.

Thus Laplace transform is basically Fourier Transform on imaginary axis in the s-plane.

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## Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems.

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