# Relation between Laplace and Fourier Transform

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

$X(jomega&space;)&space;&space;int_-infty&space;infty&space;&space;x(t)&space;e-jomega&space;tdt----EQN(I)&is-pending-load=1$

Fourier Transform exists only if $int_-infty&space;infty&space;&space;left&space;|&space;x(t)&space;right&space;|dt<&space;infty&is-pending-load=1$

we know that $s=\sigma&space;_&space;jomega&is-pending-load=1$

$X(S)&space;&space;int_-infty&space;infty&space;&space;x(t)&space;e-s&space;tdt&is-pending-load=1$

$X(S)&space;&space;int_-infty&space;infty&space;&space;left&space;|&space;x(t)e-sigma&space;t&space;right&space;e-jomega&space;tdt----EQN(II)&is-pending-load=1$

if we compare Equations (I) and (II) both are equal when  $sigma&space;=0&is-pending-load=1$.

i.e, $X(S)&space;=X(j\omega)|&space;right&space;|_s=j\omega&space;&is-pending-load=1$.

This means that Laplace Transform is same as Fourier transform when $s=j\omega&is-pending-load=1$.

Fourier Transform is nothing but the special case of Laplace transform where  $s=j\omega&is-pending-load=1$indicates the imaginary axis in complex-s-plane.

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

## Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems.

Insert math as
$${}$$