Relation between Laplace and Fourier Transform

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

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

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

we know that $s=\sigma + j\omega$

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

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

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

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

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. View all posts by Lakshmi Prasanna Ponnala