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.


1 Star2 Stars3 Stars4 Stars5 Stars (No Ratings Yet)

Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.