4.1.2. Relation with the Fourier transform

In case we take \(s=j\omega\) we obtain the continuous time Fourier transform (of a causal signal \(x(t)\)):

\[X(s) \big|_{s=j\omega} = X(\omega)\]

Please note that we need this somewhat sloppy notation to distinguish the Laplace transform \(X(s)\) from the Fourier transform \(X(\omega)\). Writing \(X(j\omega)=X(\omega)\) would be totally confusing. This practice of using the argument of a function to distinguish it from other functions has penetrated deeply into the signal processing community and is a fact we have to live with.

The domain of the Laplace transform \(X(s)\) is its region of convergence being a subset of the complex plane. Along the imaginary axis we find the Fourier transform.

Now consider \(X\) to be the Laplace transform of \(x\), i.e.:

\[X(s) = \int_{0}^{\infty} x(t) e^{-st} dt\]

Setting \(s=\sigma+j\omega\) we have:

\[X(\sigma+j\omega) = \int_{0}^{\infty} x(t) e^{-(\sigma+j\omega) t} dt = \int_{0}^{\infty} e^{-\sigma t} x(t) e^{-j\omega t} dt\]

Indeed for \(\sigma=0\) we find the Fourier transform of \(x(t)\) but for other values of \(\sigma\) we find that the Laplace transform equals the Fourier transform of \(x(t)e^{-\sigma t}\).