Connection Between the Fourier and Laplace Transforms

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B. P. Lathi & Roger Green. (2018). Chapter 7: Continuous-Time Signal Analysis. Signal Processing and Linear Systems (3nd ed., pp. 700-701). Oxford University Press.

The (bilateral) Laplace transform of a signal \(x(t)\) is \[ X(s)=\int_{-\infty}^{\infty} x(t) e^{-s t} d t \]

Setting \(s=j \omega\) in this equation yields \[ X(j \omega)=\int_{-\infty}^{\infty} x(t) e^{-j \omega t} d t \] where \(X(j \omega)=\left.X(s)\right|_{s=j \omega}\). But, the right-hand-side integral defines \(X(\omega)\), the Fourier transform of \(x(t)\). Does this mean that the Fourier transform can be obtained from the corresponding Laplace transform by setting \(s=j \omega\) ? In other words, is it true that \(X(j \omega)=X(\omega)\) ?

Yes and no. Yes, it is true in most cases. For example, when \(x(t)=e^{-a t} u(t)\), its Laplace transform is \(1 /(s+a)\), and \(X(j \omega)=1 /(j \omega+a)\), which is equal to \(X(\omega)\) (assuming \(a<0)\). However, for the unit step function \(u(t)\), the Laplace transform is \[ u(t) \Longleftrightarrow \frac{1}{s} \quad \operatorname{Re} s>0 \]

The Fourier transform is given by \[ u(t) \Longleftrightarrow \frac{1}{j \omega}+\pi \delta(\omega) \]

Clearly, \(X(j \omega) \neq X(\omega)\) in this case. To understand this puzzle, consider the fact that we obtain \(X(j \omega)\) by setting \(s=j \omega\) in Eq. (7.24). This implies that the integral on the right-hand side of Eq. (7.24) converges for \(s=j \omega\), meaning that \(s=j \omega\) (the imaginary axis) lies in the ROC for \(X(s)\). The general rule is that only when the ROC for \(X(s)\) includes the \(\omega\) axis, does setting \(s=j \omega\) in \(X(s)\) yield the Fourier transform \(X(\omega)\), that is, \(X(j \omega)=X(\omega)\). This is the case of absolutely integrable \(x(t)\). If the ROC of \(X(s)\) excludes the \(\omega\) axis, \(X(j \omega) \neq X(\omega)\). This is the case for exponentially growing \(x(t)\) and also \(x(t)\) that is constant or is oscillating with constant amplitude.

The reason for this peculiar behavior has something to do with the nature of convergence of the Laplace and the Fourier integrals when \(x(t)\) is not absolutely integrable. \({ }^{\dagger}\)

This discussion shows that although the Fourier transform may be considered as a special case of the Laplace transform, we need to circumscribe such a view. This fact can also be confirmed by noting that a periodic signal has the Fourier transform, but the Laplace transform does not exist.