In a previous post we explored some relationships between the Laplace and the Fourier transforms. In this post we address a common question: “when is the Fourier transform of a signal, $X(j\omega)$, equal to its Laplace transform, $X(s)$, evaluated at $s = j \omega$?” The short answer is: when the region of convergence of the corresponding Laplace transform contains the imaginary axis.

There are two more cases worth mentioning. The first is when the region of convergence is a subset of the open half-plane. In this case the corresponding signal must necessarily diverge, which implies that it does not admit a Fourier transform. The second is when the region of convergence is the entire open half-plane.

Take for example the case of the exponential signal

$$x(t)=e^{-\alpha t} 1(t)$$

whose Laplace transform is

$$X(s)=\int_{0}^{\infty} e^{-\alpha t} e^{-s t} \, dt= \int_{0}^{\infty} e^{-(s + \alpha) t} \, dt = \frac{1}{s + \alpha}, \quad \operatorname{Re}(s) > -\alpha.$$

If $\alpha > 0$ is real then the region of convergence contains the imaginary axis and, as expected,

$$X(j \omega) = \frac{1}{j \omega + \alpha}$$

is the corresponding Fourier transform. The case when $\alpha < 0$ does not admit a Fourier transform since the integral

$$\int_{0}^{\infty} e^{-\alpha t} e^{-j \omega t} \, dt$$

diverges for all possible values of $\omega$. Yet, if one attempts to invert the function $X(j\omega)$ when $\alpha < 0$ one would obtain the function

$$x(t) = -e^{-\alpha t} 1(-t)$$

which is a function different than the one that we started with! Clearly the Fourier transform is not a good tool to deal with unbounded signals, such as the unbounded impulse response of unstable system models.

Examples of signals that have the entire open half-plane as its Laplace transform region of convergence are sines, cosines, the step, and all other signals with poles on the imaginary axis. For all such signals, the Laplace and Fourier transforms are also significantly different.

Continuing the above example with $\alpha = 0$, that is $x(t)=1(t)$, one needs to be careful when evaluating

$$\int_{0}^{\infty} e^{-j \omega t} \, dt.$$

Formally

$$\int_{0}^{\infty} e^{-j \omega t} \, dt= \lim_{\rho \rightarrow \infty} \int_{0}^{\rho} e^{-j \omega t} \, dt = \lim_{\rho \rightarrow \infty} \frac{1 – e^{-j \rho \omega}}{j \omega},$$

which does not have a clear limit. The usual path is to start with $\alpha > 0$ and take the limit as $\alpha \rightarrow 0$. Borrowing from above

$$X(j \omega) = \frac{1}{j \omega + \alpha} = \frac{\alpha}{\omega^2+\alpha^2} – \frac{j \omega}{\omega^2+\alpha^2}$$

and evaluating the limits

$$\lim_{\alpha \rightarrow 0} \frac{\alpha}{\omega^2+\alpha^2} = \pi \delta(\omega), \qquad \lim_{\alpha \rightarrow 0} – \frac{j \omega}{\omega^2+\alpha^2}=\frac{1}{j \omega}$$

one obtains

$$X(j \omega) = \frac{1}{j \omega} + \pi \delta(\omega).$$

The convergence of the first term to an impulse is the result of its area being equal to $\pi$ independently of the value of $\alpha > 0$ (verify!) and $\lim_{\alpha \rightarrow 0} \frac{\alpha}{\omega^2+\alpha^2} = 0$ for all $\omega \neq 0$. Once again $X(j\omega)$ is not equal to the Laplace transform evaluated at $s = j \omega$. Note that the impulse is the Fourier transform of the constant signal $x(t) = 1$, which has Fourier transform $X(\omega) = 2 \pi \delta(\omega)$, so it’s presence on the Fourier transform of $x(t)=1(t)$ should not be that surprising.

A similar pattern holds for other signals for which the Laplace transform has imaginary poles. For example

$$x(t) = \sin(\beta t) 1(t)$$

has as Laplace transform

$$X(s) = \frac{\beta}{s^2 + \beta^2}, \quad \operatorname{Re}(s) > 0$$

and Fourier transform

$$X(j \omega) = \frac{\beta}{\beta^2- \omega^2}+\frac{\pi}{2} \delta(\omega + \beta)\, – \frac{\pi}{2} \delta(\omega – \beta)$$

which contains additional impulses at the frequencies $\beta$ and $-\beta$.

In closing, one might wonder whether such differences can lead to different answers when using the Laplace versus the Fourier transform to, say calculate the response of a stable linear system to a given input. To make it concrete let

$$g(t)=\alpha e^{-\alpha t} 1(t), \quad \alpha > 0, \quad u(t)=1(t).$$

Using the Laplace transform

$$G(s)=\frac{\alpha}{s + \alpha}, \quad U(s) = \frac{1}{s}$$

and

$$Y(s) = G(s) U(s) = \frac{\alpha}{s(s+\alpha)}=\frac{1}{s} – \frac{1}{s+\alpha}$$

or

$$y(t)=(1 – e^{-\alpha t})1(t).$$

Contrast that with the Fourier transform

$$G(j\omega)=\frac{\alpha}{j\omega + \alpha}, \quad U(j\omega) = \frac{1}{j\omega} + \pi \delta(\omega)$$

and

$$\begin{aligned}

Y(j\omega) = G(j\omega) U(j\omega) &= \frac{1}{j\omega} -\frac{1}{j\omega+\alpha} + \frac{\alpha \pi}{j\omega + \alpha} \delta(\omega) \\&= \frac{1}{j\omega} + \pi \delta(\omega) -\frac{1}{j\omega+\alpha} + \frac{j\omega\pi}{j\omega + \alpha} \delta(\omega)

\end{aligned}$$

so that

$$y(t)=(1 – e^{-\alpha t})1(t)$$

after noticing that

$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{j\omega\pi}{j\omega + \alpha} \delta(\omega) e^{j \omega t} \, d\omega = 0.$$

So the answers are the same, as one would expect, but which tool does the job *better*?

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