# Fourier transform of exponent: Delta pulse or hyperbola?

Why do some tables say that Laplace (or Fourier?) inverse of exponential is a time-shifted delta pulse

\begin{align} \delta (t) &\overset{\mathcal F}{\Longleftrightarrow} 1\\ \delta (t-t_0) &\overset{\mathcal F}{\Longleftrightarrow} e^{-j2\pi f t_0}\\ 1 &\overset{\mathcal F}{\Longleftrightarrow} \delta (f)\\ e^{j2\pi f_0 t} &\overset{\mathcal F}{\Longleftrightarrow}\delta (f-f_0) \end{align}

And table: \begin{array}{c|c} f(t)=\mathcal L^{-1}\left\{\mathcal F(s)\right\}&\mathcal F(s)=\mathcal L\left\{f(t)\right\}\\\hline e^{at}&\displaystyle\frac{1}{s-a} \end{array}

whereas others say that it is a (pole) hyperbola rather than a delta-pulse?

$$e^{at}\overset{\mathcal F}{\Longleftrightarrow}\frac{1}{s-a}$$

As pointed out by Peter K., it is important to distinguish between Laplace and Fourier transforms. The first few transform pairs in your question are Fourier transform pairs, whereas the last pair is a correspondence of the unilateral Laplace transform:

$$F(s)=\int_{0}^{\infty}f(t)e^{-st}dt$$

In the last transform pair in your question the $\mathcal{F}$ symbol is wrong because it would imply that it is a general Fourier transform pair. To be precise, the (Laplace transform) correspondence is actually

$$e^{at}u(t)\Longleftrightarrow \frac{1}{s-a}\tag{1}$$

because with the unilateral Laplace transform we only consider causal time functions, which satisfy $f(t)=0$ for $t<0$. Note that this is not necessarily the case with Fourier transforms.

Let's now consider the Fourier transform of the (causal) time function in (1):

$$F(j\omega)=\int_{-\infty}^{\infty}e^{at}u(t)e^{-j\omega t}dt=\int_{0}^{\infty}e^{-(j\omega - a)t}dt=\frac{1}{j\omega-a},\quad \textrm{for Re}\{a\}<0\tag{2}$$

Comparing (1) and (2) we see that, for $Re\{a\}<0$, the two transforms are indeed the same for $s=j\omega$. For $Re\{a\}>0$ the Fourier transform does not exist because the region of convergence of the Laplace transform $F(s)$ does not contain the imaginary axis.

Let us finally consider the time function $f(t)=e^{j\omega_0 t}$. If we were to multiply it with the step function $u(t)$, its unilateral Laplace transform exists according to equation (1) with $a=j\omega_0$. However, if we consider the function for $-\infty<t<\infty$, then its (bilateral) Laplace transform does not exist, whereas its Fourier transform does exist. It is given by a delta impulse in the frequency domain as shown in your table.

How can we now make sense of the relation between the Fourier and the (unilateral) Laplace transform? We must consider three cases:

1. The region of convergence of the Laplace transform $F(s)$ contains the $j\omega$ axis: then the Fourier transform is simply the Laplace transform evaluated at $s=j\omega$.

2. The region of convergence of the Laplace transform $F(s)$ does not contain the $j\omega$ axis: then the Fourier transform does not exist.

3. The region of convergence is $Re\{s\}>0$ but there are singularities on the $j\omega$ axis: both transforms exist but they have different forms. The Fourier transform has additional delta impulses. Consider the function $f(t)=e^{j\omega_0 t}u(t)$. From (1), its Laplace transform is given by

$$F(s)=\frac{1}{s-j\omega_0}$$ However, due to the singularity on the $j\omega$ axis, its Fourier transform is

$$F(j\omega)=\pi\delta(\omega-\omega_0)+\frac{1}{j\omega-j\omega_0}$$

It might seem that the Laplace transform is more general than the Fourier transform (when looking at the second point above), but this is actually not the case. In system theory, there are many important functions which are not causal, e.g. the impulse responses of ideal band-limiting (brick-wall) filters. For these functions the Laplace transform does not exist, but their Fourier transform exists. The same is of course true for sinusoidal functions defined for $-\infty<t<\infty$.

• Wolfram Alpha Link: FourierTransform[E^(I a t) Piecewise[{{1, t >= 0}}, 0], t, ω] Use the non-unitary fourier transform convention to get the same result as Matt L. Sep 2, 2021 at 7:36

The first table is a table of Fourier transform pairs. The second table is a table of one Laplace transform pair.

If things were all correct, then you should be able to set $s = j\omega$ in the Laplace transform pair to get the Fourier transform pair:

$$e^{j\omega_0 t} \leftrightarrow \frac{1}{j(\omega - \omega_0)}$$

However, there is a problem with this: it has a singularity at $\omega = 0$.

What this means is that the usual definition of functions breaks down. See this great answer to a related (duplicate?) question from Dilip.

• It is actually because things are correct that we can't always just set $s=j\omega$. :) Apr 1, 2014 at 9:12