I have a spatio-temporal signal $f(x, t)$ that propagates at a constant velocity $v$. To represent that propagation I'm reading in multiple papers that we can represent it in the Fourier domain with the delta function, where the propagation is formalized by $\delta(x - vt)$. The papers say that it can be proven that the Fourier transform of that delta function is $\delta(k,\omega) = \delta(kv + \omega)$ where $\omega$ is the time angular frequency [$s^{-1}$], $k$ the spatial frequency [$m^{-1}$] and $v = \frac{\omega}{k}$. I could not find that proof anywhere, so I'm trying to get as close as possible to a proof. Is the following reasoning true for showing the integration property of $\delta$ in the Fourier domain?

Starting from the Fourier transform of $\delta(x - vt)$: $$ F(k,\omega ) = \int_{-\infty}^\infty \int_{-\infty}^\infty \delta(x - vt)e^{-i(kx + \omega t)}\, \mathrm{d}x \, \mathrm{d}t $$

I make a change of variable $\tau$ = $x - vt$ with $\mathrm{d}x = \mathrm{d}\tau$, which gives:

\begin{eqnarray} F(k,\omega ) &=& \int_{-\infty}^\infty \int_{-\infty}^\infty \delta(\tau)e^{-i(k(\tau + vt) + \omega t)} \,\mathrm{d}\tau \,\mathrm{d}t \\ &=& \int_{-\infty}^\infty e^{-i(kv + \omega)t}\,\mathrm{d}t \end{eqnarray}

From the basic definition of the Fourier transform of a signal $f(t)$ given by $F(\omega ) =\int_{-\infty}^\infty f(t) e^{-i\omega t} \,\mathrm{d}t$, the above equation gives $F(k, \omega ) = \delta(kv+\omega)$.

If incomplete or wrong, what's the right way to show the final result?

[UPDATE] I should rephrase my problem. From the answers below regarding the relationship between a given Fourier transform and its inverse mapping applied to the delta function, fixing for the missing $2\pi$ in my above formulations, I would get:

$$ \mathscr{F}^{-1}(\delta(\omega)) = \frac{1}{2\pi}\int_{-\infty}^\infty \delta(\omega) e^{i\omega t}d\omega = \frac{1}{2\pi} $$

leading to the delta function expressed as:

$$ \delta(\omega) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-i \omega t}dt $$

and if $\omega$ changes to $kv + \omega$, I get to the final result: $$ \frac{1}{2\pi}\int_{-\infty}^\infty e^{-i (kv + \omega) t}dt = \delta(kv + \omega) $$

The paper goes a bit further, saying that if we formalize the 1D propagation of some signal at velocity $r_x$ as

$$ f(x,t) = f_0(x)\delta(x - r_x t) $$ with x a 1D spatial dimension, and $f_0(x)$ the spatial distribution of the signal at time $t=0$, then it's Fourier transform is (here I'll use ordinary space-time frequencies $u$ and $w$ like in the paper):

\begin{eqnarray} F(u,w) &=& \iint_{-\infty}^\infty f(x,t)e^{-i2\pi(ux + wt)} dx dt \\ &=& F_0(u) \ \delta(ur_x + w) \end{eqnarray}

where $F_0(u)$ is the Fourier transform of $f_0(x)$.

The previous proof makes me see that the Fourier transform of $\delta(x - r_x t)$ taken "in isolation" would be equal to $\delta(u r_x + w)$, but I do not understand how the last integral ends up equal exactly to the product of the two fourier transforms.

  • 1
    $\begingroup$ There's gonna be a $\frac1{2\pi}$ factor in there because you're using angular frequency. And I think there is a time-reversal problem. $\endgroup$ Jun 8, 2023 at 3:57
  • 2
    $\begingroup$ @robertbristow-johnson: That factor is not only related to the use of angular frequency but also to the scaling definition of the Fourier transform. With unitary scaling, the 2-D Fourier transform of $\delta(x-vt)$ would indeed be $\delta(kv+\omega)$. $\endgroup$
    – Matt L.
    Jun 8, 2023 at 15:10
  • 2
    $\begingroup$ oh, that's right. shows you how often in my life I do either 1. 2-D Fourier transform or 2. use angular frequency. I'm a big believer in using ordinary frequency $f$ in the continuous Fourier Transform and taking advantage of the simplicity of scaling factors that come with it. With that $\frac{1}{\sqrt{N}}$ thing in the DFT, I am not yet completely decided which convention is best. $\endgroup$ Jun 8, 2023 at 15:49

2 Answers 2


The Fourier Transform, using angular frequency:

$$ F(\omega) \triangleq \int\limits_{-\infty}^{\infty} f(t) \, e^{-i\omega t} \,\mathrm{d}t $$

has inverse mapping:

$$ f(t) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} F(\omega) \, e^{+i\omega t} \,\mathrm{d}\omega $$


$$ e^{-i\omega \tau} = \int\limits_{-\infty}^{\infty} \delta(t-\tau) \, e^{-i\omega t} \,\mathrm{d}t $$


$$\begin{align} \delta(t-\tau) &= \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} e^{-i\omega \tau} \, e^{+i\omega t} \,\mathrm{d}\omega \\ \\ &= \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} e^{-i\omega (\tau-t)} \,\mathrm{d}\omega \\ \end{align}$$

Now simply swap the roles of the two symbols $t$ and $\omega$.

$$ \delta(\omega-\tau) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} e^{-i (\tau-\omega) t} \,\mathrm{d}t $$

change the sign of $\omega$

$$ \delta(-\omega-\tau) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} e^{-i (\tau+\omega) t} \,\mathrm{d}t $$

and set $\tau \leftarrow kv$

$$ \delta(-\omega-kv) = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty} e^{-i (kv+\omega) t} \,\mathrm{d}t $$

and finally multiply both sides by $2\pi$ and recognize that $\delta(\cdot)$ is an even-symmetry function.

$$ 2 \pi \, \delta(\omega+kv) = \int\limits_{-\infty}^{\infty} e^{-i (kv+\omega) t} \,\mathrm{d}t $$

  • $\begingroup$ ooooooo current arr bee jayy - if Hugh Hefner did signal processing $\endgroup$ Jun 8, 2023 at 15:10
  • $\begingroup$ man, you look smashing. I've always liked the smile. $\endgroup$ Jun 8, 2023 at 15:11
  • $\begingroup$ @PeterK. took the pic of both of us. I just cropped it. $\endgroup$ Jun 8, 2023 at 15:20
  • $\begingroup$ I very much appreciate this detailed derivation. The absence of the $2\pi$ in the original paper added another bit of confusion, I was unclear which properties of the delta function were being used. Here's the original paper where I first came across this problem (in 3D: 2D + time): ntrs.nasa.gov/citations/19830015902 For simplicity I just turned the problem in 2D: 1D + time. $\endgroup$
    – Wall-E
    Jun 8, 2023 at 20:51
  • $\begingroup$ @robertbristow-johnson I updated my problem to consider the most important result I wish to prove, which is the Fourier transform of the product of that delta function with an initial signal to describe its propagation, and not just the propagation of delta itself. $\endgroup$
    – Wall-E
    Jun 14, 2023 at 0:41

Your derivation is correct. Assuming the definition of the Fourier transform that you used and that is common in signal processing, there's just that factor $2\pi$ missing, as already pointed out in RBJ's answer.

It is well-known, and easy to show using the correspondence $\mathcal{F}\{\delta(t)\}=1$, that


where I've tried to use innocent names for the variables. Eq. $(1)$ can also be found in this list of properties of the Dirac delta impulse.

From $(1)$ it is obvious that


which, apart from the factor $2\pi$, agrees with your result.

Note that in different fields, the Fourier transform is defined differently. This can affect the scaling of the transform and the sign of the exponent. So if you define the 2-D Fourier transform as

$$F(k,\omega)={ \color{red} {\frac{1}{2\pi}} }\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,t)e^{-i(kx+\omega t)}dxdt\tag{3}$$

then you will find that the Fourier transform of $\delta(x-vt)$ equals $\delta(kv+\omega)$. The definition of the scaling of the Fourier transform could be the reason for the discrepancy between the results.

If you define the 2-D Fourier transform as in $(3)$, the inverse transform also needs to be scaled accordingly:

$$f(x,t)={ \color{red} {\frac{1}{2\pi}} }\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}F(k,\omega)e^{i(kx+\omega t)}dkd\omega\tag{4}$$

With definitions $(3)$ and $(4)$, the two Dirac impulses $\delta(x-vt)$ and $\delta(kv+\omega)$ are a transform pair.

The scaling convention used in $(3)$ and $(4)$ makes the (2-dimensional) Fourier transform unitary (see the table here).

  • $\begingroup$ "You're derivation" how is it just me who never in one's life made this error. If odds are fixed, the outcome's exponentially less likely to persist, so lasting till retirement will be impressive. Except I'll never retire voluntarily, and my lifespan shall only potentially end by heat death of the $\text{multiverse}$ - where $\text{multiverse} \supseteq \text{universe}$, making the lower bound on the upper bound $T_\text{retirement_age} /T_\text{current_age} \geq 10^{89}$ - which will prove that the non-zero exponential base, hence the original odds, are very impressively low. $\endgroup$ Jun 8, 2023 at 14:11
  • 5
    $\begingroup$ @OverLordGoldDragon: Glad you never make that error. Corrected. But you could have used fewer words to point it out ... $\endgroup$
    – Matt L.
    Jun 8, 2023 at 14:46
  • 1
    $\begingroup$ I realize that the paper I looked at was using ordinary frequency. I rewrote the problem with angular frequency while forgetting to adapt the definition of the Fourier transform. That made me miss the $2\pi$ factor. $\endgroup$
    – Wall-E
    Jun 8, 2023 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.