Fourier transform of a propagating Dirac delta

Question

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.

Answer 1

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:

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

Since

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

Then

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

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.

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

Answer 2

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

$$\int_{-\infty}^{\infty}e^{-ixy}dx=\int_{-\infty}^{\infty}e^{ixy}dx=2\pi\delta(y)\tag{1}$$

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

$$\int_{-\infty}^{\infty}e^{-i(kv+\omega)t}dt=2\pi\delta(kv+\omega)\tag{2}$$

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).