# Representation of 1D motion in the frequency domain with Delta function

Given a time-dependent pattern in a 2D function $$b(x,t)$$ that follows a path with a constant velocity $$v$$, which I define with the use of the delta function as: $$b(x,t)=b_0(x)\delta(x-v t)$$ where $$b_0(x)$$ is the function at time $$t=0$$.

I'm confronted to a frequency-domain representation of it with the fourier transform: $$B(k,w) = \iint_{-\infty}^{\infty} b(x,t)e^{-i2\pi(kx + wt)}dx \ dt$$ with $$k,w$$ the 1D space & time frequencies. I'm seeing the following result that I cannot prove:

$$B(k,w) = B_0(k) \delta(w + k v)$$

I tried starting by simply replacing $$b(x,t)$$ with its delta representation:

$$B(k,w) = \iint_{-\infty}^{\infty} b_0(x)e^{-i2\pi(kx + wt)} \delta(x-v t) dx \ dt$$

I tried first to have a more explicit separation of space and time with

$$B(k,w) = \int_{-\infty}^{\infty} b_0(x)e^{-i2\pi kx} dx \int_{-\infty}^{\infty} \delta(x-v t) e ^{-i2\pi w t} dt$$

That would get me at least half-way there:

$$B(k,w) = B_0(k) \int_{-\infty}^{\infty} e ^{-i2\pi w t} \delta(x-v t) dt$$

So this problem seems to come down to proving that the Fourier transform of $$\delta(x-vt)$$ acts as $$\delta(w + kv)$$, which eludes me.

[Update 1]: I cannot separate the way I did as $$\delta(x - vt)$$ depends on both $$x$$ and $$t$$.

So I shoud let $$B(k, w)$$ written as: $$B(k,w) = \iint_{-\infty}^{\infty} b_0(x) \delta(x-v t) e ^{-i2\pi(kx + w t)} dx \ dt$$

[Update 2]: In the latter expression, if I apply the fundamental property of the Dirac distribution that for a test function $$f(x)$$ continuous at $$x=a$$ it does $$\int_{-\infty}^{\infty} f(x)\delta(x-a) dx = f(a)$$, could I consider $$a = v t$$ and get:

$$\begin{eqnarray} B(k,w) &=& \int_{-\infty}^{\infty} b_0(vt) e ^{-i2\pi(kvt + w t)} \ dt \\ &=& \int_{-\infty}^{\infty} b_0(vt) e ^{-i2\pi t(kv + w)} dt \end{eqnarray}$$

So rearranging the argument of the exponential shows an effect of $$\delta(kv+w)$$ but I don't see the formal way to get the final result of $$B_0(k) \delta(kv+w)$$.