So, in my previous question https://math.stackexchange.com/questions/582224/where-does-this-formula-for-prediction-of-a-multiple-wave-come-from, I get that using this picture:
we have so far written the time it takes for a multiple to travel in the form:
$$T_2 = T_1 + \left( \frac{1}{v} \frac{2h}{\cos (\theta)} \right)$$
where
$$\cos(\theta) = \frac{h}{\sqrt{\frac{x^2}{4} + h^2}}.$$
AFter substituting in for $\cos(\theta)$, this can then be re-written as
$$T_2 = T_1 + \sqrt{\tau^2 + \frac{x^2}{v^2}}$$
where $\tau = \frac{4h^2}{v^2}$. From here though, it then says that we can represent this in the frequency domain as:
$$R h(\omega) e^{2 \pi i f (T_1 + \sqrt{\tau^2 + \frac{x^2}{v^2}}}$$
where $R$ is some constant including decay and $h(\omega)$ is the source wavelet.
I don't understand how they have managed to transform the previous equation into this one. Where has the exponential come from?
Here's what I can make of it.
You have
$$t_2 = t_1 + \left( \frac{1}{v} \frac{2h}{\cos (\theta)} \right)$$.
The part in parentheses is the additional time that it takes for the wave to go through the red path.
Then, you express this in terms of $x$ and $h$:
$$\cos(\theta) = \frac{h}{\sqrt{\frac{x^2}{4} + h^2}}.$$
so that
$$t_2 = t_1 + \sqrt{\tau^2 + \frac{x^2}{v^2}}$$
where $\tau = \frac{4h^2}{v^2}$. So far, so good.
Now, what it looks like is that you want to express $h(t)$ for this delayed wave in the frequency domain. Then
$$h(t + \sqrt{\tau^2 + \frac{x^2}{v^2}}) = \frac{1}{\sqrt{2\pi}}H(\omega)e^{+i\omega t_2}$$
where $\omega = 2\pi f$ is the angular frequency and $H(\omega)$ is the Fourier transform of $h(t)$.
Something like that?