first of all, sorry for this basic question in signal processing, I'm pretty new in this field and have been stuck in a situation a little beyond my expertise.

The idea is to derive the instantaneous wavenumber direction as an estimation of the direction normal of a 2D field by using a directional Hilbert transform in one of the directions. 

As far as I can recall correctly, all I have to do is to do a **1D Fourier** transform of each data vector, **apply a $\pi/2$ phase-shift** in the wavenumber domain and then apply an **inverse Fourier transformation** to get the Hilbert transform.

So the Hilbert transform is defined as:
$$
H(x(t)) = \frac{1}{\pi} PV \int_{-\infty}^{\infty}\frac{x(\tau)}{\tau-t}d\tau
$$
where $PV$ denotes the principal value of the integral and $H$ is the Hilbert transform of a vector $x(t)$. This is supposed to be equal to the convolution $\frac{1}{\pi t} * x(t)$ and

$$
F(H(x(t))) = F(\frac{1}{\pi t}) F(x(t)).
$$

So the Fourier transformation and multiplication of the two factors on the right side is equal to the Fourier transformation of the Hilbert transformation of my vector, right?

My question now is how to apply the phase shift on the Hilbert transform? is it simply multiplying the product on the right side by $\pi/2$ in the wavenumber domain?

Thanks a lot for any help!