A Fourier domain look

It is also possible to implement the transformation to full-aperture angle in the Fourier domain. Although this transformation does not take into account the lateral and vertical variations of $\mathcal{S}$, it's still an interesting exercise. We can link this theory with Fourier transform by knowing that:  

$$\frac{\partial t}{\partial m_\xi} = \frac{k_{m_\xi}}{\omega}... ...\; \frac{\partial t}{\partial h_\xi} = \frac{k_{h_\xi}}{\omega}$$ (8)

From equations (8), it is well known that

$$\frac{\partial z_\xi}{\partial h_\xi} = \frac{k_{h_\xi}}{k_{... ...{\partial z_\xi}{\partial m_\xi} = \frac{k_{m_\xi}}{k_{z_\xi}}.$$ (9)

Therefore, the Fourier equivalent for equations (7) is

$$\begin{eqnarray} -\tan{\gamma} &=& \frac{\mathcal{S}^2 -1 \pm \sqrt{ (1-\mathca... ...} \mathcal{S}- \frac{k_{m_\xi}}{k_{z_\xi}} \mathcal{S}^2 \right]}.\end{eqnarray}$$ (10)

Equations (10) can be used to transform SODCIGs into full-aperture ADCIGs through stretching of the offset and midpoint axes, but this is only valid for a velocity-ratio function $\mathcal{S}$ that is constant along the image ($I(z_\xi,m_\xi,h_\xi)$).