In the paper,
Sheen, David M., Douglas L. McMakin, and Thomas E. Hall. "Three-dimensional millimeter-wave imaging for concealed weapon detection." Microwave Theory and Techniques, IEEE Transactions on 49.9 (2001): 1581-1592.
the author decomposes the spherical wave into a superposition of plan-wave components \begin{equation} e^{-j2k\sqrt{(x-x')^2+(y-y')^2+z_0^2}} = \iint_{R^2} e^{jk_{x'}(x'-x) + jk_{y'}(y'-y) + j k_z z_0} dk_{x'}dk_{y'} \end{equation}
$e^{-j2k\sqrt{(x-x')^2+(y-y')^2+z_0^2}}$ represents a spherical wave from $(x',y',z_0)$. k_{x'} and k_{y'} are the Spatial frequency corresponding to x' and y'. $k = \frac{\omega}{c}, (2k)^2 = k_x^2 + k_y^2 + k_z^2$.
How is the formula deducted?
The physical process is as follows. As shown in the Figure. The source(transceiver position) is placed at $(x',y',z_0)$, and there is an object(our target) at position $(x,y,0)$. The target has a reflectivity function $f(x,y,z)$, the ratio of reflected field to incident field.
There is a receiver also at $(x',y',z_0)$. Its signal would be $s(x',y') = \iint f(x,y,z=0) e^{-j2k\sqrt{(x-x')^2 + (y-y')^2 + z_0^2}} dxdy$.
$\nabla^2 f +k^2f =-\delta(x-x') \delta (y-y') \delta(z)$ is the Helmholtz equation with a point source at $(x',y',0)$ with well-known solution
$$\frac{e^{-jkR}}{4\pi R}$$
where $R=\sqrt{(x-x')^2+(y-y')^2+z^2}$.
Now, apply the Fourier Transform to both sides of the Helmholtz equation to obtain
$$-(k_x^2+k_y^2+k_z^2-k^2)F=-e^{jk_xx'+jk_yy'}$$
whereupon solving for $F$ and taking the inverse Fourier Transform reveals that
$$f=\frac{1}{(2\pi)^3}\int \frac{e^{-jk_x(x-x')-jk_y(y-y')-jk_zz}}{k^2-k_x^2-k_y^2-k_z^2}dk_xdk_ydk_z$$