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Use ${\hat{\psi }}_{\mathrm{sc}}(\nu )=\int {\psi }_{\mathrm{sc}}(x,0)e^{-i\nu x}dx$, inverse factor $1/(2\pi )$, and $k=\omega /c$. Fourier transform the homogeneous Helmholtz equation givesAn upward radiating field has no incoming component from $z=+\infty$. For $q(\nu )=\sqrt{k^2-{\nu }^2}$, take the nonnegative real root when $|\nu |<k$ and the positive imaginary root when $|\nu |>k$. Consequently the outgoing angular spectrum isThe propagating terms go upward; the evanescent terms decay with increasing $z$. Grazing values are understood by a radiation limit.

There is a reference-plane qualification in the question. For an arbitrary rough surface the full line $z=0$ need not lie in the medium: already $f(x)=h>0$ puts it in vacuum. In that case the stated physical trace is undefined without continuation. The exact representation is obtained instead on a plane $z=z_{\ast }>\sup f$,One can formally set ${\hat{\psi }}_{\mathrm{sc}}={\hat{\psi }}_{\ast }{e}^{-iq{z}_{\ast }}$, but convergence of that continued spectrum down to the rough boundary is an additional assumption. This is the reference-plane validity of a rough-surface angular spectrum. The first boxed formula is exact in an appropriate homogeneous upper half-space; it is not automatically an exact representation at every point of a completely arbitrary rough domain. Small-height perturbation theory in part (c) uses the mean plane order by order.