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The angular spectrum is the transverse Fourier transform of a monochromatic field on a reference plane. In a uniform medium, each transverse wavevector component propagates with a longitudinal wavenumber satisfying the Helmholtz equation. Components beyond the propagating wavenumber range are evanescent waves. The outgoing angular spectrum chooses the longitudinal branch that propagates or decays away from the plane.

Taylor expansion of the Dirichlet boundary condition gives the first-order boundary trace $-h{\partial }_z{\psi }^{[0]}$ at the mean plane. The outgoing angular spectrum then propagates it. A zero-mean surface gives zero mean for this linear rough correction, while the complete scattered field still contains its nonzero flat reflection.

The outgoing angular spectrum propagates prescribed scalar boundary data into a half-space. For $e^{-i\omega t}$ time dependence, $\beta (q)=\sqrt{k^2-q^2}$ has nonnegative real part on propagating modes and nonnegative imaginary part on evanescent modes. Its trace at the boundary is $g$, and each mode solves the Helmholtz equation. The spectral representation can be interpreted through distributions for nondecaying plane-wave data.

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.

Fix time dependence $e^{-i\omega t}$ and use pressure amplitude for $\psi$. Write the incident acoustic plane wave as ${\psi }_i=A{e}^{i\alpha x-i\beta z}$, where $\alpha =k\cos \theta$, $\beta =k\sin \theta >0$. “Perfectly reflecting” specifies zero absorbed energy but does not uniquely select a boundary condition: a rigid wall has a Neumann boundary condition, whereas a pressure-release surface has a Dirichlet boundary condition. We first give the rigid-wall result, and then the alternative pressure-release result.

For the rigid surface the normal-velocity condition, without an immaterial normalizing factor, isExpand $\psi ={\psi }^{[0]}+{\psi }^{[1]}+{\psi }^{[2]}+\cdots$ in the height amplitude, scaling a fixed differentiable profile. The flat total field is ${\psi }^{[0]}=A{e}^{i\alpha x}(e^{-i\beta z}+e^{i\beta z})$. Taylor expansion givesLet $\hat{h}(q)=\int h(x)e^{-iqx}dx$ and $\beta (q)=\sqrt{k^2-q^2}$ on the outgoing sheet: it is positive real for propagating modes and positive imaginary for evanescent modes. Fourier transforming this boundary derivative gives $2A(k^2-\alpha q)\hat{h}(q-\alpha )$. The outgoing angular spectrum therefore yields the small-height Neumann scattering fieldFor an infinite stationary surface the Fourier quantities may be interpreted as random distributions or through a finite illumination window and its limiting mean. The correction is linear in $h$ and $h^{\prime }$, so zero mean height and stationarity giveOnly the rough correction has zero mean at first order; the flat reflected wave remains.

The first nontrivial coherent correction depends on correlation, not just the r.m.s. height. Define $C_h(s)=⟨h(x+s)h(x)⟩$, its power spectrum of surface height $S_h(q)=\int C_h(s)e^{-iqs}ds$, and ${\sigma }^2=(2\pi {)}^{-1}\int S_h(q)dq$. At second order the flat-field third normal derivative and mixed derivative vanish at zero, leavingFor each first-order outgoing Fourier component, its derivative data are $g_1(q)=2A(k^2-\alpha q)\hat{h}(q-\alpha )$, its second normal derivative is $i\beta (q)g_1(q)$, and its tangential derivative is $q{g}_1(q)/\beta (q)$. Averaging their products with $h,h^{\prime }$ gives the mean derivativeDividing by the specular outgoing derivative $i\beta$ therefore gives oblique coherent reflection from a stationary rigid rough surface:This second-order expression requires the relevant spectral moments to exist and the boundary expansion to remain controlled. Small $|kh|$ alone does not control arbitrarily large slopes or high-frequency roughness. The given stationarity and $\sigma$ determine the first-order mean above, but do not determine this spectrum-weighted correction numerically.

For a pressure-release surface, instead impose $\psi (x,h)=0$. The flat reflected coefficient is $-1$, and Taylor expansion gives ${\psi }^{[1]}(x,0)=2i\beta Ah{e}^{i\alpha x}$. Thus small-height Dirichlet scattering givesThe zero-mean first correction again vanishes on averaging. At second order ${\psi }^{[2]}(x,0)=-h{\partial }_z{\psi }^{[1]}(x,0)$, givingIf the roughness spectrum is concentrated near $q=\alpha$ after the incident tangential shift, the additional long-correlation approximation yields $r_N\simeq 1-2{\beta }^2{\sigma }^2$ and $r_D\simeq -1+2{\beta }^2{\sigma }^2$. A local Gaussian random-height phase model would then give $r_N=e^{-2{\beta }^2{\sigma }^2}$ and $r_D=-e^{-2{\beta }^2{\sigma }^2}$, but Gaussian statistics and that local approximation were not supplied. They must not be inferred from stationarity and r.m.s. height alone.

Take $e^{-i\omega t}$ time dependence and measure angles from the upward normal. Write $p_i=k\sin {\theta }_i$, $q_i=k\cos {\theta }_i>0$ and choose incident amplitude $A_i$. Separate the flat Dirichlet reflection from the rough correction:Taylor expansion of the boundary condition about $z=0$ givesSince ${\partial }_z{\psi }^{[0]}(x,0)=-2i{q}_i{A}_i{e}^{i{p}_{i}x}$, the induced trace isUse $\hat{h}(Q)=\int h(x)e^{-iQx}dx$ and $\beta (\xi )=\sqrt{k^2-{\xi }^2}$ with the outgoing branch: $\beta >0$ for $|\xi |<k$ and $\mathrm{Im}\beta >0$ outside. The outgoing angular spectrum solves the Helmholtz equation with that trace, giving the first-order rough-surface scattered fieldIf “scattered field” includes the specular reflection, add $-A_i{e}^{i{p}_{i}x+i{q}_{i}z}$ to this rough correction. Fourier transforms of an infinite random surface are generalized objects; finite-window transforms give an ordinary integral and are also needed for its far-field intensity. The expansion assumes a fixed regular surface shape scaled to small height. Highly oscillatory profiles can require additional control of slopes and spectral moments; small height alone is not a uniform error estimate for every profile.

Work above the surface, with $e^{-i\omega t}$ time dependence. Define the scattered field by $\psi ={\psi }_i+{\psi }_s$, so it includes the reflection from a flat surface. For small-height Dirichlet scattering, write $h=\epsilon \eta$ and expandThe zeroth-order total field ${\psi }^{[0]}={\psi }_i+{\psi }_s^{[0]}$ satisfies the Dirichlet boundary condition ${\psi }^{[0]}(x,0)=0$. Taylor expansion at the perturbed boundary givesThus the first-order rough-surface scattered field has mean-plane boundary data $g_1(x)=-h(x){\partial }_z{\psi }^{[0]}(x,0)$.

Use the outgoing angular spectrum to solve this boundary-value problem. For $\hat{g}(q)=\int g(x)e^{-iqx}dx$, letThe branch ensures upward propagation or upward evanescent decay. Each component solves the Helmholtz equation, and $\mathcal{E}g$ has trace $g$ at $z=0$. ConsequentlyThis gives the scattered field through first order by adding the two contributions. If one reserves “rough scattered field” for the non-specular correction, it is ${\psi }_s^{[1]}$ alone; the convention here keeps the flat reflection as well.

The expansion is in height for a fixed sufficiently regular profile. The condition $|kh|\ll 1$ controls the incident wave's height expansion, but very short spatial scales can create large evanescent normal derivatives. The surface regularity and relevant spectral moments must also control the subsequent boundary expansions; small amplitude alone is not a uniform guarantee for arbitrarily fine roughness.

First-order Dirichlet roughness has mean-plane trace $2i{q}_i{A}_{i}h(x)e^{i{p}_{i}x}$, where $p_i^2+q_i^2=k^2$ and $q_i>0$. Stationary phase applied to the outgoing angular spectrum evaluates the far-field amplitude at tangential wavenumber $k\sin {\theta }_s$. For a stationary surface illuminated over length $D$, the mean squared height transform divided by $D$ tends to its power spectrum of surface height. Thus the diffuse intensity samples the spectrum at the tangential momentum transfer. An infinite stationary surface has no finite total far-field amplitude without a window or an intensity-density convention.