I am working with a surface that's both spatially and temporally periodic.

It's defined by

$$\zeta(x,y,t) = \sum_{mnl = -\infty}^\infty P(m,n,l)e^{ia(mx+ny) -iwlt} $$

where $P(m,n,l)$ is the coefficient of the $m,n,l$th Fourier component, $a=2\pi / L$ (where $L$ is the spatial period in both the $x$ and $y$ directions) and $w=2\pi/T$ where $T$ is the temporal period of the surface.

In the paper I'm reading (First-Order Theory and Analysis of MF/HF/VHF Scatter from the Sea - D.E.Barrick, January 1972), we "define the average spatial-temporal spectrum $W(p,q,\omega)$ of the surface height in terms of the Fourier coefficients as $$ W(p,q,\omega) = \frac{1}{\pi^3} \iiint \langle \zeta(x_1,y_1,t_1)\zeta(x_2,y_2,t_2) \rangle e^{ip\tau_x + iq\tau_y - i\omega \tau}\,d\tau_x\,d\tau_y\,d\tau" $$

where $p=am$, $q=an,$ $\omega=wl$, $\tau_x = x_2-x_1$, $y_2-y_1$, $\tau = t_2-t_1$ and $\langle \rangle$ denotes a statistical average.

I gather from the Wiener-Khinchin theorem that the Fourier transform of the autocorrelation function of the surface gives the spectrum. However the factor of $\frac{1}{\pi^3}$ in their definition of $W(p,q,\omega)$ confuses me as I am unaware of any version of the Fourier transform where $\frac{1}{\pi}$ is used.

So my question is where does the $\frac{1}{\pi^3}$ come from?

Any ideas or suggestions are most welcome.

Thanks for reading, Rachael

  • $\begingroup$ Rice ("Reflection of Electromagnetic Waves from Slightly Rough Surfaces" 1951) also uses this definition. $\endgroup$ Nov 21 '16 at 12:26
  • $\begingroup$ what do you mean with "a statistical average" ? each $P(m,n,l)$ is a random variable ? $\endgroup$
    – reuns
    Nov 21 '16 at 16:43
  • $\begingroup$ And if $g(x)$ is $2\pi$ periodic (and say $C^1$) then $g(x) = \sum_{n=-\infty}^\infty c_n(g)e^{inx}$ where $c_n(g) = \color{red}{\frac{1}{2\pi}}\int_0^{2\pi} g(x) e^{-inx}dx$, and in dimension $3$ it give a factor $\frac{1}{8\pi^3}$. For remembering this scaling factor you need to look at the Hilbert space theory. Let $e_n(x) = C e^{2 \pi i n x / T}$ where $C = \frac{1}{\sqrt{T}}$ such that $\|e_n\|_{L^2([0,T])} = 1$, and since $e_n$ is an orthonormal basis of $L^2([0,T])$, you have that $g \in L^2([0,T]) \implies g = \sum_n \langle g,e_n \rangle e_n$ $\endgroup$
    – reuns
    Nov 21 '16 at 16:46

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