# Formula for PSD across an axis of a 2D output

Consider a 2D stationary input $$e(x,y)$$ and a 2D real convolution function $$h(x,y)$$. Let $$S=h*e$$ be the result of the convolution of $$e$$ by $$h$$.

If needed, we may assume $$e$$ is isotropic (spectrum analysis across any axis of $$e$$ is the same).

1) How can I mathematically express the power spectral density (PSD) along a given axis of $$S$$, say the PSD of the 1D function $$S_{y=0} : x\mapsto S(x,0)$$ ?

(In 1D, we have the formula $$PSD_S(f) = \left|H(f)\right|^2PSD_e(f)$$. I'm hoping for / expecting something similar here, probably needing to integrate H over something...)

2) If $$h(x,y)$$ is circular ($$h(x,y)=h_r(r)$$ with $$r=\sqrt{x^2+y^2}$$), do things get any simpler ? We may assume E is isotropic (spectrum analysis across any axis of E is the same).

3) If I now am working with discrete signals, can you express things as well (probably with FFTs) ?

Edit : I think I'm starting to get it. "Coarsely", if I try to think what frequencies would be "seen" on $$x\mapsto S(x,0)$$, then you realize all 2D frequencies $$f_x+if_y$$ "apply"/"have an effect" on the $$y=0$$ line, but shifted by $$\cos(angle(f_x+if_y))$$.

Since $$\left|f_x+if_y\right|\cos(angle(f_x+if_y))=f_x$$ (for $$f_x>=0$$), I would expect to integrate over $$f_y$$, and the result would be something like

$$PSD_{S_{y=0}}(f)=2\int_{0}^{+\infty}\left|\mathfrak{F}\{h\}(f,f_y)\right|^2PSD_e(f,f_y)df_y$$

with $$\mathfrak{F}(h)$$ being the 2D Fourier transform of $$h$$.

Would that be right ?