0
$\begingroup$

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 ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.