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Background: I'm working on a 2D Moving-Average (MA) estimator for image processing stuff and I'm trying to follow along from a paper that defines the algorithm for estimation.

The main equation for the parameter estimation is: $$\eta_i = (1/N_i) \sum_{u \in \Omega_i}|Y(u)|^2 $$

where $\Omega_i = \{u \in \Omega_u | ||Au|| = \rho_i\} $

However, $\Omega_u$ is defined earlier as $\Omega_u = \{(u_1, u_2) \in \Omega | 0 \le u_1 \lt N, 0 \le u_2 \le N/2 , (u_1, u_2) \neq (0,0)\}$. Since we're dealing with images, $N$ is the image size, which in my case is 128 for a 128x128 image. $u_1, u_2$ are both indices in the frequency domain after the DFT is used on the original image.

I'm hung up on what the set $\Omega_i$ actually entails and consequently, what the I'm summing over. Is this saying there are two sets like $\Omega_{i1}$ and $\Omega_{i2}$?

My current interpretation is that $u_1 = \{0...127\}, u_2 = \{0...64\}$ where $u_1$ is the "row" of $Y$ and $u_2$ is the "column" of $Y$. Then in the algorithm, I'm summing across the row of $Y$ for all $u_2$'s. Yet this leads to drastically different results for $\eta_i$ than what the paper has.

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  • $\begingroup$ Could you share a link to the paper? $\endgroup$
    – Royi
    Commented May 11, 2023 at 12:05
  • $\begingroup$ $\Omega_i$ is related to $\rho_i$ via $A$: it includes the frequency points such that their image by the linear transformation $A$ has the given norm. This corresponds to a pair of parallel lines. The number and values of the $\rho_i$ must be specified elsewhere. $\endgroup$
    – user67664
    Commented May 11, 2023 at 12:31
  • $\begingroup$ @Royi here is the link on IEEE: ieeexplore.ieee.org/document/730388 $\endgroup$
    – Brian
    Commented May 11, 2023 at 18:16
  • $\begingroup$ @YvesDaoust so $\Omega_I$ shouldn't be considered to be related to $\Omega_u$? I'm interpreting the set as containing "pairs" $u_1$, $u_2$. Also yes, $\rho_i$ is defined elsewhere as $\{\rho_i\} = \{||Au|| | u \in \Omega_u\}$. So it's just the reverse of what $\Omega_i$ is being defined as/ $\endgroup$
    – Brian
    Commented May 11, 2023 at 18:18
  • $\begingroup$ You did not supply information for the question to be answered then. $\endgroup$
    – user67664
    Commented May 11, 2023 at 18:30

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