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.

  • $\begingroup$ 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.