references【前置知識】

Theoretical characterizations of images that can be totally refined using the transducer oscillation algorithm

這篇文章據說就是用解方程做的，但是根本找不到文章本體x

Improving image resolution using subpixel motion

用local search+退火做的

Generalized Sampling Expansion（比較有用）

數學上證明了：對於\(f(t)\)，如果傅立葉變換滿足：

\[\begin{gathered} F(\omega)=0, \quad \text { for } \quad|\omega| \geqslant \sigma \\ \int_{-\infty}^{\infty}|f(t)|^{2} d t=\frac{1}{2 \pi} \int_{-\sigma}^{\sigma}|F(\omega)|^{2} d \omega<\infty \end{gathered} \]

當用m個線性系統\(H_{1}(\omega), \cdots, H_{m}(\omega)\)

去處理\(f(t)\)得到：

\[g_{k}(t)=\frac{1}{2 \pi} \int_{-\sigma}^{\sigma} F(\omega) H_k(\omega) e^{j \omega t} d \omega \]

然後對\(g_k(t)\)進行取樣得到\(g_k(nT)\)，其中\(T=\frac{2 \pi}{c} \quad c=\frac{2 \sigma}{m}\)，則\(f(t)\)可以用\(g_k(nT)\)來表示。過程為：

先列一個方程組：

\[\begin{aligned} &H_{1}(\omega) Y_{1}(\omega, t)+\cdots+H_{m}(\omega) Y_{m}(\omega, t)=1 \\ &\ldots \\ &H_{1}(\omega+r c) Y_{1}(\omega, t)+\cdots+H_{m}(\omega+r c) Y_{m}(\omega, t)=e^{j r c t} \\ &\ldots \\ &H_{1}(\omega+m c-c) Y_{1}(\omega, t)+\cdots+H_{m}(\omega+m c-c) Y_{m}(\omega, t)=e^{j(m-1) c t} \end{aligned} \]

where \(t\)

is any number and \(\omega\) is between \(-\sigma\) and \(-\sigma+c\).

這樣可以定義m個函式：\(Y_{1}(\omega, t), \cdots, Y_{m}(\omega, t)\)

然後可以用它們來表示\(f(t)\)：

\[\begin{array}{r} f(t)=\sum_{n=-\infty}^{\infty}\left[g_{1}(n T) y_{1}(t-n T)+\right. \left.\cdots+g_{m}(n T) y_{m}(t-n T)\right] ,\\ where \ \ \ \ \ y_{k}(t)=\frac{1}{c} \int_{-\sigma}^{-\sigma+c} Y_{k}(\omega, t) e^{j \omega t} d \omega, \quad k=1, \cdots, m \end{array} \]

論文字體

重建過程

(1) Determining the model parameters, i.e., the PSF and the relative subpixel shifts in the ensemble \(\left(g_{k}\right)\).
(2) Registering and merging the ensemble over a finer grid to obtain \(\hat{g}\).
(3) Deconvolving the combined picture \((\hat{g})\) with the restoration filter \(\left(y_{I}\right)\) to obtain an improved picture \((\hat{f})\).
如何得到 \(\hat{g}\)？通過下面的分析：

the sampling

\(h_{I}\)和\(h_{II}\)是filter的分解，分別代表了shift和blur，由於是linear space invariant operator，a和b是等價的。對channel的response用傅立葉變換：

\[\begin{aligned} H_{k}(\omega) &=\frac{1}{j \omega}\left(e^{j \omega A / 2}-e^{-j \omega A / 2}\right) e^{-j \omega x_{k}} \\ &=\underbrace{A \frac{\sin (\omega A / 2)}{(\omega A / 2)}}_{H_{l}(\omega)} \cdot \underbrace{e^{-j \omega A / k}}_{H_{I I}(\omega, k)} \end{aligned} \]

A是detector長度。

reference中取樣定理的channel：

k個線性channel的output（\(g_k\)）sampled at a rate of \(\omega=2 \sigma / K\)（\(1/K\)of the Nyquist rate）。如何重建：passing the sampled signals through \(K\) linear filters whose impulse responses are \(y_{k}\). The output \(\hat{f}\) is equal to \(f\), and

\[\hat{f}(x)=\sum_{k=1}^{K} \hat{f}_{k}(x)=\sum_{k=1}^{K} \sum_{m=-\infty}^{\infty} g_{k}(m D) \cdot y_{k}(x-m D) \]\[\begin{aligned} &H(\omega)={\left[\begin{array}{ccc} H_{1}(\omega) & \cdots & H_{K}(\omega) \\ H_{1}(\omega+\Omega) & \cdots & H_{K}(\omega+\Omega) \\ \vdots & & \vdots \\ H_{1}(\omega+(K-1) \Omega) & \cdots & H_{K}(\omega+(K-1) \Omega) \end{array}\right] } \end{aligned} \]\[y_{k}(k)=\frac{1}{\sigma} \sum_{\nu=1}^{K} \int_{I_{1}} y_{k \nu} e^{j(\omega+(\nu-1) \sigma) x} d \omega \]

where \(y_{k \nu}\) is the \(k \nu^{\text {th }}\) element of the inverse matrix of \(H(\omega)\)
【變數對應取樣定理】

我的理解：這裡解出的就是\(\hat{g}\)。
論文後面就直接說，shift的closed form表示式：

\[\begin{aligned} &\hat{g}(x)=\sum_{k=1}^{x} \sum_{n=-x}^{\infty} g\left(m D+x_{k}\right) \cdot \frac{\Pi_{n=1}^{K} \sin \left(\Omega\left[\left(x-m D-x_{n}\right) / 2\right]\right)}{\left(\Omega\left[\left(x-m D-x_{k}\right) / 2\right]\right) \cdot \prod_{n=1 \atop n \neq k}^{K} \sin \Omega\left[\left(x_{k}-x_{n}\right) / 2\right]} \end{aligned} \]

也可以寫作（是通過另一個定理還是直接寫，沒看懂）：

\[\begin{aligned} \hat{g}(x)=& \sum_{k=1}^{\infty} \sum_{m=-\infty}^{\infty} g_{k}(m D) \varphi_{k m}\left(x, m, x_{1}, \ldots ., x_{k}, \ldots, x_{K}\right) \end{aligned} \]

為了計算方便，regroup：

\[\hat{g}(x)=\sum_{k=1}^{K} \sum_{m=-\infty}^{\infty} g\left(m D+x_{k}\right) \frac{\sin \Omega\left[\left(x-m D-x_{k}\right) / 2\right]}{\Omega\left[\left(x-m D-x_{k}\right) / 2\right]} \cdot \frac{\prod_{n=1 \atop n \neq k}^{K} \sin \left(\Omega\left[\left(x-m D-x_{n}\right) / 2\right]\right)}{\left.\prod_{n=1 \atop n \neq k}^{K} \sin \Omega\left[\left(x_{k}-x_{n}\right) / 2\right]\right)} . \]

則：\(\varphi_{mk}(x)=\frac{\sin \Omega\left[\left(x-m D-x_{k}\right) / 2\right]}{\Omega\left[\left(x-m D-x_{k}\right) / 2\right]} \cdot \frac{\prod_{n=1 \atop n \neq k}^{K} \sin \left(\Omega\left[\left(x-m D-x_{n}\right) / 2\right]\right)}{\left.\prod_{n=1 \atop n \neq k}^{K} \sin \Omega\left[\left(x_{k}-x_{n}\right) / 2\right]\right)}\)

\(sinc= \frac{\sin \Omega\left[\left(x-m D-x_{k}\right) / 2\right]}{\Omega\left[\left(x-m D-x_{k}\right) / 2\right]}\), \(\varphi_{m k}^{*}(x)=\frac{\prod_{n=1 \atop n \neq k}^{K} \sin \left(\Omega\left[\left(x-m D-x_{n}\right) / 2\right]\right)}{\left.\prod_{n=1 \atop n \neq k}^{K} \sin \Omega\left[\left(x_{k}-x_{n}\right) / 2\right]\right)}\)

我的理解：只是為了方便計算變形了一下。

這篇文章大概的思路瞭解了，但是沒有解決最關心的問題：為什麼這個問題要到頻域上做，和退火那個做法的根本出發點區別在哪裡。不是從數學上知道“噢有這麼一個定理可以在頻域上做”，而是更有邏輯的直觀的“為什麼頻域上更好”。以及有沒有別的正交空間上也有這樣的好的性質呢？

說實話我傅立葉變換這節學得很差...其實最近看這些都挺難理解挺痛苦的...

順便我二刷也沒看懂前面說的一些看起來很關鍵的想法和後面的解法有什麼關係。大概是這樣的內容：
bandwidth is limited by sampling rate（單位面積的detector數量），限制：intersample distance cannot be smaller than the detector size. -> shifted picture

temporal和special bandwidth的trade-off：對於resolving power超過傳統限制的光學系統，fundamental invariant是系統傳遞的光學影象的N個自由度，並不僅僅與spatial bandwidth有關（是product of the spatial x, y and temporal dimensions）。因此可以通過減小temporal bandwidth來增大spatial bandwidth。

這篇文章考慮的sample theorem：Papoulis considers a \(\sigma\)-Band-Limited (BL) signal that passes through \(K\) linear channels and then sampled at \(1/K\) the Nyquist rate. 這裡的linear channel對應的是這篇文章中得到低解析度影象的過程和重建過程。

把低解析度影象merge到一張大的圖的過程可以認為是週期非均勻取樣，有插值方法。