## Abstract

In this paper we present a new approach providing super resolved imaging at the center of the field of view and yet allowing seeing the remaining of the original field of view with the original resolution. This operation resembles optical zooming while the zoomed and the non zoomed images are obtained simultaneously. This is obtained by taking a single snap shot and using a single imaging lens. The technique utilizes a special static/still coding element and a post processing algorithmic, without any mechanical movements.

© 2005 Optical Society of America

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### Equations (19)

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(1)
$${G}_{1}\left(v\right)\bullet {G}_{k}\left(v\right)=\delta \left[1,k\right]$$
(2)
$$I\left(\nu \right)=\sum _{k=-1}^{1}{S}_{k}\left(\nu \right)\times {G}_{k}\left(\nu \right)\phantom{\rule{1.5em}{0ex}}\nu \in [\frac{-1}{3{\nu}_{max}},\frac{1}{3{\nu}_{max}}]$$
(3)
$$\mathrm{CCD}\left(x\right)={\sum}_{n=-\infty}^{\infty}\delta \left(x-\mathrm{n\Delta x}\right)$$
(4)
$$\mathrm{CDM}\tilde{A}\left(\nu \right)={\sum}_{n=-1}^{1}{G}_{n}\left(\nu -\mathrm{n\Delta \nu}\right)$$
(5)
$$D\left(v\right)=\left[S\left(\nu \right){\sum}_{n=-1}^{1}{G}_{n}\left(\nu -\mathrm{n\Delta \nu}\right)\right]\ast \left[{\sum}_{n=-\infty}^{\infty}\delta \left(\nu -n\frac{2\pi}{\mathrm{\Delta x}}\right)\right]$$
(6)
$$D\left(\nu \right)=\lfloor S\left(\nu \right){\sum}_{n=-1}^{1}{G}_{n}\left(\nu -\mathrm{n\Delta \nu}\right)\rfloor \ast \lfloor {\sum}_{n=-\infty}^{\infty}\delta \left(\nu -\mathrm{n\Delta \nu}\right)\rfloor $$
(7)
$$\phantom{\rule{.7em}{0ex}}={\sum}_{n=-\infty}^{\infty}S\left(\nu -\mathrm{n\Delta \nu}\right)\left[{\sum}_{k=-1}^{1}{G}_{n}\left(\nu -\left(n+k\right)\mathrm{\Delta \nu}\right)\right]$$
(8)
$$R\left(\nu \right)=D\left(\nu \right)\mathrm{CDM}\tilde{A}\left(\nu \right)=\left\{\sum _{n=-\infty}^{\infty}S\left(\nu -\mathrm{n\Delta}\nu \right)\left[\sum _{k=-1}^{1}{G}_{n}\left(\nu -\left(n+k\right)\mathrm{\Delta \nu}\right)\right]\right\}\left[\sum _{m=-1}^{1}{G}_{m}\left(\nu -\mathrm{m\Delta \nu}\right)\right]$$
(9)
$$\phantom{\rule{.2em}{0ex}}=\sum _{n=-\infty}^{\infty}S\left(\nu -\mathrm{n\Delta \nu}\right){G}_{n}\left(\nu -\mathrm{n\Delta \nu}\right)=S\left(\nu \right)\mathrm{CDM}\tilde{A}\left(\nu \right)\underset{\mathrm{downsampling}}{\to}S\left(\nu \right)$$
(10)
$${i}_{L}\left(x\right)=({s}_{0}*{g}_{0})\left(x\right)\bullet \mathrm{rect}\left(\frac{x}{{L}_{T}}\right)$$
(11)
$$\mathrm{rect}\left(\frac{x}{{L}_{T}}\right)=\{\begin{array}{c}1\mid x\mid \le \frac{{L}_{T}}{2}\\ 0\phantom{\rule{.5em}{0ex}}\mathrm{otherwise}\end{array}$$
(12)
$${g}_{0}\left(x\right)=\sum _{n=-1}^{1}{a}_{n}\times \delta \left(x-\frac{n{L}_{T}}{2}\right)$$
(13)
$${r}_{1}\left(x\right)={a}_{0}{f}_{1}\left(x\right)+{a}_{-1}{f}_{4}\left(x\right)$$
(14)
$${r}_{2}\left(x\right)={a}_{0}{f}_{2}\left(x\right)+{a}_{-1}{f}_{5}\left(x\right)$$
(15)
$${r}_{3}\left(x\right)={a}_{0}{f}_{3}\left(x\right)+{a}_{-1}{f}_{6}\left(x\right)$$
(16)
$${r}_{4}\left(x\right)={a}_{0}{f}_{4}\left(x\right)+{a}_{1}{f}_{1}\left(x\right)$$
(17)
$${r}_{5}\left(x\right)={a}_{0}{f}_{5}\left(x\right)+{a}_{1}{f}_{2}\left(x\right)$$
(18)
$${r}_{6}\left(x\right)={a}_{0}{f}_{6}\left(x\right)+{a}_{1}{f}_{3}\left(x\right)$$
(19)
$$\left[\begin{array}{c}{r}_{1}\left(x\right)\\ {r}_{2}\left(x\right)\\ {r}_{3}\left(x\right)\\ {r}_{4}\left(x\right)\\ {r}_{5}\left(x\right)\\ {r}_{6}\left(x\right)\end{array}\right]=\left[\begin{array}{cccccc}{a}_{0}& 0& 0& {a}_{-1}& 0& 0\\ 0& {a}_{0}& 0& 0& {a}_{-1}& 0\\ 0& 0& {a}_{0}& 0& 0& {a}_{-1}\\ {a}_{1}& 0& 0& {a}_{0}& 0& 0\\ 0& {a}_{1}& 0& 0& {a}_{0}& 0\\ 0& 0& {a}_{1}& 0& 0& {a}_{0}\end{array}\right]\phantom{\rule{.2em}{0ex}}\left[\begin{array}{c}{f}_{1}\left(x\right)\\ {f}_{2}\left(x\right)\\ {f}_{3}\left(x\right)\\ {f}_{4}\left(x\right)\\ {f}_{5}\left(x\right)\\ {f}_{6}\left(x\right)\end{array}\right]$$