## Abstract

A high performance lensless optical security system based on the discrete Fresnel transform is presented. Two phase-only masks are generated with what we believe to be a novel and efficient algorithm. Their position coordinates and the wavelength are used as encoding parameters in the encryption process. Compared with previous studies, the main advantage of this proposed encryption system is that it does not need any iterative algorithms to produce the masks, and that makes it very efficient and easy to implement without losing the encryption security.

© 2006 Optical Society of America

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### Equations (7)

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(1)
$$u({x}_{1},{y}_{1})=\int \int \phantom{\rule{0.2em}{0ex}}\mathrm{exp}\left[\phantom{\rule{0.1em}{0ex}}j{\psi}_{2}({x}_{2},{y}_{2})\right]h({x}_{1},{y}_{1};{x}_{2},{y}_{2};{z}_{1},{z}_{2};\lambda )\mathrm{d}{x}_{2}\mathrm{d}{y}_{2},$$
(2)
$$h({x}_{1},{y}_{1};{x}_{2},{y}_{2};{z}_{2};\lambda )=\frac{\mathrm{exp}(\phantom{\rule{0.1em}{0ex}}j2\pi {z}_{2}\u2215\lambda )}{j\lambda {z}_{2}}\mathrm{exp}\{-\frac{j\pi}{\lambda {z}_{2}}[{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}]\}$$
(3)
$$u({x}_{1},{y}_{1})={\mathrm{FrT}}_{\lambda}\{\mathrm{exp}\left[\phantom{\rule{0.1em}{0ex}}j{\psi}_{2}({x}_{2},{y}_{2})\right];{z}_{2}\},$$
(4)
$${u}^{\prime}({x}_{1},{y}_{1})={\mathrm{IFrT}}_{\lambda}\{g({x}_{0},{y}_{0});{z}_{1}\}.$$
(5)
$${\psi}_{1}({x}_{1},{y}_{1})=\mathrm{arg}\left\{\frac{{u}^{\prime}({x}_{1},{y}_{1})}{u({x}_{1},{y}_{1})}\right\},$$
(6)
$$\widehat{g}({x}_{0},{y}_{0})={\mathrm{FrT}}_{\lambda}\{u({x}_{1},{y}_{1})\mathrm{exp}\left[\phantom{\rule{0.1em}{0ex}}j{\psi}_{1}({x}_{1},{y}_{1})\right];{z}_{1}\}={\mathrm{FrT}}_{\lambda}[u({x}_{1},{y}_{1})\frac{{u}^{\prime}({x}_{1},{y}_{1})}{u({x}_{1},{y}_{1})};{z}_{1}]={\mathrm{FrT}}_{\lambda}[{u}^{\prime}({x}_{1},{y}_{1});{z}_{1}]={\mathrm{FrT}}_{\lambda}\{{\mathrm{IFrT}}_{\lambda}[g({x}_{0},{y}_{0});{z}_{1}];{z}_{1}\}=g({x}_{0},{y}_{0}).$$
(7)
$$\rho =\frac{E\left\{[g-E\left(g\right)][\mid \widehat{g}\mid -E\left(\mid \widehat{g}\mid \right)]\right\}}{{\left(E\left\{{[g-E\left(g\right)]}^{2}\right\}E\left\{{[\mid \widehat{g}\mid -E\left(\mid \widehat{g}\mid \right)]}^{2}\right\}\right)}^{1\u22152}},$$