## Abstract

The kinogram-based single-phase decryption technique is experimentally demonstrated.
Only one phase spatial light modulator is used to simultaneously display the encrypted information and the decrypting key. The intensity decrypted image is obtained by Fourier transforming the phase decrypted information. We investigate the effect of the binary and multiphase keys on the security level of the encrypted information. The accepted displacement of the decrypting key within the system is determined. The influence of the optical system bandwidth and noise on the decryption quality is also investigated.

© 2007 Optical Society of America

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### Equations (9)

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(1)
$$E\left(\upsilon ,\nu \right)=F\left(\upsilon ,\nu \right){K}_{e}\left(\upsilon ,\nu \right)\text{,}$$
(2)
$${K}_{e}\left(\upsilon ,\nu \right)=\mathrm{exp}\left[i2\pi R\left(\upsilon ,\nu \right)\right]\text{.}$$
(3)
$$D\left(\upsilon ,\nu \right)=E\left(\upsilon ,\nu \right){K}_{d}\left(\upsilon ,\nu \right),=F\left(\upsilon ,\nu \right){K}_{e}\left(\upsilon ,\nu \right){K}_{d}\left(\upsilon ,\nu \right),=F\left(\upsilon ,\nu \right)\text{.}$$
(4)
$$g\left(x,y\right)={\left|F{T}^{-1}\left\{D\left(\upsilon ,\nu \right)\right\}\right|}^{2}\approx f\left(x,y\right)\text{.}$$
(5)
$${g}_{mn}={\left|RS\text{\hspace{0.17em} sinc}\left(\frac{m}{M},\frac{n}{N}\right){\displaystyle \sum _{k=-M/2}^{M/2-1}{\displaystyle \sum _{l=N/2}^{N/2-1}{F}_{kl}\times \mathrm{exp}\left[i2\pi \left(\frac{mk}{M}+\frac{nl}{N}\right)\right]}}\right|}^{2}\text{.}$$
(6)
$$e={\left[\frac{1}{AB}\text{\hspace{0.17em}}{\displaystyle \sum _{m=-A/2}^{A/2-1}{\displaystyle \sum _{n=-B/2}^{B/2-1}{\left|{f}_{mn}-\alpha {g}_{mn}\right|}^{2}}}\right]}^{1/2}\text{,}$$
(7)
$$\alpha =\frac{{\displaystyle \sum _{m=-A/2}^{A/2-1}{\displaystyle \sum _{n=-B/2}^{B/2-1}{f}_{mn}{g}_{mn}}}}{{\displaystyle \sum _{m=-A/2}^{A/2-1}{\displaystyle \sum _{n=-B/2}^{B/2-1}{{g}_{mn}}^{2}}}}\text{.}$$
(8)
$${f}_{n1}\left(x,y\right)={\left|F{T}^{-1}\left\{D\left(\upsilon ,\nu \right)\mathrm{exp}\left[j2\pi n\left(x,y\right)\right]\right\}\right|}^{2},$$
(9)
$${f}_{n2}\left(x,y\right)=\frac{1}{2}-\frac{1}{2}\text{\hspace{0.17em}}\mathrm{cos}\left\{\pi \left[f\left(x,y\right)+2n\left(x,y\right)\right]\right\}\text{,}$$