Optical Security with Double Random Fractional Fourier Domain Encoding

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Utilizing double random fractional Fourier domain encoding for optical security involves encryption and decryption methods based on the fractional Fourier transform of various orders, involving specific mathematical operations and notations. The process includes transforming the input function, encrypting it with fractional phases, decrypting with two different methods, and applying product and convolution theorems to the encrypted function.


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  1. Double random fractional Fourier- domain encoding for optical security

  2. Outline Fractional Fourier transform Encryption Decryption of two method Optical implementation Result 2

  3. Fractional Fourier transform of ath- order ???,? ??? ?? ??? ? Where ???,? = ??? = = ??exp ?? ?2cot? 2?? csc? + ? 2cot? ?? ??? sin? 4 ? = ??/2 , ?? ??= sin? 1/2exp , 2 3

  4. Some case and property of ath-order fractional Fourier transform = ? ? ??= ? ?0?,? = (? + ? ) ? 2= ? ? 2?,? ?1= Fourier transform ? 1= inverse Fourier transform ????= ??+? ?4?= ?0= ? 4

  5. Notation Original signal : ? ?0 is a real valued function ??? = ?? ? = exp[??1( )],? = exp[??2( )] are two statistically independent random phase. ?1( ) and ?2( ) are two statistically independent white sequences uniformly distributed in interval [0,2?]. 5

  6. Encryption ?? ??(??) ? ?0? ?0 ?? ? ????? ?? = ?(??) ??(??) a th order fractional transform transform ?? ? ?? (? ?)? fractional transform Input function ? Encrypted function ?? fractional Phase ? Phase ? 6

  7. Decryption of first method ?? ? ??(??)= ?(??) ??(??) ????? ?? = ???? ?? = ????? ??? ?? = ???? ? ? ,? ?0 is a phase function ???? ???0 = ? ?0? ?0 a th order fractional transform transform (? ?)? fractional ?? ?? ? Encrypted function ?? Decrypted function ?? fractional transform Phase ? 7

  8. Decryption of second method ?? ? ?? ?? by {??[?( )]} = ? ?[? ( )] ???? = [ ???] ????? ?? = [ ???] ? ?? = [????] ? ??? ?? = [????] [????] ?? ???? = [? ?0] [? ?0] ,? ?0 is a phase function a th order fractional transform transform (? ?)? fractional ?? [ ?] [??] Encrypted function Conjugate ?? ?? Decrypted function ?? fractional transform Phase ? 8

  9. Encrypted function ??(??) Apply product and convolution theorems for the fractional Fourier transform The encrypted function ???? is a stationary white noise. ? ???? ???? = ??(?)2?? ?(?? ?? ) 9

  10. Optical Engineering , 2000 , 39 (11) :2853-2859 10

  11. Optical implementation The distance : ?1= ??tan(?1/2), ?2= ??tan(?2/2) The focal length of lens : ?1= ??/sin(?1), ?2= ??/sin(?2) ?1= a /2, ?2= (? ?) /2,??is a scale factor. 11 Optical Engineering , 2000 , 39 (11) :2853-2859

  12. Numerical Simulation Result (a) primary image to be encrypted (b) encrypted image with orders (0.75, 0.9) and (1.25,1.1) (c) decrypted image with the orders (0.75, 0.9) and (1.25, 1.1) (d) decrypted image with wrong orders (0.7, 0.85) and (1.2, 1.05) Optical Engineering , 2000 , 39 (11) :2853-2859 12

  13. References [1] Unnikrishnan, G., and K. Singh. "Double random fractional Fourier domain encoding for optical security." Optical Engineering 39.11(2000):2853- 2859. [2] Almeida, L. B. "Product and Convolution Theorems for the Fractional Fourier Transform." Signal Processing Letters IEEE 4.1(1997):15-17. [3] Ozaktas, Haldun M., and O. Ayt r. "Fractional Fourier domains." Signal Processing 46.1(1995):119-124. 13

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