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Double random fractional Fourier-domain encoding for optical security

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Presentation on theme: "Double random fractional Fourier-domain encoding for optical security"— Presentation transcript:

1 Double random fractional Fourier-domain encoding for optical security
鄭任傑

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

3 Fractional Fourier transform of a’th-order
𝐹 𝑎 𝑓 𝑥 = 𝑓 𝑎 𝑥 = −∞ ∞ 𝐵 𝑎 𝑥,𝑥′ 𝑓 𝑎 𝑥′ 𝑑 𝑥 ′ Where 𝐵 𝑎 𝑥,𝑥′ = 𝐴 𝜙 exp 𝑖𝜋 𝑥 2 cot 𝜙 −2𝑥 𝑥 ′ csc 𝜙 + 𝑥 ′ 2 cot 𝜙 , 𝐴 𝜙 = sin 𝜙 −1/2 exp 𝑖𝜋 𝑠𝑔𝑛 sin 𝜙 4 − 𝑖𝜙 2 , 𝜙=𝑎𝜋/2

4 Some case and property of a’th-order fractional Fourier transform
𝐵 0 𝑥,𝑥′ =δ 𝑥− 𝑥 ′ ⇒ 𝐹 𝑂 =𝐼 𝐵 ±2 𝑥,𝑥′ =δ(𝑥+𝑥′)⇒ 𝐹 ±2 =𝑃 𝐹 1 = Fourier transform 𝐹 −1 = inverse Fourier transform 𝐹 𝑎 𝐹 𝑏 = 𝐹 𝑎+𝑏 𝐹 4𝑘 = 𝐹 0 =𝐼

5 Notation Original signal : 𝑓 𝑥 0 is a real valued function
𝐹 𝑎 𝑓 ∙ = 𝑓 𝑎 ∙ 𝛼 ∙ = exp [𝑗 𝑛 1 (∙)] ,𝛽 ∙ = exp [𝑗 𝑛 2 (∙)] are two statistically independent random phase. 𝑛 1 (∙) and 𝑛 2 (∙) are two statistically independent white sequences uniformly distributed in interval [0,2𝜋].

6 Encryption 𝑓 𝑥 0 𝛼 𝑥 0 𝐹 𝑎 𝑔 𝑒 ( 𝑥 𝑎 )
𝑓 𝑥 0 𝛼 𝑥 0 𝐹 𝑎 𝑔 𝑒 ( 𝑥 𝑎 ) 𝑔 𝑒 𝑥 𝑎 𝛽 𝑥 𝑎 = ℎ 𝑒 ( 𝑥 𝑎 ) 𝐹 𝑏−𝑎 𝜉 𝑒 ( 𝑥 𝑏 ) 𝑔 𝑒 a’th order fractional transform 𝑎 𝑡ℎ fractional transform ℎ 𝑒 (𝑏−𝑎) 𝑡ℎ fractional transform Input function 𝑓 Encrypted function 𝜉 𝑒 Phase 𝛼 Phase 𝛽

7 Decryption of first method
𝜉 𝑒 ( 𝑥 𝑏 ) 𝐹 𝑎−𝑏 𝑔 𝑑 ( 𝑥 𝑎 )= ℎ 𝑒 ( 𝑥 𝑎 ) 𝑔 𝑑 𝑥 𝑎 𝛽 ∗ 𝑥 𝑎 = ℎ 𝑒 𝑥 𝑎 𝛽 ∗ 𝑥 𝑎 = 𝑔 𝑒 𝑥 𝑎 𝛽 𝑥 𝑎 𝛽 ∗ 𝑥 𝑎 = 𝑔 𝑒 𝑥 𝑎 𝑔 𝑒 𝑥 𝑎 𝐹 −𝑎 𝑓 𝑑 𝑥 0 =𝑓 𝑥 0 𝛼 𝑥 ,𝛼 𝑥 0 is a phase function (𝑎−𝑏) 𝑡ℎ fractional transform a’th order fractional transform 𝑔 𝑒 ℎ 𝑒 −𝑎 𝑡ℎ fractional transform Encrypted function 𝜉 𝑒 Decrypted function 𝑓 𝑑 𝛼()= Phase 𝛽 ∗

8 Decryption of second method
𝜉 𝑒 ∗ 𝑥 𝑏 𝐹 𝑏−𝑎 𝑔 𝑑 𝑥 𝑎 = [ ℎ 𝑒 𝑥 𝑎 ] ∗ by {𝐹 𝑎 [𝑓(∙)]} ∗ = 𝐹 −𝑎 [ 𝑓 ∗ (∙)] 𝑔 𝑑 𝑥 𝑎 𝛽 𝑥 𝑎 = [ ℎ 𝑒 𝑥 𝑎 ] ∗ 𝛽 𝑥 𝑎 = [ 𝑔 𝑒 𝑥 𝑎 ] ∗ 𝛽 ∗ 𝑥 𝑎 𝛽 𝑥 𝑎 = [ 𝑔 𝑒 𝑥 𝑎 ] ∗ [ 𝑔 𝑒 𝑥 𝑎 ] ∗ 𝐹 𝑎 𝑓 𝑑 𝑥 𝑎 = [𝑓 𝑥 0 ] ∗ [𝛼 𝑥 0 ] ∗ ,𝛼 𝑥 0 is a phase function (𝑏−𝑎) 𝑡ℎ fractional transform a’th order fractional transform [ℎ 𝑒 ] ∗ [𝑔 𝑒 ] ∗ 𝑎 𝑡ℎ fractional transform Encrypted function Conjugate 𝜉 𝑒 ∗ 𝑥 𝑏 Decrypted function 𝑓 𝑑 Phase 𝛽

9 Encrypted function 𝜉 𝑒 ( 𝑥 𝑏 )
Apply product and convolution theorems for the fractional Fourier transform The encrypted function 𝜉 𝑒 𝑥 𝑏 is a stationary white noise. 𝐸 𝜉 𝑒 𝑥 𝑏 𝜉 𝑒 𝑥 𝑏 ′ ∗ = 𝑓 𝑏 (𝑢) 2 𝑑𝑢 ×𝛿( 𝑥 𝑏 − 𝑥 𝑏 ′)

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

11 Optical implementation
The distance : 𝑧 1 = 𝑓 𝑠 tan ( 𝜃 1 /2) , 𝑧 2 = 𝑓 𝑠 tan ( 𝜃 2 /2) The focal length of lens : 𝐹 1 = 𝑓 𝑠 / sin ( 𝜃 1 ) , 𝐹 2 = 𝑓 𝑠 / sin ( 𝜃 2 ) 𝜃 1 =aπ/2, 𝜃 2 =(𝑏−𝑎)π/2, 𝑓 𝑠 is a scale factor. 《Optical Engineering》 , 2000 , 39 (11) :

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) :

13 References [1] Unnikrishnan, G., and K. Singh. "Double random fractional Fourier domain encoding for optical security." Optical Engineering 39.11(2000): [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. Aytür. "Fractional Fourier domains." Signal Processing 46.1(1995):


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