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3D object watermarking by a 3D hidden object

Sherif Kishk and Bahram Javidi

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Abstract

In this paper we present a method to watermark a 3D object with another hidden 3D object using digital holography. The watermark or the hidden information is a 3D object that is embedded in the digital hologram of a 3D host object. The digital holograms are obtained optically by phase shift interferometery. The hologram of the hidden 3D object is double phase encoded before embedding it to the host 3D object hologram. Then, the watermarked hologram is double phase encoded again using different set of codes. The resultant watermarked hologram is very secure because of the multi-key nature of the watermarking process. We discuss the effect of distortion caused by hologram quantization and occlusion of some of the hologram pixels. We present tests to illustrate the effect of using a window of the hologram to reconstruct the hidden 3D object and the host 3D object. Both mathematical analysis and simulations are presented to illustrate the system performance. To the best of our knowledge, this is the first report of embedding a 3D objects within another 3D object.

©2003 Optical Society of America

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Figures (14)
Fig. 1.
Fig. 1. Phase shifting interferometer

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Fig. 2.
Fig. 2. Block diagram of the proposed system (a) Transmitter (b) Receiver

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Fig. 3.
Fig. 3. 3D object reconstructed at different distances from the output plane

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Fig. 4.
Fig. 4. The original host and hidden 3D objects.

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Fig. 5.
Fig. 5. (a) The autocorrelation for the watermarked hologram without using the second double phase encoding process. (b) The autocorrelation for the watermarked hologram when using the second double phase encoding process.

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Fig. 6.
Fig. 6. The Reconstructed 3D objects from the double phase encoded watermarked hologram without distortions (a) the 3D host object(b) The 3D hidden object

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Fig. 7.
Fig. 7. The reconstructed 3D host object using uniform quantization (a) 8 bits (b) 2 bits

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Fig. 8.
Fig. 8. The reconstructed 3D hidden object using uniform quantization (a) 8 bits (b) 2 bits

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Fig. 9.
Fig. 9. The reconstructed 3D objects using different portions of the digital holograms using quantized watermarked hologram. (a,b) 1/16 of the hologram area and 8 bits quantization (c,d) 1/4 of the hologram area and 4 bits quantization (e,f) 1/2 of the hologram area and 2 bits quantization

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Fig. 10.
Fig. 10. (a) The transmitted hologram having 50% occlusion and 4 bits quantization (b) The transmitted hologram having 75% occlusion and 4 bits quantization (c) The reconstructed 3D host object using the hologram in a (d) The reconstructed 3D hidden object using the hologram in a (e) The reconstructed 3D host object using the hologram in b (f) The reconstructed 3D hidden object using the hologram in b.

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Fig. 11.
Fig. 11. The effect of number of quantization levels and occluding parts of the transmitted double phase encoded watermarked hologram when using a uniform quantizer

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Fig. 12.
Fig. 12. The reconstructed 3D objects using an optimum quantizer with 4 bits quantization and 25% of the holograms (a) host object (b) hidden object

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Fig. 13.
Fig. 13. The error when using an optimum quantizer compared to the error when using a uniform quantizer, 25% of the hologram is used.

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Fig. 14.
Fig. 14. Effect of blind decoding on the hidden object (a) only the hidden object spatial domain phase code is unknown. (b) only the hidden object Fourier domain phase code is unknown.

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Tables (2)

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Table 1. error when using a uniform quantizer

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Table 2. Error when using only 25% of the hologram and a uniform quantizer

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Equations (48)

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D ( x , y ) = A ( x , y ) exp ( j ϕ ( x , y ) )
R ( x , y ; α ) = A R exp ⌊ j ( ϕ R + α ) ⌋
I ( x , y ; α ) = A 2 ( x , y ) + A R 2 + 2 A ( x , y ) A R cos ⁡ [ ϕ ( x , y ) − ϕ R − α ] .
ϕ O ( x , y ) = arctan ⁡ [ I 4 ( x , y ) − I 2 ( x , y ) I 1 ( x , y ) − I 3 ( x , y ) ] ,
A O ( x , y ) = 1 4 { [ I 1 ( x , y ) − I 3 ( x , y ) ] 2 + [ I 2 ( x , y ) − I 4 ( x , y ) ] 2 } 1 2
H O ( x , y ) = A O ( x , y ) exp [ j ϕ O ( x , y ) ]
H d ( x , y ) = { H ( x , y ) ψ 1 ( x , y ) } ⊗ IFT [ ψ 2 ( ξ , γ ) ]
σ 2 = 1 N . M ∑ y = 0 M − 1 ∑ x = 0 N − 1 H ( x , y ) H * ( x , y )
H w ( x , y ) = { [ H host ( x , y ) + α [ H hidden ( x , y ) ψ 1 ( x , y ) ⊗ ψ 2 ( x , y ) ] ] ψ ′ 1 ( x , y ) } ⊗ ψ ′ 2 ( x , y )
σ w 2 = 1 2 . N . M ∑ y = 0 M − 1 ∑ x = 0 N − 1 H hidden ( x , y ) H hidden * ( x , y )
H ˜ hidden ( x , y ) = α H hidden ( x , y ) + IFT { H ̂ host ( ξ , γ ) ψ * 2 ( ξ , γ ) } ψ * 1 ( x , y )
σ w 2 = 1 N . M ∑ y = 0 M − 1 ∑ x = 0 N − 1 H host ( x , y ) H host * ( x , y )
O ˜ d ( u , v ; d 1 ) = exp [ − j π λ d 1 ( Δ u 2 u 2 + Δ v 2 v 2 ) ] ∑ x = 0 N − 1 ∑ y = 0 M − 1 { H ˜ hidden ( x , y )
exp [ − j π λ d 1 ( Δ x 2 x + Δ y 2 y ) ] exp [ j 2 π ( x u N + y v M ) ] }
Δ u = λ d 1 N Δ x , and
Δ v = λ d 1 M Δ y
O ˜ d ( u , v ; d 1 ) = α O d ( u , v ; d 1 ) + exp [ − j π λ d 1 ( Δ u 2 u 2 + Δ v 2 v 2 ) ] .
∑ x = 0 N − 1 ∑ y = 0 M − 1 { ( IFT { H ˜ host ( ξ , γ ) ψ ′ * 2 ( ξ , γ ) } ψ ′ 1 * ( x , y ) )
exp [ − j π λ d 1 ( Δ x 2 x + Δ y 2 y ) ] exp [ j 2 π ( x u N + y v M ) ] }
O ˜ h ( u , v ; d 2 ) = O h ( u , v ; d 2 ) + exp [ − j π λ d 2 ( Δ u 2 u 2 + Δ v 2 v 2 ) ]
∑ x = 0 N − 1 ∑ = 01 M − 1 { α [ H hidden ( x , y ) ψ * 1 ( x , y ) ⊗ ψ * 2 ( x , y ) ]
exp [ − j π λ d 2 ( Δ x 2 x + Δ y 2 y ) ] exp [ j 2 π ( x u N + y v M ) ] }
σ host 2 = α 2 N . M ∑ y = 0 M − 1 ∑ x = 0 N − 1 H hidden ( x , y ) H hidden * ( x , y )
H ˜ w ( x , y ) = H w ( x , y ) + Δ H w ( x , y )
O ˜ d ( u , v ; d 1 ) = α O d ( u , v ; d 1 ) + exp [ − jπ λ d 1 ( Δ u 2 u 2 + Δ v 2 v 2 ) ] .
{ ∑ x = 0 N − 1 ∑ y = 0 M − 1 ( IFT { H ˜ host ( ξ , γ ) ψ ′ 2 * ( ξ , γ ) } ψ ′ 1 * ( x , y ) ) .
exp [ − j π λ d 1 ( Δ x 2 x + Δ y 2 y ) ] exp [ j 2 π ( x u N + y v M ) ] +
∑ x = 0 N − 1 ∑ y = 0 M − 1 Δ H w ( x , y ) exp [ − j π λ d ( Δ x 2 x + Δ y 2 y ) ] exp [ j 2 π ( x u N + y v M ) ] } .
O ˜ d ( u , v ; d 2 ) = O d ( u , v ; d 2 ) + exp [ − j π λ d 2 ( Δ u 2 u 2 + Δ v 2 v 2 ) ] .
{ ∑ u = 0 N − 1 ∑ v = 0 M − 1 α [ H hidden ( x , y ) ψ 1 ( x , y ) ⊗ ψ 2 ( x , y ) ] .
exp [ − j π λ d 2 ( Δ x 2 x + Δ y 2 y ) ] exp [ j 2 π ( x u N + y v M ) ] +
∑ x = 0 N − 1 ∑ y = 0 M − 1 Δ H w ( x , y ) exp [ − jπ λ d 2 ( Δ x 2 x + Δ y 2 y ) ] exp [ j 2 π ( x u N + y v M ) ] }
σ host 2 = α 2 N . M ∑ y = 0 M − 1 ∑ x = 0 N − 1 H hidden ( x , y ) H hidden * ( x , y ) + Δ 2 12
σ w 2 = 1 N . M ∑ y = 0 M − 1 ∑ x = 0 N − 1 H host ( x , y ) H host * ( x , y ) + Δ 2 12
H w o ( x , y ) = H w ( x , y ) ( 1 − W ( x , y ) ) = H w ( x , y ) − H w ( x , y ) W ( x , y )
O d ( u , v ; d ) = ∫ − ∞ ∞ − exp [ j 2 π λ ( z − d ) ] λ ( z − d ) exp [ j π λ ( z − d ) ( u 2 + v 2 ) ] ·
∫ − ∞ ∞ ∫ O ( x , y , z ) exp [ j π λ ( z − d ) ( x 2 + y 2 ) ]
exp [ j 2 π λ ( z − d ) ( x u + y v ) ] dx dy dz
error = 1 P ∑ p = 1 P 1 N × M ∑ y = 0 M − 1 ∑ x = 0 N − 1 [ O d ( u , v ; d ( p ) ) − O ( u , v ; d ( p ) ) ] 2 O ( u , v ; d ( p ) )
X ( u ) ≡ exp [ − j π λ d ( Δ u 2 u 2 ) ] · ∑ x = 0 N − 1 ( IFT { H ̂ host ( ξ ) ψ ′ 2 ( ξ ) } ψ ′ 1 ( x ) ) exp [ − j π λ d ( Δ x 2 x ) ] exp [ j 2 π ( x u N ) ] .
R x ( u , u ′ ) = E ⌊ X * ( u ) X ( u ′ ) ⌋
R x ( u , u ′ ) = E [ exp [ j π λ d ( Δ u 2 u 2 ) ] ∑ x = 0 N − 1 ∑ ξ = 0 N − 1 1 N H ̂ host * ( ξ ) ψ ′ 2 * ( ξ ) ψ ′ 1 * ( x ) exp [ j π λ d ( Δ x 2 x ) ] exp [ − j 2 π ( x u N ) ] . exp [ − j π λ d ( Δ u 2 u ′ 2 ) ] ∑ x ′ = 0 N − 1 ∑ ξ ′ = 0 N − 1 1 N H ̂ host ( ξ ′ ) ψ ′ 2 ( ξ ′ ) ψ ′ 1 ( x ′ ) exp [ − j π λ d ( Δ x 2 x ′ ) ] exp [ j 2 π ( x ′ u ′ N ) ] ]
R x ( u , u ′ ) = 1 N 2 E [ exp [ j π λ d Δ u 2 ( u 2 − u ′ 2 ) ] ∑ x = 0 N − 1 ∑ x ′ = 0 N − 1 ∑ ξ = 0 N − 1 ∑ ξ ′ N − 1 H * ( ζ ) H ( ξ ′ ) ψ ′ 2 ( ξ ) ψ ′ 2 ( ξ ' ) ψ ′ 1 ( x ) ψ ′ 1 ( x ′ ) exp [ j π λ d Δ x 2 ( x − x ′ ) ] exp [ − j 2 π ( x u N − x ′ u ′ N ) ] ]
R x ( u , u ′ ) = 1 N 2 exp [ j π λ d Δ u 2 ( u 2 − u ′ 2 ) ] ∑ x = 0 N − 1 ∑ x ′ = 0 N − 1 ∑ ξ = 0 N − 1 ∑ ξ ′ N − 1 H * ( ζ ) H ( ξ ′ ) E n 2 [ ψ ′ 2 ( ξ ′ ) ψ ′ 2 ( ξ ′ ) ] E n 1 [ ψ ′ 1 ( x ) ψ ′ 12 ( x ′ ) ]
exp [ j π λ d Δ x 2 ( x − x ′ ) ] exp [ − j 2 π ( x u N − x ′ u ′ N ) ]
R ( u , u ′ ) = 1 N 2 exp [ j π λ d Δ u 2 ( u 2 − u ′ 2 ) ] ∑ x = 0 N − 1 ∑ ζ = 0 N − 1 ∣ H ( ξ ) ∣ 2 exp [ − j 2 π x ( u N − u ′ N ) ] .
R ( u , u ′ ) = 1 N exp [ j π λ d Δ u 2 ( u 2 − u ′ 2 ) ∑ ξ = 0 N − 1 ∣ H ( ξ ) ∣ 2 δ ( u − u ′ ) ]
R ( u , u ′ ) = 1 N ∑ ξ = 0 N − 1 ∣ H ( ξ ) ∣ 2 δ ( u − u ′ )
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