On robust watermark detection for optimum multichannel compressive transmission

Abstract

An improved detection model for spread spectrum image watermarking is developed assuming that compressive measurements of watermarked image are transmitted over multiple channels. A closed form expression of detection threshold in log-likelihood ratio model is derived followed by developing the two optimization problems. First one is to find the optimal number of compressed sensing measurements as the product of the number of channels and the number of samples per channel (bandwidth) under the constraint of a detection reliability and watermarked image power. The second one finds an optimal set of watermarked image measurements (that lead to watermarked image power) under the constraint of a detection reliability. Extensive simulation results are reported to highlight the efficacy of the watermark detector for both the optimization problems. An improvement of $\sim$ 5.65% in detection performance is observed for 82.35% CS measurements and for a given probability of false alarm value 0.1 on multiplicative (fading) degradation having power 0.7.

Appendices

Appendix A

1.1 Parameter estimation of the detector

Let $\mu _{\mathcal {H}_i}$ and $\sigma _{\mathcal {H}_i}^2$ be the mean and the variance of $t(\mathbf {Y})$ under $\mathcal {H}_i~ (i=0,1)$. Then the detector distribution parameters are derived as,

$$\begin{aligned} \mu _{\mathcal {H}_0}&= E[t(\mathbf {Y}) / \mathcal {H}_0] \\&= E[\mathbf {Y}^T \mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P} / \mathcal {H}_0] \\&= E[(\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {x} + \varvec{\eta })^T \mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P}] \\&\approx 0 \\ \mu _{\mathcal {H}_1}&= E[t(\mathbf {Y}) / \mathcal {H}_1] \\&= E[\mathbf {Y}^T \mathbf {W}\varvec{\mathbf {H}\Theta }\mathbf {P} / \mathcal {H}_1] \\&= E[(\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {x} + \varvec{\eta } + \alpha \mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P})^T \mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P}] \\&=\alpha \Vert \mathbf {W} \Vert _2^2\sigma _h^2 E[\Vert \varvec{\Theta } \mathbf {P} \Vert _2^2] =\alpha \frac{LK}{N} \Vert \mathbf {W} \Vert _2^2\sigma _h^2 \sigma _\mathbf {P}^2 \\ \sigma _{\mathcal {H}_0}^2&=E[\lbrace t(\mathbf {Y} / \mathcal {H}_0) \rbrace ^2] = E[\lbrace \mathbf {Y}^T \mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P} / \mathcal {H}_0 \rbrace ^2]~[\because \mu _{\mathcal {H}_0}=0] \\&=E[(\mathbf {Y}^T \mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P})^T (\mathbf {Y}^T \mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P})] \\&=E[(\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P})^T \mathbf {Y}\mathbf {Y}^T (\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P})] \\&=\frac{LK}{N} \mathbf {C} \Vert \mathbf {W} \Vert _2^2\sigma _h^2\sigma _\mathbf {P}^2 \end{aligned}$$

$$\begin{aligned} \sigma _{\mathcal {H}_1}^2&=E[\lbrace t(\mathbf {Y} / \mathcal {H}_1) - E(t(\mathbf {Y} / \mathcal {H}_1))\rbrace ^2] \\&=E[\lbrace \mathbf {Y}^T \mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {P} - \alpha (\mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P})^T (\mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P}) \rbrace ^2] \\&=E[\lbrace (\mathbf {Y} - \alpha \mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P})^T (\mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P}) \rbrace ^2] \\&=E[\lbrace (\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {x} + \varvec{\eta })^T (\mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P}) \rbrace ^2] ~[from ~Eq.~4] \\&=E[\lbrace (\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {x} + \varvec{\eta })^T (\mathbf {W}\varvec{\mathbf {H}\Theta } \mathbf {P}) \rbrace ^T \lbrace (\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {x} + \varvec{\eta })^T (\mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P}) \rbrace ] \\&=E[(\mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P})^T (\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {x} + \varvec{\eta })(\mathbf {W}\mathbf {H}\varvec{\Theta }\mathbf {x} + \varvec{\eta })^T (\mathbf {W}\mathbf {H}\varvec{\Theta } \mathbf {P})] \\&=\frac{LK}{N} \mathbf {C} \Vert \mathbf {W} \Vert _2^2\sigma _h^2\sigma _\mathbf {P}^2 \end{aligned}$$

Appendix B

1.1 Proof of $E[\Vert \varvec{\Theta } \mathbf {P} \Vert _2^2]=\frac{LK}{N}\sigma _\mathbf {P}^2$

Let $\varvec{\Theta }$ be taken as $\varvec{\Theta }=[\varvec{\phi }^1 \vdots \varvec{\phi }^2 \vdots \ldots \vdots \varvec{\phi }^L]^T$ and each $\varvec{\phi }^i \in \mathcal {R}^{K\times N}$, thus the matrix $\varvec{\Theta } \in \mathcal {R}^{LK\times N}$. The watermark $\mathbf {P} \in \mathcal {R}^N$, thus the product $\varvec{\Theta }\mathbf {P}$ is a vector in $\mathcal {R}^{LK}$. Since each $\varvec{\phi }^i \in \mathcal {N}(0,1/N)$, thus

$$\begin{aligned} E[\Vert \varvec{\Theta }\mathbf {P} \Vert _2^2]=&E[\sum _{i=1}^{LK} \left( \varvec{\Theta } \mathbf {P} \right) ^2] \\ =&\sum _{i=1}^{LK} E[\left( \varvec{\Theta } \mathbf {P} \right) ^2] \\ =&(LK) E[<\varvec{\Theta } \mathbf {P}, \varvec{\Theta } \mathbf {P}>] \\ =&(LK)\sigma _\mathbf {P}^2 E[\varvec{\Theta }^T \varvec{\Theta }] \\&[\because E[\mathbf {P}^T \mathbf {P}]=\sigma _\mathbf {P}^2] \end{aligned}$$

Here,

$$\begin{aligned} \varvec{\Theta }^T \varvec{\Theta }&=[(\varvec{\phi }^1)^T \vdots (\varvec{\phi }^2)^T \vdots \ldots \vdots (\varvec{\phi }^L)^T] \begin{bmatrix} \varvec{\phi }^1 \\ \ldots \\ \varvec{\phi }^2 \\ \ldots \\ \vdots \\ \ldots \\ \varvec{\phi }^L \end{bmatrix} \\&=\begin{bmatrix} (\varvec{\phi }^1)^T \varvec{\phi }^1 \ldots (\varvec{\phi }^1)^T \varvec{\phi }^L \\ \vdots \\ (\varvec{\phi }^L)^T \varvec{\phi }^1 \ldots (\varvec{\phi }^L)^T \varvec{\phi }^L \end{bmatrix} \\ \text {Now,}~~ E[(\varvec{\phi }^i)^T \varvec{\phi }^j ]&= {\left\{ \begin{array}{ll} 0~for~i \ne j \\ 1/N~for~i=j \end{array}\right. } \\ \therefore E[\varvec{\Theta }^T \varvec{\Theta }]&= \begin{bmatrix} 1/N&\\&\ddots&\\&1/N \end{bmatrix}=\frac{1}{N}\mathbf {I}_N \end{aligned}$$

Hence,

$$\begin{aligned} E[\Vert \varvec{\Theta }\mathbf {P} \Vert _2^2]&=\left( LK/N\right) \sigma _\mathbf {P}^2 \end{aligned}$$