A triple-layer steganography scheme for low bit-rate speech streams

Abstract

With the wide application of low bit-rate codecs in speech communication systems, low bit-rate speech streams have become new cover media of great potential for steganography. In this paper, through analyzing the pitch period prediction process in G.729 codec, the pitch parameter of the second speech subframe is found suitable for performing embedding. Then a novel triple-layer steganography method is proposed for low bit-rate speech streams. In this method, modification directions (adding or subtracting one) of the pitch parameter are selected adaptively in order to achieve a high embedding efficiency. Based on the “Hamming + 1” scheme, we use the matrix encoding method twice to increase the hiding capacity. Experimental results show that while keeping a good perceived quality of the synthetic speech, the proposed method has a good real-time performance and a satisfactory steganography security.

Appendix

Assume that there are 2^k–1 speech frames. In the first layer, k secret bits can be embedded in 2^k–1 pitch parameters using the ME method. After calculating the value of r, there are two cases:

Case 1: r ≠ 0, P (Case 1) = (2^k–1)/2^k. In the third layer, we can embed another k secret bits into 2^k–1 L3-0 bits using the ME method. After calculating the value of h, there are also two cases:

Case 1.1: h ≠ 0, P (Case 1.1) = (2^k–1)/2^k. By controlling the modification direction of p ^r ₂, make the result of Eq. (12) not equal to the secret bit to be embedded, namely λ ≠ β. Then flip the L3-0 _h bit. The total number of cover bits changed is 2.

Case 1.2: h = 0, P (Case 1.2) = 1/2^k. By controlling the modification direction of p ^r ₂, make the result of Eq. (12) equal to the secret bit to be embedded, namely λ = β. The total number of cover bits changed is 1.

Case 2: r = 0, P (Case 2) = 1/2^k. According to the values of λ and β, there are also two cases:

Case 2.1: λ = β, P (Case 2) = 1/2.

Case 2.1.1: h ≠ 0, P (Case 2.1.1) = P (Case 1.1). Flip the L3-0 _f and L3-0 _g bits. The total number of cover bits changed is 2.

Case 2.1.2: h = 0, P (Case 2.1.2) = P (Case 1.2). All the three layers have been completed, so the total number of cover bits changed is 0.

Case 2.2: λ ≠ β, P (Case 2.2) = 1/2.

Case 2.2.1: h ≠ 0, P (Case 2.2.1) = P (Case 1.1). Flip the L3-0 _h bit. The total number of cover bits changed is 1.

Case 2.2.2: h = 0, P (Case 2.2.2) = P (Case 1.2). Flip the L3 ‐ 0 _h ', L3 ‐ 0 _f ' and L3 ‐ 0 _g ' bits. The total number of cover bits changed is 3.

In conclusion, when embedding 2 k + 1 secret bits, the average number of cover bits changed is:

$$ \begin{array}{l}D=P\left(\mathrm{Case}\ 1\right)\times \left[P\left(\mathrm{Case}\ 1.1\right)\times 2+P\left(\mathrm{Case}\ 1.2\right)\times 1\right]\\ {}\kern1.5em +P\left(\mathrm{Case}\ 2\right)\times P\left(\mathrm{Case}\ 2.1\right)\times \left[P\left(\mathrm{Case}\ \mathrm{2.1.1}\right)\times 2+P\left(\mathrm{Case}\ \mathrm{2.1.2}\right)\times 0\right]\\ {}\kern1.5em +P\left(\mathrm{Case}\ 2\right)\times P\left(\mathrm{Case}\ 2.2\right)\times \left[P\left(\mathrm{Case}\ \mathrm{2.2.1}\right)\times 1+P\left(\mathrm{Case}\ \mathrm{2.2.2}\right)\times 3\right]\\ {}\kern0.75em =\frac{4^{k+1}-3\cdot {2}^k+2}{2^{2k+1}}\end{array} $$

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Thus the bit-change rate (the average rate of being changed per cover bit) of the proposed method is:

$$ C=\frac{D}{N}=\frac{4^{k+1}-3\cdot {2}^k+2}{4\cdot \left({2}^{3k}-{2}^{2k}\right)} $$

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Where N = 2 · (2^k–1) is the total number of cover bits. Therefore, we can figure out the embedding rate and the embedding efficiency:

$$ R=\frac{2k+1}{N}=\frac{2k+1}{2\cdot \left({2}^k-1\right)} $$

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$$ E=\frac{R}{C}=\frac{\left(2k+1\right)\cdot {2}^{2k+1}}{4^{k+1}-3\cdot {2}^k+2} $$

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