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Secure Communication Using Chaos Synchronization 23-11<br />

It can be known from (23.16) and (23.17) that the original signal can be recovered only when<br />

two identical chaotic circuits in both the encrypter and the decrypter are synchronized. Similarly, an<br />

intruder can know the original message only when he knows the parameters and the structure of the<br />

circuits and the synchronization impulses.<br />

Remark 23.2: The magnifying glass is used to transform the chaotic state variables that act<br />

as key sequence before XOR with the plaintext. Assuming that there is a small mismatch that<br />

results in perturbations Δx i (t) = σ i (i = 1, 2, 3), then the signal getting through the amplifier gives<br />

⎛<br />

1/<br />

2<br />

⎢ 3<br />

⎛<br />

⎞ ⎥<br />

k( t) = ⎢K ( xi( t) + i) mod( )<br />

⎝<br />

⎜<br />

i<br />

⎠<br />

⎟ ⎥<br />

⎣⎢<br />

⎦⎥ +<br />

⎞<br />

2<br />

⎝<br />

⎜ ∑ σ λ 256 . Since the parameter K is a large number, it can<br />

= 1<br />

⎠<br />

⎟<br />

be seen that the signal is enlarged many times, which implies that the parameters mismatch can be<br />

enlarged. Thus, even a minor mismatch in the parameters will produce a large decryption error, resulting<br />

in a decryption key sequence that is not the same as the encryption key signal. So, one cannot<br />

recover the plaintext signal. The security of the chaotic <strong>communication</strong> system is thus improved.<br />

Before transmitting the ciphertext through a digital channel, it is necessary to packetize the ciphertext.<br />

To simplify the operation, a packetization algorithm is designed, in which the packet length is<br />

fixed. Since the lengths of the impulsive intervals are piecewise constant, it is difficult to find the synchronization<br />

impulse. The security of the system can then be improved. To maintain the simplicity of<br />

the presented secure system, the length of the packets is the same as the length of the first impulsive<br />

interval. The length of other impulsive interval is determined by the length of the first impulsive interval<br />

and the parameters of the Chua’s circuits.<br />

23.3.4 Synchronization Time Estimation<br />

It is noticeable that the time-varying impulsive intervals do not affect the synchronization of two identical<br />

chaotic circuits embedded in the encrypter and the decrypter, and the time required for synchronization<br />

can be estimated. A scheme in [LLWS03] is presented to achieve this objective.<br />

From the estimation, it is remarked that the value of K will influence the synchronization stable time<br />

of two identical chaotic <strong>systems</strong> and the desired precision of the system. A larger K will require longer<br />

synchronization time. On the other hand, the security of the system is higher with a larger K. A trade-off<br />

should be obtained in practice.<br />

23.3.5 Implementation<br />

Text transmission is taken to illustrate the performance of the proposed chaotic secure <strong>communication</strong><br />

system. In order to make the presented encryption system widely used in the computer, 256 characters ASCII<br />

code is used, which includes 128 standard ASCII codes and other 128 that are known as extended ASCII.<br />

Thus, the ciphertext in the simulation has not only the standard characters but also some local symbols and<br />

marks. When the ciphertext can be decrypted by the correct key, the message is exactly recovered. Otherwise,<br />

the recovered message is spread in the extended ASCII code.<br />

Choose the parameters of Chua’s circuit as k = 1, α = 9.35159085, β = 14.790313805, γ = 0.016072965,<br />

m 0 = −1.138411196, and m 1 = −0.723451121. For synchronization to be ensured, matrix B is selected as<br />

B = diag(−1.05, −1, −1) and the lengths of impulsive time intervals are chosen as T 2h−1 = 5 × 10 −3 .s and<br />

T 2h = 2 × 10 −3 .s where h = 1, 2, …. Then starting with τ 0 = 0, the impulsive instants can be determined.<br />

The initial condition is given by [x 1 (0); x 2 (0); x 3 (0)] = [−2.12; −0.05; 0.8] and [x˜1(0); x˜ 2(0); x˜3(0)] = [−0.2;<br />

−0.02; 0.1], respectively. As the initial error is e(0) = [−1.92; −0.03; 0.7], so the encrypter and decrypter<br />

are initially not synchronized. Choose the magnifying glass as K = 1 × 10 3 , λ = 12, and the quantization<br />

step is q = 5 × 10 −8 .<br />

© <strong>2011</strong> by Taylor and Francis Group, LLC

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