V0.9.5.0
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# Chryptography with Chaotic Systems
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## Deterministic Chaotic Systems
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These are `non-linear` differential equations that are
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***very*** sensitive to initial conditions. One of the most
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famous is the `Lorentz-Attractor`.
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Since we cannot get back to these initial conditions, these are
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***unpredictable systems***.
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>[!NOTE]
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> $s(x)$ is a sync message added to the other `system` to get it
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> produce the same `output` as the `first one`.
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>
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> Here it is represented its effect in `red`.
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>
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> In `yellow` you see a simple gain factor coming from the same
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> $s(x)$
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Let's say we have 2 of these `systems`:
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$$
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\begin{align*}
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x(t); z(t) &\triangleq \text{chaotic systems}\\
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e(t) = x(t) - z(t) \rightarrow 0
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&\triangleq \text{synchronization} \\
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s(x) = f(x) + Kx &\triangleq \text{synchronization signal}
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\end{align*}\\
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\begin{cases}
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\dot{x}(t) = Ax(t) + Bf(x) + c\\
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\dot{z}(t) = Az(t) + Bf(z) + c
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\textcolor{red}{+ Bf(x) - Bf(z)}\\
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\dot{e}(t) = \dot{x}(t) - \dot{z}(t)
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\end{cases}\\
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\begin{align*}
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\dot{e}(t) &= \dot{x}(t) - \dot{z}(t) = \\
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&= Ax(t) + Bf(x) + c - Az(t) - Bf(z) - c = \\
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&= A(x(t) - z(t))+ B(f(x) - f(z)) = \\
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&= Ae(t)+ B(f(x) - f(z)) = \\
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&= Ae(t)+ B(f(x) - f(z)\textcolor{red}{-f(x) + f(z)}) = \\
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&= Ae(t) \longrightarrow e(t) \rightarrow 0
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\text{ if } eig(A)\in \R^- \\
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&= Ae(t) \textcolor{yellow}{-BKx + BKz} = (A - BK)e
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\end{align*}
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$$
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At the end of all, you can basically add a message into
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$\hat{s}(x) = s(x) + m(t)$ and after getting the new value $z(t)$
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I can extract from here the message:
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$$
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\hat{s}(x) - f(z) -Kz = f(x) + Kx + m(t) - f(x) - Kx = m(t)
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$$
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@ -0,0 +1,38 @@
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# Smith Predictor
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Once you have your system with `pure delays`, desing $G_c$
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as if $G_p$ ha no `delay` at all. The resulting system
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should be something similar to: $\frac{G_cG_p}{1 + G_cG_p}$.
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Now consider the real delay of the system and desing $\hat{G}_c$
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so that the new $\hat{G}(s) = G(s)e^{-sT}$:
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$$
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\begin{align*}
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\frac{\hat{G}_cG_pe^{-sT}}{1 + \hat{G}_cG_pe^{-sT}}
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&= \frac{G_cG_p}{1 + G_cG_p}e^{-sT} \rightarrow \\
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\rightarrow
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\frac{\hat{G}_c}{1 + \hat{G}_cG_pe^{-sT}}
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&= \frac{G_c}{1 + G_cG_p} \rightarrow \\
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\rightarrow
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\hat{G}_c + \hat{G}_c G_cG_p
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&= G_c + G_c\hat{G}_cG_pe^{-sT} \rightarrow \\
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\rightarrow
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\hat{G}_c \left[
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1 + G_cG_p(1 - e^{-sT})
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\right]
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&= G_c \rightarrow \\
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\rightarrow
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\hat{G}_c
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&= \frac{ G_c}{ 1 + G_cG_p(1 - e^{-sT})}
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\end{align*}
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$$
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@ -47,4 +47,6 @@ If $| rank_k(p) - rank_{k-1}(p)| < \epsilon$ we will stop
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iterating.
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At time $k = 0$ all pages have the same importance that is
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$rank_0(p) = \frac{1}{n}$
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$rank_0(p) = \frac{1}{n}$
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## Economic System
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