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Christian Risi 2025-01-17 20:15:01 +01:00
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# Chryptography with Chaotic Systems
## Deterministic Chaotic Systems
These are `non-linear` differential equations that are
***very*** sensitive to initial conditions. One of the most
famous is the `Lorentz-Attractor`.
Since we cannot get back to these initial conditions, these are
***unpredictable systems***.
>[!NOTE]
> $s(x)$ is a sync message added to the other `system` to get it
> produce the same `output` as the `first one`.
>
> Here it is represented its effect in `red`.
>
> In `yellow` you see a simple gain factor coming from the same
> $s(x)$
Let's say we have 2 of these `systems`:
$$
\begin{align*}
x(t); z(t) &\triangleq \text{chaotic systems}\\
e(t) = x(t) - z(t) \rightarrow 0
&\triangleq \text{synchronization} \\
s(x) = f(x) + Kx &\triangleq \text{synchronization signal}
\end{align*}\\
\begin{cases}
\dot{x}(t) = Ax(t) + Bf(x) + c\\
\dot{z}(t) = Az(t) + Bf(z) + c
\textcolor{red}{+ Bf(x) - Bf(z)}\\
\dot{e}(t) = \dot{x}(t) - \dot{z}(t)
\end{cases}\\
\begin{align*}
\dot{e}(t) &= \dot{x}(t) - \dot{z}(t) = \\
&= Ax(t) + Bf(x) + c - Az(t) - Bf(z) - c = \\
&= A(x(t) - z(t))+ B(f(x) - f(z)) = \\
&= Ae(t)+ B(f(x) - f(z)) = \\
&= Ae(t)+ B(f(x) - f(z)\textcolor{red}{-f(x) + f(z)}) = \\
&= Ae(t) \longrightarrow e(t) \rightarrow 0
\text{ if } eig(A)\in \R^- \\
&= Ae(t) \textcolor{yellow}{-BKx + BKz} = (A - BK)e
\end{align*}
$$
At the end of all, you can basically add a message into
$\hat{s}(x) = s(x) + m(t)$ and after getting the new value $z(t)$
I can extract from here the message:
$$
\hat{s}(x) - f(z) -Kz = f(x) + Kx + m(t) - f(x) - Kx = m(t)
$$

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# Smith Predictor
Once you have your system with `pure delays`, desing $G_c$
as if $G_p$ ha no `delay` at all. The resulting system
should be something similar to: $\frac{G_cG_p}{1 + G_cG_p}$.
Now consider the real delay of the system and desing $\hat{G}_c$
so that the new $\hat{G}(s) = G(s)e^{-sT}$:
$$
\begin{align*}
\frac{\hat{G}_cG_pe^{-sT}}{1 + \hat{G}_cG_pe^{-sT}}
&= \frac{G_cG_p}{1 + G_cG_p}e^{-sT} \rightarrow \\
\rightarrow
\frac{\hat{G}_c}{1 + \hat{G}_cG_pe^{-sT}}
&= \frac{G_c}{1 + G_cG_p} \rightarrow \\
\rightarrow
\hat{G}_c + \hat{G}_c G_cG_p
&= G_c + G_c\hat{G}_cG_pe^{-sT} \rightarrow \\
\rightarrow
\hat{G}_c \left[
1 + G_cG_p(1 - e^{-sT})
\right]
&= G_c \rightarrow \\
\rightarrow
\hat{G}_c
&= \frac{ G_c}{ 1 + G_cG_p(1 - e^{-sT})}
\end{align*}
$$

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@ -47,4 +47,6 @@ If $| rank_k(p) - rank_{k-1}(p)| < \epsilon$ we will stop
iterating.
At time $k = 0$ all pages have the same importance that is
$rank_0(p) = \frac{1}{n}$
$rank_0(p) = \frac{1}{n}$
## Economic System