Generating a 50-scroll chaotic attractor at 66 MHz by using FPGAs

Abstract

Multi-scroll chaotic oscillators exhibit more rich dynamics and higher unpredictability than double-scroll ones. However, in electronics, the challenge is yet generating as many scrolls as the device capabilities allow it. For instance, experiments realized during the last 4 years showed the generation of 12 scrolls using commercially available amplifiers and 5 scrolls with an integrated circuit fabricated with technology of 0.5 $\upmu$m. The generation of these very few number of scrolls is due to the limitations of the electronic devices, like voltage range and frequency response. In this manner, to cope with the problem on generating many scrolls, this article shows the usefulness of field-programmable gate arrays (FPGAs) to generate 50 scrolls at 66 MHz. Computer arithmetic is detailed as well as realizations of single-constant multiplication blocks with fixed-point notation both to accelerate the processing speed and to reduce FPGA resources. Finally, experimental results show good agreement with numerical simulations for the generation of 10, 11, 18, 20, 30, 40 and 50 scrolls using the Altera Cyclone IV GX FPGA DE2i-150.

Appendices

Appendix 1

Appendix 2

A one-step method for solving a dynamical system like (5) can be described in a general way by (6), where $\phi (t,y,h)$ is known as increment function and $\lambda =\left( \frac{\partial f}{\partial y} \right) _n$ [38, 39]

$$\begin{aligned}&\frac{\mathrm{d}y}{\mathrm{d}t}=f(t,y)=\lambda y,\quad y(t_0)=y_0,\quad t\in [t_0,b] \end{aligned}$$

(5)

$$\begin{aligned}&y_{n+1}=y_n+h\phi (t_n,y_n,h),\quad n=0,1,\ldots , N-1 \end{aligned}$$

(6)

Assuming that (5) has a unique solution in $[t_0,b]$ and that $y(t)\in C^{(p+1)}$ $[t_0,b]$ for $p\ge 1$, then the solution y(t) can be expanded (Taylor’s series) at any point $t_n$ like in (7). Considering (8) and that $h \phi (t_n,y_n,h)$ is obtained from $h\phi (t_n,y(t_n),h)$, then the approximated value for $y_n$ instead of the exact value of $y(t_n)$ is obtained from Taylor’s series method of order p and described by (9) to approximate $y(t_{n+1})$ [39].

$$\begin{aligned}&y(t_{n+1})=y(t_n)+hy^{\prime }(t_n)+\frac{h^2}{2!}y^{\prime \prime }(t_n)+\cdots \end{aligned}$$

(7)

$$\begin{aligned}&h\phi (t_n,y(t_n),h)=hy^{\prime }(t_n)+\frac{h^2}{2!}y^{\prime \prime }(t_n)\nonumber \\&\quad +\cdots +\frac{h^p}{p!}y^{(p)}(t_n) \end{aligned}$$

(8)

$$\begin{aligned}&y_{n+1}=y_n+h\phi (t_n,y_n,h),\quad n=0, 1, 2, \ldots , N-1\nonumber \\ \end{aligned}$$

(9)

Therefore, if $p=1$, one gets Euler’s method, as described by (10)

$$\begin{aligned} y_{n+1}=y_n+hf(t_n,y_n),\quad n=0, 1, 2, \ldots , N-1 \end{aligned}$$

(10)

From an stability analysis considering (5) and assuming that $\partial f/ \partial y$ is relatively invariant in the region of interest, then the solution to (5) is given by (11), and considering that $t=t_0+nh$ one gets (12),

$$\begin{aligned} y(t)= & {} y(t_0)\mathrm{e}^{\lambda (t-t_0)} \end{aligned}$$

(11)

$$\begin{aligned} y(t_n)= & {} y(t_0)\mathrm{e}^{\lambda n h}=y_0(\mathrm{e}^{\lambda h})^n \end{aligned}$$

(12)

When applying a one-step method to (5), the solution is given by (13), where $c_1$ is a constant and $E(\lambda h) \approx \mathrm{e}^{\lambda h}$ [39].

$$\begin{aligned} y_n=c_1(E(\lambda h))^n \end{aligned}$$

(13)

Considering the set of equations for the chaotic oscillator shown in (14), it has a unique equilibrium point at (0, 0, 0), then its characteristic equation is given by (15),

$$\begin{aligned} \dot{x}= & {} y\nonumber \\ \dot{y}= & {} z\nonumber \\ \dot{z}= & {} -ax-by-cz+d_1f(x;_1,h_1,p_1,q_1) \end{aligned}$$

(14)

$$\begin{aligned} \lambda ^3+c\lambda ^2+b\lambda +a=0 \end{aligned}$$

(15)

Solving (15), one gets: $\lambda _1=-\eta$ and $\lambda _{2,3}=\alpha _1\pm \beta i$, where $\lambda _1<0$, $\alpha _1>0$ and $\beta \ne 0$. Then, (14) has a negative eigenvalue and one pair of complex eigenvalues with positive real part [8]. As a result, a numerical method solving (14) will simulate chaotic behavior under the conditions of $\lambda _1, \alpha _1, \beta$. This result is extended to a multi-scroll chaotic oscillator because in (14), $f(x;_1,h_1,p_1,q_1)$ is a piecewise-linear (PWL) function consisting of slopes, offsets and saturation values. For example, simulating the generation of 2 scrolls, the PWL function is given by (16), where $\alpha =16.5e-3$, $m=60,606$, $a=b=c=d_1=0.7$. In this case: $\lambda _1=-0.8410142, \lambda _{2,3}=0.07050\pm 0.90201i$.

$$\begin{aligned} f(x) = \left\{ \begin{array}{l@{\quad }l} 1 &{} \text{ if } x>\alpha \\ \frac{x}{m} &{} \text{ if } |x|\le \alpha \\ -1 &{} \text{ if } x<-\alpha \end{array} \right. \end{aligned}$$

(16)

Since (14) has at least one $\lambda >0$, then Euler’s method is relatively stable if $\mid E(\lambda h) \mid \le \mathrm{e}^{\lambda h}$ for $\lambda >0$. Therefore, applying (10) for solving (14) will lead to the true solution if for an arbitrary initial condition $y_0$,

$$\begin{aligned} \displaystyle \lim _{h \rightarrow 0}{y_n=y(t)} \text{ for } t\in [t_0,b],\quad t_n=t \end{aligned}$$

(17)

Finally, considering the sampling theorem, $\tau _{\min }=\frac{1}{f}$, the condition is $h\le \frac{\tau _{\min }}{2}$ in order to to converge the numerical method to the true solution when $h\rightarrow 0$, i.e., better when h is relatively low.