Flipflop stationary state

exa11
Variables and parameters:

y1,y2: Voltages in inverter 1 ( $y_1=U_1,\ y_2=V_1$)

y3,y4: Voltages in inverter 2 ( $y_3=U_2,\ y_4=V_2$)

$\lambda$: voltage supply


System of equations:

\begin{displaymath}U_1\left({1\over R_l}+{1\over r}\right)-V_1{1\over r}-\lambda{1\over R_l} &= 0\cr \end{displaymath}


\begin{displaymath}I(U_1,U_2)-U_1{1\over r}+V_1\left({1\over R}+{1\over r}\right) &=0\cr \end{displaymath}


\begin{displaymath}U_2\left({1\over R_l}+{1\over r}\right)-V_2{1\over r}-\lambda{1\over R_l} &=0\cr \end{displaymath}


\begin{displaymath}I(U_2,U_1)-U_2{1\over r}+V_2\left({1\over R}+{1\over r}\right) &=0\cr\end{displaymath}



\begin{displaymath}I(a,b):=\left\{0\qquad\qquad\qquad\qquad\qquad\ & \hbox{ for }b\leq V_0\cr \end{displaymath}


\begin{displaymath}\beta(b-V_0)^2(1+\delta a)\qquad\quad & \hbox{ for }V_0<b<V_0+a\car \beta a[2(b-V_0)-a](1+\delta a)\ & \hbox{ elsewhere }\cr\end{displaymath}





\begin{displaymath}R_l=4.5 * 10^5,\ R=10^{15},\ r=60\cr \end{displaymath}


\begin{displaymath}V_0=1,\ \beta=8.5\cdot 10^{-6},\ \delta=0.02\cr\end{displaymath}


Comments Demos

Figure 1
Two identical inverters coupled to a flipflop

Figure 2
Bifurcation diagram y1 versus $\lambda$.

Figure 3
(y1,y3) phase diagram for $\lambda=5$.

Figure 4
(y1,y3) phase diagram for $\lambda=1$.

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