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According to Kennedy's Robust op-amp realization of Chua's circuit(1992), the differential equations satisfied by several physical quantities in Chua's circuit are

$$\begin{aligned} C_{1} \frac{d v_{C_{1}}}{d t} &=G\left(v_{C_{2}}-v_{C_{1}}\right)-g\left(v_{C_{1}}\right) \\ C_{2} \frac{d v_{C_{2}}}{d t} &=G\left(v_{C_{1}}-v_{C_{2}}\right)+i_{L} \\ L \frac{d i_{L}}{d t} &=-v_{C_{2}} \end{aligned}$$

Where $G=1/R$. The circuit diagram is given at the bottom, where the quantity represented by each letter is indicated clearly.

When Chua invented his circuit, he deliberately designed it in such a way that the three equilibrium points were unstable. But I don't know how he did this, although I can analyze the stability using eigenvalues.

For small $x$, $f(x)=-mx$ for some positive $m$. So the equations above are linear for samll $x$. In order to analyze the stability of the zero solution $v_{C_1}=v_{C_2}=i_L=0$, I write $\mathbf x=[v_{C_1},v_{C_2},i_L]^T$ and $$ M=\begin{bmatrix}-a & b & 0\\ c & -c & d \\ 0 & -f & 0 \end{bmatrix} $$ where all letters are positive constants, and $a=1/RC_1,b=1/RC_1+m/C_1$ and so on. So the differential equations above can be written as $\dot{\mathbf x}=M\mathbf x$.

The characteristics equation $\det(M-\lambda I)=0$ is $\lambda^3+(a+c)\lambda^2+(ac-bc+fd)\lambda+fda=0$.

If the trivial solution $\mathbf x=0$ is a stable solution, then according to a theorem(see here), the real parts of all eigenvalues of $M$ have to be negative. Thus it is not hard to deduce that the coefficients of the characteristic equation must be all positive. Therefore $ac-bc+fd>0$. Substituting the expressions for $a,b,\ldots,f$ in terms of $R,C_1,...$ and rearranging gives $m<\frac{RC_1}{L}$.

Questions:

  1. So does it mean that when $m\geq\frac{RC_1}{L}$ the zero solution is always unstable? Does this result help us to choose the values of parameters?
  2. Intuitively, for the trivial solution to be unstable, we want the Chua's diode to develop power at a faster rate than the resistor consuming power. If we ignore the capacitors for a moment, then this means that $m<1/R$. However, the $R$ in $m\geq\frac{RC_1}{L}$ is on the numerator rather than the denominator. why are the two inequalities ($m\geq\frac{RC_1}{L}$ and $m<1/R$) so different? this seems quite counter-intuitive. Also, intuitively, for the trivial solution to be unstable, we want the Chua's diode to develop power at a faster rate than the resistor consuming power. If we ignore the capacitors for a moment, then this means that $m<1/R$. However, the $R$ in $m\geq\frac{RC_1}{L}$ is on the numerator rather than the denominator. why are the two inequalities ($m\geq\frac{RC_1}{L}$ and $m<1/R$) so different? this seems quite counter-intuitive.

Also, can I improve my results? This is just a sufficient condition for the equilibrium to be unstable. Can I somehow make it a necessary condition as well?

Or are there any better methods I can use instead of evaluating eigenvalues?

EDIT: It appears that the condition $m\geq\frac{RC_1}{L}$ should be also sufficient if all eigenvalues are real.

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