Solutions of Circuits with Fractional, Nonlinear Elements by Means of a SubIval Solver

Abstract

The work concerns the application of a solver based on SubIval (the subinterval-based method for computations of the fractional derivative in initial value problems). A general form of a system of equations is considered. The system is one that can be formulated for a circuit with fractional elements, nonlinear elements and elements that are both fractional and nonlinear. A few details are given on the computations that need to be performed in each time step. An example with fractional, nonlinear elements is introduced to display the usefulness of the solver. The results obtained through the solver are compared with ones obtained through a harmonic balance methodology (steady state solution). Finally, a measure of the accuracy of the results is introduced and computed for the selected example.

A Formulation of State Equations

The vector of state variables takes the form:

$$\begin{aligned} \varvec{x}(t) = [ \begin{array}{l l l l} u_\kappa (t)&i_L(t)&q(t)&\psi (t) \end{array} ]^\mathrm {T}, \end{aligned}$$

(12)

with the derivative orders:

$$\begin{aligned} \varvec{\alpha } = [ \begin{array}{l l l l} \kappa&1&\beta&\gamma \end{array} ]^\mathrm {T}. \end{aligned}$$

(13)

The vector of additional variables is given by:

$$\begin{aligned} \varvec{y}(t) = [ \begin{array}{llll llll llll} V_1(t)&V_2(t)&V_3(t)&V_4(t)&u_\mathrm {NL}(t)&u_L(t)&u_q(t)&u_\psi (t)&i_\mathrm {NL}&i_\kappa (t)&i_q(t)&i_\psi (t) \end{array} ]^\mathrm {T}. \end{aligned}$$

(14)

The vector of source time functions is:

$$\begin{aligned} \varvec{v}(t) = [ \begin{array}{ll} E(t)&J(t) \end{array} ]^\mathrm {T}. \end{aligned}$$

(15)

The current balances contribute to the matrices marked by $\varvec{M}_\mathrm {I}$, $\varvec{M}_\mathrm {II}$ and $\varvec{T}$. For the node with the potential $V_1$:

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 1, 1} = \frac{1}{R_1}+\frac{1}{R_2}, \\ \varvec{M}_{\mathrm {I} \; 1, 2} = -\frac{1}{R_2}, \\ \varvec{M}_{\mathrm {I} \; 1, 11} = 1, \\ \varvec{T}_{1, 1} = \frac{1}{R_1}, \end{array} \end{aligned}$$

(16)

for the $V_2$ node:

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 2, 2} = \frac{1}{R_2}+\frac{1}{R_3}, \\ \varvec{M}_{\mathrm {I} \; 2, 1} = -\frac{1}{R_2}, \\ \varvec{M}_{\mathrm {I} \; 2, 3} = -\frac{1}{R_3}, \\ \varvec{M}_{\mathrm {I} \; 2, 9} = 1, \\ \varvec{M}_{\mathrm {I} \; 2, 12} = 1, \end{array} \end{aligned}$$

(17)

while for the node marked by $V_3$:

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 3, 3} = \frac{1}{R_3}, \\ \varvec{M}_{\mathrm {II} \; 3, 2} = 1, \\ \varvec{M}_{\mathrm {I} \; 3, 10} = 1, \\ \varvec{T}_{3, 2} = -1 \end{array} \end{aligned}$$

(18)

and for the $V_4$ node:

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 4, 4} = \frac{1}{R_4}, \\ \varvec{M}_{\mathrm {II} \; 4, 2} = -1. \end{array} \end{aligned}$$

(19)

The relations between the marked voltages and the node potentials contribute to $\varvec{M}_\mathrm {I}$ and $\varvec{M}_\mathrm {II}$. From the relation between $V_1$ and $u_q$:

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 5, 7} = -1, \\ \varvec{M}_{\mathrm {I} \; 5, 1} = 1, \end{array} \end{aligned}$$

(20)

between $V_2$ and $u_\mathrm {NL}$

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 6, 5} = -1, \\ \varvec{M}_{\mathrm {I} \; 6, 2} = 1, \end{array} \end{aligned}$$

(21)

between $V_2$ and $u_\psi$

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 7, 8} = -1, \\ \varvec{M}_{\mathrm {I} \; 7, 2} = 1, \end{array} \end{aligned}$$

(22)

while for $u_L$ and its terminal potentials:

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {I} \; 8, 6} = -1, \\ \varvec{M}_{\mathrm {I} \; 8, 3} = 1, \\ \varvec{M}_{\mathrm {I} \; 8, 4} = -1 \end{array} \end{aligned}$$

(23)

and for the relation between $u_\kappa$ and $V_3$:

$$\begin{aligned} \begin{array}{l} \varvec{M}_{\mathrm {II} \; 9, 1} = -1, \\ \varvec{M}_{\mathrm {I} \; 9, 3} = 1. \end{array} \end{aligned}$$

(24)

The nonlinear dependency $i_\mathrm {NL}(u_\mathrm {NL})$ describing the response of the nonlinear resistor $R_\mathrm {NL}$ is given in $\varvec{F}_\mathrm {NL}(\varvec{w}(t))$ by:

$$\begin{aligned} f_\mathrm {NL\;1}( w_{(\varvec{i}_\mathrm {arg})_1} ) = i_\mathrm {NL}(u_\mathrm {NL}), \end{aligned}$$

(25)

hence the index of this nonlinear function’s argument:

$$\begin{aligned} (\varvec{i}_\mathrm {arg})_1 = 5. \end{aligned}$$

(26)

The left-hand side of this equation introduces:

$$\begin{aligned} \varvec{M}_{\mathrm {I} \; 10, 9} = 1. \end{aligned}$$

(27)

In the same manner the nonlinear function of $C_q$ introduces:

$$\begin{aligned} f_\mathrm {NL\;2}( w_{(\varvec{i}_\mathrm {arg})_2} ) = u_q(q), \end{aligned}$$

(28)

with:

$$\begin{aligned} (\varvec{i}_\mathrm {arg})_2 = 15 \end{aligned}$$

(29)

and:

$$\begin{aligned} \varvec{M}_{\mathrm {I} \; 11, 7} = 1. \end{aligned}$$

(30)

For the nonlinear function of $L_\psi$ one obtains:

$$\begin{aligned} f_\mathrm {NL\;3}( w_{(\varvec{i}_\mathrm {arg})_3} ) = \psi (i_\psi ), \end{aligned}$$

(31)

with:

$$\begin{aligned} (\varvec{i}_\mathrm {arg})_3 = 12 \end{aligned}$$

(32)

and:

$$\begin{aligned} \varvec{M}_{\mathrm {I} \; 12, 16} = 1. \end{aligned}$$

(33)

The differential equation describing the fractional capacitor $C_\kappa$ introduces:

$$\begin{aligned} \varvec{M}_{\mathrm {III} \; 1, 10} = -\frac{1}{C_\kappa }, \end{aligned}$$

(34)

for the coil L:

$$\begin{aligned} \varvec{M}_{\mathrm {III} \; 2, 6} = -\frac{1}{L}, \end{aligned}$$

(35)

while for the fractional, nonlinear capacitor $C_q$:

$$\begin{aligned} \varvec{M}_{\mathrm {III} \; 3, 11} = -1 \end{aligned}$$

(36)

and for the fractional, nonlinear coil $L_\psi$:

$$\begin{aligned} \varvec{M}_{\mathrm {III} \; 4, 8} = -1. \end{aligned}$$

(37)