Electrical circuits RC and RL involving fractional operators with bi-order

Abstract

This article describes electrical series circuits RC and RL using the concept of derivative with two fractional orders $\alpha$ and $\beta$ in Liouville–Caputo sense. The fractional equations consider derivatives in the range of $\alpha ,\beta \in (0;1]$ . Numerical solutions are presented considering different source terms introduced in the fractional equation. This new approach considers electrical elements with two different properties. In addition, we prove that if $\alpha =0$ , the fractional derivative with Mittag-Leffler kernel in Liouville–Caputo sense is recovered, and when $\beta =0$ , the Liouville–Caputo fractional derivative is recovered.

Introduction

Fractional calculus (FC) allows to research the non-local response of multiple phenomena,^1–6 particularly in electrical circuits, several authors studied the behavior of capacitors, coils, memristors, and meminductors.^7–20 These elements involve irreversible dissipative effects (ohmic friction) and nonlinear effects due to the electric and magnetic fields.^17,21,22 Rousan et al. ²³ suggested a fractional order differential equation to study an LC and RC circuit. Obeidat et al. ²⁴ studied the evolution of current in a simple source-wire circuit. Ertik et al. ²⁵ considered the RC electrical circuit and showed charging and discharging processes of different capacitors, and experimental results showed similar behaviors that obtained within the fractional calculus approach. In 2015, a new definition of fractional operator without singular kernel and regular is presented, ²⁶ the Caputo–Fabrizio fractional operator. Gómez-Aguilar et al. ²⁷ presented some electrical circuits in terms of Caputo–Fabrizio fractional operator; they obtained numerical simulations of these circuits by applying the numerical Laplace transform algorithm, and more works related to this fractional operator are given by Atangana and Alkahtani, ²⁸ Atangana and Nieto, ²⁹ Atangana and Alkahtani, ³⁰ Gómez-Aguilar et al., ³¹ and Alsaedi et al.^32,33 Recently, Atangana and Baleanu proposed a new definition with non-local and nonsingular kernel based on the Mittag-Leffler function, this definition has all the benefits of the fractional derivatives of Liouville–Caputo and Caputo–Fabrizio types.^34–38

The fractional operators mentioned above have been widely used. However, none of these fractional operators is capable of handling the concept of heterogeneity with great success. Using the Mittag-Leffler law, new derivatives with two fractional orders were developed by A Atangana ³⁹ in the paper; in this work, the author introduces fractional operators with two orders in Riemann and Liouville–Caputo sense; these operators allow to describe more complex problems with different layers and with different properties, for instance, in thermal science where the heat is flowing within a medium with two different properties.

The aim of this contribution is to present an alternative representation of the electrical series circuits RC and RL using the concept of derivative with two fractional orders $\alpha$ and $\beta$ . The orders considered for the fractional equations are $\alpha ,\beta \in (0;1]$ .

The article is organized as follows: second section presents the new definitions of fractional operators with bi-order and with Mittag-Leffler kernel, third section discusses the fractional electrical series circuits, and the fourth section presents the conclusions.

Basic concepts

In the following, we present the definition of fractional operators with bi-order ³⁹ in Liouville–Caputo sense

$$\begin{array}{l}{}_0{}^{AC}{D}_t^{\alpha ,\beta }\{f(t)\}=\frac{B(\beta )}{n-\beta }\cdot \frac{1}{\Gamma (n-\alpha )}\hfill \\ {\displaystyle \underset{0}{\overset{t}{\int }}\frac{d^n}{d{t}^n}f(t){(t-\tau )}^{1-\alpha -n}{E}_{\beta }\{-\frac{\beta }{n-\beta }{(t-\tau )}^{\beta +\alpha }\}}d\tau ,\hfill \\ n-1<\beta <n,n-1<\alpha <n\hfill \end{array}$$

(1)

The Laplace transform of equation (1) produces

$$\begin{array}{c}\mathcal{L}\left[{}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }\{f(t)\}\right](s)=\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\hfill \\ \{\mathrm{sF}(s)-f(0)\}\mathcal{L}\left\{t^{-\alpha }{E}_{\beta }(-\frac{\beta }{1-\beta }{t}^{\beta +\alpha })\right\}\hfill \end{array}$$

(2)

where

$$\begin{array}{c}\mathcal{L}\{t^{-\alpha }{E}_{\beta }(-\frac{\beta }{1-\beta }{t}^{\beta +\alpha })\}={s^{\alpha -1}}_2{\Psi }_1\hfill \\ \left[\begin{array}{cc}(1,1),(1-\alpha ,\alpha +\beta )\hfill & \hfill \\ \multicolumn{1}{c}{}& -(\frac{\beta }{1-\beta }·\frac{1}{s^{\alpha +\beta }})\hfill \\ \multicolumn{1}{c}{(1,\beta )}& \hfill \end{array}\right]\hfill \end{array}$$

(3)

the function ${}_a{\Psi }_b$ is Wright’s generalized hyper-geometric function. ³⁹

Substituting Wright’s function (equation (3)) in equation (2), we have

$$\begin{array}{c}\mathcal{L}[{}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }f(t)](s)\hfill \\ =\frac{B(\beta )}{1-\beta }·\frac{s}{\Gamma (1-\alpha )}{}_2{\Psi }_1\hfill \\ \left[\begin{array}{cc}(1,1),(1-\alpha ,\alpha +\beta )\hfill & \hfill \\ \multicolumn{1}{c}{}& -(\frac{\beta }{1-\beta }·\frac{1}{s^{\alpha +\beta }})\hfill \\ \multicolumn{1}{c}{(1,\beta )}& \hfill \end{array}\right]F(s)\hfill \\ -\frac{B(\beta )}{1-\beta }·\frac{s^{\alpha -1}}{\Gamma (1-\alpha )}{}_2{\Psi }_1\hfill \\ \left[\begin{array}{cc}(1,1),(1-\alpha ,\alpha +\beta )\hfill & \hfill \\ \multicolumn{1}{c}{}& -(\frac{\beta }{1-\beta }·\frac{1}{s^{\alpha +\beta }})\hfill \\ \multicolumn{1}{c}{(1,\beta )}& \hfill \end{array}\right]f(0)\hfill \end{array}$$

(4)

For this operator, the inverse Laplace transform is defined as follows

$${}_0{}^{AC}{D}_t^{\alpha \mathrm{,}\beta }f(t)={}_0{}^{AR}{D}_t^{\alpha ,\beta }f(t)-\frac{B(\beta )}{1-\beta }\cdot \frac{1}{\Gamma (1-\alpha )}\cdot f(0)t^{-\alpha }{E}_{\beta }\{-t^{\beta +\alpha }\}$$

(5)

For the fractional operator with bi-order (equation (1)), when $\alpha =0$ , the Atangana–Baleanu fractional derivative in Liouville–Caputo sense (ABC) is recovered. This operator is defined as follows^34–38

$${}_0{}^{ABC}{D}_t^{\beta }f(t)=\frac{B(\beta )}{1-\beta }{\displaystyle \underset{0}{\overset{t}{\int }}\stackrel{.}{f}(\tau )E_{\beta }\left[-\frac{\beta }{1-\beta }{(t-\tau )}^{\beta }\right]}d\tau \mathrm{,}0<\beta \le 1$$

(6)

where $d^{\beta }/d{t}^{\beta }{=}_0^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }$ is an ABC fractional derivative with respect to $t$ , $B(\beta )$ is a normalization function.

The Laplace transform of equation (6) is defined as follows^34–38

$$\begin{array}{c}\mathcal{L}[{}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }f(t)](s)=\frac{B(\beta )}{1-\beta }\mathcal{L}\hfill \\ [\underset{a}{\overset{t}{\int }}\stackrel{·}{f}(\tau )E_{\beta }[-\frac{\beta }{1-\beta }{(t-\tau )}^{\beta }]d\tau ]\hfill \\ =\frac{B(\beta )}{1-\beta }\frac{s^{\beta }\mathcal{L}[f(t)](s)-s^{\beta -1}f(0)}{s^{\beta }+\frac{\beta }{1-\beta }}\hfill \end{array}$$

(7)

For the fractional operator with bi-order (equation (1)), when $\beta =0$ , the Liouville–Caputo fractional derivative (C) is recovered. This operator is defined as follows

$${}_0{}^c{\mathcal{D}}_t^{\alpha }f(t)=\frac{1}{\Gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}{(t-\tau )}^{-\alpha }\frac{d}{\mathrm{dt}}f(\tau )d\tau ,0<\alpha \le 1$$

(8)

Electrical circuits

To keep the physical dimensionality of the fractional differential equation with bi-order (equation (1)), a parameter $\sigma$ is introduced, ⁴⁰ this parameter has the dimension of seconds. In this case, we have

$$\frac{d}{\mathrm{dt}}\to \frac{1}{{\sigma }^{1-\alpha ,\beta }}{·}_0^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta },0<\alpha ,\beta \le 1$$

(9)

$$\frac{d^2}{d{t}^2}\to \frac{1}{{\sigma }^{2(1-\alpha ,\beta )}}{·}_0^{\mathrm{AC}}{\mathcal{D}}_t^{2(\alpha ,\beta )},0<\alpha ,\beta \le 1$$

(10)

when $(\alpha =\beta =1)$ , the expressions mentioned above are recovered in the classical case.

RC electrical circuit

Applying Kirchhoff’s laws, the equation of the RC electrical series circuit is given as follows

$${\mathcal{D}}_{t}V(t)+\frac{1}{\mathrm{RC}}V(t)=\frac{1}{\mathrm{RC}}E(t)$$

(11)

where C is the capacitance, R is the resistance, and the source voltage is $E(t)$ .

Considering equation (1), the fractional equation with bi-order for this circuit is given as follows

$${}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }V(t)=\tau E(t)-\tau V(t)$$

(12)

where

$$\tau =\frac{{\sigma }^{1-\alpha ,\beta }}{\mathrm{RC}}={\tau }_0·{\sigma }^{1-\alpha ,\beta }$$

(13)

τ is the fractional time constant with bi-order and ${\tau }_0=1/\mathrm{RC}$ is the classical time constant. Now, we obtain the numerical approximation of equation (12) considering different source terms.

First case

Unit step source, $E(t)=u(t)$ , $V(0)=V_0$ , $(V_0>0)$ , equation (12) can be written as follows

$${}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }V(t)=\tau u(t)-\tau V(t)$$

(14)

The numerical approximation of equation (14) is given as follows

$$\begin{array}{c}\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{\mathrm{dV}(\nu )}{\mathrm{dt}}{(t-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\}d\nu +\tau V(t)=\tau u(t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{V(t)-V(t+\Delta t)}{2\Delta t}{(t-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\}d\nu +\tau V(t)=\tau u(t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\underset{t_i}{\overset{t_{i+1}}{\int }}\frac{V(t_{i+1})-V(t_i)}{2\Delta t}{(t_n-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t_n-\nu )}^{\beta +\alpha }\}d\nu \hfill \\ +\tau \left[\frac{V(t_{n+1})-V(t_n)}{2}\right]=\tau u(t)\hfill \end{array}$$

(15)

then

$$\begin{array}{c}V(t)=\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\frac{V(t_{i+1})-V(t_i)}{2\Delta t}[{(t_n-t_{i+1})}^{1-\alpha }{E}_{\beta ,2-\alpha }\{-\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\}\hfill \\ -{(t_n-t_i)}^{1-\alpha }{E}_{\beta ,2-\alpha }\{-\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\}]+\tau \left[\frac{V(t_{n+1})-V(t_n)}{2}\right]-\tau u(t)\hfill \end{array}$$

(16)

where $E_{\beta }$ is the Mittag-Leffler function.

If $\alpha =0$ in equation (1), we recover ABC fractional derivative equation (6). Considering this operator, we have

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }V(t)=\tau E(t)-\tau V(t)$$

(17)

where

$$\tau =\frac{{\sigma }^{1-\beta }}{\mathrm{RC}}={\tau }_0·{\sigma }^{1-\beta }$$

(18)

τ is the fractional time constant and ${\tau }_0=1/\mathrm{RC}$ is the classical time constant. For the unit step source, $E(t)=u(t)$ , $V(0)=V_0$ , $(V_0>0)$ , equation (17) can be written as follows

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }V(t)=\tau u(t)-\tau V(t)$$

(19)

Applying the Laplace transform to equation (19), we have

$$\mathcal{L}\{{}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }V(t)\}(s)=\tau \frac{1}{s}-\tau \tilde{V}(s)$$

(20)

where $\tilde{V}(s)=\mathcal{L}\{V(t)\}(s)$ .

Considering

$$\mathcal{L}\{{}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }V(t)\}(s)=\frac{B(\beta )}{1-\beta }·\frac{s^{\beta }\tilde{V}(s)-s^{\beta -1}V(0)}{s^{\beta }+\frac{\beta }{1-\beta }}$$

(21)

substituting equation (21) into equation (20), we have

$$\frac{B(\beta )}{1-\beta }·\frac{s^{\beta }\tilde{V}(s)}{s^{\beta }+\frac{\beta }{1-\beta }}+\tau \tilde{V}(s)=\tau \frac{1}{s}+\frac{B(\beta )}{1-\beta }·\frac{s^{\beta -1}{V}_0}{s^{\beta }+\frac{\beta }{1-\beta }}$$

(22)

the expression for the voltage is

$$\begin{array}{c}\tilde{V}(s)=\frac{\tau }{B(\beta )+\tau (1-\beta )}\hfill \\ [\frac{(1-\beta )s^{\beta -1}}{s^{\beta }+\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}}+\frac{\beta }{s^{\beta }+\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}}·\frac{1}{s}]\hfill \\ +\frac{B(\beta )V_0}{B(\beta )+\tau (1-\beta )}·\frac{s^{\beta -1}}{s^{\beta }+\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}}\hfill \end{array}$$

(23)

Taking the inverse Laplace transform of equation (23), we obtain the following solution

$$\begin{array}{c}V(t)=\frac{\tau (1-\beta )}{B(\beta )+\tau (1-\beta )}{E}_{\beta ,1}[-\left(\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}\right)t^{\beta }]\hfill \\ +\frac{B(\beta )V_0}{B(\beta )+\tau (1-\beta )}{E}_{\beta ,1}[-\left(\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}\right)t^{\beta }]\hfill \\ +\frac{\tau (\beta )}{B(\beta )+\tau (1-\beta )}\hfill \\ \underset{0}{\overset{t}{\int }}{(t-\tau )}^{\beta -1}{E}_{\beta ,\beta }[-\left(\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}\right){(t-\tau )}^{\beta }]d\tau \hfill \end{array}$$

(24)

Second case

Periodic source, $E(t)=\sin (\phi t)$ , $V(0)=V_0$ , $(V_0>0)$ , equation (12) can be written as follows

$${}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }V(t)=\tau \sin (\phi t)-\tau V(t)$$

(25)

where τ is given by equation (13).

The numerical approximation of equation (25) is given as follows

$$\begin{array}{c}\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{\mathrm{dV}(\nu )}{\mathrm{dt}}{(t-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\}d\nu +\tau V(t)=\tau \sin (\phi t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{V(t)-V(t+\Delta t)}{2\Delta t}{(t-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\}d\nu +\tau V(t)=\tau \sin (\phi t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\underset{t_i}{\overset{t_{i+1}}{\int }}\frac{V(t_{i+1})-V(t_i)}{2\Delta t}{(t_n-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t_n-\nu )}^{\beta +\alpha }\}d\nu \hfill \\ +\tau \left[\frac{V(t_{n+1})-V(t_n)}{2}\right]=\tau \sin (\phi t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\frac{V(t_{i+1})-V(t_i)}{2\Delta t}\underset{t_n-t_i}{\overset{t_n-t_i+1}{\int }}{\mu }^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{\mu }^{\beta +\alpha }\}d\mu \hfill \\ +\tau \left[\frac{V(t_{n+1})-V(t_n)}{2}\right]=\tau \sin (\phi t)\hfill \end{array}$$

(26)

then

$$\begin{array}{c}V(t)=\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\frac{V(t_{i+1})-V(t_i)}{2\Delta t}[{(t_n-t_{i+1})}^{1-\alpha }{E}_{\beta ,2-\alpha }\left\{\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\right\}\hfill \\ -(t_n-t_i{)}^{1-\alpha }{E}_{\beta ,2-\alpha }\{-\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\}]+\tau \left[\frac{V(t_{n+1})-V(t_n)}{2}\right]-\tau \sin (\phi t)\hfill \end{array}$$

(27)

where $E_{\beta }$ is the Mittag-Leffler function.

Considering $\alpha =0$ in equation (1), we recover ABC fractional derivative (equation (6)). Considering this operator, we have

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }V(t)=\tau E(t)-\tau V(t)$$

(28)

where τ is given by equation (18). For the periodic source, $E(t)=\sin (\phi t)$ , $V(0)=V_0$ , $(V_0>0)$ , equation (39) can be written as follows

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }V(t)=\tau \sin (\phi t)-\tau V(t)$$

(29)

Applying the Laplace transform to equation (29), we have

$${\mathcal{L}}_t\{{}_a{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }V(t)\}(s)=\tau \left(\frac{\phi }{s^2+{\phi }^2}\right)-\tau \tilde{V}(s)$$

(30)

where $\tilde{V}(s)={\mathcal{L}}_t\{V(t)\}(s)$ .

The expression for the voltage is as follows

$$\begin{array}{c}\tilde{V}(s)=\frac{\tau }{B(\beta )+\tau (1-\beta )}\hfill \\ [\frac{s^{\beta }(1-\beta )}{s^{\beta }+\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}}·\frac{\phi }{s^2+{\phi }^2}+\frac{\beta }{s^{\beta }+\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}}·\frac{\phi }{s^2+{\phi }^2}]\hfill \\ +\frac{B(\beta )V_0}{B(\beta )+\tau (1-\beta )}·\frac{s^{\beta -1}}{s^{\beta }+\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}}\hfill \end{array}$$

(31)

Taking the inverse Laplace transform of equation (31), we obtain the following solution

$$+\frac{B(\beta )V_0}{B(\beta )+\tau (1-\beta )}{E}_{\beta ,1}[-\frac{{(\tau t)}^{\beta }}{B(\beta )+\tau (1-\beta )}]$$

(32)

RL electrical circuit

Applying Kirchhoff’s laws, the equation of the RL electrical series circuit is given as follows

$${\mathcal{D}}_{t}I(t)+\frac{R}{L}I(t)=\frac{1}{L}E(t)$$

(33)

where L is the inductance, R is the resistance, and the source voltage is E(t).

Considering equation (1), the fractional equation with bi-order for this circuit is given as follows

$${}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }I(t)=\left(\frac{\eta }{R}\right)E(t)-\eta I(t)$$

(34)

where

$$\eta =\frac{{\sigma }^{1-\alpha ,\beta }}{{\eta }_0}$$

(35)

$\eta$ is the fractional time constant with bi-order, and ${\eta }_0=L/R$ is the classical time constant. Now, we obtain the numerical approximation of equation (34) considering different source terms.

First case

Unit step source, $E(t)=u(t)$ , $I(0)=I_0$ , $(I_0>0)$ , equation (34) can be written as follows

$${}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }I(t)=\left(\frac{\eta }{R}\right)u(t)-\eta I(t)$$

(36)

The numerical approximation of equation (36) is given as follows

$$\begin{array}{c}\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{\mathrm{dI}(\nu )}{\mathrm{dt}}{(t-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\}d\nu +\eta I(t)=\left(\frac{\eta }{R}\right)u(t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{I(t)-I(t+\Delta t)}{2\Delta t}{(t-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\}d\nu +\eta I(t)=\left(\frac{\eta }{R}\right)u(t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\underset{t_i}{\overset{t_{i+1}}{\int }}\frac{I(t_{i+1})-I(t_i)}{2\Delta t}{(t_n-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t_n-\nu )}^{\beta +\alpha }\}d\nu \hfill \\ +\eta \left[\frac{I(t_{n+1})-I(t_n)}{2}\right]=\left(\frac{\eta }{R}\right)u(t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\frac{I(t_{i+1})-I(t_i)}{2\Delta t}\underset{t_n-t_i}{\overset{t_n-t_i+1}{\int }}{\mu }^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{\mu }^{\beta +\alpha }\}d\mu \hfill \\ +\eta \left[\frac{I(t_{n+1})-I(t_n)}{2}\right]=\left(\frac{\eta }{R}\right)u(t)\hfill \end{array}$$

(37)

then

$$\begin{array}{cc}\multicolumn{1}{c}{I(t)}& =\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\frac{I(t_{i+1})-I(t_i)}{2\Delta t}[{(t_n-t_{i+1})}^{1-\alpha }{E}_{\beta ,2-\alpha }\{-\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\}\hfill \\ -(t_n-t_i{)}^{1-\alpha }{E}_{\beta ,2-\alpha }\{-\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\}]+\eta \left[\frac{I(t_{n+1})-I(t_n)}{2}\right]-\left(\frac{\eta }{R}\right)u(t)\hfill \end{array}$$

(38)

where $E_{\beta }$ is the Mittag-Leffler function.

Considering $\alpha =0$ in equation (1), we recover ABC fractional derivative (equation (6)). Considering this operator we have

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }I(t)=\left(\frac{\eta }{R}\right)E(t)-\eta I(t)$$

(39)

where

$$\eta =\frac{{\sigma }^{1-\beta }}{{\eta }_0}$$

(40)

$\eta$ is the fractional time constant and ${\eta }_0=L/R$ is the classical time constant. Now, we obtain the numerical approximation of equation (39). For the unit step source, $E(t)=u(t)$ , $I(0)=I_0$ , $(I_0>0)$ , equation (34) can be written as follows

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }I(t)=\left(\frac{\eta }{R}\right)u(t)-\eta I(t)$$

(41)

Applying the Laplace transform to equation (41), we have

$$\mathcal{L}\{{}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }I(t)\}(s)=\left(\frac{\eta }{R}\right)\frac{1}{s}-\eta \tilde{I}(s)$$

(42)

where $\tilde{I}(s)=\mathcal{L}\{I(t)\}(s)$ .

Considering

$$\mathcal{L}\{{}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }I(t)\}(s)=\frac{B(\beta )}{1-\beta }·\frac{s^{\beta }\tilde{I}(s)-s^{\beta -1}I(0)}{s^{\beta }+\frac{\beta }{1-\beta }}$$

(43)

substituting equation (43) into equation (42) we have

$$\frac{B(\beta )}{1-\beta }·\frac{s^{\beta }\tilde{I}(s)}{s^{\beta }+\frac{\beta }{1-\beta }}+\left(\frac{\eta }{R}\right)\tilde{I}(s)=\left(\frac{\eta }{R}\right)\frac{1}{s}+\frac{B(\beta )}{1-\beta }·\frac{s^{\beta -1}{I}_0}{s^{\beta }+\frac{\beta }{1-\beta }}$$

(44)

the expression for the current is as follows

$$\begin{array}{c}\tilde{I}(s)=\frac{\eta }{R[B(\beta )+\eta (1-\beta )]}\hfill \\ [\frac{(1-\beta )s^{\beta -1}}{s^{\beta }+\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}}+\frac{\beta }{s^{\beta }+\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}}·\frac{1}{s}]\hfill \\ +\frac{B(\beta )V_0}{B(\beta )+\tau (1-\beta )}·\frac{s^{\beta -1}}{s^{\beta }+\frac{\tau \beta }{B(\beta )+\tau (1-\beta )}}\hfill \end{array}$$

(45)

Taking the inverse Laplace transform of equation (45), we obtain the following solution

$$\begin{array}{c}I(t)=\frac{\eta (1-\beta )}{R[B(\beta )+\eta (1-\beta )]}{E}_{\beta ,1}[-\left(\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}\right)t^{\beta }]\hfill \\ +\frac{B(\beta )V_0}{B(\beta )+\eta (1-\beta )}{E}_{\beta ,1}[-\left(\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}\right)t^{\beta }]\hfill \\ +\frac{\eta (\beta )}{R[B(\beta )+\eta (1-\beta )]}\hfill \\ \underset{0}{\overset{t}{\int }}{(t-\tau )}^{\beta -1}{E}_{\beta ,\beta }[-\left(\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}\right){(t-\tau )}^{\beta }]d\tau \hfill \end{array}$$

(46)

Second case

Periodic source, $E(t)=\sin (\phi t)$ , $I(0)=I_0$ , $(I_0>0)$ , equation (34) can be written as follows

$${}_0{}^{\mathrm{AC}}{\mathcal{D}}_t^{\alpha ,\beta }I(t)=\left(\frac{\eta }{R}\right)\sin (\phi t)-\eta I(t)$$

(47)

where $\eta$ is given by equation (35).

The numerical approximation of equation (47) is given as follows

$$\begin{array}{c}\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{\mathrm{dI}(\nu )}{\mathrm{dt}}{(t-\nu )}^{-\alpha }{E}_{\beta }\{-\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\}d\nu +\eta I(t)=\left(\frac{\eta }{R}\right)\sin (\phi t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}\frac{I(t)-I(t+\Delta t)}{2\Delta t}{(t-\nu )}^{-\alpha }{E}_{\beta }-\left\{\frac{\beta }{1-\beta }{(t-\nu )}^{\beta +\alpha }\right\}d\nu +\eta I(t)=\left(\frac{\eta }{R}\right)\sin (\phi t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\underset{t_i}{\overset{t_{i+1}}{\int }}\frac{I(t_{i+1})-I(t_i)}{2\Delta t}{(t_n-\nu )}^{-\alpha }{E}_{\beta }-\left\{\frac{\beta }{1-\beta }{(t_n-\nu )}^{\beta +\alpha }\right\}d\nu \hfill \\ +\eta \left[\frac{I(t_{n+1})-I(t_n)}{2}\right]=\left(\frac{\eta }{R}\right)\sin (\phi t)\hfill \\ \frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\frac{I(t_{i+1})-I(t_i)}{2\Delta t}\underset{t_n-t_i}{\overset{t_n-t_i+1}{\int }}{\mu }^{-\alpha }{E}_{\beta }-\left\{\frac{\beta }{1-\beta }{\mu }^{\beta +\alpha }\right\}d\mu \hfill \\ +\eta \left[\frac{I(t_{n+1})-I(t_n)}{2}\right]=\left(\frac{\eta }{R}\right)\sin (\phi t)\hfill \end{array}$$

(48)

then

$$\begin{array}{c}I(t)=\frac{B(\beta )}{1-\beta }·\frac{1}{\Gamma (1-\alpha )}\sum _{i=0}^n\frac{I(t_{i+1})-I(t_i)}{2\Delta t}[{(t_n-t_{i+1})}^{1-\alpha }{E}_{\beta ,2-\alpha }-\left\{\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\right\}\hfill \\ -{(t_n-t_i)}^{1-\alpha }{E}_{\beta ,2-\alpha }\{-\frac{\beta }{1-\beta }{(t_n-t_{i+1})}^{\beta +\alpha }\}]+\eta \left[\frac{I(t_{n+1})-I(t_n)}{2}\right]-\left(\frac{\eta }{R}\right)\sin (\phi t)\hfill \end{array}(49)$$

(49)

where $E_{\beta }$ is the Mittag-Leffler function.

Considering $\alpha =0$ in equation (1), we recover ABC fractional derivative (equation (6)). Considering this operator, we have

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }I(t)=\left(\frac{\eta }{R}\right)E(t)-\eta I(t)$$

(50)

where $\eta$ is given by equation (40). For the periodic source, $E(t)=\sin (\phi t)$ , $I(0)=I_0$ , $(I_0>0)$ , equation (50) can be written as follows

$${}_0{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }I(t)=\left(\frac{\eta }{R}\right)\sin (\phi t)-\eta I(t)$$

(51)

Applying the Laplace transform to equation (51), we have

$$L_t\{{}_a{}^{\mathrm{ABC}}{\mathcal{D}}_t^{\beta }I(t)\}(s)=\left(\frac{\eta }{R}\right)\left(\frac{\phi }{s^2+{\phi }^2}\right)-\eta \tilde{I}(s)$$

(52)

where $\tilde{I}(s)={\mathcal{L}}_t\{I(t)\}(s)$ .

The expression for the current is as follows

$$\begin{array}{c}\tilde{I}(s)=\frac{\eta }{R[B(\beta )+\tau (1-\beta )]}[\frac{s^{\beta }(1-\beta )}{s^{\beta }+\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}}·\frac{\phi }{s^2+{\phi }^2}\hfill \\ +\frac{\beta }{s^{\beta }+\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}}·\frac{\phi }{s^2+{\phi }^2}]\hfill \\ +\frac{B(\beta )V_0}{B(\beta )+\eta (1-\beta )}·\frac{s^{\beta -1}}{s^{\beta }+\frac{\eta \beta }{B(\beta )+\eta (1-\beta )}}\hfill \end{array}$$

(53)

Taking the inverse Laplace transform of (53), we obtain the following solution

$$I(t)=\frac{\eta {(1-\beta )}^2}{R[B(\beta )+\eta (1-\beta )]}\underset{0}{\overset{t}{\int }}\frac{d}{d\gamma }\underset{0}{\overset{\gamma }{\int }}\delta (\phi )E_{\beta ,1}[-(\frac{\eta \beta (\gamma -\phi )}{B(\beta )+\eta (1-\beta )}]d\phi ·\sin (\phi (t-\gamma ))d\gamma +\frac{B(\beta )V_0}{B(\beta )+\eta (1-\beta )}{E}_{\beta ,1}[-\frac{{(\eta t)}^{\beta }}{B(\beta )+\eta (1-\beta )}]$$

(54)

Examples

Consider the electrical series circuit RC with $R=10\Omega$ , $C=0.1\mathrm{F}$ , and $V_0=10\mathrm{V}$ . Figure 1(a)–(d) shows numerical simulations for the voltage across the capacitor considering unit step source with bi-order operator, equation (16) and unit step source with ABC operator, equation (24) and periodic source with bi-order operator (27) and periodic source with ABC operator, equation (32), for different particular cases of $\alpha$ and $\beta$ .

Figure 1.

Numerical simulation for RC electrical series circuit: (a) fractional operator with bi-order, equation (16); (b) fractional operator of type ABC, equation (24); (c) fractional operator with bi-order, equation (27); and (d) fractional operator of type ABC, equation (32).

Consider the electrical series circuit RL with $R=10\Omega$ , $L=10\mathrm{mH}$ , and $I_0=10\mathrm{A}$ . Figure 2(a)–(d) shows numerical simulations for the current across the inductor considering unit step source with bi-order operator, equation (38) and unit step source with ABC operator, equation (46) and periodic source with bi-order operator (49), and periodic source with ABC operator, equation (54), for different particular cases of $\alpha$ and $\beta$ .

Figure 2.

Numerical simulation for RL electrical series circuit: (a) fractional operator with bi-order, equation (38); (b) fractional operator of type ABC, equation (46); (c) fractional operator with bi-order, equation (49); and (d) fractional operator of type ABC, equation (54).

Conclusion

In this article, we describe electrical series circuits RC and RL using the concept of derivative with two fractional orders $\alpha$ and $\beta$ in Liouville–Caputo sense. The definitions developed by Atangana are the extensions of Liouville–Caputo and Riemann–Liouville derivatives with Mittag-Leffler kernel. These fractional operators have been widely used; however, none of these fractional operators is capable of handling the concept of heterogeneity and structures with different scales with great success. In the literature, the fractional calculus is applied to various electrical circuit problems as a useful mathematical tool. In the electrical circuits, the experimental results do not comply with standard theoretical calculations; this is due to the effects of ohmic friction and temperature (dissipative effects related to resistance), and these losses are not taken into account in the standard approach. Since the temperature of these circuits increases with time, an unstable ohmic friction occurs during the process. This behavior is nonlinear and non-local in time (resistance value increasing nonlinearly with increasing temperature is not a constant parameter). In order to make a more realistic description, time-fractional derivative with bi-order can be used in the calculations; in this context, the results showed that there exists a close relationship between the temperature and fractional operator with bi-order. The case of $(\alpha =1,\beta =1)$ corresponds to local behavior in time or classical case; in this case, $R$ is a constant parameter and dissipative effects are neglected, whereas the case $(\alpha \ne 1,\beta \ne 1)$ corresponds to non-local behavior in time. In this alternative representation, considering fractional operators with bi-order is possible to describe more complex systems with different properties represented by $\alpha$ and $\beta$ . These types of problems cannot be portrayed with existing derivatives with fractional order. Alternative methods of analysis involving novel integral transforms^41–43 are goals of future works.