New equivalent resistance formula of $m\times n$ rectangular resistor network represented by Chebyshev polynomials

Abstract

In the process of exploring the field of circuits, obtaining the exact solution of the equivalent resistance between two nodes in a resistor network has become an important problem. This paper aims to introduce Chebyshev polynomial of the second kind to improve the equivalent resistance formula of $m\times n$ rectangular resistor network, thereby improving the calculation efficiency. Additionally, the discrete sine transform of the first kind (DST-I) is utilized to solve the modeling equation. Under the condition of applying the new equivalent resistance formula, several equivalent resistance formulas with different parameters are given, and three-dimensional views are used to illustrate them. Six comparison tables are provided to showcase the advantages of the improved explicit formula in terms of computational efficiency, as well as the relationship between resistivity and the maximum size of the resistor network that the formula can effectively handle. This may provide more convenient and effective technical support for research and practice in electronic engineering and other related fields.

Introduction

The resistor network is an important research direction in the field of circuit analysis and design. These play a vital role in various electronic systems and applications, and are widely used in signal processing, power systems, communication systems and so on. Accurate analysis and efficient calculation of the equivalent resistance in a resistor network is essential for understanding its behavior and optimizing performance. Tan’s^{1,2,3,4,5,6,7,8,9,10} innovative work in establishing various resistor network models has provided significant theoretical support and has far-reaching implications for scientific researchers. By improving the efficiency of calculating the equivalent resistance formula, researchers can better solve complex scientific and technical problems. Therefore, in order to enhance the numerical processing of the formula, this paper re-expresses the original formula using Chebyshev polynomials to save calculation time.

According to research, many practical problems have been solved by establishing resistor network models^{11,12,13,14,15,16,17,18} and neural network models^{20,21,22,23,24,25,26,27}. In the past few decades, researchers have extensively studied resistor networks, focusing on research directions such as electromigration phenomenon, graph theory, studies of impedance network, infinite network, finite network and the other Laplace matrix (LM) methods^{17,18,19,28,29,30,31,32,33,34,35,36,37,38,39,40,41,43,44,46}. The neural network proposed by Shi et al.^20,21,22,23 has similarities with the resistor network in processing and analyzing complex systems.

In recent years, the Recursion-Transform (RT) method proposed by Tan^{1,2,3,4,5,6,7,8,9,10,47,48,49,50,51,52} has attracted extensive attention in the research of resistor networks as a novel computational method. The RT method calculates the equivalent resistance by establishing a tridiagonal matrix and using matrix transformation and operation. Currently, there are many research results on tridiagonal matrices^{53,54,55,56,57,58,59,60,61,62,63}. The traditional $\hbox {Green}^{\prime }$s function method^41,42,43 is usually employed to solve infinite resistor network problems or those with periodic boundary conditions by calculating equivalent resistance through the construction of a point source in the network. The $\hbox {Green}^{\prime }$s function approach relies on network symmetry and Fourier transforms, making it suited for obtaining analytical solutions in infinite or periodic networks. However, for finite networks, the resistance formulas derived from the Green’s function method are less suitable for numerical computation, as the integral convergence slows with increasing grid points, thereby complicating the calculations. In contrast, the RT method is applicable to finite network models of various sizes and complexities, offering greater flexibility in engineering applications. In 2015, the method is further improved, Tan² studies the hard problem of two-point resistance on irregular $m \times n$ spider webs with an arbitrary longitude. Additionally, Tan conducted research and analysis on spherical³ and sector⁸ network models. The RT method has become an important technical mean to study various topological resistor networks^{1,2,3,4,5,6,7,8,9,10,47,48,49,50,51,52}, which has the potential to bring new breakthroughs and development opportunities to the field of electronic engineering and other fields.

This paper is organized as follows: In Sect. 2, the original equivalent resistance formula of rectangular resistor network is given. In Sect. 3, a new formula of equivalent resistance expressed by Chebyshev polynomials is given. In Sect. 4, the derivation of the new formula is introduced in detail. In Sect. 5, the equivalent resistance formulas and their three-dimensional diagrams for several special cases are presented. In Sect. 6, the efficiency of the original formula and the new formula for calculating equivalent resistance is analyzed. In Sect. 7, the paper is concluded.

Original equivalent resistance formula

In this section, the equivalent resistance formula for an $m \times n$ resistor network with an arbitrary boundary, derived by Tan⁷, is provided, along with the key equations necessary for solving it.

In 2016, Tan⁷ proposed an $m \times n$ rectangular resistor network, as shown in Fig. 1. The resistance in the vertical and horizontal directions are $r_{0}$ and r, where $r_1$ is the right boundary resistor, m and n are the number of resistors between two nodes on each vertical line and horizontal line, respectively. $r_1$ is an arbitrary resistor on the right boundary. Various geometric structures can be obtained by adjusting the right boundary. For example, when $r_1=0$, a fan-shaped network model is obtained, and when $r_1=r_{0}$, a regular rectangular network model is formed. The nodes in the resistor network are represented by coordinates (x, y). Where $d_{1}(x, y_{1})$ and $d_{2}(x, y_{2})$ are two arbitrary nodes on the common vertical axis of the $m \times n$ resistor network. A part of the rectangular resistor network is selected for analysis and study using Kirchhoff’s law. The schematic diagram of the partial resistor network is shown in Fig. 2, which represents all current distributions and parameters in the resistor network.

Fig. 1

An $m \times n$ rectangular resistor network, except the right boundary resistor in the vertical direction is $r_1$, its horizontal and vertical resistors are r and $r_{0}$, respectively.

Fig. 2

Partial resistor network with current directions and parameters.

The equivalent resistance $R_{m \times n}(d_{1},d_{2})$ between two arbitrary nodes $d_{1}(x, y_{1})$ and $d_{2}(x, y_{2})$ in an $m \times n$ rectangular resistor network is shown below

$$\begin{aligned} R_{m \times n}(d_{1},d_{2}) = \frac{r_{0}}{m+1}\sum ^{m}_{i=1}\frac{(C_{1,i}-C_{2,i})^{2}}{1-\cos \theta _{i}}\times \left( \frac{\left[ \Delta F^{(i)}_{n-x}+(h_1-1)\Delta F^{(i)}_{n-x-1}\right] \Delta F^{(i)}_{x}}{F^{(i)}_{n+1}+(h_1-1)F^{(i)}_{n}}\right) , \end{aligned}$$

(1)

where

$$\begin{aligned} h_1 = r_1/r_0,~C_{k,i} = \cos (y_{k}+\frac{1}{2})\theta _{i},\ \theta _{i} =\frac{i\pi }{m+1}, \nonumber \\ F^{(i)}_{k} = \left( \lambda ^{k}_{i}-{\bar{\lambda }}^{k}_{i}\right) /(\lambda _{i}-{\bar{\lambda }}_{i}),\ \Delta F^{(i)}_{k} = F^{(i)}_{k+1}-F^{(i)}_{k}, \end{aligned}$$

(2)

$$\begin{aligned} \begin{aligned}&\lambda _{i} = 1+h-h\cos \theta _{i}+\sqrt{(1+h-h\cos \theta _{i})^{2} - 1},\\&{\bar{\lambda }}_{i} = 1+h-h\cos \theta _{i}-\sqrt{(1+h-h\cos \theta _{i})^{2} - 1}, ~~h=r/r_{0}. \end{aligned} \end{aligned}$$

(3)

Tan analyzed and studied the resistor network, and established a resistor network model based on Kirchhoff’s law. The general matrix equation is given below.

$$\begin{aligned} {\textbf{I}}_{k+1} = {\textbf{A}}_{m}{\textbf{I}}_{k} - {\textbf{I}}_{k-1} - J {\textbf{H}}_{x}\delta _{k,x}, \end{aligned}$$

(4)

the function $\delta _{k,x}$ is defined as $\delta _{k,x}={\left\{ \begin{array}{ll}1,& \ x =k\\ 0,& \ x\ne k \end{array}\right. }$ , ${\textbf{I}}_{k}$ and ${\textbf{H}}_{x}$ are the $m \times 1$ column matrices which can be described as

$$\begin{aligned} {\textbf{I}}_{k} = \left[ I^{(1)}_{k},~I^{(2)}_{k},~...,I^{(m)}_{k}\right] ^{T}\ (0 \le k \le n), \\ ({\textbf{H}}_{x})_{j} = h(-\delta _{j,y_{1}}+\delta _{j,y_{1}+1}+\delta _{j,y_{2}}-\delta _{j,y_{2}+1}), \end{aligned}$$

where $({\textbf{H}}_{x})_{j}$ is the element of ${\textbf{H}}_{x}$ when injecting current J at $d_1(x, y_1)$ and exiting at $d_2(x, y_2)$ ,

$$\begin{aligned} {\textbf{A}}_{m}= \left( \begin{array}{cccccc} 2+2h & -h & 0& \cdots & 0\\ -h & 2+2h & -h & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & -h & 2+2h & -h\\ 0 & \cdots & 0 & -h & 2+2h \end{array} \right) _{m\times m}, \end{aligned}$$

(5)

where $h = r/r_0$.

New formula of equivalent resistance represented by Chebyshev polynomials

For the equivalent resistance formula (1), Eq. (2) is an explanation of the symbols in formula (1), which involves complex exponential operations and has high computational complexity. In order to improve the calculation efficiency of equivalent resistance, this section introduces the improved equivalent resistance formula using the Chebyshev polynomial of the second kind.

Let the current J be input at $d_{1}(x, y_{1})$ and output at $d_{2}(x, y_{2})$, the equivalent resistance between two nodes in the $m \times n$ resistor network is given by

$$\begin{aligned} R_{m \times n}(d_{1},d_{2}) = \frac{r_{0}}{m+1}\sum ^{m}_{j=1}\frac{(S_{1,j}-S_{2,j})^{2}}{1-\cos \theta _{j}}\times \left( \frac{[\Delta B^{(j)}_{n-x}+(h_1-1)\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{B^{(j)}_{n+1}+(h_1-1)B^{(j)}_{n}}\right) , \end{aligned}$$

(6)

where

$$\begin{aligned} \Delta B^{(j)}_{k}=B^{(j)}_{k+1}-B^{(j)}_{k}, \end{aligned}$$

(7)

$$\begin{aligned} S_{q,j} = \cos \frac{(q+\frac{1}{2})j\pi }{m+1},\ q=y_{1},\ y_{2}, \nonumber \\ \theta _{j} =\frac{j\pi }{m+1}, \end{aligned}$$

(8)

$$\begin{aligned} \begin{aligned} B^{(j)}_{k} = B^{(j)}_{k}(\cosh \mu _{j}) = \frac{\sinh (k\mu _{j})}{\sinh \mu _{j}},\ \cosh \mu _{j}=\frac{\sigma _{j}}{2},\\ k=n-x+1,~n-x,~n-x-1,~x+1,~x,~n+1,~n,~~j=1,~2,~\ldots ,~m. \end{aligned} \end{aligned}$$

(9)

$$\begin{aligned} \sigma _{j}=2+\frac{2r}{r_{0}}-\frac{2r}{r_{0}}\cos \frac{j\pi }{m+1}. \end{aligned}$$

(10)

Derivation of the new equivalent resistance formula

In this section, Chebyshev polynomial of the second kind is adopted to signify the Horadam sequence⁶⁴, which improves the calculation efficiency. And the discrete sine transform is introduced to obtain the solution of the model equations, the equivalent resistance formula is re-derived.

Horadam sequence represented by Chebyshev polynomials

Horadam sequence contains the following conditions:

$$\begin{aligned} W_k=dW_{k-1}-qW_{k-2},~~W_0=A,~~W_1=B, \end{aligned}$$

(11)

where $k\in {\textbf{N}},~~k\ge 2,~A,~B,~d,~q\in {\textbf{C}}$, ${\textbf{N}}$ is the set of all natural numbers and ${\textbf{C}}$ is the set of all complex numbers.

Horadam sequence⁶⁵ represented by Chebyshev polynomial of the second kind is

$$\begin{aligned} W_{k} =(\sqrt{q})^k\left( \frac{B}{\sqrt{q}}U_{k-1}\left( \frac{d}{2\sqrt{q}}\right) - AU_{k-2}\left( \frac{d}{2\sqrt{q}}\right) \right) , \end{aligned}$$

(12)

where

$$\begin{aligned} U_{k} = \frac{\sin (k+1)\eta }{\sin \eta },~~\cos \eta =\frac{d}{2\sqrt{q}},~~\eta \in {\textbf{C}}, \end{aligned}$$

(13)

is the Chebyshev polynomial of the second kind⁶⁶.

Equation (13) contains complex numbers, since $\frac{d}{2\sqrt{q}}>1$, which in this study can be described as

$$\begin{aligned} B_{k}=U_{k} = \frac{\sinh (k+1)\mu }{\sinh \mu },~~\cosh \mu =\frac{d}{2\sqrt{q}},~~\mu \in {\textbf{R}}, \end{aligned}$$

(14)

where $i\mu =\eta$, i is the imaginary unit.

Discrete sine transform

Let

$$\begin{aligned} {\mathbb {S}}_m^{I}=\sqrt{\frac{2}{m+1}}\left( \sin \frac{jk\pi }{m+1} \right) _{k,j=1}^m. \end{aligned}$$

(15)

The matrix ${\mathbb {S}}_m^{I}$ is a well-known discrete sine transform of the first kind (DST-I)^67,68. ${\mathbb {S}}_m^{I}$ is an orthogonal matrix, and the inverse and transpose of ${\mathbb {S}}_m^{I}$ are still itself, i.e.

$$\begin{aligned} ({\mathbb {S}}_m^{I})^{-1}=({\mathbb {S}}_m^{I})^T={\mathbb {S}}_m^{I}. \end{aligned}$$

(16)

For Eq. (5), perform the following orthogonal diagonalization

$$\begin{aligned} ({\mathbb {S}}_m^{I})^{-1}{\textbf{A}}_{m}{\mathbb {S}}_m^{I}=\textrm{diag}(\sigma _1,~\sigma _2,~\ldots ,~\sigma _m), \end{aligned}$$

(17)

therefore,

$$\begin{aligned} {\textbf{A}}_{m}={\mathbb {S}}_m^{I}\textrm{diag}(\sigma _1,~\sigma _2,~\ldots ,~\sigma _m)({\mathbb {S}}_m^{I})^{-1}, \end{aligned}$$

(18)

where

$$\begin{aligned} \sigma _j=2+2h-2h\cos \frac{(j-1)\pi }{m},~~j=1,~2,~\ldots ,~m. \end{aligned}$$

(19)

From Eq. (17), it is known that the matrix ${\textbf{A}}_{m}$ is similar to $\textrm{diag}(\sigma _1,~\sigma _2,~\ldots ,~\sigma _m)$, so $\sigma _j$ is the eigenvalue of ${\textbf{A}}_{m}$.

By left-multiplying Eq. (17) by ${\mathbb {S}}_m^{I}$, we obtain the following equation

$$\begin{aligned} {\textbf{A}}_{m}{\mathbb {S}}_m^{I}={\mathbb {S}}_m^{I}\textrm{diag}(\sigma _1,~\sigma _2,~\ldots ,~\sigma _m), \end{aligned}$$

i.e.,

$$\begin{aligned} {\textbf{A}}_{m}\left( {\mathcal {T}}^{(1)},~{\mathcal {T}}^{(2)},~\ldots ,~{\mathcal {T}}^{(m)}\right) =\left( {\mathcal {T}}^{(1)},~{\mathcal {T}}^{(2)},~\ldots ,~{\mathcal {T}}^{(m)}\right) \textrm{diag}(\sigma _1,~\sigma _2,~\ldots ,~\sigma _m), \end{aligned}$$

(20)

where ${\mathcal {T}}^{(j)}=\left( {\mathcal {T}}^{(j)}_1,~{\mathcal {T}}^{(j)}_2,~\ldots ,~{\mathcal {T}}^{(j)}_m \right)$,

$$\begin{aligned} {\mathcal {T}}^{(j)}_k=\sqrt{\frac{2}{m+1}}\sin \frac{jk\pi }{m+1},~~k=1,~2,~\ldots ,~m,~j=1,~2,~\ldots ,~m. \end{aligned}$$

Equation (20) can be expressed as

$$\begin{aligned} {\textbf{A}}_{m}{\mathcal {T}}^{(j)}=\sigma _j{\mathcal {T}}^{(j)},~~j=1,~2,~\ldots ,~m. \end{aligned}$$

(21)

Based on Eq. (21), the eigenvector ${\mathcal {T}}^{(j)}= \left( {\mathcal {T}}^{(j)}_1,~{\mathcal {T}}^{(j)}_2,~\ldots ,~{\mathcal {T}}^{(j)}_m \right)$ corresponding to $\sigma _j$ is obtained.

Let

$$\begin{aligned} \sqrt{\frac{m+1}{2}}{\mathbb {S}}_m^{I}{\textbf{I}}_{k} = {\textbf{L}}_{k}, \end{aligned}$$

(22)

where the $m\times 1$ column matrix ${\textbf{L}}_{k}$ is

$$\begin{aligned} {\textbf{L}}_{k} = \left[ L^{(1)}_{k},L^{(2)}_{k},...,L^{(m)}_{k}\right] ^{T} (0 \le k \le n). \end{aligned}$$

According to Eqs. (16) and (22), it can be obtained as follows

$$\begin{aligned} {\textbf{I}}_{k} = \sqrt{\frac{2}{m+1}}{\mathbb {S}}_m^{I}{\textbf{L}}_{k}. \end{aligned}$$

(23)

Considering the boundary conditions of the rectangular resistor network, the following current equations are established based on Kirchhoff’s law

$$\begin{aligned} {\textbf{I}}_{1} = [{\textbf{A}}_{m}-{\textbf{E}}_{m}]{\textbf{I}}_{0}, \end{aligned}$$

(24)

$$\begin{aligned} {\textbf{I}}_{n-1} = [{\textbf{A}}_{m}-(2-h_1){\textbf{E}}_{m}]{\textbf{I}}_{n}, \nonumber \\ h_1{\textbf{I}}_{n} +{\textbf{I}}_{n-2}={\textbf{A}}_{m}{\textbf{I}}_{n-1}, \end{aligned}$$

(25)

where $h_1=\frac{r_1}{r_0}$ , ${\textbf{A}}_{m}$ is given by Eq.(5) and ${\textbf{E}}_{m}$ is the $m\times m$ identity matrix.

Equations (4), (24) and (25) are multiplied by $\sqrt{\frac{2}{m+1}}{\mathbb {S}}_m^{I}$ on the left, and then combine with Eq. (22) to obtain the following equations

$$\begin{aligned} L^{(j)}_{k+1} = \sigma _{j}L^{(j)}_{k}-L^{(j)}_{k-1}-Jh\delta _{k,x}\zeta _j, \end{aligned}$$

(26)

$$\begin{aligned} L^{(j)}_{1} = (\sigma _{j}-1)L^{(j)}_{0}, \nonumber \\ L^{(j)}_{n-1} = (\sigma _{j}+h_1-2)L^{(j)}_{n}, \end{aligned}$$

(27)

$$\begin{aligned} h_1L^{(j)}_{n}+L^{(j)}_{n-2}= \sigma _{j}L^{(j)}_{n-1}, \end{aligned}$$

(28)

where

$$\begin{aligned} \begin{aligned} \zeta _j=2\sin (\frac{1}{2}\theta _j)\left[ \left[ \cos (y_1+\frac{1}{2})\theta _j-\cos (y_2+\frac{1}{2})\theta _j \right] \right] . \end{aligned} \end{aligned}$$

(29)

Solving the matrix equations

The homogeneous equation of Eq. (26) is expressed as follows

$$\begin{aligned} L^{(j)}_{k+1} = \sigma _{j}L^{(j)}_{k}-L^{(j)}_{k-1}, \end{aligned}$$

let $W_0=L_x$, $W_1=L_{x+1}$, $d=\sigma _{j}$ and $q=1$ in Eq. (11), combine Eqs. (12), (13) and (14) to get the following equation

$$\begin{aligned} L^{(j)}_{k} = L^{(j)}_{x+1}B^{(j)}_{k-x}-L^{(j)}_{x}B^{(j)}_{k-x-1}, \end{aligned}$$

(30)

where

$$\begin{aligned} \begin{aligned} B^{(j)}_{k} = B^{(j)}_{k}(\cosh \mu _{j}) = \frac{\sinh (k\mu _{j})}{\sinh \mu _{j}},\ \cosh \mu _{j}=\frac{\sigma _{j}}{2},\\ \end{aligned} \end{aligned}$$

(31)

$\sigma _{j}$ is defined by Eq. (19).

Next, consider the solution of Eq. (26) with the current input at $d_{1}(x_{1}, y_{1})$ and output at $d_{2}(x_{2}, y_{2})$. According to Eq. (30), the piecewise solutions of Eq. (26) are obtained as follows

$$\begin{aligned} L^{(j)}_{k} = L^{(j)}_{1}B^{(j)}_{k}-L^{(j)}_{0}B^{(j)}_{k-1},~~0 \le k \le x, \end{aligned}$$

(32)

$$\begin{aligned} L^{(j)}_{x+1} = \sigma _{j}L^{(j)}_{x}-L^{(j)}_{x-1}-Jh\zeta _j, \end{aligned}$$

(33)

$$\begin{aligned} L^{(j)}_{k} = L^{(j)}_{x+1}B^{(j)}_{k-x}-L^{(j)}_{x}B^{(j)}_{k-x-1},~~x \le k \le n, \end{aligned}$$

(34)

where $B^{(j)}_{k}$ is defined by Eq. (31).

Based on Eqs. (27), (28), (32), (33) and (34), the expression of $L^{(j)}_{x}$ can be described as

$$\begin{aligned} L^{(j)}_{x} = Jh\frac{[(\sigma _j-2) B^{(j)}_{n-x}+h_1\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{(\sigma _j-2)(\Delta B^{(j)}_{n}+h_1B^{(j)}_{n})}\zeta _j. \end{aligned}$$

(35)

From Eqs. (15), (23) and (35), the sum of currents between two nodes can be expressed as

$$\begin{aligned} \sum ^{y_2}_{j=y_1+1}I^{(j)}_{x} = \frac{J}{m+1}\sum ^{m}_{j=1}\frac{(S_{1,j}-S_{2,j})^{2}}{1-\cos \theta _{j}}\times \left( \frac{[(\sigma _j-2) B^{(j)}_{n-x}+h_1\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{\Delta B^{(j)}_{n}+h_1B^{(j)}_{n}}\right) . \end{aligned}$$

(36)

According to Ohm’s law, the equivalent resistance formula between two nodes is described as

$$\begin{aligned} R_{m \times n}(d_{1},d_{2}) =\frac{1}{J}|U_{d_1}-U_{d_2}|=\frac{1}{J}\left( \sum ^{y_2}_{j=y_1+1}I^{(j)}_{x}\right) r_0, \end{aligned}$$

(37)

due to Eqs. (36) and (37), the explicit formula (6) for the equivalent resistance between two nodes can be obtained.

Demonstrating the equivalent resistance formulas for some special cases

Formula (6) is a general conclusion for rectangular resistor networks that includes all cases. The influence of different variables on the explicit equivalent resistance formula is analyzed from two aspects as follows, and 3D views are used to demonstrate them.

Influence of current input node on equivalent resistance

This part gives examples of the change of equivalent resistance when the current input node is different.

Case 1. Assume that the current J is input at node $d_{1}(x_{1}, y_{1})(x_1=x, y_1=0)$, and the current flows out of the resistor network at node $d_{2}(x_{2}, y_{2})(x_2=x, y_2=y)$, the equivalent resistance between nodes $d_1$ and $d_2$ can be written as

$$\begin{aligned} R_{m \times n}(\{x,0\},\{x,y\}) = \frac{r_{0}}{m+1}\sum ^{m}_{j=1}\frac{[\cos \frac{\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{[\Delta B^{(j)}_{n-x}+(h_1-1)\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{B^{(j)}_{n+1}+(h_1-1)B^{(j)}_{n}}\right) , \end{aligned}$$

where $\Delta B^{(j)}_{k}$, $\theta _j$ and $B^{(j)}_{k}$ are defined by Eqs. (7), (8) and (8), respectively.

When $m = n = 80,\ J = 10,\ y_{1} = 0,\ r_{0}=r=1$ and $r_{1}=11$ , in other words, $h_1=11$, the following formula is obtained

$$\begin{aligned} R_{80 \times 80}(\{x,0\},\{x,y\}) = \frac{1}{81}\sum ^{80}_{j=1}\frac{\left[ \cos \frac{\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j \right] ^{2}}{1-\cos \theta _{j}}\times \left( \frac{(\Delta B^{(j)}_{80-x}+10\Delta B^{(j)}_{79-x})\Delta B^{(j)}_{x}}{B^{(j)}_{81}+10B^{(j)}_{80}}\right) , \end{aligned}$$

(38)

where

$$\begin{aligned} \Delta B^{(j)}_{80-x} =B^{(j)}_{81-x}-B^{(j)}_{80-x}, \end{aligned}$$

(39)

$$\begin{aligned} \Delta B^{(j)}_{79-x} =B^{(j)}_{8-x}-B^{(j)}_{79-x}, \end{aligned}$$

(40)

$$\begin{aligned} \theta _{j} =\frac{j\pi }{m+1}, \end{aligned}$$

(41)

$$\begin{aligned} B^{(j)}_{k} = B^{(j)}_{k}(\cosh \mu _{j}) = \frac{\sinh (k\mu _{j})}{\sinh \mu _{j}},\ \cosh \mu _{j}=1+\frac{r}{r_{0}}-\frac{r}{r_{0}}\cos \frac{j\pi }{m+1}, \end{aligned}$$

(42)

$$\begin{aligned} k=81-x,~80-x,~79-x,~x+1,~x,~81,~80,~~j=1,~2,~\ldots ,~80. \end{aligned}$$

A three-dimensional view of Eq. (38) is shown in Fig. 3.

Fig. 3

The 3D equivalent resistance distribution diagram of $R_{80\times 80}(\{x,0\},\{x,y\})$ in Eq. (38).

Case 2. If the current J is input at node $d_{1}(x_{1}, y_{1})(x_1=x, y_1=30)$ in the resistor network and output at node $d_{2}(x_{2}, y_{2})(x_2=x, y_2=y)$, the equivalent resistance formula between $d_1$ and $d_2$ can be characterized as

$$\begin{aligned} R_{m \times n}(\{x,30\},\{x,y\}) = \frac{r_{0}}{m+1}\sum ^{m}_{j=1}\frac{\left[ \cos \frac{61\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j \right] ^{2}}{1-\cos \theta _{j}}\times \left( \frac{[\Delta B^{(j)}_{n-x}+(h_1-1)\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{B^{(j)}_{n+1}+(h_1-1)B^{(j)}_{n}}\right) , \end{aligned}$$

(43)

where $\Delta B^{(j)}_{k}$, $\theta _j$ and $B^{(j)}_{k}$ are defined by Eqs. (7), (8) and (8), respectively.

In a resistor network of size $80\times 80$, when $J = 10,\ y_1=30,\ h_1=11$ and $r_0=r=1$, Eq. (43) is defined as

$$\begin{aligned} R_{80 \times 80}(\{x,30\},\{x,y\}) =\frac{1}{81}\sum ^{80}_{j=1}\frac{[\cos \frac{61\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{(\Delta B^{(j)}_{80-x}+10\Delta B^{(j)}_{79-x})\Delta B^{(j)}_{x}}{B^{(j)}_{81}+10B^{(j)}_{80}}\right) , \end{aligned}$$

(44)

where $\Delta B^{(j)}_{80-x}$, $\Delta B^{(j)}_{79-x}$, $\theta _j$ and $B^{(j)}_{k}$ ($k=81-x,~80-x,~79-x,~x+1,~x,~81,~80,~~j=1,~2,~\ldots ,~80$) are the same as Eqs. (39), (40), (41) and (42), respectively.

A three-dimensional view of Eq. (44) is shown in Fig. 4.

Fig. 4

The 3D equivalent resistance distribution diagram of $R_{80\times 80}(\{x,30\},\{x,y\})$ in Eq. (44).

Case 3. Assume that the current J is input into the resistor network, $d_{1}(x_{1}, y_{1})(x_1=x, y_1=80)$ is the input node of the current and $d_{2}(x_{2}, y_{2})(x_2=x, y_2=y)$ is the output node, then the equivalent resistance formula between these two nodes can be expressed as

$$\begin{aligned} R_{m \times n}(\{x,80\},\{x,y\}) = \frac{r_{0}}{m+1}\sum ^{m}_{j=1}\frac{[\cos \frac{161\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{[\Delta B^{(j)}_{n-x}+(h_1-1)\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{B^{(j)}_{n+1}+(h_1-1)B^{(j)}_{n}}\right) , \end{aligned}$$

(45)

where $\Delta B^{(j)}_{k}$, $\theta _j$ and $B^{(j)}_{k}$ are defined by Eqs. (7), (8) and (8), respectively.

When $J = 10,\ y_{1} = 80,\ r_{1}=11$ and $r_{0}=r=1$, Eq. (45) is described in the resistor network of size $80\times 80$ as

$$\begin{aligned} R_{80 \times 80}(\{x,80\},\{x,y\}) = \frac{1}{81}\sum ^{80}_{j=1}\frac{[\cos \frac{161\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{(\Delta B^{(j)}_{80-x}+10\Delta B^{(j)}_{79-x})\Delta B^{(j)}_{x}}{B^{(j)}_{81}+10B^{(j)}_{80}}\right) , \end{aligned}$$

(46)

where $\Delta B^{(j)}_{80-x}$, $\Delta B^{(j)}_{79-x}$, $\theta _j$ and $B^{(j)}_{k}$ ($k=81-x,~80-x,~79-x,~x+1,~x,~81,~80,~~j=1,~2,~\ldots ,~80$) are the same as Eqs. (39), (40), (41) and (42), respectively.

A three-dimensional view of Eq. (46) is shown in Fig. 5.

Fig. 5

The 3D equivalent resistance distribution diagram of $R_{80\times 80}(\{x,80\},\{x,y\})$ in Eq. (46).

Effect of resistivity h ($h=\frac{r}{r_0}$) on equivalent resistance

The following discusses the values of the equivalent resistance between nodes $d_1$ and $d_2$ on each vertical axis of the resistor network when the value of $y_1$ at the current input node $d_1$ remains constant and the resistivity h is different. The resistivity here is the ratio of r to $r_0$, denoted by h, i.e., $h=\frac{r}{r_0}$.

Case 4. If the node $d_{1}(x_{1}, y_{1})(x_1=x, y_1=40)$ in the resistor network is used as the input node of the current J, and $d_{2}(x_{2}, y_{2})(x_2=x, y_2=y)$ is used as the output node, then the equivalent resistance formula between the two nodes $d_1$ and $d_2$ is written as

$$\begin{aligned} R_{m \times n}(\{x,40\},\{x,y\}) = \frac{r_{0}}{m+1}\sum ^{m}_{j=1}\frac{[\cos \frac{81\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{[\Delta B^{(j)}_{n-x}+(h_1-1)\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{B^{(j)}_{n+1}+(h_1-1)B^{(j)}_{n}}\right) , \end{aligned}$$

where $\Delta B^{(j)}_{k}$, $\theta _j$ and $B^{(j)}_{k}$ are defined by Eqs. (7), (8) and (8), respectively.

When $m = n = 80,\ J = 10,\ y_{1} = 40,\ r_1=11,\ r_{0}=1$ and $r=0.1,$ the following formula is obtained

$$\begin{aligned} R_{80 \times 80}(\{x,40\},\{x,y\}) = \frac{1}{81}\sum ^{80}_{j=1}\frac{[\cos \frac{161\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{(\Delta B^{(j)}_{80-x}+10\Delta B^{(j)}_{79-x})\Delta B^{(j)}_{x}}{B^{(j)}_{81}+10B^{(j)}_{80}}\right) , \end{aligned}$$

(47)

where $\Delta B^{(j)}_{80-x}$, $\Delta B^{(j)}_{79-x}$, $\theta _j$ and $B^{(j)}_{k}$ ($k=81-x,~80-x,~79-x,~x+1,~x,~81,~80,~~j=1,~2,~\ldots ,~80$) are the same as Eqs. (39), (40), (41) and (42), respectively.

A three-dimensional view of Eq. (47) is shown in Fig. 6.

Fig. 6

The 3D equivalent resistance distribution diagram of $R_{80\times 80}(\{x,40\},\{x,y\})$ in Eq. (47).

Case 5. Assume that the current $J=10$ flows from a fixed input node $d_{1}(x_{1}, y_{1})(x_1=x, y_1=40)$ to a fixed output node $d_{2}(x_{2}, y_{2})(x_2=x, y_2=y)$ in a rectangular resistor network.

In this case, given $h_1=11,\ h=0.01$, in other words, $r_1=11,\ r_0=1,\ r=0.01$ the equivalent resistance formula between these two nodes is described as

$$\begin{aligned} R_{m \times n}(\{x,40\},\{x,y\}) = \frac{r_{0}}{m+1}\sum ^{m}_{j=1}\frac{[\cos \frac{81\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{[\Delta B^{(j)}_{n-x}+(h_1-1)\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{B^{(j)}_{n+1}+(h_1-1)B^{(j)}_{n}}\right) , \end{aligned}$$

(48)

where $\Delta B^{(j)}_{k}$, $\theta _j$ and $B^{(j)}_{k}$ are defined by Eqs. (7), (8) and (8), respectively.

When the size of the resistor network is $80\times 80$, ie $m=80$, $\ n=80$, Eq. (48) is represented as

$$\begin{aligned} R_{80 \times 80}(\{x,40\},\{x,y\}) = \frac{1}{81}\sum ^{80}_{j=1}\frac{[\cos \frac{161\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{(\Delta B^{(j)}_{80-x}+10\Delta B^{(j)}_{79-x})\Delta B^{(j)}_{x}}{B^{(j)}_{81}+10B^{(j)}_{80}}\right) , \end{aligned}$$

(49)

where $\Delta B^{(j)}_{80-x}$, $\Delta B^{(j)}_{79-x}$, $\theta _j$ and $B^{(j)}_{k}$ ($k=81-x,~80-x,~79-x,~x+1,~x,~81,~80,~~j=1,~2,~\ldots ,~80$) are the same as Eqs. (39), (40), (41) and (42), respectively.

A three-dimensional view of Eq. (49) is shown in Fig. 7.

Fig. 7

The 3D equivalent resistance distribution diagram of $R_{80\times 80}(\{x,40\},\{x,y\})$ in Eq. (49).

Case 6. Suppose the current $J=10$ flows into resistor network through $d_{1}(x_{1}, y_{1})(x_1=x, y_1=40)$ and out at $d_{2}(x_{2}, y_{2})(x_2=x, y_2=y)$, at this time, the equivalent resistance formula between $d_1$ and $d_2$ can be expressed as

$$\begin{aligned} R_{m \times n}(\{x,40\},\{x,y\}) = \frac{r_{0}}{m+1}\sum ^{m}_{j=1}\frac{[\cos \frac{81\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{[\Delta B^{(j)}_{n-x}+(h_1-1)\Delta B^{(j)}_{n-x-1}]\Delta B^{(j)}_{x}}{B^{(j)}_{n+1}+(h_1-1)B^{(j)}_{n}}\right) , \end{aligned}$$

where $\Delta B^{(j)}_{k}$, $\theta _j$ and $B^{(j)}_{k}$ are defined by Eqs. (7), (8) and (8), respectively.

Let $m = n = 80,\ h_1=11,\ h=0.001$, that is , $r_1=11,\ r_0=1,\ r=0.001$ the following formula is obtained

$$\begin{aligned} R_{80 \times 80}(\{x,40\},\{x,y\}) = \frac{1}{81}\sum ^{80}_{j=1}\frac{[\cos \frac{161\theta _j}{2}-\cos (y+\frac{1}{2})\theta _j]^{2}}{1-\cos \theta _{j}}\times \left( \frac{(\Delta B^{(j)}_{80-x}+10\Delta B^{(j)}_{79-x})\Delta B^{(j)}_{x}}{B^{(j)}_{81}+10B^{(j)}_{80}}\right) , \end{aligned}$$

(50)

where $\Delta B^{(j)}_{80-x}$, $\Delta B^{(j)}_{79-x}$, $\theta _j$ and $B^{(j)}_{k}$ ($k=81-x,~80-x,~79-x,~x+1,~x,~81,~80,~~j=1,~2,~\ldots ,~80$) are the same as Eqs. (39), (40), (41) and (42), respectively.

A three-dimensional view of Eq. (50) is shown in Fig. 8.

Fig. 8

The 3D equivalent resistance distribution diagram of $R_{80\times 80}(\{x,40\},\{x,y\})$ in Eq. (50).

Calculation efficiency of different equivalent resistance formulas

In this section, examples are shown that demonstrate the computational efficiency of two equivalent resistor formulas. In the $m\times n$ rectangular resistor network, $d_{1}(x,y_{1})$ and $d_{2}(x,y_{2})$ represent the current input and output nodes, respectively. In the experiment, the $y_1$ value of the input node is fixed, and each $y_2$ node on each vertical axis is traversed. The CPU processing times $t_1$ and $t_2$ represent the time required to calculate the equivalent resistance using formula (1) and formula (6), respectively, and demonstrate the calculation efficiency of the two different formulas.

These experiments are done on an Intel Core i7-12700H laptop with 2.30 GHz CPU and NVIDIA GeForce RTX 3060 GPU. In the following tables, the calculation time is in seconds, “$m\times n$” denotes the scale of resistor network, “*” and the dashed empty bar indicate computer memory overflow.

When $y_1=30$, $r/r_0=1$, the CPU time spent calculating the equivalent resistance by formula (1) and formula (6), respectively is shown in Fig. 9.

Fig. 9

CPU time to calculate equivalent resistance using formula (1) and formula (6), respectively.

When $y_1=30$, $r/r_0=0.1$, the CPU time spent calculating the equivalent resistance by formula (1) and formula (6), respectively is shown in Fig. 10.

Fig. 10

CPU time to calculate equivalent resistance using formula (1) and formula (6), respectively.

When $y_1=30$, $r/r_0=0.01$, the CPU time spent calculating the equivalent resistance by formula (1) and formula (6), respectively is shown in Fig. 11.

Fig. 11

CPU time to calculate equivalent resistance using formula (1) and formula (6), respectively.

When $y_1=50$, $r/r_0=1$, the CPU time spent calculating the equivalent resistance by formula (1) and formula (6), respectively is shown in Fig. 12.

Fig. 12

CPU time to calculate equivalent resistance using formula (1) and formula (6), respectively.

When $y_1=50$, $r/r_0=0.1$, the CPU time spent calculating the equivalent resistance by formula (1) and formula (6), respectively is shown in Fig. 13.

Fig. 13

CPU time to calculate equivalent resistance using formula (1) and formula (6), respectively.

When $y_1=50$, $r/r_0=0.01$, the CPU time spent calculating the equivalent resistance by formula (1) and formula (6), respectively is shown in Fig. 14.

Fig. 14

CPU time to calculate equivalent resistance using formula (1) and formula (6), respectively.

It can be clearly seen from the above six visualization charts that the computational efficiency of the improved formula (6) is higher than that of formula (1), and as the scale of the resistor network increases, the advantages of formula (6) become more obvious. As the resistivity decreases, the size of the data that can be processed using the equivalent resistance formula increases.

Conclusion

This paper uses Chebyshev polynomial of the second kind to improve the equivalent resistance formula of the $m \times n$ rectangular resistor network. Some special and interesting equcations of the resistor network, such as Eqs. (38), (44), (46), (47), (49) and (50) were introduced. To provide a visual representation, their three-dimensional views were plotted using MATLAB. Finally, several comparison tables were provided to show the calculation efficiency of two equivalent resistance formulas. The design philosophy and formulas presented in this study will inspire further research in fields such as neural networks and other related areas.