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# Time-domain Formulation of a Multi-layer Plane Circuit Coupled with Lumped-parameter Circuits using Maxwell Equations

## Abstract

We calculate electromagnetic phenomena in the multi-layer plane circuit starting from the Maxwell equations. We present a numerical method of potential and current density in two-dimensional conductors, where their time developments are treated as phenomena of wave propagation. We treat the plane conductors by dividing them into small finite-volume elements, similar to the case of the partial element equivalent circuit method, and the transport equations are then solved by the finite-difference time-domain method. Furthermore, we develop a calculation method for the boundary in a multi-layer plane by applying the method we have used in multi-transmission lines. We formulate the boundary conditions of a multi-layer plane coupled with lumped-parameter circuits and introduce an algorithm to reduce calculation costs that are largely associated with the two-dimensional extension from the multi-transmission-line case. We perform calculations of the wave propagation of potential, current density, and charge density in the time domain for a simple plane circuit. These calculations are presented as supplementary materials of the present paper.

## Introduction

Electromagnetic noise is troublesome because it is difficult to understand the fundamental phenomena associated with noise. In recent years, circuit design that reduces the influence of electromagnetic noise has become increasingly important for reliable performance with the trend toward high-frequency and low-voltage electronics1. This trend is likely to result in electromagnetic noise, causing the malfunction of electric devices, and can be a major bottleneck in future circuit design2. There are many symptomatic treatments, such as filters, to suppress the electromagnetic noise3,4. However, these treatments require the experience of skilled engineers because the causes of electromagnetic noise phenomena are not clearly understood and cannot be described quantitatively using equivalent circuit modeling and impedance analysis in the frequency domain5,6. A fundamental understanding of the mechanisms of electromagnetic noise is needed for noiseless circuit design.

In previous studies, we elucidated the mechanism of electromagnetic noise phenomena in line circuit systems using the multi-conductor transmission line (MTL) theory derived from the Maxwell equations7. We calculated common-mode noise for a configuration of three-transmission-line theory and developed a noiseless circuit design8,9. We have also proposed a numerical calculation algorithm for hybrid systems, where the distributed- and lumped-parameter circuits are connected10,11.

For further analysis of electromagnetic noise in an actual circuit, such as a printed circuit board (PCB), we have extended the MTL theory to two- and three-dimensional conductors and formulated transport equations of potential and current. In the numerical calculation, we realized calculation of a multi-layer plane (MLP) circuit with a discontinuous ground using the finite-difference time-domain (FDTD) method12. Although we applied our algorithm to an MLP circuit to connect lumped-parameter circuits at boundaries, there are still major problems to be solved. First, we have to consider the directions of current according to the locations of the boundaries in the MLP. Second, the number of connection points of the lumped-parameter circuit greatly increases as compared to the case of the MTL theory. For these reasons, applying the previous calculation algorithm to arbitrary shapes of MLPs is difficult due to complicated boundary conditions and the associated heavy computer burden.

In the present paper, we would like to extend the MTL theory to the case of two- and three-dimensional conductors. To this end, we derive the transport equations in the form of finite-volume elements of potential and current density from the Maxwell equations. Using these equations, we are able to visualize the transitional phenomena as a function of time, which is important in order to understand the electromagnetic noise phenomena. In addition, we would like to resolve problems of complex calculations at boundaries of MLPs. For this purpose, we formulate the boundary equations to consider any direction of current at the boundaries. We use Kirchhoff’s current law (KCL) to consider the connective relation between lumped- and distributed-parameter circuits. We also use Kirchhoff’s voltage law (KVL) and branch constitutive equation (BCE) to consider the lumped-parameter circuit conditions.

In the present paper, a calculation algorithm for a two-dimensional conductor is presented in order to clarify the mechanism of electromagnetic noise occurring in the MLP circuit. We assume that two- or three-dimensional MLP circuit systems are connected with lumped-parameter circuits at the boundaries (an example is shown in Fig. 1). The proposed method provides information of transient phenomena of the potential and current density in the planar circuit system. In Section 2, two- and three-dimensional telegraphic equations expressed in terms of potential and current density are derived. Then, coupled differential equations are expressed in the form of discretized finite-volume elements for numerical calculation. In Section 3, we introduce the numerical method at the boundary. In Section 4, we describe how to reduce boundary calculation costs by considering the fact that most of the boundary of the MLP is open, by which the calculation cost of the boundary conditions can be significantly reduced.

## Derivation of Coupled Difference Equations of Potential and Current Density in MLP Circuits

In previous studies, we derived equations of potential and current in a multi-transmission-line (MTL) system from the Maxwell equations7. We have reported discretized equations in an MLP circuit system and compared the results of simulations and experiments12. In the present paper, we discuss the derivation of discretized equations in detail in order to further understand the calculations. We start with the retarded scalar $$U({\bf{r}},t)$$ and vector A$$({\bf{r}},t)$$ potentials obtained by applying the Lorenz gauge from the Maxwell equations.

$$U({\bf{r}},t)=\frac{1}{4\pi \varepsilon }\,\int \,{\rm{d}}{\bf{r}}{\rm{^{\prime} }}\frac{q({\bf{r}}{\rm{^{\prime} }},t-\frac{|{\bf{r}}-{\bf{r}}{\rm{^{\prime} }}|}{c})}{|{\bf{r}}-{\bf{r}}{\rm{^{\prime} }}|},$$
(1)
$${\boldsymbol{A}}({\bf{r}},t)=\frac{\mu }{4\pi }\,\int \,{\rm{d}}{\bf{r}}{\rm{^{\prime} }}\frac{{\boldsymbol{j}}({\bf{r}}{\rm{^{\prime} }},t-\frac{|{\bf{r}}-{\bf{r}}{\rm{^{\prime} }}|}{c})}{|{\bf{r}}-{\bf{r}}{\rm{^{\prime} }}|},$$
(2)

where $$q({\bf{r}},t)$$ and j$$({\bf{r}},t)$$ are the charge and current densities in the conductors. We assume that, in addition to $$U({\bf{r}},t)$$ and A$$({\bf{r}},t)$$, both Ohm’s law and charge conservation hold in the planar conductors.

$$-\nabla U({\bf{r}},t)-\frac{\partial \,{\boldsymbol{A}}({\bf{r}},t)}{\partial t}=\rho {\boldsymbol{j}}({\bf{r}},t),$$
(3)
$$\frac{\partial }{\partial t}q({\bf{r}},t)-\nabla \cdot {\boldsymbol{j}}({\bf{r}},t)=0.$$
(4)

Here, $$\rho$$ denotes the resistance in the plane conductors.

These four equations satisfy the necessary and sufficient conditions for solving the scalar potential, vector potential, charge density, and current density in conductors. Here, we realize a numerical instability in the presence of the delay time in Eqs.. (1) and (2). This stability problem is a large challenge and a long-standing problem for circuit theories such as the partial element equivalent circuit (PEEC) method13. At present, we avoid this numerical difficulty by neglecting the delay time in the coupled differential equations14. For numerical calculation, we use finite-volume elements by taking an average in the small finite volume. In the present paper, we assume a multi-layer planar conductor circuit, as shown in Fig. 2, and we use a cubic finite-volume element to divide the conductor planes into finite-volume elements. We refer to one of the divided elements as cell $$i(k,l)$$, where i is a conductor plane, k is the x coordinate, l is the y coordinate, and the mesh dimensions are Δx and Δy. Here, we consider the distribution in two directions (x and y) because the thickness of the plane conductor, denoted Δz, is sufficiently thin compared with Δx and Δy. The volume of cell $$i(k,l)$$ is represented by $${V}_{i(k,l)}=\Delta x\Delta y\Delta z$$. We can express all variables using quantities of cell elements as:

$${U}_{i(k,l)}(t)=\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,U({\bf{r}},t){\rm{d}}{\bf{r}},\,{{\boldsymbol{A}}}_{i(k,l)}(t)=\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,{\boldsymbol{A}}({\bf{r}},t){\rm{d}}{\bf{r}},$$
(5)
$${q}_{i(k,l)}(t)=\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,q({\bf{r}},t){\rm{d}}{\bf{r}},\,{{\boldsymbol{j}}}_{i(k,l)}(t)=\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,{\boldsymbol{j}}({\bf{r}},t){\rm{d}}{\bf{r}}.$$
(6)

We apply the averaging procedure to cell $$i(k,l)$$ for Eqs. (1) and (2):

$${U}_{i(k,l)}(t)=\sum _{i^{\prime} ,k^{\prime} ,l^{\prime} }\,{P}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}{q}_{i^{\prime} (k^{\prime} ,l^{\prime} )}(t).$$
(7)
$${{\boldsymbol{A}}}_{i(k,l)}(t)=\sum _{i^{\prime} ,k^{\prime} ,l^{\prime} }\,{L}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}{{\boldsymbol{j}}}_{i^{\prime} (k^{\prime} ,l^{\prime} )}(t).$$
(8)

Equations (1) and (2) can be expressed as the sum of cell elements. Here, P and L represent the local potential coefficient and the local inductance coefficient of the conductors:

$${P}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}=\frac{1}{4\pi \varepsilon }\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,{\int }_{{V}_{i^{\prime} (k^{\prime} ,l^{\prime} )}}\,\frac{1}{|{\bf{r}}-{\bf{r}}^{\prime} |}{\rm{d}}{\bf{r}}{\rm{d}}{\bf{r}}^{\prime} ,$$
(9)
$${L}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}=\frac{\mu }{4\pi }\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,{\int }_{{V}_{i^{\prime} (k^{\prime} ,l^{\prime} )}}\,\frac{1}{|{\bf{r}}-{\bf{r}}^{\prime} |}{\rm{d}}{\bf{r}}{\rm{d}}{\bf{r}}^{\prime} .$$
(10)

The local coefficients P and L have two sets of subscripts, which represent the relation between cells $$i(k,l)$$ and $$i^{\prime} (k^{\prime} ,l^{\prime} )$$. Next, we derive spatial difference equations by taking the average of Eqs. (3) and (4) in cell $$i(k,l)$$. Here, Eq. (3) includes the spatial differentiation, which is the inclination in the x, y, and z directions in cell $$i(k,l)$$. For instance, spatial differentiation of any function $$f(x,y,z)$$ in the x direction can be converted into the cell spatial difference using the following relation:

$$\begin{array}{c}\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,\frac{{\rm{\partial }}}{{\rm{\partial }}x}f(x,y,z){\rm{d}}{\bf{r}}\\ \,=\,\frac{1}{\Delta x\Delta y\Delta z}{\int }_{(k-1/2)\Delta x}^{(k+1/2)\Delta x}\,dx\,{\int }_{(l-1/2)\Delta y}^{(l+1/2)\Delta y}\,dy\Delta z\frac{[f(x+\frac{\Delta x}{2},y)-f(x-\frac{\Delta x}{2},y)]}{\Delta x}\\ \,=\,\frac{{f}_{i(k+\frac{1}{2},l)}-{f}_{i(k-\frac{1}{2},l)}}{\Delta x}.\end{array}$$
(11)

We express spatial differentiation with respect to x as a central difference shifted from x by half a cell, and take the average in the x direction. The average spatial differentiation in cell $$i(k,l)$$ is converted into the cell spatial difference between cells, which is positively or negatively shifted by half a cell in the x direction from cell $$i(k,l)$$. Furthermore, the divergence of the space is included in Eq. (4) and can be expressed using the inner product:

$$\begin{array}{c}\frac{1}{{V}_{i(k,l)}}\,{\int }_{{V}_{i(k,l)}}\,{\rm{\nabla }}\cdot {\bf{f}}(x,y,z)\,{\rm{d}}{\bf{r}}\\ \,=\,\frac{1}{{V}_{i(k,l)}}\,\,{\int }_{{V}_{i(k,l)}}\,[\frac{{\rm{\partial }}}{{\rm{\partial }}x}{f}_{x}(x,y,z)+\frac{{\rm{\partial }}}{{\rm{\partial }}y}{f}_{y}(x,y,z)+\frac{{\rm{\partial }}}{{\rm{\partial }}z}{f}_{z}(x,y,z)]{\rm{d}}{\bf{r}}\\ \,=\,\frac{{f}_{xi(k+\frac{1}{2},l)}-{f}_{xi(k-\frac{1}{2},l)}}{\Delta x}+\frac{{f}_{yi(k,l+\frac{1}{2})}-{f}_{yi(k,l-\frac{1}{2})}}{\Delta y}.\end{array}$$
(12)

Here, we do not consider the thin z direction of a rectangular cell for simplicity. The average divergence in cell $$i(k,l)$$ is converted into the sum of the spatial difference in each direction. Finally, Ohm’s law (3) and the continuous Eq. (4) can be given in terms of the cell spatial differences:

$$\frac{{U}_{i(k+1,l)}(t)-{U}_{i(k,l)}(t)}{\Delta x}=-\,\sum _{i^{\prime} ,k^{\prime} ,l^{\prime} }\,{L}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}\frac{\partial {j}_{xi^{\prime} (k^{\prime} +\frac{1}{2},l^{\prime} )}(t)}{\partial t}-{\rho }_{i}{j}_{xi(k+\frac{1}{2},l)}(t),$$
(13)
$$\frac{{U}_{i(k,l+1)}(t)-{U}_{i(k,l)}(t)}{\Delta y}=-\,\sum _{i^{\prime} ,k^{\prime} ,l^{\prime} }\,{L}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}\frac{\partial {j}_{yi^{\prime} (k^{\prime} ,l^{\prime} +\frac{1}{2})}(t)}{\partial t}-{\rho }_{i}{j}_{yi(k,l+\frac{1}{2})}(t),$$
(14)
$$\frac{\partial }{\partial t}{q}_{i(k,l)}(t)-\frac{{j}_{xi(k+\frac{1}{2},l)}-{j}_{xi(k-\frac{1}{2},l)}}{\Delta x}-\frac{{j}_{yi(k,l+\frac{1}{2})}-{j}_{yi(k,l-\frac{1}{2})}}{\Delta y}=0.$$
(15)

Here, Eqs. (13) and (14) express the vector components of Eq. (3). The difference of the half integer of space between the potential and the current densities comes from Eqs. (11) and (12). We used the scalar potential and the current density as variables in circuit theory and eliminated the vector potential and charge density cell elements from Eqs. (7), (8), (13), (14) and (15). Finally, the transport equations are expressed in the form of recurrence formulas by discretizing the time direction:

$$\begin{array}{rcl}\frac{{U}_{i(k,l)}^{m+1}-{U}_{i(k,l)}^{m}}{\Delta t} & = & -\,\mathop{\sum }\limits_{i^{\prime} ,k^{\prime} ,l^{\prime} }^{N,{N}_{x},{N}_{y}}\,{P}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}\frac{{j}_{xi^{\prime} (k^{\prime} +\frac{1}{2},l^{\prime} )}^{m+\frac{1}{2}}-{j}_{xi^{\prime} (k^{\prime} -\frac{1}{2},l^{\prime} )}^{m+\frac{1}{2}}}{\Delta x}\\ & & -\,\mathop{\sum }\limits_{i^{\prime} ,k^{\prime} ,l^{\prime} }^{N,{N}_{x},{N}_{y}}\,{P}_{i(k,l)i^{\prime} (k^{\prime} ,l^{\prime} )}\frac{{j}_{yi^{\prime} (k^{\prime} ,l^{\prime} +\frac{1}{2})}^{m+\frac{1}{2}}-{j}_{yi^{\prime} (k^{\prime} ,l^{\prime} -\frac{1}{2})}^{m+\frac{1}{2}}}{\Delta y},\end{array}$$
(16)
$$\begin{array}{rcl}\frac{{U}_{i(k+1,l)}^{m+1}-{U}_{i(k,l)}^{m+1}}{\Delta x} & = & -\,\mathop{\sum }\limits_{i^{\prime} ,k^{\prime} ,l^{\prime} }^{N,{N}_{x},{N}_{y}}\,{L}_{i(k+\frac{1}{2},l)i^{\prime} (k^{\prime} +\frac{1}{2},l^{\prime} )}\frac{{j}_{xi^{\prime} (k^{\prime} +\frac{1}{2},l^{\prime} )}^{m+\frac{3}{2}}-{j}_{xi^{\prime} (k^{\prime} +\frac{1}{2},l^{\prime} )}^{m+\frac{1}{2}}}{\Delta t}\\ & & -\,{\rho }_{i(k+\frac{1}{2},l)}\frac{{j}_{xi(k+\frac{1}{2},l)}^{m+\frac{3}{2}}+{j}_{xi(k+\frac{1}{2},l)}^{m+\frac{1}{2}}}{2},\end{array}$$
(17)
$$\begin{array}{rcl}\frac{{U}_{i(k,l+1)}^{m+1}-{U}_{i(k,l)}^{m+1}}{\Delta y} & = & -\,\mathop{\sum }\limits_{i^{\prime} ,k^{\prime} ,l^{\prime} }^{N,{N}_{x},{N}_{y}}\,{L}_{i(k,l+\frac{1}{2})i^{\prime} (k^{\prime} ,l^{\prime} +\frac{1}{2})}\frac{{j}_{yi^{\prime} (k^{\prime} ,l^{\prime} +\frac{1}{2})}^{m+\frac{3}{2}}-{j}_{yi^{\prime} (k^{\prime} ,l^{\prime} +\frac{1}{2})}^{m+\frac{1}{2}}}{\Delta t}\\ & & -\,{\rho }_{i(k,l+\frac{1}{2})}\frac{{j}_{yi(k,l+\frac{1}{2})}^{m+\frac{3}{2}}+{j}_{yi(k,l+\frac{1}{2})}^{m+\frac{1}{2}}}{2}.\end{array}$$
(18)

These equations are difference equations for the potential and current density in space and time. We can obtain the potential in the future time from Eq. (16), and the values of current density in the x and y directions are obtained from Eqs. (17) and (18), respectively. Moreover, if the current densities can be calculated, we can also quantify the vector potential and charge density using Eqs. (8) and (15), respectively. The suffix m at the upper right represents the discretized time. The difference of the half integer of time between the potential and current densities comes from the FDTD method. Here, N represents the number of plane, and Nx and Ny represent the numbers of divisions in the x and y directions, respectively. The calculation of boundaries expressed by gray cells in Fig. 3 is explained in the next section.

## Method of Boundary Calculation of MLP with a Lumped-Parameter Circuit

In this section, we explain the method of connecting the two-dimensional distributed-parameter circuit and the lumped-parameter circuit at any boundary position. We use the node potential and branch current in the lumped-parameter circuit and the cell potential and cell current density in the distributed-parameter circuit as unknown variables. In order to treat these circuits, we use KVL and the BCE in the lumped-parameter circuit and Eqs. (16), (17) and (18), which are cell constitutive equations (CCE), in the distributed-parameter circuit. Assuming that the node potential is equivalent to the cell potential and branch current is equivalent to the cell current, we can express current conservation by KCL, which is the connection relationship between the lumped- and distributed-parameter circuits, as shown in Fig. 4.

We formulated the time-domain incident-potential equation (IPE), which is a coupled equation of a lumped-parameter circuit and an MTL, in a previous study11. As for the lumped-parameter circuit, we can use time-domain IPE by using the incidence matrix of directed graph A and time-domain impedance matrix Z:

$$(\begin{array}{cc}{{\bf{A}}}_{l}^{T} & -{{\bf{Z}}}_{l}\\ {\bf{0}} & {{\bf{A}}}_{l}\end{array})(\begin{array}{c}{{\bf{U}}}_{l}^{m+1}\\ {{\bf{I}}}_{l}^{m+1}\end{array})=(\begin{array}{cc}-\epsilon {{\bf{A}}}_{l}^{T} & \delta {{\bf{Z}}}_{l}\\ {\bf{0}} & {{\bf{A}}}_{l}\end{array})(\begin{array}{c}{{\bf{U}}}_{l}^{m}\\ {{\bf{I}}}_{l}^{m}\end{array})+(\begin{array}{c}{{\bf{V}}}_{s}^{m+1}+{{\bf{V}}}_{s}^{m}\\ -{{\bf{A}}}_{s}{{\bf{I}}}_{s}^{m+1}\end{array})$$
(19)

Here, the upper right subscript m of U and I refers to discretized time. Unknown variables at a future time are moved to the left-hand side to solve the IPE. Here, Al is an incidence matrix of size $$p\times q$$, where p is associated with the number of nodes, and q is the number of branches. In addition, Al can express the relationship of connections between nodes and branches, and its entries are 0, 1, or −1. If the component of the α-th row and the β-th column is 1, then the α-th node is connected to the β-th branch, and the current direction is outward from the node. If the component of the α-th row and the β-th column is −1, then the current direction is inward to the node. If the component of the α-th row and the β-th column is 0, then the nodes and branches are not connected. In addition, Zl is a time-domain impedance matrix of size $$p\times p$$, which has diagonal elements depending on β-th lumped-parameter circuit elements. From a previous study10,11, the time-domain impedance elements are derived as follows. The time-domain impedance elements for the resistor, capacitor, inductor, and independent voltage source are R, $$\Delta t/(2C)$$, $$2L/(\Delta t)$$, and 0, respectively. Vector Vs is for an independent voltage source, and vector Is is for an independent current source. Then, $$\epsilon$$ and δ are $$p\times p$$ diagonal matrices, which are used to change signs depending on the β-th lumped-parameter circuit elements:

$${\epsilon }_{\beta \beta }=\{\begin{array}{cc}-1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{c}}{\rm{a}}{\rm{p}}{\rm{a}}{\rm{c}}{\rm{i}}{\rm{t}}{\rm{a}}{\rm{n}}{\rm{c}}{\rm{e}}\\ 1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{o}}{\rm{t}}{\rm{h}}{\rm{e}}{\rm{r}}\,{\rm{e}}{\rm{l}}{\rm{e}}{\rm{m}}{\rm{e}}{\rm{n}}{\rm{t}}{\rm{s}}\end{array},\,{\delta }_{\beta \beta }=\{\begin{array}{cc}-1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{i}}{\rm{n}}{\rm{d}}{\rm{u}}{\rm{c}}{\rm{t}}{\rm{a}}{\rm{n}}{\rm{c}}{\rm{e}}\\ 1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{o}}{\rm{t}}{\rm{h}}{\rm{e}}{\rm{r}}\,{\rm{e}}{\rm{l}}{\rm{e}}{\rm{m}}{\rm{e}}{\rm{n}}{\rm{t}}{\rm{s}}\end{array}.$$
(20)

Next, we derive the time-domain IPE in the same form as Eq. (19) for the distributed-parameter circuit. We consider boundary equations of lumped-parameter circuits and the MLP circuit at boundaries A through D shown in Fig. 5. Since the size of the current density cell is halved at these boundaries and results in a complicated numerical calculation of the impedance matrix, we replace the boundary current density with the time average of the current density flowing in a potential cell $$i(k,l)$$ at the boundary, which is expressed as follows:

$$\begin{array}{rcl}{j}_{xi(k\pm \frac{1}{2},l)}^{m+\frac{1}{2}} & = & \frac{{j}_{xi(k,l)}^{m+1}+{j}_{xi(k,l)}^{m}}{2},\\ {j}_{yi(k,l\pm \frac{1}{2})}^{m+\frac{1}{2}} & = & \frac{{j}_{yi(k,l)}^{m+1}+{j}_{yi(k,l)}^{m}}{2}.\end{array}$$
(21)

This replacement of current densities (21) at the boundaries applies to $$k-1/2$$ with $$k=1$$ (Boundary A) and $$k+\mathrm{1/2}$$ with $$k={N}_{x}$$ (Boundary B) for jx and $$l-1/2$$ with $$l=1$$ (Boundary D) and $$l+1/2$$ with $$l={N}_{y}$$ (Boundary C) for jy.

We use the recurrence formula of the potential in Eq. (16) as a CCE for the boundary equation. Unknown variables in the distributed-parameter circuit are the boundary potential and the current density flowing in the potential cell and are extracted from the summation in Eq. (16), which is brought to the left-hand side:

$$\begin{array}{c}{U}_{i(k,l)}^{m+1}-\mathop{\sum }\limits_{{i}^{{\rm{^{\prime} }}},k{\rm{^{\prime} }},l{\rm{^{\prime} }}}^{{\rm{B}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}}\,\mathop{\sum }\limits_{\alpha }^{x,y}\,{\gamma }_{\alpha i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}\frac{\Delta t}{\Delta \alpha }{P}_{i(k,l)i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}{j}_{\alpha i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}^{m+1}\\ \begin{array}{ccc} & = & {U}_{i(k,l)}^{m}+\mathop{\sum }\limits_{i{\rm{^{\prime} }},k{\rm{^{\prime} }},l{\rm{^{\prime} }}}^{{\rm{B}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}}\,\mathop{\sum }\limits_{\alpha }^{x,y}\,{\gamma }_{\alpha i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}\frac{\Delta t}{\Delta \alpha }{P}_{i(k,l)i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}{j}_{\alpha i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}^{m}\\ & & -\,\mathop{\sum }\limits_{i{\rm{^{\prime} }},k{\rm{^{\prime} }},l{\rm{^{\prime} }}}^{{\rm{B}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}}\,\mathop{\sum }\limits_{\alpha }^{x,y}\,{\gamma }_{\alpha i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}{\Gamma }_{\alpha i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}\frac{\Delta t}{\Delta \alpha }{P}_{i(k,l)i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}{j}_{\alpha i{\rm{^{\prime} }}(k{\rm{^{\prime} }}+{\beta }_{x},l{\rm{^{\prime} }}+{\beta }_{y})}^{m+\frac{1}{2}}\\ & & -\,\mathop{\sum }\limits_{i{\rm{^{\prime} }},k{\rm{^{\prime} }},l{\rm{^{\prime} }}}^{{\rm{N}}{\rm{o}}\,{\rm{b}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}}\,\frac{\Delta t}{\Delta x}{P}_{i(k,l)i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}({j}_{xi{\rm{^{\prime} }}(k{\rm{^{\prime} }}+\frac{1}{2},l{\rm{^{\prime} }})}^{m+\frac{1}{2}}-{j}_{xi{\rm{^{\prime} }}(k{\rm{^{\prime} }}-\frac{1}{2},l{\rm{^{\prime} }})}^{m+\frac{1}{2}})\\ & & -\,\mathop{\sum }\limits_{i{\rm{^{\prime} }},k{\rm{^{\prime} }},l{\rm{^{\prime} }}}^{{\rm{N}}{\rm{o}}\,{\rm{b}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}}\,\frac{\Delta t}{\Delta y}{P}_{i(k,l)i{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }})}({j}_{yi{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }}+\frac{1}{2})}^{m+\frac{1}{2}}-{j}_{yi{\rm{^{\prime} }}(k{\rm{^{\prime} }},l{\rm{^{\prime} }}-\frac{1}{2})}^{m+\frac{1}{2}}).\end{array}\end{array}$$
(22)

We introduced factors $${\gamma }_{xi(k,l)}$$ and $${\gamma }_{yi(k,l)}$$ to represent the position and the direction (x or y) of the current density at boundary cell $$i(k,l)$$. These factors are shown in Table 1, and the position of cell $$i(k,l)$$ corresponds to boundaries A through D, as shown in Fig. 5. For instance, the case of $${\gamma }_{xi(k,l)}=1$$ and $${\gamma }_{yi(k,l)}=0$$ means that the input current flows in the x direction at boundary A. The factors $${\Gamma }_{xi(k,l)}$$ and $${\Gamma }_{yi(k,l)}$$ are introduced in Table 1 to consider the shift in the current density by half in Eq. (21). This shift is taken care of by the β factor, where $${\beta }_{x}=\frac{1}{2}$$ and $${\beta }_{y}=0$$ at boundary A, $${\beta }_{x}=-\,\frac{1}{2}$$ and $${\beta }_{y}=0$$ at boundary B, $${\beta }_{y}=\frac{1}{2}$$ and $${\beta }_{x}=0$$ at boundary D, and $${\beta }_{y}=-\,\frac{1}{2}$$ and $${\beta }_{x}=0$$ at boundary C.

We can express the CCE as a matrix equation, as used in Eq. (19):

$${{\bf{A}}}_{d}^{T}{{\bf{U}}}_{d}^{m+1}-{{\bf{Z}}}_{d}{{\bf{j}}}_{d}^{m+1}={{\bf{A}}}_{d}^{T}{{\bf{U}}}_{d}^{m}+{{\bf{Z}}}_{d}{{\bf{j}}}_{d}^{m}-{\tilde{{\bf{U}}}}_{d}^{m+1}.$$
(23)

Here, matrix Ad is a unit matrix because each cell is not connected to other cells at boundaries. Vector Ud and jd are the cell potential and the cell current density, respectively. Moreover, Zd is the time-domain cell impedance matrix expressed as $${{\bf{Z}}}_{d}={\sum }_{\alpha =x,y}\,{\gamma }_{\alpha }\frac{\Delta t}{\Delta \alpha }{\bf{P}}$$. We write $$\tilde{{\bf{U}}}$$ for all terms in the second, third, and fourth summation terms in the right-hand side of Eq. (22) and treat this term as a source term in the CCE.

Next, we can use KCL to derive the connection relation between distributed- and lumped-parameter circuits using the incidence matrix:

$$({\gamma }_{x}{{\bf{S}}}_{x}+{\gamma }_{y}{{\bf{S}}}_{y}){{\bf{A}}}_{d}{{\bf{j}}}_{d}^{m+1}+{{\bf{A}}}_{l}{{\bf{I}}}_{l}^{m+1}=-\,{{\bf{A}}}_{s}{{\bf{I}}}_{s}^{m+1}.$$
(24)

Here, matrices Sx and Sy are diagonal matrices of size $$p\times p$$, where p is the number of boundaries, and the elements are determined by the cross section of the β-th boundary cell ($${S}_{x\beta \beta }=\Delta z\Delta y$$ and $${S}_{y\beta \beta }=\Delta z\Delta x$$).

Finally, we can express the time domain IPE for the coupled lumped- and distributed-parameter circuit by combining Eqs. (19), (23) and (24). We express the variables of the boundary, which are Ul, Ud, Il, and jd, as follows:

$${\bf{U}}=(\begin{array}{c}{{\bf{U}}}_{d}\\ {{\bf{U}}}_{l}\end{array}),\,{\bf{I}}=(\begin{array}{c}{{\bf{I}}}_{l}\\ {{\bf{j}}}_{d}\end{array}).$$
(25)

The incidence matrices of lumped- and distributed-parameter circuits are joined horizontally:

$${\bf{A}}=({{\bf{A}}}_{l}\,{{\bf{A}}}_{d})$$
(26)

The time-domain impedance matrices in both lumped- and distributed-parameter circuits are arranged diagonally:

$${\bf{Z}}=(\begin{array}{cc}{{\bf{Z}}}_{l} & {\bf{0}}\\ {\bf{0}} & {{\bf{Z}}}_{d}\end{array})$$
(27)

The voltage sources are joined vertically:

$${{\bf{E}}}^{m+1}=(\begin{array}{c}{{\bf{V}}}_{s}^{m+1}+{{\bf{V}}}_{s}^{m}\\ {\mathop{{\bf{U}}}\limits^{ \sim }}_{d}^{m+1}\end{array})$$
(28)

Finally, we can summarize all of the conditions at boundaries in the same form as Eq. (19):

$$(\begin{array}{cc}{{\bf{A}}}^{T} & -{\bf{Z}}\\ {\bf{0}} & {\gamma }_{S}{\bf{A}}\end{array})(\begin{array}{c}{{\bf{U}}}^{m+1}\\ {{\bf{I}}}^{m+1}\end{array})=(\begin{array}{cc}-\epsilon {{\bf{A}}}^{t} & \delta {\bf{Z}}\\ {\bf{0}} & {\bf{0}}\end{array})(\begin{array}{c}{{\bf{U}}}^{m}\\ {{\bf{I}}}^{m}\end{array})+(\begin{array}{c}{{\bf{E}}}^{m+1}\\ -{{\bf{A}}}_{s}{{\bf{I}}}_{s}^{m+1}\end{array})$$
(29)

Here, the matrices $$\epsilon$$ and δ are diagonal matrices with the signs chosen according to the β-th element of the lumped-parameter circuit and the distributed-parameter circuit, as in Eqs. (30) and (31).

$${\epsilon }_{\beta \beta }=\{\begin{array}{cc}-1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{c}}{\rm{a}}{\rm{p}}{\rm{a}}{\rm{c}}{\rm{i}}{\rm{t}}{\rm{a}}{\rm{n}}{\rm{c}}{\rm{e}}\,{\rm{a}}{\rm{n}}{\rm{d}}\,{\rm{d}}{\rm{i}}{\rm{s}}{\rm{t}}{\rm{r}}{\rm{i}}{\rm{b}}{\rm{u}}{\rm{t}}{\rm{e}}{\rm{d}}\,{\rm{p}}{\rm{a}}{\rm{r}}{\rm{a}}{\rm{m}}{\rm{e}}{\rm{t}}{\rm{e}}{\rm{r}}\,{\rm{c}}{\rm{e}}{\rm{l}}{\rm{l}},\\ 1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{t}}{\rm{h}}{\rm{e}}\,{\rm{o}}{\rm{t}}{\rm{h}}{\rm{e}}{\rm{r}}{\rm{s}}.\end{array}$$
(30)
$${\delta }_{\beta \beta }=\{\begin{array}{cc}-1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{i}}{\rm{n}}{\rm{d}}{\rm{u}}{\rm{c}}{\rm{t}}{\rm{a}}{\rm{n}}{\rm{c}}{\rm{e}},\\ 1 & {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{t}}{\rm{h}}{\rm{e}}\,{\rm{o}}{\rm{t}}{\rm{h}}{\rm{e}}{\rm{r}}{\rm{s}}.\end{array}$$
(31)

The matrix γS is also a diagonal matrix to fit the unit of current in KCL and consider the direction of current in the β-th distributed-parameter circuit.

$${\gamma }_{{S}_{\beta \beta }}=\{\begin{array}{c}1\\ {\gamma }_{x\beta \beta }{S}_{x\beta \beta }+{\gamma }_{y\beta \beta }{S}_{y\beta \beta }\end{array}\,\begin{array}{c}{\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{l}}{\rm{u}}{\rm{m}}{\rm{p}}{\rm{e}}{\rm{d}}\,{\rm{p}}{\rm{a}}{\rm{r}}{\rm{a}}{\rm{m}}{\rm{e}}{\rm{t}}{\rm{e}}{\rm{r}}\,{\rm{e}}{\rm{l}}{\rm{e}}{\rm{m}}{\rm{e}}{\rm{n}}{\rm{t}}{\rm{s}},\\ {\rm{f}}{\rm{o}}{\rm{r}}\,{\rm{d}}{\rm{i}}{\rm{s}}{\rm{t}}{\rm{r}}{\rm{i}}{\rm{b}}{\rm{u}}{\rm{t}}{\rm{e}}{\rm{d}}\,{\rm{p}}{\rm{a}}{\rm{r}}{\rm{a}}{\rm{m}}{\rm{e}}{\rm{t}}{\rm{e}}{\rm{r}}\,{\rm{C}}{\rm{e}}{\rm{l}}{\rm{l}}.\end{array}$$
(32)

In the Supplementary Information, we show the calculation results of the time variation of potentials and charge densities in a simple two-layer circuit system, as shown in Fig. 4. Comparing the simulation results of “Supplementary Movie 1.mov” and “Supplementary Movie 2.mov”, the potentials of the bottom layer plane, which are usually used as the ground, largely depend on the position of the upper layer plane.

## Calculation Algorithm of Boundary Conditions to Reduce Calculation Cost

In this section, we explain the appropriate calculation algorithm for an MLP circuit. In the previous algorithm shown in Table 2, we calculated all boundaries using Eq. (29), although the size of the matrix is proportional to the number of cells in the MLP circuit. The calculation is burdensome in the case of a complicated circuit. In the new algorithm, we divide the boundary into two cases, considering that most of the boundaries in the MLP circuit are open. One is the case with a lumped-parameter circuit (W), and the other is the case without a lumped-parameter circuit (WO), as shown in Table 2. We can reduce the size of the matrix in Eq. (29) considerably and obtain a simplified boundary equation in the case of no lumped-parameter circuit.

In the case without a lumped-parameter circuit, the KCL at the boundary is derived from Eq. (29):

$$({\gamma }_{x}{{\bf{S}}}_{x}+{\gamma }_{y}{{\bf{S}}}_{y}){{\bf{A}}}_{d}{{\bf{j}}}_{d}^{m+1}=0.$$
(33)

The current density in the distributed-parameter circuit flowing perpendicular to the boundary is always zero when a lumped-parameter circuit is not connected. As for the condition of the distributed-parameter circuit, the following equation can be obtained by substituting Eq. (33) into Eq. (22):

$$\begin{array}{ccc}{U}_{i(k,l)}^{m+1} & = & {U}_{i(k,l)}^{m}\\ & & +\,\mathop{\sum }\limits_{{i}^{{\rm{^{\prime} }}},{k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}}}^{{\rm{B}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}({\rm{W}})}\,\mathop{\sum }\limits_{\alpha }^{x,y}\,{\gamma }_{\alpha {i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}\frac{\Delta t}{\Delta \alpha }{P}_{i(k,l){i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}({j}_{\alpha {i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}^{m+1}+{j}_{\alpha {i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}^{m})\\ & & -\,\mathop{\sum }\limits_{{i}^{{\rm{^{\prime} }}},{k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}}}^{{\rm{B}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}({\rm{W}},{\rm{W}}{\rm{O}})}\,\mathop{\sum }\limits_{\alpha }^{x,y}\,{\Gamma }_{\alpha {i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}\frac{\Delta t}{\Delta \alpha }{P}_{i(k,l){i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}\,{j}_{\alpha {i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}}+{\beta }_{x},{l}^{{\rm{^{\prime} }}}+{\beta }_{y})}^{m+\frac{1}{2}}\\ & & -\,\mathop{\sum }\limits_{{i}^{{\rm{^{\prime} }}},{k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}}}^{{\rm{N}}{\rm{o}}\,{\rm{b}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}}\,\frac{\Delta t}{\Delta x}{P}_{i(k,l){i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}({j}_{x{i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}}+\frac{1}{2},{l}^{{\rm{^{\prime} }}})}^{m+\frac{1}{2}}-{j}_{x{i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}}-\frac{1}{2},{l}^{{\rm{^{\prime} }}})}^{m+\frac{1}{2}})\\ & & -\,\mathop{\sum }\limits_{{i}^{{\rm{^{\prime} }}},{k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}}}^{{\rm{N}}{\rm{o}}\,{\rm{b}}{\rm{o}}{\rm{u}}{\rm{n}}{\rm{d}}{\rm{a}}{\rm{r}}{\rm{y}}}\,\frac{\Delta t}{\Delta y}{P}_{i(k,l){i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}})}({j}_{y{i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}}+\frac{1}{2})}^{m+\frac{1}{2}}-{j}_{y{i}^{{\rm{^{\prime} }}}({k}^{{\rm{^{\prime} }}},{l}^{{\rm{^{\prime} }}}-\frac{1}{2})}^{m+\frac{1}{2}}).\end{array}$$
(34)

Here, $${j}_{\alpha i^{\prime} (k^{\prime} ,l^{\prime} )}^{m+1}$$ is a known value obtained by Eq. (29). Equation (16) is the same as Eq. (34) when the summation term is divided into boundary and no-boundary terms. In other words, the potential at the boundary where the lumped circuit is not connected can be calculated in the same manner as the no-boundary case when the current density flowing perpendicular to the boundary is set to zero. In this case, we can significantly decrease the size of the boundary matrix in Eq. (29). The calculation algorithms are listed in Table 2.

## Discussion with Numerical Results

We performed numerical calculation using the present method and the corresponding experiment. We then compared the numerical results with the experimental results together with the results calculated by the PEEC method15,16. We used the equations and space discretization methods used in the PEEC method. There are differences in the calculation method of boundary conditions and the updating time step between our formulation and the PEEC method. We calculated the boundaries by connecting the lumped-parameter circuit and the transport equations derived from the Maxwell equations, as discussed in the present paper, while the PEEC method converts all finite-volume cells into equivalent circuit models and solves both the boundary and no-boundary parts at the same time. As for time discretization, we used the leapfrog time discretization method, which is the concept of the FDTD method, for updating the potential and the current density17, while the PEEC method uses a lumped-parameter circuit simulation algorithm18,19.

We consider a two-layer plane circuit with two bends used in the simulation and experiment (see Supplementary Information, Figs 1 and 2). We compare our simulation results with sub-nanosecond time domain reflectometry (TDR) experiments and PEEC method simulation. We can see good agreement between the numerical results of the proposed method and the experimental results (see Supplementary Information, Fig. 3). We perform simulation with both the proposed method and the PEEC method with the same numerical calculation parameters, such as the space and time steps. The proposed method can reproduce the reflected wave with higher accuracy because of the calculation method for boundary conditions and leapfrog time updating.

In addition, we present electromagnetic noise simulation in a simple circuit, which has a micro-strip line configuration, as shown in Fig. 4. The ground noise voltage, which causes external radiation and common mode noise, is generated in various directions according to the position of the signal line20 (see Supplementary Information, Movies 1 and 2).

At the moment, the delay time in the scalar and vector potentials in the charge and current densities has been dropped in the numerical calculations. We are currently studying the inclusion of the delay time effect in the present formulation14. In addition, multi-layer-plane conductors exist in a uniform medium, and the skin effect is not considered. In order to precisely consider the change of dielectric properties in the conductors and surrounding materials as well as the skin effect, it is necessary to extend our calculation to three dimensions. If three-dimensional calculation is realized, then the propagation speed will change inside and outside of the conductor plane, which means that we can consider the variation of dielectric properties in each plane, as well as the skin effect. We hope to consider these factors in the near future.

This calculation method enables us to visualize transient basic electromagnetic phenomena, such as propagation, reflection, and coupling with other conductors. We can also clarify the relationship between complex circuit shapes and electromagnetic phenomena observed in high-frequency filters, antennas, and meta-materials21,22,23. This calculation method will be important not only in elucidating electromagnetic noise phenomena but also in developing future advanced circuit design.

## Summary

In the present paper, we introduced a method to quantify the transient phenomena of the potential and current density in a multi-layer plane (MLP) circuit. The theoretical equations were obtained from the Maxwell equations. In addition, we used Ohm’s law and continuity equations in order to describe the transport phenomena. In the numerical method, we use the spatial difference method used in the PEEC method and the central difference approximation used in the FDTD method. We also developed a calculation method for the boundary in the MLP and proposed an appropriate calculation algorithm in order to greatly reduce the calculation cost. We applied the present theoretical method for a two-dimensional circuit system and demonstrated the power of the newly developed theory.

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Kitora, S., Jinno, S., Toki, H. & Abe, M. Coupling of Differential and Common Modes of Two-Line Circuits in the Multi-Conductor Transmission-Line Theory Including Radiation. In Progress in Electromagnetics Research Symposium 2018-Augus, 570–574 (2018).

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## Acknowledgements

The present study was supported in part by Grants-in-Aid for Scientific Research (18K19023, 16H03872, 19J14259,19H05789) from the Ministry of Education, Culture, Sports, Science, and Technology of Japan (MEXT) and by the Japan Science Society through a Sasakawa Scientific Research Grant (29-201). The present study was conducted in collaboration with the Nichicon Corporation. S.J. and S.K. acknowledge the support of the Faculty of Engineering Science of Osaka University.

## Author information

Authors

### Contributions

S.J. derived the coupled difference equations of potential and current density, developed the method of boundary calculation in a multi-layer plane circuit and wrote the main manuscript text. S.K. reviewed the derivation of equations. H.T. reviewed and revised the equations. M.A. reviewed and organized the manuscript text and composition.

### Corresponding author

Correspondence to Souma Jinno.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Supplementary information

### 41598_2019_53288_MOESM2_ESM.mov

Supplementary Movie1

### 41598_2019_53288_MOESM3_ESM.mov

Supplementary Movie2

## Rights and permissions

Reprints and Permissions

Jinno, S., Kitora, S., Toki, H. et al. Time-domain Formulation of a Multi-layer Plane Circuit Coupled with Lumped-parameter Circuits using Maxwell Equations. Sci Rep 9, 17891 (2019). https://doi.org/10.1038/s41598-019-53288-x

• Accepted:

• Published:

• ### Origin of common-mode noise in two-dimensional finite-size circuit and reduction of the noise using a symmetric three-line circuit

• Souma Jinno
• , Shuji Kitora
• , Hiroshi Toki
•  & Masayuki Abe

International Journal of Circuit Theory and Applications (2020)


text:
# Spherical Triangles: Area and mapping to Euclidean space

If we take a sphere of radius 1 and travel a quarter-circumference south from the north pole, turn 90 degrees, travel another quarter circumference, then return North, we form a triangle with an angle of 270$$^\text{o}$$. According to Mathworld, the area of such a triangle would be given by $$\Delta=R^{2}[(A+B+C)-\pi]$$ With A, B and C the angles of the triangle.

Now if we consider lines of latitude and longitude (I understand these always intersect at $$90^\text{o}$$), and shift the base of the triangle closer to the north pole but so that it remains on a line of latitude, this would suggest that the base of the triangle still meets the two longitudinal lines at 90$$^\text{o}$$. This would suggest that the area stays the same, which seems very wrong.

I am also unsure of the maximum sum of angles of a spherical triangle. On the same Mathworld page it states:

The sum of the angles of a spherical triangle is between $$\pi$$ and $$3\pi$$ radians (180 degrees and 540 degrees; Zwillinger 1995, p. 469)

However, considering a very small triangle (s.t. the sum of interior angles tends to 180$$^\text{o}$$, and taking this as the outside of the triangle (and the rest of the sphere as the inside) this would create a 'triangle' with angles summing 900$$^\text{o}$$. My guess would be that this might not count as a triangle because it perhaps cannot be mapped appropriately to Euclidaen space. So then, what would be an appropriate mapping to Euclidean space? Would it result in a 'triangle' with curved sides?

## 1 Answer

A spherical triangle must be formed by arcs of great circles, hence you cannot take a parallel as side, apart from the equator.

And of the two parts into which a sphere is divided by the sides of a spherical triangle, only the smaller (and convex) one is considered "the inner part" of the triangle. That explains the 540° limit.

• Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle. – Lubin Dec 19 '18 at 18:14

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# Finding area

I follow the Five Triangles ‘blog for cool math problems.  A recent one proved particularly nice.

At first I wasn’t sure this situation was invariant.  I didn’t see how fixing three triangle areas guaranteed a fixed quadrilateral area.  Not seeing an immediate general solution approach, I reasoned that if there was a solution, it worked for multiple overall configurations.  If it worked in general, then it must also work for any particular case I chose, so I made the cevians perpendicular.  That made each of the given area triangles right.  I modeled that by constructing the overall triangle with the cevian intersection at the origin and the legs of the given area triangles along the coordinate axes.

There are many ways to do this, but I reasoned that if there was a single answer, then any one of them would work.  A right triangle with legs of length 8 and 5 would have area 20.  Constructing that triangle in GeoGebra fixed the lengths of the legs of the other two triangles and the hypotenuses of the area 8 & 15 triangles intersected at a Quadrant II point.  Here’s my construction.

I  overlayed a polygon to create the quadrilateral and measured its area directly.  For fun, I also wrote algebraic equations for lines CB and DA, found the coordinates of point F by solving the 2×2 linear system, used that to derive the area of $\Delta BDF$, and determined the area of the quadrilateral from that.

While I realized that this approach was just a single case of the given problem, it absolutely convinced me that the solution was unique.  Once the area 20 triangle was defined (whether or not the triangle was right), a side and the area of each of the other two given triangles is known.  That meant the heights of the triangles would be determined and thereby the location of the quadrilateral’s fourth vertex.  So, I knew without a doubt that the unknown area was $27 cm^2$, but I didn’t know a general solution.

Chronology of the General Solution

While I worked more on the problem, I also pitched it to my Twitter network and asked a colleague at my school, Tatiana Yudovina, if she was interested in the problem.  Next is Tatiana’s initial solution, followed by my generic Geogebra construction, and a much shorter solution Tatiana created.  My conclusion takes the problem to a more generic state and raises some potential extensions.

Tatiana’s First Solution:

Leveraging the fact that triangles with the same base have equivalent height and area ratios, she created a system of equations that solved to eventually determine the quadrilateral’s area.

My Generic GeoGebra Solution:

While Tatiana was working on her algebraic answer, I was creating  a dynamic version on GeoGebra.  I built the area 20 triangle by first drawing a segment AB and measuring its length, a.  That meant the height of this triangle, h, was given by $\frac{1}{2} a \cdot h =20\longrightarrow h=\frac{40}{a}$. Then I constructed a perpendicular line to AB and used the “Segment with Fixed Length” tool and defined the length using the generic length of h as defined above to create segment AC.  This worked because GeoGebra defined the length of AB as a variable as shown below.

I used the “Compass” tool to create a circle with radius AC through the perpendicular line created earlier. Point D is the intersection of the circle and the normal line.  I then constructed a perpendicular to AD through D and placed a random point E on this new line.  Point E was the requisite height above AB to guarantee that $\Delta ABE$ always had area 20 which I confirmed by drawing the triangle and computing its area.

I hid AC, the circle, and both normals.  Segment AB was a completely independent object, and point E was free to move along the second “height” normal.  I measured AE and repeated the previous construction to create the area 15 triangle. Because BE was part of a cevian, I drew line BE to determine point J on the normal defining the final vertex of the area 15 triangle.

Again, I hid all of my constructions and repeated the process to create the final vertex, K, of the area 8 triangle off side BE of the area 20 triangle.  Extending segments AJ and BK defined point L, the final vertex of the quadrilateral.  Laying a quadrilateral in the figure let me compute its area.  Moving points A, B, and E around the screen and seeing the areas remain fixed is pretty compelling evidence that the quadrilateral’s area is always 27, and Tatiana’s proof showed why.  You can play with my final construction on GeoGebra Tube here.

Then Tatiana emailed me a much shorter proof.

Tatiana’s Short Solution:

Reversing the logic of her first solution, Tatiana reasoned that equivalent-altitude triangles had equal base and area ratios.

And the sum of X and Y gave the quadrilateral’s area.

Conclusion:

This problem was entertaining both in the solution and the multiple ways we found it.  Creating the dynamic construction gave  insights into the critical features of the problem.

Here are some potential extensions I developed for this problem.  I haven’t fully explored any of them yet, hoping some of my geometry students this year might take up the exploration challenge.  I’d love to hear if any of my readers have any further suggestions.

1. It might be interesting to create an even more dynamic construction with the areas of the three given triangles defined by sliders.
2. Can the quadrilateral’s area be expressed as a closed-form function of the areas of the three given triangles.
3. What happens on the boundaries of this problem?  That is, what happens if one of the side triangles was a degenerate with area 0? What would happen to the quadrilateral? Would would be the corresponding affect on the area formula from extension 2?
4. Extending 3 even further, if both given side triangles were degenerates with area 0, it seems that the area formula from extension 2 should collapse to the area of the final given non-zero triangle, but does it?

Thanks again, Five Triangles, for another great problem!


text:
Properties

 Label 147.3.b.f Level $147$ Weight $3$ Character orbit 147.b Analytic conductor $4.005$ Analytic rank $0$ Dimension $4$ CM no Inner twists $2$

Related objects

Newspace parameters

 Level: $$N$$ $$=$$ $$147 = 3 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$3$$ Character orbit: $$[\chi]$$ $$=$$ 147.b (of order $$2$$, degree $$1$$, minimal)

Newform invariants

 Self dual: no Analytic conductor: $$4.00545988610$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: 4.0.65856.1 Defining polynomial: $$x^{4} + 14 x^{2} + 21$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 21) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$

$q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + \beta_{1} q^{2} + ( \beta_{1} + \beta_{2} - \beta_{3} ) q^{3} + ( -3 + 2 \beta_{2} ) q^{4} + ( 1 - \beta_{2} + 2 \beta_{3} ) q^{5} + ( -3 - \beta_{1} + 2 \beta_{2} + \beta_{3} ) q^{6} + ( 2 - 5 \beta_{1} - 2 \beta_{2} + 4 \beta_{3} ) q^{8} + ( -6 - \beta_{1} + 2 \beta_{2} - 2 \beta_{3} ) q^{9} +O(q^{10})$$ $$q + \beta_{1} q^{2} + ( \beta_{1} + \beta_{2} - \beta_{3} ) q^{3} + ( -3 + 2 \beta_{2} ) q^{4} + ( 1 - \beta_{2} + 2 \beta_{3} ) q^{5} + ( -3 - \beta_{1} + 2 \beta_{2} + \beta_{3} ) q^{6} + ( 2 - 5 \beta_{1} - 2 \beta_{2} + 4 \beta_{3} ) q^{8} + ( -6 - \beta_{1} + 2 \beta_{2} - 2 \beta_{3} ) q^{9} + ( -7 - \beta_{2} ) q^{10} + 2 \beta_{1} q^{11} + ( 6 - 7 \beta_{1} - \beta_{2} + \beta_{3} ) q^{12} + ( 9 - \beta_{2} ) q^{13} + ( 9 - \beta_{1} + 2 \beta_{2} + 4 \beta_{3} ) q^{15} + ( 9 - 4 \beta_{2} ) q^{16} + ( -2 + 2 \beta_{1} + 2 \beta_{2} - 4 \beta_{3} ) q^{17} + ( 15 - 8 \beta_{1} - 2 \beta_{2} + 2 \beta_{3} ) q^{18} + ( -3 + 5 \beta_{2} ) q^{19} + ( 3 - 4 \beta_{1} - 3 \beta_{2} + 6 \beta_{3} ) q^{20} + ( -14 + 4 \beta_{2} ) q^{22} + ( -2 + 8 \beta_{1} + 2 \beta_{2} - 4 \beta_{3} ) q^{23} + ( 33 + 3 \beta_{1} - 6 \beta_{2} + 3 \beta_{3} ) q^{24} + ( -3 - 10 \beta_{2} ) q^{25} + ( -1 + 12 \beta_{1} + \beta_{2} - 2 \beta_{3} ) q^{26} + ( -3 - 5 \beta_{1} - 8 \beta_{2} - \beta_{3} ) q^{27} + ( -2 - 2 \beta_{1} + 2 \beta_{2} - 4 \beta_{3} ) q^{29} + ( -3 - 5 \beta_{1} - 8 \beta_{2} + 8 \beta_{3} ) q^{30} + ( -34 - 2 \beta_{2} ) q^{31} + ( 4 + \beta_{1} - 4 \beta_{2} + 8 \beta_{3} ) q^{32} + ( -6 - 2 \beta_{1} + 4 \beta_{2} + 2 \beta_{3} ) q^{33} + 6 \beta_{2} q^{34} + ( 24 + 13 \beta_{1} - 8 \beta_{2} - 10 \beta_{3} ) q^{36} + ( 4 + 14 \beta_{2} ) q^{37} + ( 5 - 18 \beta_{1} - 5 \beta_{2} + 10 \beta_{3} ) q^{38} + ( -3 + 11 \beta_{1} + 8 \beta_{2} - 8 \beta_{3} ) q^{39} + ( -21 - 15 \beta_{2} ) q^{40} + ( -8 + 22 \beta_{1} + 8 \beta_{2} - 16 \beta_{3} ) q^{41} + ( -40 - 6 \beta_{2} ) q^{43} + ( 4 - 18 \beta_{1} - 4 \beta_{2} + 8 \beta_{3} ) q^{44} + ( 33 - 2 \beta_{1} + 13 \beta_{2} - 4 \beta_{3} ) q^{45} + ( -42 + 18 \beta_{2} ) q^{46} + ( -4 - 8 \beta_{1} + 4 \beta_{2} - 8 \beta_{3} ) q^{47} + ( -12 + 17 \beta_{1} + 5 \beta_{2} - 5 \beta_{3} ) q^{48} + ( -10 + 27 \beta_{1} + 10 \beta_{2} - 20 \beta_{3} ) q^{50} + ( -24 - 6 \beta_{3} ) q^{51} + ( -41 + 21 \beta_{2} ) q^{52} + ( -16 + 10 \beta_{1} + 16 \beta_{2} - 32 \beta_{3} ) q^{53} + ( 30 + 23 \beta_{1} - \beta_{2} - 17 \beta_{3} ) q^{54} + ( -14 - 2 \beta_{2} ) q^{55} + ( 15 - 13 \beta_{1} + 2 \beta_{2} - 2 \beta_{3} ) q^{57} + ( 28 - 2 \beta_{2} ) q^{58} + ( 1 + 26 \beta_{1} - \beta_{2} + 2 \beta_{3} ) q^{59} + ( 39 + \beta_{1} - 2 \beta_{2} + 8 \beta_{3} ) q^{60} + ( 39 - 7 \beta_{2} ) q^{61} + ( -2 - 28 \beta_{1} + 2 \beta_{2} - 4 \beta_{3} ) q^{62} + ( 1 - 18 \beta_{2} ) q^{64} + ( 6 + 2 \beta_{1} - 6 \beta_{2} + 12 \beta_{3} ) q^{65} + ( 12 - 22 \beta_{1} - 10 \beta_{2} + 10 \beta_{3} ) q^{66} + ( -6 - 8 \beta_{2} ) q^{67} + ( -2 - 10 \beta_{1} + 2 \beta_{2} - 4 \beta_{3} ) q^{68} + ( -42 - 6 \beta_{1} + 12 \beta_{2} ) q^{69} + ( 6 + 18 \beta_{1} - 6 \beta_{2} + 12 \beta_{3} ) q^{71} + ( -9 + 36 \beta_{1} + 36 \beta_{2} - 18 \beta_{3} ) q^{72} + ( 8 - 26 \beta_{2} ) q^{73} + ( 14 - 38 \beta_{1} - 14 \beta_{2} + 28 \beta_{3} ) q^{74} + ( -30 + 17 \beta_{1} - 13 \beta_{2} + 13 \beta_{3} ) q^{75} + ( 79 - 21 \beta_{2} ) q^{76} + ( -45 - 11 \beta_{1} + 22 \beta_{2} + 8 \beta_{3} ) q^{78} + ( 32 - 36 \beta_{2} ) q^{79} + ( -3 + 8 \beta_{1} + 3 \beta_{2} - 6 \beta_{3} ) q^{80} + ( -15 + 20 \beta_{1} - 22 \beta_{2} + 4 \beta_{3} ) q^{81} + ( -98 + 52 \beta_{2} ) q^{82} + ( 9 + 18 \beta_{1} - 9 \beta_{2} + 18 \beta_{3} ) q^{83} + ( 42 + 18 \beta_{2} ) q^{85} + ( -6 - 22 \beta_{1} + 6 \beta_{2} - 12 \beta_{3} ) q^{86} + ( -12 + 4 \beta_{1} - 8 \beta_{2} - 10 \beta_{3} ) q^{87} + ( 42 - 24 \beta_{2} ) q^{88} + ( 16 - 42 \beta_{1} - 16 \beta_{2} + 32 \beta_{3} ) q^{89} + ( 39 + 2 \beta_{1} - 13 \beta_{2} + 22 \beta_{3} ) q^{90} + ( 10 - 64 \beta_{1} - 10 \beta_{2} + 20 \beta_{3} ) q^{92} + ( -6 - 30 \beta_{1} - 36 \beta_{2} + 36 \beta_{3} ) q^{93} + ( 84 - 12 \beta_{2} ) q^{94} + ( 12 - 10 \beta_{1} - 12 \beta_{2} + 24 \beta_{3} ) q^{95} + ( 33 - 5 \beta_{1} + 10 \beta_{2} + 17 \beta_{3} ) q^{96} + ( 2 - 8 \beta_{2} ) q^{97} + ( 30 - 16 \beta_{1} - 4 \beta_{2} + 4 \beta_{3} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + 2q^{3} - 12q^{4} - 14q^{6} - 20q^{9} + O(q^{10})$$ $$4q + 2q^{3} - 12q^{4} - 14q^{6} - 20q^{9} - 28q^{10} + 22q^{12} + 36q^{13} + 28q^{15} + 36q^{16} + 56q^{18} - 12q^{19} - 56q^{22} + 126q^{24} - 12q^{25} - 10q^{27} - 28q^{30} - 136q^{31} - 28q^{33} + 116q^{36} + 16q^{37} + 4q^{39} - 84q^{40} - 160q^{43} + 140q^{45} - 168q^{46} - 38q^{48} - 84q^{51} - 164q^{52} + 154q^{54} - 56q^{55} + 64q^{57} + 112q^{58} + 140q^{60} + 156q^{61} + 4q^{64} + 28q^{66} - 24q^{67} - 168q^{69} + 32q^{73} - 146q^{75} + 316q^{76} - 196q^{78} + 128q^{79} - 68q^{81} - 392q^{82} + 168q^{85} - 28q^{87} + 168q^{88} + 112q^{90} - 96q^{93} + 336q^{94} + 98q^{96} + 8q^{97} + 112q^{99} + O(q^{100})$$

Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} + 14 x^{2} + 21$$:

 $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$\nu^{2} + 7$$$$)/2$$ $$\beta_{3}$$ $$=$$ $$($$$$\nu^{3} + \nu^{2} + 13 \nu + 5$$$$)/4$$
 $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$2 \beta_{2} - 7$$ $$\nu^{3}$$ $$=$$ $$4 \beta_{3} - 2 \beta_{2} - 13 \beta_{1} + 2$$

Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/147\mathbb{Z}\right)^\times$$.

 $$n$$ $$50$$ $$52$$ $$\chi(n)$$ $$-1$$ $$1$$

Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
50.1
 − 3.50592i − 1.30710i 1.30710i 3.50592i
3.50592i −0.822876 2.88494i −8.29150 1.24197i −10.1144 + 2.88494i 0 15.0457i −7.64575 + 4.74789i −4.35425
50.2 1.30710i 1.82288 + 2.38267i 2.29150 7.37953i 3.11438 2.38267i 0 8.22359i −2.35425 + 8.68663i −9.64575
50.3 1.30710i 1.82288 2.38267i 2.29150 7.37953i 3.11438 + 2.38267i 0 8.22359i −2.35425 8.68663i −9.64575
50.4 3.50592i −0.822876 + 2.88494i −8.29150 1.24197i −10.1144 2.88494i 0 15.0457i −7.64575 4.74789i −4.35425
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 inner

Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 147.3.b.f 4
3.b odd 2 1 inner 147.3.b.f 4
7.b odd 2 1 21.3.b.a 4
7.c even 3 2 147.3.h.c 8
7.d odd 6 2 147.3.h.e 8
21.c even 2 1 21.3.b.a 4
21.g even 6 2 147.3.h.e 8
21.h odd 6 2 147.3.h.c 8
28.d even 2 1 336.3.d.c 4
35.c odd 2 1 525.3.c.a 4
35.f even 4 2 525.3.f.a 8
56.e even 2 1 1344.3.d.b 4
56.h odd 2 1 1344.3.d.f 4
63.l odd 6 2 567.3.r.c 8
63.o even 6 2 567.3.r.c 8
84.h odd 2 1 336.3.d.c 4
105.g even 2 1 525.3.c.a 4
105.k odd 4 2 525.3.f.a 8
168.e odd 2 1 1344.3.d.b 4
168.i even 2 1 1344.3.d.f 4

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
21.3.b.a 4 7.b odd 2 1
21.3.b.a 4 21.c even 2 1
147.3.b.f 4 1.a even 1 1 trivial
147.3.b.f 4 3.b odd 2 1 inner
147.3.h.c 8 7.c even 3 2
147.3.h.c 8 21.h odd 6 2
147.3.h.e 8 7.d odd 6 2
147.3.h.e 8 21.g even 6 2
336.3.d.c 4 28.d even 2 1
336.3.d.c 4 84.h odd 2 1
525.3.c.a 4 35.c odd 2 1
525.3.c.a 4 105.g even 2 1
525.3.f.a 8 35.f even 4 2
525.3.f.a 8 105.k odd 4 2
567.3.r.c 8 63.l odd 6 2
567.3.r.c 8 63.o even 6 2
1344.3.d.b 4 56.e even 2 1
1344.3.d.b 4 168.e odd 2 1
1344.3.d.f 4 56.h odd 2 1
1344.3.d.f 4 168.i even 2 1

Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{3}^{\mathrm{new}}(147, [\chi])$$:

 $$T_{2}^{4} + 14 T_{2}^{2} + 21$$ $$T_{5}^{4} + 56 T_{5}^{2} + 84$$ $$T_{13}^{2} - 18 T_{13} + 74$$

Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$1 - 2 T^{2} + 5 T^{4} - 32 T^{6} + 256 T^{8}$$
$3$ $$1 - 2 T + 12 T^{2} - 18 T^{3} + 81 T^{4}$$
$5$ $$1 - 44 T^{2} + 1034 T^{4} - 27500 T^{6} + 390625 T^{8}$$
$7$ 1
$11$ $$1 - 428 T^{2} + 74630 T^{4} - 6266348 T^{6} + 214358881 T^{8}$$
$13$ $$( 1 - 18 T + 412 T^{2} - 3042 T^{3} + 28561 T^{4} )^{2}$$
$17$ $$1 - 988 T^{2} + 407046 T^{4} - 82518748 T^{6} + 6975757441 T^{8}$$
$19$ $$( 1 + 6 T + 556 T^{2} + 2166 T^{3} + 130321 T^{4} )^{2}$$
$23$ $$1 - 1444 T^{2} + 980166 T^{4} - 404090404 T^{6} + 78310985281 T^{8}$$
$29$ $$1 - 2972 T^{2} + 3611558 T^{4} - 2102039132 T^{6} + 500246412961 T^{8}$$
$31$ $$( 1 + 68 T + 3050 T^{2} + 65348 T^{3} + 923521 T^{4} )^{2}$$
$37$ $$( 1 - 8 T + 1382 T^{2} - 10952 T^{3} + 1874161 T^{4} )^{2}$$
$41$ $$1 - 1292 T^{2} + 2832038 T^{4} - 3650883212 T^{6} + 7984925229121 T^{8}$$
$43$ $$( 1 + 80 T + 5046 T^{2} + 147920 T^{3} + 3418801 T^{4} )^{2}$$
$47$ $$1 - 6148 T^{2} + 19144326 T^{4} - 30000278788 T^{6} + 23811286661761 T^{8}$$
$53$ $$1 + 20 T^{2} - 13350138 T^{4} + 157809620 T^{6} + 62259690411361 T^{8}$$
$59$ $$1 - 3676 T^{2} + 15964266 T^{4} - 44543419036 T^{6} + 146830437604321 T^{8}$$
$61$ $$( 1 - 78 T + 8620 T^{2} - 290238 T^{3} + 13845841 T^{4} )^{2}$$
$67$ $$( 1 + 12 T + 8566 T^{2} + 53868 T^{3} + 20151121 T^{4} )^{2}$$
$71$ $$1 - 10588 T^{2} + 78813510 T^{4} - 269058878428 T^{6} + 645753531245761 T^{8}$$
$73$ $$( 1 - 16 T + 5990 T^{2} - 85264 T^{3} + 28398241 T^{4} )^{2}$$
$79$ $$( 1 - 64 T + 4434 T^{2} - 399424 T^{3} + 38950081 T^{4} )^{2}$$
$83$ $$1 - 13948 T^{2} + 141899946 T^{4} - 661948661308 T^{6} + 2252292232139041 T^{8}$$
$89$ $$1 - 11468 T^{2} + 120945830 T^{4} - 719528019788 T^{6} + 3936588805702081 T^{8}$$
$97$ $$( 1 - 4 T + 18374 T^{2} - 37636 T^{3} + 88529281 T^{4} )^{2}$$


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# The inside dimensions of a rectangular wooden box are 6 inches by 8 in

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The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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Updated on: 15 Apr 2015, 02:47
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Question Stats:

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The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-inside-dimensions-of-a-rectangular-wooden-box-are-128053.html

Originally posted by hb05sv on 17 Mar 2010, 17:30.
Last edited by Bunuel on 15 Apr 2015, 02:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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17 Mar 2010, 19:02
1
hb05sv wrote:
Can someone show me how to solve the following question?

The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?
A)3
B)4
c)5
D)6
E)8

Correct Ans is B.

let the face on which the cylinder is placed is 6 by 8. In this case the volume will be pi* 3^2*10 = 90pi (here r =3inches)

If the cylinder is placed on the face having dimensions 8 by 10 then volume in that case will be pi* 4^2* 6 = 96pi (here r = 4inches)

If the cylinder is placed on the face having dimensions 6 by 10 then volume in that case will be pi* 3^2* 8 = 72pi (here r = 3inches)

so for r = 4inches the cylinder will have maximum area.
so B - 4
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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17 Mar 2010, 19:26
kp1811 wrote:
let the face on which the cylinder is placed is 6 by 8. In this case the volume will be pi* 3^2*10 = 90pi (here r =3inches)

If the cylinder is placed on the face having dimensions 8 by 10 then volume in that case will be pi* 4^2* 6 = 96pi (here r = 4inches)

If the cylinder is placed on the face having dimensions 6 by 10 then volume in that case will be pi* 3^2* 8 = 72pi (here r = 3inches)

so for r = 4inches the cylinder will have maximum area.
so B - 4

Nice explanation -

To be more specific, logically, for any face down of the cube, the smaller length only can be the diameter of the cylindrical canister. So either ways, it is 6 or 8. The height will be either the shortest or the longest dimension accordingly.

Volume of cylinder is Pi * r squared * h. The greater the value of r^2*h, the greater the volume.
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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18 Mar 2010, 14:23
BarneyStinson, Can you explain why logically the diameter of the cylinder will be one of the shortest?
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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18 Mar 2010, 16:04
hb05sv wrote:
BarneyStinson, Can you explain why logically the diameter of the cylinder will be one of the shortest?

Because if the diameter were wider say 8", it won't fit into a box that has one side 6", when the face down is 8" X 6".
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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20 Aug 2010, 06:22
1
kp1811 wrote:
hb05sv wrote:
Can someone show me how to solve the following question?

The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?
A)3
B)4
c)5
D)6
E)8

Correct Ans is B.

let the face on which the cylinder is placed is 6 by 8. In this case the volume will be pi* 3^2*10 = 90pi (here r =3inches)

If the cylinder is placed on the face having dimensions 8 by 10 then volume in that case will be pi* 4^2* 6 = 96pi (here r = 4inches)

If the cylinder is placed on the face having dimensions 6 by 10 then volume in that case will be pi* 3^2* 8 = 72pi (here r = 3inches)

so for r = 4inches the cylinder will have maximum area.
so B - 4

Sorry guyz but dont understand your ways

My approach is;

pi approx. = 3,14

The box volume is 480 so;

Cylinder max. volume should be pi*5^2*8 = 200pi means more then 600 so cant be, If the radius cant be 5 so it should be 4;

pi*4^2*10= slightly more then 480 so CANT BE

pi*4^2*6 < 480 it means The radius should be "4".

B
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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21 Aug 2010, 15:17
1
Since we want Max Volume for cylinder hence Max Vol can only max if the Radius is maximum

We can always have Cylinder with radius 3 in this Box but we need max radius if we take (6 and 8) or (6 and 10) or (10 and 8)

So let us pick 5 as radius so the diameter will be 10 but other two sides are 8 and 6 which cause a cylinder with 10 radius out of the box. Hence 5 is not the answer
Anything above 5 i.e 6 and 8 are gone
Now we come to our last option 4 if we take sides 10 and 8 as the base we can surely incorporate cylinder inside the box hence our answer is B i.e. 4.

Try to give it a thought because I have not used any calculation to solve this question.
And don't forget in Gmat Exam we need to conserve all our energies because after Quants exam the beast awaits......VERBAL!!!!!!
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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31 Aug 2010, 10:15
And don't forget in Gmat Exam we need to conserve all our energies because after Quants exam the beast awaits......VERBAL!!!!!!

And remember to reserve some extra energy for AWA:)
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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31 Aug 2010, 23:34
Calculate r^2 * h for all cases you gat the answer
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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01 Jun 2011, 04:08
6*8*10-hence:
side 6*8 r=3 h=10
side 6*10 r=3 h=8
side 8*10 r=4 h=6

pir^2 will be bigger at side 8*10
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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13 Jun 2011, 00:15
l = 10,w = 8 and h = 6
gives max volume.

r = 4.
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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23 Nov 2011, 08:40
1
The above mentioned solutions are correct, but from my perspective the question could be misleading (at least for non-natives).
The question states "a cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on (ANY) one of its six faces".
Since the canister has a fixed volume and we cannot be sure on which face the wooden box will stand, we would have to assume that it could also stay on the smallest possible area, namely 6 by 8.
If so, the correct solution, 4, would be wrong, since the diameter would exceed the side length 6.
Sorry if I confused you guys, but I personally dislike such questions, since they inhabit the potential for incorrect choices only due to the unclear phrasing of the question.

My 2 cents.
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Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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15 Apr 2015, 02:48
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hb05sv wrote:
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

$$volume_{cylinder}=\pi{r^2}h$$

If the cylinder is placed on 6*8 face then it's maximum radius is 6/2=3 and $$volume==\pi*{3^2}*10=90\pi$$;
If the cylinder is placed on 6*10 face then it's maximum radius is 6/2=3 and $$volume==\pi*{3^2}*8=72\pi$$;
If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and $$volume==\pi*{4^2}*6=96\pi$$;

So, the maximum volume is for $$r=4$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-inside-dimensions-of-a-rectangular-wooden-box-are-128053.html
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Posts: 52911
Re: The inside dimensions of a rectangular wooden box are 6 inches by 8 in  [#permalink]

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15 Apr 2015, 02:48
Bunuel wrote:
hb05sv wrote:
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

$$volume_{cylinder}=\pi{r^2}h$$

If the cylinder is placed on 6*8 face then it's maximum radius is 6/2=3 and $$volume==\pi*{3^2}*10=90\pi$$;
If the cylinder is placed on 6*10 face then it's maximum radius is 6/2=3 and $$volume==\pi*{3^2}*8=72\pi$$;
If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and $$volume==\pi*{4^2}*6=96\pi$$;

So, the maximum volume is for $$r=4$$.

Similar questions to practice:
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a-rectangular-box-has-dimensions-12-10-8-inches-what-is-the-28790.html
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a-rectangular-box-has-dimensions-12-10-8-inches-what-is-the-28790.html
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# Why does mechanical equilibrium depend only on potential energy?

As far as I understand, for a system to be considered in equilibrium, the sum of the forces that is applied to it must be $0$:

$\vec F = 0$

which is

$\partial \frac{E_p}{\partial x}\bigg\rvert _{x=x_0} = 0$

But it only depends on the potential energy, what about kinetic energy?

• The derivative of kinetic energy with respect to position is zero anyway... After all, kinetic energy only depends on velocity. – Physicist137 Jun 7 '17 at 1:25

Equilibrium here just means that if a particle is placed at the location $x_0 = 0$ with $0$ kinetic energy, it will stay at that location. Imagine another location, say $x_1$, such that $\nabla V(x_1)\not = 0$, if a particle is placed there, it will start moving, if the kinetic energy is zero initially.
A simple example of this is a pendulum. Imagine a bar with ends points labeled $A$ and $B$ suspended from one of its end-points ($A$). If the bar is placed vertically with $B$ lower than $A$, the bar will stay at that configuration forever, unless you add kinetic energy. This is also the case if you place $B$ vertically above $A$. The difference between these two configurations is the type of equilibrium (stable vs unstable), but both of them are equilibrium points.


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# Tangent – Explanation & Examples

In the context of a right triangle, we can simply define the tangent function or any other trigonometric function using the terms hypotenuse, opposite, and adjacent in a right-angled triangle. Sounds interesting? Yes, it is. But, how can we define the tangent function using a right-angled triangle?

The tangent function is defined by determining the ratio of the length of the side opposite a reference angle (acute angle) of a right triangle to the length of the adjacent side of a right triangle.

After studying this lesson, we are expected to learn the concepts driven by these questions and be qualified to address accurate, specific, and consistent answers to these questions.

• What is a tangent function?
• How can we determine the formula for tangent function from a right-angled triangle?
• How can we solve actual problems using trigonometric functions?

This lesson aims to clear up any confusion you might have about the concepts involving the tangent function.

## What is tangent?

In the context of a triangle, the tangent function is the ratio of the opposite side to the adjacent side. For an angle $\alpha$, the tangent function is denoted by $\tan \alpha$. In other words, the tangent is a trigonometric function of any given angle.

The following figure 5-1 represents a typical right triangle. The lengths of the three legs (sides) of the right triangle are named $a$, $b$, and $c$. The angles opposite the legs of lengths $a$, $b$, and $c$ are named $\alpha$, $\beta$, and $\gamma$. The tiny square with the angle $\gamma$ shows that it is a right angle.

Use the diagram in Figure 5-1 to determine the tangent function from the perspective of the angle $\alpha$.

Looking at Figure 5-1, we can determine the tangent function from the right-angled triangle if we divide the length of the side opposite the reference angle $\alpha$ (acute angle) by the length of the adjacent side.

The following figure 5-2 represents a tangent function.

Looking at Figure 5-2, we can identify that the side of length $a$ is the opposite side that lies exactly opposite the reference angle $\alpha$, and the side of length $b$ is the adjacent side that lies right next to the reference angle $\alpha$. Thus,

Opposite = $a$

Adjacent = $b$

Therefore, the tangent of an angle $\alpha$ is

${\displaystyle \tan \alpha ={\frac {a}{b}}}$

Therefore, we conclude that the tangent function is the ratio of the opposite side to the adjacent side.

Tangent function from the perspective of the angle $\beta$

We should be cautious when we apply the terms opposite and adjacent because the meaning of these terms is dependent on the reference angle we are using.

The following figure 5-3 represents a typical right triangle from the perspective of the angle $\beta$.

You can observe that now the roles of the sides have been shifted.

Looking at Figure 5-3, it is clear now the length of the side $a$ is right next to the reference angle $\beta$, and the length of the side $b$  lies exactly opposite the reference angle $\beta$. Thus, in relation to the angle measuring $\beta$, now we have

Adjacent = $a$

Opposite = $b$

While the hypotenuse $c$ remains the same. This is why the hypotenuse is very special in a right-angled triangle.

The following figure 5-4 represents a tangent function from the perspective of the angle $\beta$.

Looking at Figure 5-4, we can identify that the side of length $b$ is the opposite side that lies exactly opposite the reference angle $\beta$, and the length of the side $a$  lies right next to the reference angle $\beta$. Thus,

Opposite = $b$

Adjacent = $a$

We know that the tangent function is the ratio of the opposite side to the adjacent side.

Therefore, the tangent of an angle $\beta$ is

${\displaystyle \tan \beta ={\frac {b}{a}}}$

## What is the formula for the tangent?

The following figure 5-5 illustrates a clear comparison of how we determined the ratios of tangent function from the perspective of both the angles$\alpha$ and $\beta$.


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# find an expression for W

• Jan 9th 2011, 04:40 AM
wilko
find an expression for W
$W=\int_{V_{1}}^{V_{2}}{3P}^{\frac{1}{2}}V^{2}\; dV$

If $PV\; =\; \mbox{C} (constant)$

Find an expression for W.
• Jan 9th 2011, 04:48 AM
alexmahone
Quote:

Originally Posted by wilko
$W=\int_{V_{1}}^{V_{2}}{3P}^{\frac{1}{2}}V^{2}\; dV$

If $PV\; =\; \mbox{C} (constant)$

Find an expression for W.

Substitute $P=\frac{C}{V}$ in the integral.


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# R - How to fix NbClust error with error message: “The TSS matrix is indefinite. There must be too many missing values.”

I would like to know how I can use clustering methods in R (in this case, Kmeans) if I have an "unkind" input matrix (I get this error log:

The TSS matrix is indefinite. There must be too many missing values. The index cannot be calculated.)

I could see that I might get this error if my matrix produces negative eigenvalues (like, here: https://stackoverflow.com/questions/20669596/nbclust-package-error), but what I'm missing is the "next step" part. I could see a suggestion was to "go back to the Data", but what should I do then? Is there any transformation or something that might help? (I'm pretty new to R and clustering in general...)

The Data I'm using are the result of a survey (which I briefly transformed and scaled via the scale function in R) so I was wondering if there were some algorithms or methods I could use in order to go on with my analysis (from literature I couldn't find great help). Or, if you think this is unfixable or simply non the best solution, do you have any other suggestion for clustering my data? What I'm willing to do is to identify some clusters of possible users/customers of some services, depending on their usual habits (e.g.: if they use many social networks they will be more likely to use chat/whatsapp/app to ask for bank account information - I have both the information of their social network usage and their ways of communicating with a "bank assistant").

The Dataset consists of 994 rows and 103 columns. Don't know if it may help, but the code is simply this:

Data2<- read.csv(...)
bDataScale <- scale(Data2)
nc <- NbClust(bDataScale, min.nc=2, max.nc=993, method="kmeans")

And I get:

Error in NbClust(bDataScale, min.nc = 2, max.nc = 993, method = "kmeans") : The TSS matrix is indefinite. There must be too many missing values. The index cannot be calculated.

Julia

P.S.: as it would be logical to expect, I get the same error also with the unscaled matrix.

The answer you linked seems to suggest that negative eigenvalues tend to crop up with larger values of max.nc. So maybe try reducing that to something reasonable? I don't know how you'd go about interpreting 993 clusters in any case.

• Yes, in fact in the first attempts I tried using less clusters (I tried putting max.cn = 5 and 15) though it didn't work properly at any case. What I did was to apply the "simple" kmeans r-method deciding by myself the number of clusters. Still... Not a satisfactory result since it is a little bit messy. Thanks anyway. – Julia_Engl Apr 6 '16 at 12:23
• If you can motivate your choice of k by gap statistic or silhouette plots, I think that'd be fine too – fanli Apr 6 '16 at 15:10


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# Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A) (i) Z = 17, A = 35 (ii) Z = 92, A = 233 (iii) Z = 4, A = 9

+1 vote
5.6k views
Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A)
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9

+1 vote
by (306k points)
selected


text:
Categories

# Hidden triangular inequality (PRMO Problem 23, 2019)

Problem

Let ABCD be a convex cyclic quadrilateral . Suppose P is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least .If {PA,PB,PC,PD} = {3,4,6,8}.What is the maximum possible area of ABCD?

Topic

Geometry

Contest

2019

Mathematical circle .

Sequential Hints

The first step is very clear, you have to some how locate the point P .

claim : The intersection of the diagonal is the required point .
Proof : suppose that there is a point P’ other than P (the intersection of diagonal) .
now try to apply triangular inequality in $\triangle$PBD & $\triangle$APC.
(do the rest yourself , and complete the proof)

so , the area of [ABCD]=$\frac{1}{2}$$\sin\theta$(PA*PB+PB*PC+PC*PD+PD*PA)
now, the area will maximum when $\sin\theta$=1
so , just put the value of PA, PB, PC,PD


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+7 (3412) 91 60 92

## Archive of Issues

Russia Orel
Year
2017
Volume
27
Issue
1
Pages
60-68
 Section Mathematics Title Methods of conformal mappings of polyhedra in $\mathbb{R}^3$ Author(-s) Radygin V.M.a, Polyanskii I.S.a Affiliations The Academy of Federal Security Guard Service of the Russian Federationa Abstract Methods necessary to solve problems of conformal mapping of polyhedra in $\mathbb{R}^3$ are developed. The results are obtained with the use of quaternion algebra and geometric representations. The direct and inverse conformal mappings are defined: those of the upper half-space onto the unit ball, those of a ball crescent onto the dihedral angle and those of dihedral and polyhedral angles onto the upper half-space. Solutions to the direct and inverse problems of conformal mapping of the polyhedrons onto the upper half-space are found using the results obtained. The solution to the direct problem of conformal mapping is based on the results of the Christoffel-Schwarz theorem. The solution of the inverse problem is obtained by the method of successive conformal mappings. In general, the one-to-one mappings obtained are based on the fact that, by the Liouville theorem, all conformal diffeomorphisms of any area in the space are the Möbius transformations. Keywords conformal mapping, polyhedron, dihedral angle, polyhedral angle, upper half-space UDC 517.54 MSC 30C20 DOI 10.20537/vm170106 Received 27 October 2016 Language Russian Citation Radygin V.M., Polyanskii I.S. Methods of conformal mappings of polyhedra in $\mathbb{R}^3$, Vestnik Udmurtskogo Universiteta. Matematika. Mekhanika. Komp'yuternye Nauki, 2017, vol. 27, issue 1, pp. 60-68. References Arkhipov N.S., Polyanskij I.S., Stepanov D.E. The barycentric method for the field analysis in a regular waveguide with arbitrary cross-section, Antenny, 2015, no. 1 (212), pp. 32-40 (in Russian). Polyansky I.S. Poisson barycentric coordinates for multivariate approximation of scalar potential within an arbitrary area (Part 1), Vestn. Sarat. Gos. Tekh. Univ., 2015, vol. 1, no. 1 (78), pp. 30-36 (in Russian). Polyansky I.S. Poisson barycentric coordinates for multivariate approximation of scalar potential within an arbitrary area (Part 2), Vestn. Sarat. Gos. Tekh. Univ., 2015, vol. 1, no. 1 (78), pp. 36-42 (in Russian). Fil'chakov P.F. Priblizhennye metody konformnykh otobrazhenii. Spravochnoe rukovodstvo (Approximate methods of conformal mappings. Handbook), Kiev: Naukova Dumka, 1964, 536 p. Radygin V.M., Polansky I.S. Modified method of successive conformal mappings of polygonal domains, Vestn. Tomsk. Gos. Univ., Mat. Mekh., 2016, no. 1 (39), pp. 25-35 (in Russian). DOI: 10.17223/19988621/39/3 Sudbery A. Quaternionic analysis, Mathematical Proceedings of the Cambridge Philosophical Society, 1979, vol. 85, issue 2, pp. 199-225. DOI: 10.1017/S0305004100055638 Ahlfors L. Möbius transformations in several dimensions, University of Minnesota, 1981. Translated under the title Preobrazovanie Mebiusa v mnogomernom prostranstve, Moscow: Mir, 1986, 112 p. Norden A.P. Teoriya poverkhnostei (Theory of surfaces), Moscow: Moscow: Gos. Izd. Tekh. Teor. Lit., 1956, 260 p. Lavrent'ev M.A., Shabat B.V. Metody teorii funktsii kompleksnogo peremennogo (Methods of the theory of functions of a complex variable), Moscow: Nauka, 1973, 736 p. Full text


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# Output voltage function (circuit-analysis)

What will the function of the output voltage be?

My work, and am I right?:

$$\text{V}_{\text{out}}(t)=\frac{\text{R}_9\text{V}_{\text{in}}(t)}{\text{R}_{10}}+\frac{1}{\text{R}_{11}\text{C}_1}\int\text{V}_{\text{in}}(t)\space\text{d}t+\text{R}_{12}\text{C}_2\cdot\frac{\partial\text{V}_{\text{in}}(t)}{\partial t}$$

• It's a PID block - what more to say? Feb 3, 2016 at 13:51
• Yes it is, but is my output voltage function right? Feb 3, 2016 at 13:52
• how have you accounted for the gain due to op amps 1 and 2? Feb 3, 2016 at 13:53
• No the gain isn't one, I named it number 1 Feb 3, 2016 at 13:54
• The gain is 0.75 by the look of it. Feb 3, 2016 at 13:56

Ok, so op amps 3, 4 and 5 form a PID block. Considering their respective output voltages as Vp, Vi and Vd:

$$V_p = -\frac{R_9}{R_{10}}V_{in}$$

$$V_i = -\frac{1}{R_{11}C_{1}}\int{V_{in}dt}$$

$$V_d = -R_{12}C_2\frac{dV_{in}}{dt}$$

Op amp 2 and resistors R3-8 are used to sum these voltages. Using superposition, one can solve this block's output (Vx). Let's say Vi and Vd are 0; R5, R7 and R8 are then in parallel and the voltage on the op amp '+' will be Vp/4.

$$V_{x(V_p)} = \left(\frac{R_3}{R_4} + 1\right)\frac{V_p}{4} = 0.75V_p$$

Therefore, with superposition, Vx is 0.75(Vp + Vi + Vd). The last op amp is just a regular negative gain circuit, and Vout = -Vx. Final answer:

$$V_{out} = 0.75\left(\frac{R_9}{R_{10}}V_{in} + \frac{1}{R_{11}C_{1}}\int{V_{in}dt} + R_{12}C_2\frac{dV_{in}}{dt}\right)$$

• Why the 0.75? Because Vi and Vd aren't 0 Feb 3, 2016 at 14:57
• Superposition is an analysis tool for linear circuits. It takes the basic property of a linear operator L: L(a+b+c) = L(a) + L(b) + L(c). Considering the circut "simmetry" between different inputs, when one solves for L(a) (special case when b and c are 0), there will be a similar answer for L(b) and L(c), and the final result will be the sum of such answers. Feb 3, 2016 at 15:09
• Oh yes I understand, but is there a way to make it 1, I've to change the value of the resistors but how? Feb 3, 2016 at 15:10
• (1 + R3/R4) should result in 4, not 3. R3 could be 30k, for example. Feb 3, 2016 at 15:12
• So, if I make R3 30k than it will be the ouput function that I supposed in my question?! Feb 3, 2016 at 15:17


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What is CCW torque?

How to determine the direction of torque. The direction of rotation can be clockwise (cw) or counterclockwise (ccw). These terms refer to the movement of hands on a clock (see Figure 5). In physics, the counterclockwise direction is defined as positive and clockwise is negative for rotational variables.

How do you calculate torque at 90 degrees?

Note that the torque is maximum when the angle is 90 degrees. A practical way to calculate the magnitude of the torque is to first determine the lever arm and then multiply it times the applied force. The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force.

What is torque in physics class 11?

Torque is the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.

How do you calculate tension and torque?

The Torque/Tension Equation is a method used to estimate the torque/tension relationship in an assembly. T = (K D P)/12 can be used to developt a torque value that will achieve a certain tension or clamp load. T = Torque (ft. -lbs.)

Why is CCW positive?

Clockwise Rotations (CW) follow the path of the hands of a clock. These rotations are denoted by negative numbers. Counterclockwise Rotations (CCW) follow the path in the opposite direction of the hands of a clock. These rotations are denoted by positive numbers.

What is r in torque formula?

The distance from the pivot point to the point where the force acts is called the moment arm, and is denoted by ‘r’. Note that this distance, ‘r’, is also a vector, and points from the axis of rotation to the point where the force acts.

What is the formula to calculate torque?

How is torque calculated? τ = F ⋅ r \tau = F\cdot r τ=F⋅rtau, equals, F, dot, r. The direction of the torque vector is found by convention using the right hand grip rule.

Why is torque maximum at 90 degrees?

At a 90 degrees angle, if the force is applied at the furthest point of a body’s centre of mass, the force applied is fully converted in angular acceleration.

Is torque equal to tension?

However, they use different means to accomplish it. “Torque” is simply a measurement of the twisting force required to spin the nut up along the threads of a bolt, whereas “Tension” is the stretch or elongation of a bolt that provides the clamping force of a joint.

How do you find torque without an angle?

The most basic way to calculate torque is to multiply the Newtons of force exerted by the meters of distance from the axis. There’s also a rotational version of this formula for 3-dimensional objects that uses the moment of inertia and angular acceleration.

What is mean by torque in physics?

torque, also called moment of a force, in physics, the tendency of a force to rotate the body to which it is applied.

What is state Hooke’s Law?

Mathematically, Hooke’s law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. The value of k depends not only on the kind of elastic material under consideration but also on its dimensions and shape.

Why torque is called moment of force?

Torque (also known as moment, or moment of force) is the tendency of a force to cause or change the rotational motion of a body. It is a twist or turning force on an object. Torque is calculated by multiplying force and distance. It is a vector quantity, meaning it has both a direction and a magnitude.

Is torque a unit of force?

Torque is a measure of how much a force acting on an object causes that object to rotate. Torque is defined as Force (F) x Distance (r) = Torque. The distance (r) is from the pivot point to the point where the force acts. The SI units of torque is a Newton-meter (N・m).

How do you calculate mass torque and distance?

To calculate load torque, multiply the force (F) by the distance away from the rotational axis, which is the radius of the pulley (r). If the mass of the load (blue box) is 20 Newtons, and the radius of the pulley is 5 cm away, then the required torque for the application is 20 N x 0.05 m = 1 Nm.

How do I calculate tension?

1. Tension formula is articulated as. T=mg+ma.
2. Tension Formula is made use of to find the tension force acting on any object. It is useful for problems.
3. Tension Solved Examples.
4. Problem 1: A 8 Kg mass is dangling at the end of a string.

Does clockwise tighten or loosen?

Typical nuts, screws, bolts, bottle caps, and jar lids are tightened (moved away from the observer) clockwise and loosened (moved towards the observer) counterclockwise in accordance with the right-hand rule.

What is the rule for 180 degree rotation?

The rule for a rotation by 180° about the origin is (x,y)→(−x,−y) .

Is anticlockwise rotation positive or negative?

An angle measured in clockwise direction is negative and an angle measured in anticlockwise direction is positive.

What is the right hand rule for torque?

Right Hand Rule for Torque To use the right hand rule in torque problems, take your right hand and point it in the direction of the position vector (r or d), then turn your fingers in the direction of the force and your thumb will point toward the direction of the torque.