# Introduction to the Method of Moments

The MoM is the default solver in Feko. A simple electrostatic example is used to convey the basics of the solver.

## The Charge Distribution of a Straight Wire at a Constant Electric Potential of 1 V.

The basic Yagi-Uda antenna shown in Figure 5 consists of a few straight wires. Consider the solution of the charge distribution of a single straight wire of length $l$ and diameter 2a shown in Figure 1.

According to 1, a linear electric charge distribution $\rho \left(r\text{'}\right)$ will create an electric potential $V\left(r\right)$ as follows:

(1)

where $r\text{'}$ represents the source coordinates and r denotes the observation coordinates, $dl\text{'}$ is the path of integration and R is the distance from any point on the source to the observation point which can also be written as
(2) $|r-r\text{'}|=\sqrt{{\left(x-x\text{'}\right)}^{2}+{\left(y-y\text{'}\right)}^{2}+{\left(z-z\text{'}\right)}^{2}}$
Note: Equation 1 is valid on the wire and in free space. This is the so-called "boundary condition" for this particular problem.

Even though the charge distribution on arbitrarily shaped objects are not generally known, the straight wire example is useful for an introduction to the MoM.

Assume the wire is charged to a constant electric potential of 1 V. For convenience, the wire is oriented parallel to the Z axis. To solve Equation 1 on a computer, the wire is divided into smaller segments and the charge distribution can be approximated as follows:

(3) $\rho \left(z\text{'}\right)=\sum _{n=1}^{N}{a}_{n}{g}_{n}\left(z\text{'}\right)$

The functions, ${g}_{n}\left(z\text{'}\right)$ , often referred to as basis functions, are chosen to accurately model the unknown quantity (here the charge on a wire segment) as well as for computational efficiency. For simplicity, constant functions over each segment are assumed. More specifically, each ${g}_{n}\left(z\text{'}\right)$ function is equal to 1 over one segment only, and zero elsewhere. The assumption of a constant function implies that the segment length should be short enough for this assumption to hold.
Note: A rule of thumb is to make segments $\frac{1}{10}$ th of a wavelength.

Therefore Equation 1 can be approximated as follows:

(4) $4\pi {\epsilon }_{0}=\sum _{n=1}^{N}{a}_{n}{\int }_{0}^{l}\frac{{g}_{n}\left(z\text{'}\right)}{\sqrt{{\left(z-z\text{'}\right)}^{2}+{a}^{2}}}dz\text{'}$

As shown in Figure 5, the wire is divided into N uniform segments where each segment is of length $\Delta =\frac{l}{N}$ .

Since Equation 1 is valid everywhere, z can be chosen to be located at fixed points, zm, on the surface of the wire segments with radii, a. This choice simplifies Equation 4 to only a function of z', allowing the calculation of the integral. Furthermore, since the wire was divided into N segments, Equation 4 can be written as one equation with N unknowns (an) as follows:

(5) $\begin{array}{l}4\pi {\epsilon }_{0}={a}_{1}{\int }_{0}^{\Delta }\frac{{g}_{1}\left(z\text{'}\right)}{R\left({z}_{m},z\text{'}\right)}dz\text{'}+{a}_{2}{\int }_{\Delta }^{2\Delta }\frac{{g}_{2}\left(z\text{'}\right)}{R\left({z}_{m},z\text{'}\right)}dz\text{'}+\dots \\ +{a}_{n}{\int }_{\left(n-1\right)\Delta }^{n\Delta }\frac{{g}_{n}\left(z\text{'}\right)}{R\left({z}_{m},z\text{'}\right)}dz\text{'}+{a}_{N}{\int }_{\left(N-1\right)\Delta }^{l}\frac{{g}_{N}\left(z\text{'}\right)}{R\left({z}_{m},z\text{'}\right)}dz\text{'}\end{array}$

An equation of N unknowns requires N equations where each equation stands linearly independent from each other. These N equations can be constructed by selecting the observation points zm in the centre of each segment of length $\Delta$ as shown in Figure 2.
Note: The selection of observation points is denoted “testing” or “sampling” $V\left(z,\rho =a\right)$ and the method is referred to as “point-matching” or “collocation”.
Performing the selection of points N times reduces Equation 5 to the following:
(6) $\begin{array}{l}4\pi {\epsilon }_{0}={a}_{1}{\int }_{0}^{\Delta }\frac{{g}_{1}\left(z\text{'}\right)}{R\left({z}_{1},z\text{'}\right)}dz\text{'}+\dots +{a}_{N}{\int }_{\left(N-1\right)\Delta }^{l}\frac{{g}_{N}\left(z\text{'}\right)}{R\left({z}_{1},z\text{'}\right)}dz\text{'}\\ ⋮\\ 4\pi {\epsilon }_{0}={a}_{1}{\int }_{0}^{\Delta }\frac{{g}_{1}\left(z\text{'}\right)}{R\left({z}_{N},z\text{'}\right)}dz\text{'}+\dots +{a}_{N}{\int }_{\left(N-1\right)\Delta }^{l}\frac{{g}_{N}\left(z\text{'}\right)}{R\left({z}_{n},z\text{'}\right)}dz\text{'}\end{array}$

Equation 6 can be more readily written in matrix form as:

(7) $\left[{V}_{m}\right]=\left[{Z}_{mn}\right]\left[{I}_{n}\right]$

In Equation 7 each Zmn term can be written as:

(8) ${Z}_{mn}={\int }_{\left(n-1\right)\Delta }^{n\Delta }\frac{1}{{\left({z}_{m}-z\text{'}\right)}^{2}+{a}^{2}}dz\text{'}$

In addition, we can write the remaining two terms:

(9) $\left[{I}_{n}\right]=\left[{a}_{n}\right]$
(10) $\left[{V}_{m}\right]=\left[4\pi \epsilon 0\right]$

The Vm matrix consists of 1 row and N columns and all entries are equal to $4\pi {\epsilon }_{0}$ . The an values are the unknown coefficients for the charge distribution. To solve Equation 7, the matrix requires inversion where
(11) $\left[{I}_{n}\right]={\left[{Z}_{mn}\right]}^{-1}\left[{V}_{m}\right]$
Note: A well-known and computationally cheaper inversion procedure, LU decomposition, is followed. The matrix is factored into an upper and lower triangular matrix. Then a process similar to Gaussian elimination is followed to solve the matrices.
Figure 3 shows the line charge density for a wire of length 1 m discretized into 50 segments.

For more complex problems, the integrals cannot be reduced to approximations such as those made here.

1 Advanced Engineering Electromagnetics, Second Edition, Constantine A. Balanis, p. 680