Introduction to numerical techniques in electromagnetics
In the earlier chapters, we considered the analytical techniques for solving
electromagnetic field problems and obtained solution in the close form. In obtaining such
closed form solution we are often required to make certain assumptions so that analysis is
simplified. However, while solving practical problems when analytical solutions become
extremely complex or intractable, we resort to non analytical methods. With the advent of
fast digital computers, numerical methods for solving EM field problems have become
very popular. These are many numerical techniques which have been developed over the
last few decades. In this section we shall briefly discuss two popular methods for solving
EM field problems.

Finite difference method (FDM)

Method of moments (MOM)
Finite difference method
The finite difference method is a powerful numerical method for solving partial
differential equations. In applying the method of finite differences a problem is defined
by:



A partial differential equation such as Poisson's equation
A solution region
Boundary and/or initial conditions.
An FDM method divides the solution domain into finite discrete points and replaces the
partial differential equations with a set of difference equations. Thus the solutions
obtained by FDM are not exact but approximate. However, if the discritisation is made
very fine, the error in the solution can be minimized to an acceptable level. Although an
electromagnetic field produces 3-D variations, for the sake of simplicity we shall restrict
our discussion to 2-D case only.
The Poisson's equation in 3-D is given by
......................................................................... (8.1)
For 2-D case, equation (8.1) simplifies to
............................................................... (8.2)
In applying the methods of finite differences, we define the solution region into a finite
number of meshes as shown in Fig 8.1.
Fig 8.1: Division of solution region into grid points
The meshes can be various shapes; we shall only consider the rectangular and square
meshes only.
First we consider a mesh configuration having five nodes and unequal arms as shown in
the Fig. 8.2.
Fig 8.2: A mesh with unequal arms
With reference to Fig 8.1, corresponds to the voltage Vij. For the five node mesh
configuration of Fig 8.2, the voltages are defined as:
............................................................................... (8.3a)
..................................................................... (8.3b)
.................................................................... (8.3c)
.................................................................... (8.3d)
.................................................................... (8.3e)
Let , , and represent the midpoint of the arms as shown in Fig 8.2. In order to
replace the Poisson equation (8.2) by difference equations, we obtain the approximate
first derivatives at the points
second derivative.
to
The first derivative at
are
and
and use these first derivatives to approximate the
.................................................................... (8.4a)
.................................................................... (8.4b)
............................................ (8.5)
In the same manner,
............................................ (8.6)
Using (8.5) and (8.6), equation (8.2) can be written as
or
........... (8.7)
If we consider the case of uniform grid, i.e.
then we can write
.............................................................. (8.8)
Further, for Laplace equation,
and equation (8.8) simplifies to
.....................................................(8.9)
Thus we see that voltage at the central node is the mean of the voltages at the other four
nodes.
With reference to Fig 8.1, equation (8.8) can be written as
....................(8.10)
Equation (8.8) & equation (8.9) can be used to solve Poisson's and Laplace's equation
respectively when uniform grids are used. These equations, along with the specified
boundary conditions can be used to solve a given problem in two ways, the same is
illustrated with the help of an example.
Example :
The figure 8.3 shows a region boundary four sides and the potential at each side is
specified.
Fig 8.3: A square region and division of solution region into grid points
We are interested to determine the potential at the internal nodes masked 1 to 4.
is assumed to be zero. We can determine the potentials at the internal nodes by using
two methods which are explained as follows.
Interactive Method:
In this method we start by setting the initial values of the potentials at the internal nodes
to be zero (or any other reasonable guessed initial values).
The potential of these nodes are modified iteratively as explained in the Fig 8.4 shown
below till the values attain a prescribed degree of accuracy.
Fig 8.4: Illustration of iteration method
In Fig 8.4, we have shown how the node values changes in first three iterations.
Band Matrix Method :
Let the voltages of the internal nodes 1 to 4 be represented by
Using equation (8.9) we can write
....................................................(8.11a)
..............................................................(8.11b)
....................................................................(8.11c)
....................................................................(8.11d)
The above equations can be written as
....................................................(8.12a)
..................................................(8.12b)
....................................................(8.12c)
....................................................(8.12d)
Equation (8.12) can be written in a matrix form
i.e.
....................................................(8.13)
The node voltages can be obtained by using
....................................................(8.14)
Method of Moments
Basic Concepts of the Method of Moments
The method of moment (MoM) is a numerical procedure for solving linear operator
equation by transforming it into a system of simultaneous linear algebraic equation,
referred to as matrix equation. Many problems in electromagnetics can be cast in the
form of integral equations. An integral equation is one in which the unknown function
appears in the integrand. The MoM provides a way to solve such integral equations.
Instead of discussing the general procedure for MoM, here we introduce the basic
concepts involved in MoM solution through some examples, both for static as well as
time harmonic electromagnetic fields.
To start with, let us consider the an example of determining the electrostatic potential due
to an isolated charged conducting plate, 2a meters on a side and lying on the
plane
with its center at the origin as shown in figure 8.5.
Fig. 8.5 : A square conducting charged plate
The plate is assumed to have zero thickness. Let
density on the plate
represent the surface charge
Further, the boundary condition on the potential is
=constant on the plate. For the
charged conducting plate, the potential
at any point
can be written as:
............................................ (8.15)
If
is known (which is often assumed in solving simple electrostatic problems, i. e.
charged density is specified), the potential function can be computed directly. But in
practical problems, often the charge distribution
is not known. The equation
(8.15) is an example of integral equation as the unknown
appears under the
integral. To solve the unknown charge distribution we apply the method of moments. The
procedure is explained below:
We subdivide the plate into
denoted by
squares of side
as shown in figure 8.5. The
square is
and
We approximate the charge distribution as:
.......................................................... (8.16)
where
Here, the functions
are called the expansion functions or basis functions and
are the coefficients. Basis functions can be of different types; here we have considered
pulse basis functions, which have unit magnitude over some domain and are zero
everywhere else. With sufficiently large, equation (8.16) closely approximates the
actual
. To solve for the charge distribution
unknown coefficients
approximately, the
are to be determined. From the given boundary condition we
note that on the surface of the plate
and this condition can be used to
determine the unknown coefficients
Using the approximate charge distribution given by equation (8.16), let us now evaluate
the equation (8.15) at the mid point
at the midpoint of
is given by:
............................................. (8.17)
The above equation can be written as
................................................................................... (8.18)
where
is the potential at the center of
due to unit charge placed on
.
For m = n
..................................... (8.19)
For
, treating the unit charge on
of
as a point charge located at the mid point
,
......................................................... (8.20)
As
, considering the potentials at all the
sub sections we can write
Or,
................................................................................(8.21)
The unknown coefficients
s can be computed as
..............................................................................(8.22)
Figure 8.6 shows the charge distribution obtained using the MoM technique for
m.
Fig. 8.6: Approximate charge density along the side of the plate
Another parameter of interest is the capacitance of the plate.
........................................................ (8.23)
With
known, the capacitance of the plate can be approximated as:
........................................................................ (8.24)
,
Based on the discussions we had so far, we summarize the steps involved in MoM
solution.
We consider an inhomogeneous equation
.................................................................................... (8.25)
where L is a linear operator, g is a known function (excitation) and f is the unknown
function to be determined.
If we consider our previous example,
We expand f in a series of functions
........................................................................... (8.26)
Where
are constant. The set
For exact solution
is called the expansion function or basis function.
, but in practice truncated to a finite value.
............................................................................(8.27)
We define a set of weighting functions or testing functions
and take the inner product of equation (8.25) with each of the
............................................................(8.28)
A scalar product
is defined to be a scalar satisfying
..........................................................................(8.29a)
in the range of L
...............................................(8.29b)
...............................................................(8.29c)
..............................................................(8.29d)
Here b and c are scalars and * indicates complex conjugation.
The inner product corresponding to our previous example is of the form
...................................................(8.30)
Similarly, the testing function for our previous example is
..............................................................(8.31)
i.e. the testing functions are Dirac delta functions. Such choice of testing function is
called point matching.
The equation (8.26) can be reduced to
............................................................................(8.32)
where
If
is non-singular we can write
............................................................................. (8.33)
and f can be calculated by using equation (8.24).
In our example we have used pulse type basis function and point matching, that is, Dirac
delta function as testing or weighting function. In general the basis functions may be subdomain basis functions (where the structure is subdivided into N non-overlapping
segments. The basis functions are defined in conjunction with limits of one or more
segments) or entire domain basis functions (which exist over the full domain). In
particular, when
, this case is known as Galerkin's method.
Method of Moment for Wire Antennas and Wire Scatterers
In the previous sections we have seen how MoM can be used to convert integral
equations in to a set of linear equations (Matrix equations), which may be solved by
numerical techniques. In this section, we consider application of MoM techniques to wire
antennas and scatterers. Antennas can be distinguished from scatterers in terms of the
location of the source. If the source is on the wire, it is regarded as antenna. When the
wire is far from the source it acts as scatterer. For the wire objects (antenna or scatterer)
we require to know the current distribution accurately. Integral equations are derived and
solved for this purpose. If the feed voltage to an antenna is known and the current
distribution corresponding to this feed is found out, other antenna parameters such as
impedance, radiation pattern etc can be calculated. Similarly, when a wave impinges
upon surface of a wire scatterer, it induces current density, which in turn is used to find
the scattered field.
Let us consider a perfectly conducting wire of length and radius such that
and , the wavelength corresponding to the operating frequency. We consider the wire
to be a hollow metal tube open at both ends. Let us assume that an incident wave
impinges on the surface of a wire. When the wire is an antenna, the incident field is
produced by the feed at the gap as shown in Fig. 8.7.
Figure 8.7 : A wire antenna
The induced current density produces an electric field
. The total electric field is
. Since the wire is assumed to be perfectly conducting, tangential
component of the total field on the surface of the wire is zero. For a cylindrical wire
placed along z-axis we can write
............................................................................... (8.34)
that is,
..........................................................................................(8.35)
We know from equation (7.4) & (7.6)
and
For a thin cylinder, the current density can be considered to be independent of
so that
..........................................................................(8.36)
where
is the surface current density at a point
on the conductor.
The current
may be assumed to be a filamentary current located parallel to z-axis at
a distance a shown in the Fig. 8.8.
Figure 8.8: Equivalent current
For the current flowing only in the z direction,
...........................................................(8.37)
and
.........................................................(8.38)
Therefore,
.....................................(8.39)
We know that,
From (8.36),
..........................................................(8.40)
where
In cylindrical coordinates,
where
is the radial distance to the observation point.. Because of the symmetry, the
observations are not function of
and
.
For
.......................................................(8.41)
where
Therefore, we can write
.
.........................................................(8.42)
where
For
If
,
is the field at the observation point
caused by a unit point source placed at
, then the field at by a source distribution
is the integral of
over the
range of occupied by the source. The function G is called the Green's function.
is the free space Green's function.
We have,
and
.............................................(8.43)
From the above two equations we can write,
................................................(8.44)
This electric field is the field due to current
[which results because of the impressed
or source field] and this field can be written as the scattered field.
Therefore,
........................................(8.45)
Since
...............................(8.46)
This equation is called the Pocklington's Integrodifferential equation.
Simplification of Pocklington's equation :
If the observation point is on the axis of the wire,
put in a more convenient form
, the preceding equation can be
...............................(8.47)
+ J. H. Richmond, “Digital Computer Solution of the Rigorous Equations for Scattering Problems”,
Proceedings IEEE, 1965, pp 796-804.
It may be noted that the axial field intensity is made to vanish on the axis of the wire, for
a slender wire such approximation is valid.
Method of Moment Formulation
We can now apply MoM formulation to above integral equation. We divide the wire in to
N segments. Let us consider pulse basis function and express the current as
Where
is the length of the segment
segment.
Let us denote the integral equation in (8.47) by
Substituting
and evaluating at
figure, we can write
Figure 8.9 : Evaluation of field
Putting
For
Or in the compact form, we can write
(middle of the mth segment) as shown in the
Multiplying the elements on both sides by
, we can write
can be computed by matrix inversion.
Source modeling
The impressed field
is required to be known on the surface of the wire. The
simplest form of excitation is a delta-gap excitation. Using the delta gap, it is assumed
that the excitation voltage at the feed terminal is constant and zero else where. Therefore,
the incident field is also constant over the feed gap and zero elsewhere. The other
commonly used source is a magnetic frill generator where the feed gap is replaced with a
circumferentially directed magnetic current density existing over an annular aperture. The
inner radius of the aperture is that of the wire and outer radius is found from the
characteristic impedance of the transmission line feeding the antenna. For this frill
genetor the electric field is found on the surface of the wire.
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