10 slides

Dr. Rakhesh Singh Kshetrimayum
10. Method of Moments
Dr. Rakhesh Singh Kshetrimayum
1/3/20141 Electromagnetic FieldTheory by R. S. Kshetrimayum

10.1 Introduction
learn how to use method of moments (MoM) to solve
electrostatic problems
advanced & challenging problems in time-varying fields
brief discussion on the basic steps of MoM
solve a simple differential equation using MoM
1/3/2014Electromagnetic FieldTheory by R. S. Kshetrimayum2
solve a simple differential equation using MoM
in order to elucidate the steps involved
MoM tfor 1-D and 2-D electrostatic problems
MoM for electrodynamic problems

10.2 Basic Steps in Method of Moments
Method of Moments (MoM) transforms
integro-differential equations into matrix systems of linear equations
which can be solved using computers
Consider the following inhomogeneous equation
( ) kuL =
where L is a linear integro-differential operator,
u is an unknown function (to be solved) and
k is a known function (excitation)
( ) kuL =
( ) 0=−⇒ kuL

For example,
(a) consider the integral equation for a line charge density
Then
0
0
( ') '
4 ( , ')
x dx
V
r x x
λ
πε
= ∫
( )
Then ( )'
u xλ=
0k V=
0
'
4 ( , ')
dx
L
r x xπε
= ∫

(b) consider the differential equation of the form
Then
2
2
2
( )
3 2
d f x
x
dx
− = +
( )u f x=
( )u f x=
2
3 2k x= +
2
2
d
L
dx
= −

To solve u, approximate it by sum of weighted known
basis functions or
expansion functions
as given below
N N
≅ = =∑ ∑
where is the expansion function,
is its unknown complex coefficients to be determined,
N is the total number of expansion functions
1 1
, 1,2,...,
N N
n n n
n n
u u I b n N
= =
≅ = =∑ ∑
nb
nI

Since L is linear, substitution of the above equation in the
integro-differential equation,
we get,
1
N
n n
n
L I b k
=
 
≈ 
 
∑
where the error or residual is given by
1n=
 
 
∑
1
N
n n
n
R k L I b
=
 
= −  
 
∑

Mathematicians name this method as Method ofWeighted
Residuals
Why?
Next step in MoM
Enforcing the boundary condition
Enforcing the boundary condition
Make inner product of the above equation with each of the
testing or
weighting functions
should make residual or error zero

By replacing u by
where n=1,2,…,N
taking inner product with a set of
weighting or
nu
mw
weighting or
testing functions
in the range of L, we have,
( )( ), 0, 1,2,...,m nw L u k m M− = =

Since In is a constant
we can take it outside the inner product and
write
M and N should be infinite theoretically
but practically it should be a finite number
( )
1
, , , 1,2,...,
N
n m n m
n
I w L b w k m M
=
= =∑

Note that a scalar product is defined to be a scalar
satisfying
, , ( ) ( )w g g w g x w x dx= = ∫
wgcwfbwcgbf ,,, +=+
gw,
b and c are scalars and * indicates complex conjugation
wgcwfbwcgbf ,,, +=+
00,* ≠> gifgg 00,* == gifgg

In matrix form
with each matrix and vector defined by
[ ][ ] [ ]VIZ =
[ ] [ ]T
NIIII ...21= [ ] 1 2, , ... ,
T
MV k w k w k w =  
[ ]
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
1 1 1 2 1
2 1 2 2 2
3 1 3 2 3
1 2
, , ,
, , ... ,
, , ,...
, , ... ,
N
N
N
M M M N
w L b w L b w L b
w L b w L b w L b
Z w L b w L b w L b
w L b w L b w L b
 
 
 
 
=  
 
 
 
 
K
OM M M

For [Z] is non-singular,
Solve the unknown matrix [I] of amplitudes of basis function
as
Galerkin’s method
[ ] [ ] [ ] [ ][ ]
1
I Z V Y V
−
= =
Galerkin’s method
Point matching or Collocation
The testing function is a delta function
nn wb =

Methods for calculating inverse of a matrix
Seldom find the inverse of matrix directly , because,
if we have ill-conditioned matrices,
it can give highly erroneous results
MATLAB command‘pinv’ finds pseudo inverse of a matrix
[ ]
1
Z
−
MATLAB command‘pinv’ finds pseudo inverse of a matrix
using the singular value decomposition
For a matrix equation of the form AX=B,
if small changes in B leads to large changes in the solution X,
then we callA is ill-conditioned

The condition number of a matrix is the
ratio of the largest singular value of a matrix to the smallest
singular value
Larger is this condition value
closer is the matrix to singularity
closer is the matrix to singularity
It is always
greater than or equal to 1
If it is close to one,
the matrix is well conditioned
which means its inverse can be computed with good accuracy

If the condition number is large,
then the matrix is said to be ill-conditioned
Practically,
such a matrix is almost singular, and
the computation of its inverse, or
the computation of its inverse, or
solution of a linear system of equations is
prone to large numerical errors
A matrix that is not invertible
has the condition number equal to infinity

Sometimes pseudo inverse is also used for finding
approximate solutions to ill-conditioned matrices
Preferable to use LU decomposition
to solve linear matrix equations
LU factorization unlike Gaussian elimination,
LU factorization unlike Gaussian elimination,
do not make any modifications in the matrix B
in solving the matrix equation

Try to solving a matrix equation
using LU factorization
First express the matrix
[ ] [ ][ ]A L U=
[ ][ ] [ ]A X B=
[ ] [ ]
11 11 12 13 1
21 22 22 23 2
31 32 33 33 3
1 2 3 1
0 0 0
0 0 0
0 0 0
... 0 0 0 ...
N
N
N
N N N N NN
l u u u u
l l u u u
L l l l and U u u
l l l l u
   
   
   
   = =
   
   
      
L L
L L
L L
M M M O M M M M O M

through the forward substitution
[ ][ ][ ] [ ] [ ][ ] [ ]L U X B L Y B∴ = ⇒ =
1
1
1
1
; , 1
i
i i ik k
b
y y b l y i
−
 
= = − > ∑
through the backward substitution
1
111
; , 1i i ik k
kii
y y b l y i
l l =
= = − > 
 
∑
[ ][ ] [ ]U X Y=
1
1
; ,
N
N
N i i ik k
k iNN ii
y
x x y u x i N
u u = +
 
= = − < 
 
∑

This is more efficient than Gaussian elimination
since the RHS remain unchanged during the whole process
The main issue here is to
find the lower and upper triangular matrices.
MATLAB command for LU factorization of a matrix A is
MATLAB command for LU factorization of a matrix A is
[L U] = lu(A)

Example 10.1
Consider a 1-D differential equation
subject to the boundary condition f(0)=f(1)=0
2
2
2
( )
3 2
d f x
x
dx
− = +
subject to the boundary condition f(0)=f(1)=0
Solve this differential equation using Galerkin’s MoM
Solution:
Note that for this case,
( )u f x=

According to the nature of the known function ,
it is natural to choose the basis function as
2
3 2k x= +
2
2
d
L
dx
= −
2
3 2k x= +
( ) n
b x x=
it is natural to choose the basis function as
However,
the boundary condition f(1)=0
can’t be satisfied with such a basis function
( ) n
nb x x=

A suitable basis function for this differential equation
taking into account of this boundary condition is
Assume N=2 (the total number of subsections on the
( ) 1
; 1,2,...,n
nb x x x n N+
= − =
Assume N=2 (the total number of subsections on the
interval [0,1])
Approximation of the unknown function
( ) ( )2 3
1 1 2 2 1 2( ) ( ) ( )f x I b x I b x I x x I x x≅ + = − + −

For Galerkin’s MoM, the weighting functions are
Choosing a square [Z] matrix where M=N=2
( ) 1
; 1,2,...,m
mw x x x m M+
= − =
( ) ( ) ( )
1 1
2
11 1 1 1 1
0 0
1
, ( ) ( ) ( ) 2
3
Z w L b w x L b x dx x x dx= = = − =∫ ∫
( ) ( ) ( )
1 1
2
12 1 2 1 2
0 0
1
, ( ) ( ) ( ) 6
2
Z w L b w x L b x dx x x x dx= = = − =∫ ∫

( ) ( ) ( )
1 1
3
21 2 1 2 1
0 0
1
, ( ) ( ) ( ) 2
2
Z w L b w x L b x dx x x dx= = = − =∫ ∫
( ) ( ) ( )
1 1
3
22 2 2 2 2
0 0
4
, ( ) ( ) ( ) 6
5
Z w L b w x L b x dx x x x dx= = = − =∫ ∫
( )
1 1
2 2
1 1 1
0 0
3
, ( ) ( ) 3 2 ( )
5
V k w k x w x dx x x x dx= = = + − =∫ ∫
( )
1 1
2 3
2 2 2
0 0
11
, ( ) ( ) 3 2 ( )
12
V k w k x w x dx x x x dx= = = + − =∫ ∫

Therefore,
[ ][ ] [ ] 1
2
1 1 3
3 2 5
1 4 11
2 5 12
I
Z I V
I
   
    
= ⇒ =    
    
     
[ ] 1
13
10I
 
  
⇒ = =
The unknown function f(x)
[ ] 1
2
10
1
3
I
I
I
  
⇒ = =   
   
  
( ) ( ) ( ) ( )2 3 2 3
1 2
13 1
( )
10 3
f x I x x I x x x x x x≅ − + − = − + −

The above function satisfies the given boundary conditions
f(0)=f(1)=0
The analytical solution for this differential equation is
2 45 3 1
( )
3 2 6
f x x x x= − −
Check whether the above solution using MoM is
different from the analytical solution obtained by direct
integration (see Fig. 10.1)
3 2 6

Fig. 10.1 Comparison of exact solution (analytical) and approximate
solution (MoM) of Example 10.1

10.3 Introductory examples from electrostatics
In electrostatics, the problem of finding the potential
due to a given charge distribution is often considered
In practical scenario, it is very difficult to
specify a charge distribution
We usually connect a conductor to a voltage source
and thus the voltage on the conductor is specified
We will consider MoM
to solve for the electric charge distribution
when an electric potential is specified
Examples 2 and 3 discuss about calculation of inverse using LU
decomposition and SVD

1-D Electrostatic case:Charge density of a straight wire
Consider a straight wire of length l and radius a (assume
a<<l),
placed along the y-axis as shown in Fig. 10.2 (a)
The wire is applied to a constant electric potential of 1V
The wire is applied to a constant electric potential of 1V
Choosing observation along the wire axis (x=z=0) i.e.,
along the y-axis
and representing the charge density on the surface of the wire
∫=
l
yyR
dyy
0
'
''
0 ),(
)(
4
1
1
λ
πε

∆

Fig. 10.2
(a) Straight wire of length l and radius a applied with a
constant potential of 1V
(b) Its segmentation: y1, y2, …, yN are observation points and
r′ shows a source point
r′ shows a source point
(c) Division of the charged strip into N sections

where
It is necessary to solve the integral equation
to find the unknown function λ(y′)
' ' ' 2 ' 2 ' 2 ' 2 2
0
( , ) ( , ) ( ) ( ) ( ) ( ) ( )
x z
R y y R r r y y x z y y a
= =
= = − + + = − +
r r
to find the unknown function λ(y′)
The solution may be obtained numerically by
reducing the integral equation into a series of linear algebraic
equations
that may be solved by conventional matrix techniques

(a)Approximate the unknown charge density λ(y′)
by an expansion of N known basis functions with unknown
coefficients
∑
=
=
N
n
nn ybIy
1
'' )()(λ
Integral equation after substituting this is
∫∫ ∑
∑
=
= ==
l
n
l N
n
n
N
n
nn
yyR
dyyb
I
yyR
dyybI
0
'
''
0 1
'
'
1
'
0
),(
)(
),(
)(
4πε

Now we have divided the wire into N uniform segments each
of length ∆ as shown in Fig. 10.2 (b)
We will choose our basis functions as pulse functions
( ) 


 ∆≤≤∆−
=
otherwise
nynfor
ybn
'
' )1(
0
1
b)Applying the testing or weighting functions
Let us apply the testing functions as delta functions
for point matching
 otherwise0
( )my y ∂ − 

Integration of any function with this delta function
will give us the function value at
Replacing observation variable y by a fixed point such as ym,
results in an integrand that is solely a function of y′
so the integral may be evaluated.
my y=
so the integral may be evaluated.
It leads to an equation with N unknowns
∫ ∫ ∫∫
∆
∆
∆
∆− ∆−
∆
+++++=
2
)1( )1(
'
''
'
''
'
''
2
2
0
'
''
1
10
),(
)(
...
),(
)(
...
),(
)(
),(
)(
4
n
n
l
N m
N
N
m
n
n
mm yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
Iπε

Solution for these N unknown constants,
N linearly independent equations are required
N equations may be produced
by choosing an observation point y on the wire and
by choosing an observation point ym on the wire and
at the center of each ∆ length element
as shown in Fig. 10.2 (c)
Result in an equation of the form of the previous equation
corresponding to each observation point

For N such observation points we have
∫ ∫ ∫∫
∆
∆
∆
∆− ∆−
∆
+++++=
2
)1( )1(
'
1
''
'
1
''
'
1
''
2
2
0
'
1
''
1
10
),(
)(
...
),(
)(
...
),(
)(
),(
)(
4
n
n
l
N
N
N
n
n
yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
Iπε
∫ ∫ ∫∫
∆ ∆∆ 2 ''''''
2
''
1 )()()()(
n l
Nn dyybdyybdyybdyyb
∫ ∫ ∫∫
∆
∆
∆
∆− ∆−
∆
+++++=
2
)1( )1(
'
2
''
'
2
''
'
2
''
2
2
0
'
2
''
1
10
),(
)(
...
),(
)(
...
),(
)(
),(
)(
4
n
n
l
N
N
N
n
n
yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
Iπε
∫ ∫ ∫∫
∆
∆
∆
∆− ∆−
∆
+++++=
2
)1( )1(
'
''
'
''
'
''
2
2
0
'
''
1
10
),(
)(
...
),(
)(
...
),(
)(
),(
)(
4
n
n
l
N N
N
N
N
n
n
NN yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
I
yyR
dyyb
Iπε

(c)We may write the above equations in matrix form as
[ ][ ] [ ]
11 12 1 1 0
21 22 2 2 0
31 32 3 3 0
...
...
...
N
N
N mn n m
Z Z Z I V
Z Z Z I V
Z Z Z I V Z I V
Z Z Z I V
     
     
     
     = ⇒ =
     
     
     
K
OM M M M M
where
1 2 0...N N NN NZ Z Z I V          
[ ] [ ]04πε=mV
1
1 1
' ' '
' 2 2 ' 2 2
0
' '
'' 2
( )
( ) ( )
( )
n
n
n n
n n
yl
n
mn
ym m
y y
m m ny ym
b y dy dy
Z
y y a y y a
dy dy
for m n
y y y yy y
−
− −
= =
− + − +
∆
≅ = ≈ ≠
− −−
∫ ∫
∫ ∫

Special care for calculating the Zmn for m=n case
since the expression for Zmn is infinite for this case
Extraction of this singularity
Substitute ' '
my y d dyξ ξ− = ⇒ = −
( )
0
2 2
2 2 2 2 00
log ( )
( ) ( )
mn
d d
Z a
a a
ξ ξ
ξ ξ
ξ ξ
∆ ∆
∆
= − = = + +
+ +
∫ ∫
2 2
ln
a
a
 ∆ + ∆ +
=  
  

Self or diagonal terms are the
most dominant elements in the [Z] matrix
Note that linear geometry of this problem
yields a matrix that is symmetric toeplitz, i.e.,
[ ]
11 12 1
12 11 1 1
1 1 1 11
...
...
. . . .
...
N
N
mn
N N
Z Z Z
Z Z Z
Z
Z Z Z
−
−
 
 
 =
 
 
 

All other rows are a rearranged version of the first row
Required to calculate the first row of the matrix only
Remaining elements can be obtained by the rearrangement
formula:
, 2, 1Z Z m n= ≥ ≥
Therefore the unknown [I] matrix could be solved as
1, 1
, 2, 1mn m n
Z Z m n− +
= ≥ ≥
[ ] [ ] [ ]mmnn VZI 1−=

Fig. 10.3 (a) Convergence plot of Z11 and Z21 (b) Plot of line charge density of
the wire (MATLAB program provided in the book)

Let us see the convergence of these two types of elements of
the Z matrix say,
Z11 and Z21
Fig. 10.3 (a) shows the convergence plot of two elements of
the Z matrix
the Z matrix
for number of sub-sections varying from 5 to 100
The graph of Z21 (dashed line) versus number of sub-sections
is a straight line
so any number of sub-sections between 5 and 100 should give
the same result

But the graph of Z11 versus number of sub-sections is
decreasing quite fast at the initial values of number of sub-
sections and
it is decreasing more slowly for larger values of number of sub-
sections
sections
This shows that at
higher values of number of sub-sections,
we will get a more convergent result
Choose the maximum number of sub-sections and
plot the line charge density as depicted in the Fig. 10.3 (b)

See the condition number of the [Z] matrix in order to see
whether the [Z] matrix is well-behaved or not
The condition number of [Z] matrix
(=7.1409) for maximum number of sub-sections is good
No problem in taking the inverse
No problem in taking the inverse
Fig. 10.3 (b) line charge density is
maximum at the two end points of the wire and
minimum at the center of the wire
2-D Electrostatic case: Charge density of a square conducting
plate discussed in the book

10.4 Some commonly used basis functions
The weighted sum of basis functions is
used to represent the unknown function in MoM
Choose a basis function that reasonably approximates
the unknown function over the given interval
Basis functions commonly used in antenna or scattering
Basis functions commonly used in antenna or scattering
problems are of two types:
entire domain functions and
sub-domain functions

10.4.1 Entire domain basis functions
The entire domain functions exist over the full domain
-l/2<x<l/2
Some examples are:
Fourier (is well known) 


 −
=
xn
xbn
2
)
1
(cos)(
Fourier (is well known)
Chebyshev (will discuss briefly)
Legendre (will discuss briefly)
where n=1,2,3,…,N.



=
l
xbn )
2
(cos)(
)
2
()( 22
l
x
Txb nn −=
)
2
()( 22
l
x
Pxb nn −=

Chebyshev's differential equation
where n is a real number
Solutions Chebyshev functions of degree n
( ) 01 2'''2 =+−− ynxyyx
Solutions Chebyshev functions of degree n
n is a non-negative integer, i.e., n=0,1,2,3,…,
the Chebyshev functions are called Chebyshev polynomials
denoted byTn(x)

A Chebyshev polynomial at one point can be
expressed by neighboring Chebyshev polynomials at the same
point
whereT0(x)=1,T1(x)=x
( ) ( ) ( )xTxxTxT nnn 11 2 −+ −=
whereT0(x)=1,T1(x)=x
Legendre's differential equation
where n is a real number
( ) ( ) 0121 '''2 =++−− ynnxyyx

Solutions of this equation are called Legendre functions of
degree n
When n is a non-negative integer, i.e., n=0,1,2,3,…,
the Legendre functions are called Legendre polynomials denoted
by Pn(x)
by Pn(x)
Legendre polynomial at one point can be
expressed by neighboring Legendre polynomials at the same point
where P0(x)=1, P1(x)=x
( ) ( ) ( ) ( ) ( )xnPxxPnxPn nnn 11 121 −+ −+=+

Disadvantage: entire domain basis function may not be
applicable of any general problem
Choose a particular basis function for a particular problem
Crucial and only experts in the area could do it efficiently
Developing a general purpose MoM based software,
Developing a general purpose MoM based software,
software for analyzing almost every problem in
electromagnetics
this is not feasible
Sub-domain basis functions could achieve this purpose

10.4.2 Sub-domain basis functions
Sub-domain basis functions exist only on one of the N
overlapping segments
into which the domain is divided
Some examples are:
Some examples are:
Piecewise constant function (pulse)
1 [ 1] [ ]
( )
0
n
x n x x n
b x
otherwise
− < <
= 


Piecewise triangular function
[ 1] [ 1]
( )
0
n
n
x x
x n x x n
b x
otherwise
∆ − −
− < < +
=  ∆


1
1
1
1
[ 1] [ ]
[ ] [ 1]
0
n
n n
n
n n
x x
x n x x n
x x
x x
x n x x n
x x
otherwise
−
−
+
+

−
− < < −

 −
= < < +
−




Piecewise sinusoidal function ( ){ }
( )
( ){ }
( ){ }
{ }
1
sin
[ 1] [ 1]
( ) sin
0
sin
[ 1] [ ]
sin
n
n
n
n n
k x x
x n x x n
b x k
otherwise
k x x
x n x x n
k x x −
 ∆ − −
 − < < +
=  ∆


 −
− < <
−

where ∆=l/N, assuming equal subintervals but it is not mandatory
and k is a constant
( ){ }
( ){ }
1
1
sin
[ ] [ 1]
sin
0
n
n n
k x x
x n x x n
k x x
otherwise
+
−

−
= < < +
−





Fig. 10.5 Sub-domain basis functions (a) Piecewise constant
function (b) Piecewise triangular function (c) Piecewise
sinusoidal function

Since the derivative of the pulse function is impulsive
we cannot employ it for MoM problems
o where the linear operator L consists of derivatives
Piecewise triangular and sinusoidal functions
may be used for such kinds of problems
Piecewise sinusoidal functions are generally used
for analysis of wire antennas since
they can approximate sinusoidal currents in the wire
antennas

10.5 Wire Antennas and Scatterers
For Piece-wise triangular and sinusoidal functions
when we have N points in an interval
we will have N-1 sub-sections and
N-2 basis functions may be used
10.5Wire Antennas and Scatterers
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
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
Wire antennas
Feed voltage to an antenna is known
and the current distribution could be calculated
and the current distribution could be calculated
other antenna parameters such as
impedance,
radiation pattern, etc.
can be calculated

Wire scatterers
Wave impinges upon surface of a wire scatterer
it induces current density
which in turn is used to generate the scattered fields
We will consider
We will consider
how to find the current distribution on a
thin wire or
cylindrical antenna
using the MoM

10.5.1 Electric field integral equation (EFIE)
On perfect electric conductor like metal
the total tangential electric field is zero
Centrally excited cylindrical antenna (Fig. 10.6)
have two kinds of electric fields viz.,
have two kinds of electric fields viz.,
incident and
scattered electric fields
t
tan tan tan tan tan0 0ot inc scat inc scat
E E E E E= ⇒ + = ⇒ = −
r r r r r

2∆
Fig. 10.6 A thin wire antenna of length L, radius a (a<<L)
and feed gap 2∆

where the is the source or impressed field and
can be computed from the
current density induced on the cylindrical wire antenna due to the
incident or
impressed field
inc
E
r
scat
E
r
impressed field
10.5.2 Hallen’s and Pocklington's Integro-differential equation
Let us consider a perfectly conducting wire of
length L and
radius a such that a<<L and λ, the wavelength corresponding to the
operating frequency

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
( )inc
E r
r r
When the wire is an antenna
the incident field is produced by the feed at the gap (see Fig.
10.6)
The impressed field is required
to be known on the surface of the wire
inc
zE

Simplest excitation
delta-gap excitation
For delta gap excitation (assumption)
excitation voltage at the feed terminal is constant and
zero elsewhere
zero elsewhere
Implies incident field
constant over the feed gap and
zero elsewhere

2V0 (from +V0 to -V0) voltage source applied
across the feed gap 2∆,
Incident field on the wire antenna can be expressed as
0
;
V
z

< ∆r
Induced current density
due to the incident or impressed electric field
produces the scattered electric field
0
;
0;
2
inc
z
z
E
L
z
< ∆ ∆= 
 ∆ < <

r
( )scat
E r
r r

The total electric field is given by
Since the wire is assumed to be perfectly conducting,
tangential component of the total electric field on the surface of
( ) ( ) ( )tot inc scat
E r E r E r= +
r r rr r r
tangential component of the total electric field on the surface of
the wire is zero
For a cylindrical wire placed along z-axis, we can write,
( ) ( ) ( ) 0;tot inc scat
z z zE r E r E r on the wire antenna= + =
r r rr r r

that is,
Find the electric field from the potential functions using
( ) ( )scat inc
z zE r E r= −
r rr r
VAjE ∇−−=
rr
ω
Lorentz Gauge condition,
VAjE ∇−−= ω
VjA 00εωµ−=•∇
r

For a thin cylinder,
current density considered to be independent of
where is the surface current density
)(
2
1
)( '' zI
a
zJ z
π
=
φ
'
where is the surface current density
at a point on the conductor
skin depth of the perfect conductor is almost zero
and therefore all the currents flow on the surface of the wire
)( 'zJ z
'z

The current may be assumed to be
a filamentary current located parallel to z-axis
at a distance a (a is a very small number) as shown in the Fig. 10.7
For the current flowing only in the z direction,
)( 'zI
From Lorentz Gauge condition for time harmonic case,
z
V
AjE zz
∂
∂
−−= ω
2 2
0 0 0 02 2
0 0
1z z zA A AV V
j V j
z z z z j z
ωµ ε ωµ ε
ωµ ε
∂ ∂ ∂∂ ∂
= − ⇒ = − ⇒ − =
∂ ∂ ∂ ∂ ∂

Fig. 10.7 Cylindrical conductor of radius a with surface current
zJ
Fig. 10.7 Cylindrical conductor of radius a with surface current
density
and its equivalence to the case of the conductor replaced by current
filament at a distance a from the z-axis
'
( )z
A
J z
m
 
 
 
( )' '
( ) 2 ( )zI z aJ z Aπ=

Therefore,
Magnetic vector potential can be expressed as
2 2 2
2 2
0 0 02 2 2
0 0 0 0 0 0
1 1 1z z z
z z z z
A A A
E j A A A
j z j z j z
ω ω µ ε β
ωµ ε ωµ ε ωµ ε
   ∂ ∂ ∂
= − + = + = +   
∂ ∂ ∂   
Putting the Jz expression from (10.20), we have,
'
0
4
0
ds
r
e
JA
rj
s
zz
π
µ
β−
∫∫=
0/ 2 2 '
' '
0
/2 0
( )
2 4
L j r
z
L
I z e
A ad dz
a r
π β
µ φ
π π
−
−
= ∫ ∫

where
For ρ =a
( ) ( ) ( )2'2'2' zzyyxxr −+−+−=
0/ 2 2
' ' '1
( ) ( )
L j r
e
A a I z d dz
π β
ρ µ φ
−
 
= =  ∫ ∫
where
' ' '
0
/ 2 0
1
( ) ( )
2 4
z
L
e
A a I z d dz
r
ρ µ φ
π π−
 
= =  
 
∫ ∫
( ) ( )2'
'
22
2
sin4 zzaar −+








==
ϕ
ρ

Therefore, we can write
where
'
2/
2/
''
0 ),()()( dzzzGzIaA
L
L
z ∫
−
== µρ
02
' '1
( , )
j r
e
G z z d
π β
φ
−
= ∫
is the field at the observation point caused by a unit
point source placed at
' '
0
1
( , )
2 4
e
G z z d
r
φ
π π
= ∫
),( 'rrG
rr
'r
r

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
We have,
r
r
)( 'rJ
r
),()( '' rrGrJ
rrr 'r
r
We have,
and








+
∂
∂
= z
z
z A
z
A
j
E 2
02
2
00
1
β
µωε
0/ 2 2
' ' '
0
/ 2 0
1
( ) ( )
2 4
L j r
z
L
e
A a I z d dz
r
π β
ρ µ φ
π π
−
−
 
= =  
 
∫ ∫

From the above two equations we can write, two equations:
(a)
This electric field is the field due to current
/22
2 ' ' '
02
0 /2
1
( ) ( , )
L
z
L
E I z G z z dz
j z
β
ωε −
 ∂
= + 
∂ 
∫
)( 'zI
[which results because of the impressed or source field] and
this field can be written as the scattered field
Therefore,
)( 'zI
/22
2 ' ' '
02
0 /2
1
( ) ( , )
L
scat
z
L
E I z G z z dz
j z
β
ωε −
 ∂
= + 
∂ 
∫

Since from the EFIE on the surface of the wire,
( ) ( )scat inc
z zE a E aρ ρ= = − =
/22
2 ' ' '
02
1
( ) ( , ) ( )
L
inc
zI z G z z dz E a
j z
β ρ
ωε
 ∂
+ = − = 
∂ 
∫
This equation is called the Hallen's Integro-differential equation
In this case, differential is outside the integral
2
0 /2L
j zωε −
∂ 
∫

(b)
'
2/
2/
''2
02
2
0
),()(
1
dzzzGzI
zj
E
L
L
z ∫
−








+
∂
∂
= β
ωε
)( 'zI
this field can be written as the scattered field
Therefore,
/2 2
2 ' ' '
02
0 /2
1
( ) ( , )
L
scat
z
L
E I z G z z dz
j z
β
ωε −
 ∂
= + 
∂ 
∫

Since from EFIE on the surface of the wire,
( ) ( )scat inc
z zE a E aρ ρ= = − =
/2 2
2 ' ' '
02
0 / 2
1
( ) ( , ) ( )
L
inc
z
L
I z G z z dz E a
j z
β ρ
ωε −
 ∂
+ = − = 
∂ 
∫
This equation is the Pocklington's Integro-differential equation
In this case, the differential has moved inside the integral
Richmond has simplified the above equation as follows:
0 / 2L
j zωε −
∂ 

(c) In cylindrical coordinates,
( )
2 2' ' '
r r r z z ρ ρ= − = − + −
r rr r
( )' ' 2 2 ' 2 2 '
2 2 cosa a a aρ ρ ρ ρ ρ ρ ρ ρ φ φ= ∴ − = + − • = + − −
r r r r
Q
Problem under analysis has cylindrical symmetry and
observation for any values of won’t make any difference
we may assume without loss of generality
hence
( ) ( )
2' 2 2 ' '
2 cosr r r a a z zρ ρ φ φ⇒ = − = + − − + −
r r
φ
0φ =
' '
φ φ φ− =

where
and the inner integration
0/ 2 2'
' '
0
/2 0
( )
2 4
L j r
z
L
I z e
A d dz
r
π β
µ φ
π π
−
−
= ∫ ∫
( ) ( )
22 2 ' '
2 cosr a a z zρ ρ φ= + − + −
02
'
j r
e
d
π β
φ
−
∫
and the inner integration
is also referred to as cylindrical wire kernel
If we assume a<<λ and is very small, we have,
Inner integrand is no more dependent on the variable
'
0
4
e
d
r
φ
π∫
( )
22 '
r z zρ≅ + −
'
φ

Therefore
Also called as thin wire approximation
with the reduced kernel
0/ 2 '
'
0
/ 2
( )
4
L j r
z
L
I z e
A dz
r
β
µ
π
−
−
= ∫
with the reduced kernel
For this case, we can write
( ) ( )
0
'
,
4
j r
e
G z z G r
r
β
π
−
≅ =

Now in the light of this simplification of the magnetic vector
potential,
we can simplify equation 10.29c (see example 10.4) as
follows:
This form of the Pocklington’s integro-differential is more
suitable for MoM formulation
since it does not involve any differentiation.
( )( ) ( )
0/ 2
2' 2 2 '
0 05
0 / 2
1
( ) 1 2 3 ( )
4
L j r
inc
z z
L
e
I z j r r a ar dz E a
j r
β
β β ρ
ωε π
−
−
 + − + = − =
 ∫

10.5.3 MoM Formulation of Pocklington's Integro-differential
equation
Applying MoM formulation to above integral equation
Divide the wire in to N segments
Consider pulse basis function and
Consider pulse basis function and
express the current as a series expansion
in the form of a staircase approximation as
where' '
1
( ) ( )
N
n n
n
I z I b z
=
= ∑ '
' 1
( )
0
n
n
for z
b z
otherwise
 ∆
= 


'
nz
Fig. 10.8Thin wire dipole is divided into N equal segments
'
nz∆
nz

is the length of the nth segment, expressed as
( 1)
2 2
L L L L
n z n
N N
− + − < ≤ − +
'
nz∆
/ 2
' ' '
( ) ( , )
L
inc
zE I z F z z dz− = ∫
where
/2
( ) ( , )z
L
E I z F z z dz
−
− = ∫
( )'2
02
2
' ,
1
),( zzG
zj
zzF








+
∂
∂
= β
ωε

Substituting I(z′) and
evaluating at z=zm (middle of the mth segment) as shown in the Fig. 10.7
for point matching with weighting functions as
where zm is the center of the segment m
( ) ( )m mw z z zδ= −
L L
and m=1,2,3,…,M, we can write,
( 0.5)
2
m
L L
z m
N
= − + −
( ) ( )∑ ∫
= ∆
=−
N
n z
mnm
i
z
n
dzzzFIzE
1
''
'
,

To overcome the singularity
for the self term or diagonal elements of the [Z] matrix
we have assumed that the source is on the surface of the wire
whereas the observation is the axis of the wire
Using mid-point integration, we have,
Using mid-point integration, we have,
where ( ) ( )∫
∆
∆≅=
'
'''' ,,
nz
nnmmmn zzzFdzzzFF
( ) ∑
=
=−
N
n
mnnm
i
z FIzE
1

For m=1,2,…,M,
11 12 1 1 1
21 22 2 2 2
... ( )
... ( )
: : : : :
... ( )
inc
N z
inc
N z
inc
F F F I E z
F F F I E z
F F F I E z
 −   
    
−     =
    
    
−      
In the compact form, we can write
Multiplying both sides by ∆z, we can write,
1 2 ... ( )inc
M M MN N z N
F F F I E z
    
−      
[ ][ ] [ ]mnmn EIF −=
[ ][ ] [ ]mn n mZ I V= −

[In] can be computed by matrix inversion
Can find the approximate current distribution on the antenna
Other important antenna parameters are
input impedance of the antenna and
the total radiated fields
the total radiated fields
which can be obtained as:
( ) ( ) ( ) ( )
0
0
2
/ 2 2
' 2 '
02
1/ 2
2
1
( )
4
input
N
L j rN
tot inc scat inc
n n
nL
V
Z
I
e
E r E r E r E r I b z dz
j z r
β
β
ωε π
−
=−
=
 ∂
= + = + + 
∂ 
∑∫
r r r rr r r r

Example 10.5
Consider a short dipole (thin wire antenna) of length 0.3λ
and radius 0.01λ.
Find the current distribution on the short dipole and input
impedance using MoM.
impedance using MoM.
Assume frequency of operation of the antenna is at 1 MHz.
Choose the number of discretizations on the thin wire as
three segments only.

For three segments discretization ( ) on the thin
wire antenna
the MoM matrix equation will be
0.1z λ∆ =
11 12 13 1 1Z Z Z I V
Z Z Z I V
     
     = −
21 22 23 2 2
31 32 33 3 3
11 12 13 1 1
21 22 23 2 2
31 32 33 3 3
Z Z Z I V
Z Z Z I V
F F F I E
z F F F I z E
F F F I E
     = −     
          
     
     ⇒ ∆ = −∆     
          

where
Simplifying the above expression of the Z (see book)
( )( ) ( )0
22 2
0 0
5
0
1 2 3
4
mnj r
mn mn mn
mn mn
mn
e j r r a ar z
Z F z
j r
β
β β
ωε π
−  + − + ∆
 = ∆ =
Simplifying the above expression of the Zmn (see book)
2 2
0
3
2
1 2 2 3 2 cos 2 sin 2
8
mn mn mn
mn
mn
mn
r r rz a a
jZ j j
r
Z
r
π π π π
λ λ λ λ λ
π λ
λ
     ∆         − + − + −                     =
 
 
 

For thin wire approximations,
( )
22
2
11 22 33
2 2
12 21 23 32
0.01 ;
0.1 ;
mn m nr z z
r a a r r
r a z r r r
ρ
λ
λ
≅ + −
∴ = = = = =
= + ∆ = = = =
[Z] symmetricToeplitz matrix (need to calculate the first row of
the matrix only)
( )
12 21 23 32
22
13 31
0.1 ;
2 0.2
r a z r r r
r a z r
λ
λ
= + ∆ = = = =
= + ∆ = =
[ ]
1 2 3
2 1 3
3 2 1
Z Z Z
Z Z Z Z
Z Z Z
 
 =  
  

About the [V] matrix, for delta gap excitation,
since a voltage 1V exists at the feed gap only
[ ]
0
1
0
V
 
 = −  
  
since a voltage 1V exists at the feed gap only
Solve the [I] matrix and get the current distribution
and the input impedance can be calculated as
0
2
2
2 1
input
N
V
Z
I I
= =

10.6 Software language for implementation
of electromagnetics codes
Choice of software languages for implementing electromagnetics
code
FORTRAN language complex numbers are a built-in
datatype
Many computational electromagnetics programmer prefer
Many computational electromagnetics programmer prefer
to use FORTRAN language
for implementing their algorithms
Earlier versions of FORTRAN were a functional language

New versions of FORTRAN are object-oriented languages
C++, another object-oriented language,
is also widely used for many numerical methods
FORTRAN and C++ are efficient in implementation
FORTRAN and C++ are efficient in implementation
since the computational time is less
MATLAB is also convenient environments since
it accepts complex numbers,
graphics are very easy to create and
many in-built functions are readily available for use

MATLAB any additional “for” loop in the program,
the time it takes to run the program increases drastically
Good to consider the advantages and disadvantages
for employment of any software language
for employment of any software language
For instance,
drawing graphics in C is somewhat involved,
but MATLAB is convenient for such things.
program written in C runs faster than MATLAB and so on

Computational electromagnetics is a topic
which you can learn only by doing
Some simulation exercises are given at the end of the chapter,
you should always write down a
MATLAB or
MATLAB or
any other software language program
in which you are comfortable and
see those results

10.7 Summary
Summarize the three steps involved in MoM:
(a) Derivation of appropriate integral equations
(b) Conversion or discretization of the integral equation into
a matrix equation using
basis or expansion functions and
basis or expansion functions and
weighting or testing functions
as well as evaluation of the matrix elements
(c) Solving the matrix equation and
obtaining the desired parameters

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