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1/101

10. Method of Moments

.

1/3/20141 Electromagnetic Field Theory by R. S. Kshetrimayum


2/101

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

1/3/2014Electromagnetic Field Theory by R. S. Kshetrimayum2

so ve a s mp e eren a equa on us ng o in order to elucidate the steps involved

MoM tfor 1-D and 2-D electrostatic problems

MoM for electrodynamic problems


3/101

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


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


4/101

10.2 Basic Steps in Method of Moments For example,

(a) consider the integral equation for a line charge density

0

0

( ') '

4 ( , ')

x dxV

r x x

=


Then ( )

'u x=

0k V=

0

'4 ( , ')

dxLr x x

=


5/101

10.2 Basic Steps in Method of Moments (b) consider the differential equation of the form

Then

2

22( ) 3 2d f x x

dx = +

=


23 2k x= +

2

2

dL

dx=


6/101

10.2 Basic Steps in Method of Moments To solve u, approximate it by sum of weighted known

basis functions or

expansion functions

as given belowN N


where is the expansion function,

is its unknown complex coefficients to be determined, Nis the total number of expansion functions

1 1, 1, 2,...,n n n

n nu u I b n N

= = = =

nb

nI


7/101

10.2 Basic Steps in Method of Moments Since L is linear, substitution of the above equation in the

integro-differential equation,

we get,

N

n nL I b k


where the error or residual is given by

1n=

1

N

n n

n

R k L I b=

=


8/101

10.2 Basic Steps in Method of Moments Mathematicians name this method as Method of Weighted

Residuals

Why?

Next step in MoM


n orc ng t e oun ary con t on Make inner product of the above equation with each of the

testing or

weighting functions should make residual or error zero


9/101

10.2 Basic Steps in Method of Moments

By replacing u by where n=1,2,,N

taking inner product with a set of

n

u

mw


we g ng or

testing functions

in the range of L, we have,

( )( ), 0, 1, 2,...,m nw L u k m M = =


10/101


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=

= =


11/101

10.2 Basic Steps in Method of Moments Note that a scalar product is defined to be a scalar

satisfying

, , ( ) ( )w g g w g x w x dx= =

gw,


b and c are scalars and * indicates complex conjugation

wgcwwcg ,,,

+=+

00,*

> gifgg 00,*

== gifgg


12/101

10.2 Basic Steps in Method of Moments In matrix form

with each matrix and vector defined by

[ ][ ] [ ]VIZ =

[ ] [ ]TNIIII ...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


13/101

10.2 Basic Steps in Method of Moments For [Z] is non-singular,

Solve the unknown matrix [I] of amplitudes of basis functionas[ ] [ ] [ ] [ ][ ]

1I Z V Y V

= =


a er in s met o

Point matching or Collocation

The testing function is a delta function

nn wb =


14/101

10.2 Basic Steps in Method of Moments 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

[ ] 1

Z


comman p nv n s pseu o nverse o a ma r x

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 call A is ill-conditioned


15/101

10.2 Basic Steps in Method of Moments 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


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


16/101

10.2 Basic Steps in Method of Moments If the condition number is large,

then the matrix is said to be ill-conditioned

Practically,

such a matrix is almost singular, and


e compu a on o s nverse, 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


17/101

10.2 Basic Steps in Method of Moments Sometimes pseudo inverse is also used for finding

approximate solutions to ill-conditioned matrices

Preferable to use LU decomposition

to solve linear matrix equations


ac or za on un e auss an e m na on,

do not make any modifications in the matrix B

in solving the matrix equation


18/101

10.2 Basic Steps in Method of Moments 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


19/101


through the forward substitution

[ ][ ][ ] [ ] [ ][ ] [ ]L U X B L Y B = =

11 1

ib = =


through the backward substitution

1

111

,i i ik k

kiil l =

[ ][ ] [ ]U X Y=

1

1; ,

NN

N i i ik k

k iNN ii

yx x y u x i N

u u = +

= =


20/101

10.2 Basic Steps in Method of Moments 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.


comman or ac or za on o a ma r x s

[L U] = lu(A)


21/101

10.2 Basic Steps in Method of MomentsExample 10.1

Consider a 1-D differential equation2

2

2

( )3 2

d f xx

dx = +


subject to the boundary condition f(0)=f(1)=0 Solve this differential equation using Galerkins MoM

Solution:

Note that for this case,

( )u f x=


22/101


According to the nature of the known function ,

23 2k x= +

2

2dL dx=

23 2k x= +


it is natural to choose the basis function as However,

the boundary condition f(1)=0

cant be satisfied with such a basis function

( ) nn

b x x=


23/101

10.2 Basic Steps in Method of Moments A suitable basis function for this differential equation

taking into account of this boundary condition is

( ) 1; 1,2, ...,nnb x x x n N +

= =


ssume = t e tota num er o su sections on t einterval [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 + = +


24/101

10.2 Basic Steps in Method of Moments For Galerkins 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

211 1 1 1 1

0 0

1, ( ) ( ) ( ) 23

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 2Z w L b w x L b x dx x x x dx= = = =


25/101


26/101

10.2 Basic Steps in Method of Moments Therefore,

[ ][ ] [ ] 1

2

1 1 3

3 2 5

1 4 112 5 12

IZ I V

I

= =

13


The unknown function f(x)

[ ]

1

2 1

3

I I = =

( ) ( ) ( ) ( )2 3 2 31 213 1

( )10 3

f x I x x I x x x x x x + = +


27/101

10.2 Basic Steps in Method of Moments 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( )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)


28/101



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


29/101

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 LUdecomposition and SVD


30/101

10.3 Introductory examples from electrostatics 1-D Electrostatic case: Charge density of a straight wire

Consider a straight wire of length l and radius a (assume

a


31/101



32/101

10.3 Introductory examples from electrostaticsFig. 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


(c) Division of the charged strip into N sections


33/101

10.3 Introductory examples from electrostatics where

It is necessary to solve the integral equation

' ' ' 2 ' 2 ' 2 ' 2 2

0( , ) ( , ) ( ) ( ) ( ) ( ) ( )x zR y y R r r y y x z y y a= == = + + = +

r r


to find the unknown function

(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


34/101

10.3 Introductory examples from electrostatics (a) Approximate the unknown charge density (y)

by an expansion of N known basis functions with unknown

coefficients

=

=

N

n

nn ybIy

1

'' )()(


Integral equation after substituting this is

=

= ==

l

n

l N

n

n

N

n

nn

yyR

dyybIyyR

dyybI

0'

''

0 1'

'

1

'

0),(

)(

),(

)(

4


35/101

10.3 Introductory examples from electrostatics 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

( )

= nynfor

ybn

'' )1(1


b) Applying the testing or weighting functions

Let us apply the testing functions as delta functions

for point matching

( )my y


36/101

10.3 Introductory examples from electrostatics 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

my y=


so e n egra may e eva ua e .

It leads to an equation with N unknowns

+++++=

2

)1( )1(

'

''

'

''

'

''2

2

0

'

''1

10

),(

)(...

),(

)(...

),(

)(

),(

)(4

n

n

l

N m

NN

m

nn

mm yyR

dyybI

yyR

dyybI

yyR

dyybI

yyR

dyybI


37/101

10.3 Introductory examples from electrostatics

Solution for these N unknown constants,

N linearly independent equations are required

N equations may be produced


y c oos ng an o serva on po n ym on e w re an

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


38/101

10.3 Introductory examples from electrostatics

For N such observation points we have

+++++=

2

)1( )1('

1

''

'1

''

'1

''

22

0'

1

''

110),(

)(...),(

)(...),(

)(),(

)(4

n

n

l

N

NNnnyyRdyybI

yyRdyybI

yyRdyybI

yyRdyybI

2 '''''''' n l


+++++=

)1( )1('2'2'2

2

20

'2

1

10 ),(...

),(...

),(),(4

n N

N

N

n

n yyR

yyI

yyR

yyI

yyR

yyI

yyR

yyI

+++++=

2

)1( )1( '

''

'

''

'

''2

2

0 '

''1

10

),(

)(...

),(

)(...

),(

)(

),(

)(4

n

n

l

N N

NN

N

nn

NN yyR

dyybI

yyR

dyybI

yyR

dyybI

yyR

dyybI


39/101

10.3 Introductory examples from electrostatics(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

= =

K

OM M M M M


where

1 2 0...N N NN NZ Z Z I V

[ ] [ ]04=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 dyZ

y y a y y ady dy

for m ny y y yy y

= =

+ +

=


40/101

10.3 Introductory examples from electrostatics 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 ' 'm

y y d dy = =


( )0

2 2

2 2 2 20

0

log ( )( ) ( )

mnd dZ a

a a

= = = + +

+ +

2 2

ln a

a

+ +=


41/101

10.3 Introductory examples from electrostatics 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 ZZ

Z Z Z

=


42/101

10.3 Introductory examples from electrostatics 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:


Therefore the unknown [I] matrix could be solved as

1, 1, ,mn m n m n +=

[ ] [ ] [ ]mmnn VZI 1

=


43/101


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


44/101

10.3 Introductory examples from electrostatics 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


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 givethe same result


45/101

10.3 Introductory examples from electrostatics 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


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)


46/101

10.3 Introductory examples from electrostatics 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


o pro em n a ng e nverse

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


47/101

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


as s unc ons common y use n an enna or sca er ng

problems are of two types:

entire domain functions and

sub-domain functions


48/101

10.4 Some commonly used basis functions10.4.1 Entire domain basis functions

The entire domain functions exist over the full domain

-l/2


49/101

10.4 Some commonly used basis functions Chebyshev's differential equation

where n is a real number

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 by Tn(x)


50/101

10.4 Some commonly used basis functions A Chebyshev polynomial at one point can be

expressed by neighboring Chebyshev polynomials at the same

point

( ) ( ) ( )xTxxTxT nnn 11 2 + =


0 , 1

Legendre's differential equation

where n is a real number

( ) 0121 '''2 =++ ynnxyyx


51/101


52/101

10.4 Some commonly used basis functions 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


,

software for analyzing almost every problem inelectromagnetics

this is not feasible

Sub-domain basis functions could achieve this purpose


53/101

10.4 Some commonly used basis functions10.4.2 Sub-domain basis functions

Sub-domain basis functions exist only on one of the N

overlapping segments

into which the domain is divided


:

Piecewise constant function (pulse)

1 [ 1] [ ]( )

0n

x n x x nb x

otherwise

<


54/101

10.4 Some commonly used basis functions Piecewise triangular function

[ 1] [ 1]( )

0

n

n

x xx n x x n

b x

otherwise

< < +

=


1

1

1

1

[ 1] [ ]

[ ] [ 1]

0

n

n n

n

n n

x xx n x x nx x

x xx n x x n

x x

otherwise

+

+

<


55/101


56/101

10.4 Some commonly used basis functions


Fig. 10.5 Sub-domain basis functions (a) Piecewise constant

function (b) Piecewise triangular function (c) Piecewisesinusoidal function


57/101

10.4 Some commonly used basis functions 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 wireantennas


58/101

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


10.5 Wire Antennas and Scatterers

Consider application of MoM techniques

to wire antennas and scatterers


59/101

10.5 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


W en t e wire is ar rom t e source

it acts as scatterer

For the wire objects (antenna or scatterer)

we require to know the current distribution accurately


60/101

10.5 Wire Antennas and Scatterers Integral equations are derived and

solved for this purpose

Wire antennas

Feed voltage to an antenna is known


an t e current istri ution cou e ca cu ate

other antenna parameters such as

impedance,

radiation pattern, etc.

can be calculated


61/101

10.5 Wire Antennas and ScatterersWire scatterers

Wave impinges upon surface of a wire scatterer

it induces current density

which in turn is used to generate the scattered fields


e w cons er

how to find the current distribution on a

thin wire or

cylindrical antenna

using the MoM


62/101

10.5 Wire Antennas and Scatterers10.5.1 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)


ave two in s o e ectric ie s 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


63/101

2


Fig. 10.6 A thin wire antenna of length L, radius a (a


64/101

10.5 Wire Antennas and Scatterers 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

incE

r

scatE

r


mpresse e

10.5.2 Hallens and Pocklington's Integro-differential equation

Let us consider a perfectly conducting wire of

length L and

radius a such that a


65/101

10.5 Wire Antennas and Scatterers 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

( )incE rr r


en e w re s an an enna

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


66/101

10.5 Wire Antennas and Scatterers Simplest excitation

delta-gap excitation

For delta gap excitation (assumption) excitation voltage at the feed terminal is constant and

1/3/2014Electromagnetic Field Theory by R. S. Kshetrimayum

66

Implies incident field constant over the feed gap and

zero elsewhere


67/101

10.5 Wire Antennas and Scatterers 2V0 (from + V0 to -V0) voltage source applied

across the feed gap 2,

Incident field on the wire antenna can be expressed as

0V

1/3/2014Electromagnetic Field Theory by R. S. Kshetrimayum

67

Induced current density

due to the incident or impressed electric field produces the scattered electric field

0;2

inc

zE Lz

=

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