10 slides

Dr. Rakhesh Singh Kshetrimayum

10. Method of Moments

Dr. Rakhesh Singh Kshetrimayum

1/3/20141 Electromagnetic Field Theory 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 Field Theory 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 dxV

r x x

λ

πε= ∫

( )


� Then ( )'u xλ=

0k V=

0

'

4 ( , ')

dxL

r x xπε= ∫


� (b) consider the differential equation of the form

� Then

22

2

( )3 2

d f xx

dx− = +

( )u f x=


( )u f x=

23 2k x= +

2

2

dL

dx= −


� 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,

� 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

10.2 Basic Steps in Method of Moments� 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 of Weighted 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=

= =∑

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= = ∫

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 =

[ ] [ ]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


� 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 call A 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∴ = ⇒ =

11

1

1; , 1

i

i i ik k

by y b l y i

− = = − > ∑


� through the backward substitution

1

111

; , 1i i ik k

kii

y y b l y il l =

= = − >

∑

[ ][ ] [ ]U X Y=

1

1; ,

NN

N i i ik k

k iNN ii

yx 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

22

2

( )3 2

d f xx

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

23 2k x= +

2

2

dL

dx= −

23 2k x= +

( ) nb 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

3Z 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= = = − =∫ ∫


( ) ( ) ( )1 1

3

21 2 1 2 1

0 0

1, ( ) ( ) ( ) 2

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

( ) ( ) ( )1 1

3

22 2 2 2 2

0 0

4, ( ) ( ) ( ) 6

5Z 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 ( )

5V k w k x w x dx x x x dx= = = + − =∫ ∫

( )1 1

2 3

2 2 2

0 0

11, ( ) ( ) 3 2 ( )

12V 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

IZ I V

I

= ⇒ =

[ ] 1

13

10I

⇒ = =


� The unknown function f(x)

[ ] 1

2

10

1

3

II

I

⇒ = =

( ) ( ) ( ) ( )2 3 2 3

1 2

13 1( )

10 3f 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 6f 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

11

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 zR 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

∫∫ ∑∑

=

= ==

ln

l N

n

n

N

n

nn

yyR

dyybI

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

nynforybn

'' )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

NN

m

nn

mm yyR

dyybI

yyR

dyybI

yyR

dyybI

yyR

dyybIπε


� 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

NN

nn

yyR

dyybI

yyR

dyybI

yyR

dyybI

yyR

dyybIπε

∫ ∫ ∫∫∆ ∆∆ 2 ''''''

2''

1 )()()()(n l

Nn dyybdyybdyybdyyb


∫ ∫ ∫∫∆

∆

∆

∆− ∆−

∆

+++++=

2

)1( )1('

2

''

'2

''

'2

''2

2

0'

2

''1

10),(

)(...

),(

)(...

),(

)(

),(

)(4

n

n

l

N

NN

nn

yyR

dyybI

yyR

dyybI

yyR

dyybI

yyR

dyybIπε

∫ ∫ ∫∫∆

∆

∆

∆− ∆−

∆

+++++=

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

dyybIπε


(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 20

' '

'' 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 a

dy dyfor 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 2 200

log ( )( ) ( )

mn

d dZ a

a a

ξ ξξ ξ

ξ ξ

∆ ∆

∆

= − = = + ++ +

∫ ∫

2 2

lna

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 ZZ

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

xbn2

)1

(cos)(


� Fourier (is well known)

� Chebyshev (will discuss briefly)

� Legendre (will discuss briefly)� where n=1,2,3,…,N.

=

lxbn )

2(cos)(

)2

()( 22l

xTxb nn −=

)2

()( 22l

xPxb 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 by Tn(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] [ ]( )

0n

x n x x nb x

otherwise

− < <=


� 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 n

x x

x xx 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 xx n x x n

b x k

otherwise

k x xx 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 xx 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



� 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 scatE 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∆

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

� impressed field

incEr

scatEr


� 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

( )incE rr 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

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

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

EL

z

< ∆ ∆= ∆ < <

r

( )scatE rr 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 scatE 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)( ''

zIa

zJ zπ

=

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

VAjE zz

∂

∂−−= ω

2 2

0 0 0 02 2

0 0

1z z zA A AV Vj V j

z z z z j zωµ 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

AJ z

m

( )' '( ) 2 ( )z

I z aJ z Aπ=


� Therefore,

� Magnetic vector potential can be expressed as

2 2 22 2

0 0 02 2 2

0 0 0 0 0 0

1 1 1z z zz z z z

A A AE j A A A

j z j z j zω ω µ ε β

ωµ ε ωµ ε ωµ ε

∂ ∂ ∂= − + = + = +

∂ ∂ ∂


� Putting the Jz expression from (10.20), we have,

'0

4

0

dsr

eJA

rj

s

zzπ

µβ−

∫∫=

0/ 2 2 '

' '

0

/ 2 0

( )

2 4

L j r

z

L

I z eA ad dz

a r

π β

µ φπ π

−

−

= ∫ ∫


� where

� For ρ =a

( ) ( ) ( )2'2'2'zzyyxxr −+−+−=

0/ 2 2

' ' '1( ) ( )

L j re

A a I z d dz

π β

ρ µ φ−

= = ∫ ∫


� where

' ' '

0

/ 2 0

1( ) ( )

2 4z

L

eA a I z d dz

rρ µ φ

π π−

= =

∫ ∫

( ) ( )2''

22

2sin4 zzaar −+

==

ϕρ


� Therefore, we can write

� where

'2/

2/

''0 ),()()( dzzzGzIaA

L

L

z ∫−

== µρ

02

' '1( , )

j re

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

eG z z d

rφ

π π= ∫

),( 'rrGrr

'rr


� 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,

rr

)( 'rJr

),()( ''rrGrJrrr '

rr


� We have,

� and

+

∂

∂= z

zz A

z

A

jE

202

2

00

1β

µωε

0/ 2 2

' ' '

0

/ 2 0

1( ) ( )

2 4

L j r

z

L

eA a I z d dz

r

π β

ρ µ φπ π

−

−

= =

∫ ∫


� From the above two equations we can write, two equations: (a)

� This electric field is the field due to current

/ 222 ' ' '

02

0 / 2

1( ) ( , )

L

z

L

E I z G z z dzj z

βωε

−

∂= +

∂ ∫

)( 'zI


� 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,

)( 'zI

/ 222 ' ' '

02

0 / 2

1( ) ( , )

L

scat

z

L

E I z G z z dzj z

βωε

−

∂= +

∂ ∫


� Since from the EFIE on the surface of the wire,

( ) ( )scat inc

z zE a E aρ ρ= = − =

/ 222 ' ' '

02

1( ) ( , ) ( )

L

inc

zI z G z z dz E a

j zβ ρ

ωε

∂+ = − =

∂ ∫


� This equation is called the Hallen's Integro-differential equation

� In this case, differential is outside the integral

2

0 / 2Lj zωε

−∂

∫


(b)

� This electric field is the field due to current � [which results because of the impressed or source field] and

'2/

2/

''202

2

0

),()(1

dzzzGzIzj

E

L

L

z ∫−

+

∂

∂= β

ωε

)( 'zI


� [which results because of the impressed or source field] and

� 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 dzj z

βωε

−

∂= +

∂ ∫


� Since from EFIE on the surface of the wire,

( ) ( )scat inc

z zE a E aρ ρ= = − =

/ 2 22 ' ' '

02

0 / 2

1( ) ( , ) ( )

L

inc

z

L

I z G z z dz E aj z

β ρωε

−

∂+ = − =

∂ ∫


� 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 / 2Lj 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 eA d dz

r

π β

µ φπ π

−

−

= ∫ ∫

( ) ( )2

2 2 ' '2 cosr a a z zρ ρ φ= + − + −

02

'j r

ed

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

'

04

ed

rφ

π∫

( )2

2 '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 eA dz

r

β

µπ

−

−

= ∫


� with the reduced kernel

� For this case, we can write

( ) ( )0

',4

j re

G z z G rr

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

/ 22' 2 2 '

0 05

0 / 2

1( ) 1 2 3 ( )

4

L j r

inc

z z

L

eI z j r r a ar dz E a

j r

β

β β ρωε 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 zb z

otherwise

∆=

'

nz


� Fig. 10.8 Thin 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 Ln 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−

− = ∫

( )'202

2' ,

1),( zzG

zjzzF

+

∂

∂= β

ωε

10.5 Wire Antennas and Scatterers� 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 Lz m

N= − + −

( ) ( )∑ ∫= ∆

=−N

n z

mnmiz

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

mnnmiz FIzE

1


� For m=1,2,…,M,

11 12 1 1 1

21 22 2 2 2

... ( )

... ( )

: : : : :

... ( )

incN 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 ... ( )incM 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 '

021/ 2

2

1( )

4

input

N

L j rNtot inc scat inc

n n

nL

VZ

I

eE r E r E r E r I b z dz

j z r

β

βωε π

−

=−

=

∂= + = + +

∂ ∑∫

r r r rr r r r


Example 10.5

� Consider a short dipole (thin wire antenna) of length 0.3λand 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)

( ) ( ) ( )022 2

0 0

5

0

1 2 3

4

mnj r

mn mn mn

mn mn

mn

e j r r a ar zZ 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 ajZ j j

rZ

r

π π π πλ λ λ λ λ

π λλ

∆ − + − + − =


� 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] symmetric Toeplitz 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 1input

N

VZ

I I= =

10.6 Software language for implementation

of electromagnetics codes

� Choice of software languages for implementing electromagneticscode

� 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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