1

EEE 431Computational Methods in

Electrodynamics

Lecture 17By

Dr. Rasime UygurogluRasime.uyguroglu@emu.edu.tr

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Charged Conducting Plate

Moment Method Solution

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Charged Conducting Plate/ MoM Solution

Consider a square conducting plate 2a meters on a side lying on the z=0 plane with center at the origin.

x

y

z

2a

2a

2b

2b

sn

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Charged Conducting Plate/ MoM Solution

Let represent the surface charge density on the plate.

Assume that the plate has zero thickness.

( , )y x

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Charged Conducting Plate/ MoM Solution

Then, V(x,y,z):

Where;

0

1 ( ', ', ')( , , ) ' '

4

a a

a a

x y zV x y z dx dy

R

1/ 22 2 2( ') ( ')R x x y y z

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Charged Conducting Plate/ MoM Solution

Integral Equation: When

This is the integral equation for

, , 0, ( , , ) ( .)x a y a z V x y z V const

0 2 2

( ', ')4 ' '

( ') ( ')

a a

a a

x yV dx dy

x x y y

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Charged Conducting Plate/ MoM Solution

Method of Moment Solution: Consider that the plate is divided into N

square subsections. Define:

And let:

1

0n

nm

on Sf

on all other S

1

( , )N

n nn

x y f

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Charged Conducting Plate/ MoM Solution

Substituting this into the integral equation and satisfying the resultant equation at the midpoint of each , we get:

( , )m mx y

mS

1

, 1, 2,3,...,N

m mn nn

V A m N

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Charged Conducting Plate/ MoM Solution

Where:

is the potential at the center of

due to a uniform charge density of unit amplitude over

1/ 22 2

0

1' '

4 ' 'n n

mn

x ym m

A dx dyx x y y

mnAmS

nS

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Charged Conducting Plate/ MoM Solution

Let : denote the side length of each

the potential at the center of

due to the unit charge density over its own surface.

22

ab

N nS

nnA nS

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Charged Conducting Plate/ MoM Solution

So,

2 20

0

0

1

4

2ln(1 2)

2(0.8814)

b b

nn

b b

A dx dyx y

b

b

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Charged Conducting Plate/ MoM Solution

The potential at the center of can simply be evaluated by treating the charge over as if it were a point charge, so,

mS

nS

2

1/ 22 20 0

4 ( ) ( )

nmn

mn m n m n

S bA m n

R x x y y

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Charged Conducting Plate/ MoM Solution

So, the matrix equation:

1

A V

A V

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Charged Conducting Plate/ MoM Solution

The capacitance:

1 1

1( , )

1 1( )

a a

a a

a a a aN N

n n n nn na a a a

qC dx dy x y

V V

C dx dy f f dxdyV V

1

1

1 1 1

1 1

1 N

n n mn nn mn

N N

n mn mnm m

C S A SV

A V A V V A

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Charged Conducting Plate/ MoM Solution

The capacitance (Cont.):

Number of sub areas

C/2a

Approx.

C/2a

Exact

1 31.5 31.5

9 37.3 36.8

16 38.2 37.7

36 39.2 38.5

mnA mnA

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Charged Conducting Plate/ MoM Solution

Harrington, Field Computation by Moment Methods

The charge distribution along the width of the plate

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Moment Method/ Review

Consider the operator equation:

Linear Operator. Known function, source. Unknown function. The problem is to find g from f.

...........................................................(1)Lf g

:L:g:f

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Moment Method/ Review

Let f be represented by a set of functions

scalar to be determined (unknown expansion coefficients.

expansion functions or basis functions.

1

.................................................(2)N

i ii

f f

1 2 3,..., ,f f f

i

if

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Moment Method/ Review

Now, substitute (2) into (1):

Since L is linear: 1

N

i ii

L f g

1

.....................................(3)N

i ii

Lf g

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Moment Method/ Review

Now define a set of testing functions or weighting functions

Define the inner product (usually an integral). Then take the inner product of (3) with each and use the linearity of the inner product:

1 2, ,..., Nw w w

1

, , , 1, 2,....., .......(4)N

i i j ji

Lf w g w j M

jw

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Moment Method/ Review

It is common practice to select M=N, but this is not necessary.

For M=N, (4) can be written as:

................(5)A g

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Moment Method/ Review

Where,

,

,

j i

i

j

A w Lf

g w g

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Moment Method/ Review

Or,

1 1 1 2 1

2 1 2 2 2

1 2

, , ... ,

, , ... ,

. . ... .

. . ... .

, , ... ,

N

N

N N N N

w Lf w Lf w Lf

w Lf w Lf w Lf

A

w Lf w Lf w Lf

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Where,

Moment Method/ Review

1

2

. ( 1)

.

N

NX

1

2

,

,( 1)

,N

w g

w gg NX

w g

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Moment Method/ Review

If is nonsingular, its inverse exists and .

Let

A

1A g

1 2

1

1

...... (1 )

................................(6)

N

N

i ii

f f f f XN

f f f

f f A g

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Moment Method/ Review

The solution (6) may be either approximate or exact, depending upon on the choice of expansion and testing functions.

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Moment Method/ Review

Summary: 1)Expand the unknown in a series of basis

functions. 2) Determine a suitable inner product and

define a set of weighting functions. 3) Take the inner products and form the matrix

equation. 4)Solve the matrix equation for the unknown.

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Moment Method/ Review

Inner Product:

Where:

, ,

, , ,

, 0

, 0

u v v u

u v w u w v w

u u if u

u u if u

, , :

, :

u v w functions

real numbers

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Moment Method/ Review

Inner product can be defined as:

,u v uvd

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Moment Method/ Review

If u and v are complex:

*, ,

, , ,

, 0

, 0

u v v u

u v w u w v w

u u if u

u u if u

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Moment Method/ Review

Here, a suitable inner product can be defined:

*,u v uv d

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Moment Method/ Review

Example: Find the inner product of u(x)=1-x and

v(x)=2x in the interval (0,1). Solution: In this case u and v are real functions.

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Moment Method/ Review

Hence:

1

0, (1 )2u v x xdx

2 31

0, 2 0.333

2 3

x xu v