ee757 numerical techniques in electromagnetics lecture 9mbakr/ece757/lecture9_2016.pdf ·...

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EE757

Numerical Techniques in Electromagnetics

Lecture 9

2 EE757, 2016, Dr. Mohamed Bakr

Differential Equations Vs. Integral Equations

Integral equations may take several forms, e.g.,

b

a

tttxKxf d)(),()(

b

a

tttxKxxf d)(),()()(

Most differential equations can be expressed as integral

equations, e.g.,

bxa xFxdd ,),(/22

CdtttFxddx

a1))(,(/ )(1 aC

CxCdtttFtxxx

a21))(,()()( )()(2 aaaC

3 EE757, 2016, Dr. Mohamed Bakr

Green’s Functions

Green’s functions offer a systematic way of converting a

Differential Equation (DE) to an Integral Equation (IE)

A Green’s function is the solution of the DE corresponding

to an impulsive (unit) excitation

Consider the differential equation L = g, where L is a

differential operator, is the unknown field and g is the

known given excitation

For this problem, the Green’s function G(r,r’) is the

solution of the DE LG =(r’) subject to the same boundary

conditions

For an arbitrary excitation we have

volumeexcitation

),()( vdGgΦ rrr

4 EE757, 2016, Dr. Mohamed Bakr

Green’s Functions: Examples

Obtain the Green’s function for the DE

subject to =f on the boundary B

The Green’s function is the solution of

G can be decomposed into a particular integral and a

homogeneous solution G=F+U with F and U satisfying

Switching to polar form we get

gΦyx )//(2222

)()(),,,(2 yyxxyxyxG

),()(2 yyxxF 02 U

yy ,xx F

,0

1

CAF 1ln

A is obtained using 1lim0

dlF

R 2A=1

R

x’, y’

5 EE757, 2016, Dr. Mohamed Bakr

Green’s Function: Examples (Cont’d)

The method of images can also be applied to obtain an

infinite series expansion of Green’s functions

Consider the case of a line charge between two

conducting planes

G(x,y,x’,y’) represents the potential at (x, y) due to a

line charge of value 1.0 c/m located at (x’, y’)

Original problem +q

y’

h-y’

6 EE757, 2016, Dr. Mohamed Bakr

Green’s Function: Examples (Cont’d)

+q y’

h-y’

-q y’

h-y’

-q h+y’

+q

2h-y’

+q

An infinite number of

charges is required to

maintain the same

boundary conditions

7 EE757, 2016, Dr. Mohamed Bakr

The potential caused by a 1 c/m line charge in an unbounded

medium is given by

Using the figure, we conclude that the Green’s function is

given by the infinite series

Special mathematical techniques are usually utilized to sum

such a slowly convergent series

2

ln4

1)( V

Green’s Function: Examples (Cont’d)

12222

2222

2222

)2()(ln)2()(ln

)2()(ln)2()(ln)1(

)()(ln)()(ln

4

1),,,(

n nhyyxxnhyyxx

nhyyxxnhyyxx

yyxxyyxx

yxyxGn

8 EE757, 2016, Dr. Mohamed Bakr

Green’s Function: Examples (Cont’d)

The Green’s function can also be expanded in terms of

the eigenfunctions of the homogeneous problem

As an example consider the wave equation

Let the eigenvalues and eigenfunctions be kj and j

The set j is an orthonormal set, i.e.,

,02

2

2

2

2

k

yxSubject to B

non0or0

022 jjj k

ji

ji dxdy

Sij

0,

,1*

9 EE757, 2016, Dr. Mohamed Bakr

Green’s Function: Examples (Cont’d)

We then expand the Green’s function in terms of the

eigenfunctions

But as the Green’s function satisfy

),()(1

yxay,xy,x,Gj

jj

)()()(22 yyxxy,xy,x,G k

)()()(1

22 yyxxkkaj

jjj

Substitute for G

),()(*

1

*22 yxds kka ij S

jijj

Multiply by and integrate *

i

kk

yxa

i

ii 22

*),(

10 EE757, 2016, Dr. Mohamed Bakr

Green’s Function: Examples (Cont’d)

Using Green’s functions, construct the solution for the

Poisson’s equation

Subject to V(0, y)=V(a, y)=V(x, o)=V(x, b)=0

Show that

),,(2

2

2

2

yxfy

V

x

V

)sin()sin(2

b

yn

a

xm

abmn

,2

22

2

22

b

n

a

mmn

b

n

a

m

b

yn

a

xm

abAmn

2

22

2

22

sinsin2

a b

ydxdyxfyxyxGyxV0 0

),(),,,(),(

11 EE757, 2016, Dr. Mohamed Bakr

Dyadic Green’s Functions

Dyadic Green’s functions are used to express the situation

where a source in one direction gives rise to fields in

different directions

In general, a dyadic Green’s function will have 9

components

kkkjkijkjjji

ikijiiG

GGGGGG

GGGz ,yxz,yx

zzzyzxyzyyyx

xzxyxx

),,,(

For a unit source in the x direction

we obtain the field E=G.J =

For a general source (arbitrary distribution and orientations)

)()()( zzyyxx iJ

kji GGG zxyxxx

vdzyxz,y,xz,y,(x,zy,xV

),,(.)),( JGE

12 EE757, 2016, Dr. Mohamed Bakr

Inner Products

The inner product of two functions is a scalar that must

satisfy the following conditions:

fggf ,,

hg,hf,hgf ,

commutative

distributive

0if0, f ff*

0iff0, f ff*

Example: 1

0

)()()(),( dxxgxfxgxf

13 EE757, 2016, Dr. Mohamed Bakr

Adjoint Operators

For an operator L, we sometimes define an adjoint operator

La defined by gL fgLf a,,

For the DE , f(0)=f(1)=0

We utilize the inner product

)(22 xgx/dfd xddL 22 /

1

0

)()()(),( dxxgxfxgxf

dxxgxd

fdgLf

1

02

2

)(, dx xd

gdf

dx

dgfg

dx

df

1

02

21

0

if g(0)=g(1)=0, we have Lgfdx f xd

gdgLf ,,

1

02

2

L=La

14 EE757, 2016, Dr. Mohamed Bakr

Method of Moments (MoM)

MoM aims at obtaining a solution to the inhomogeneous

equation Lf = g, where L is a known linear operator, g is a

known excitation and f is unknown

Let f be expanded in a series of known basis functions f1, f2,

, fN

Substituting in the equation we get

We define a set of N weighting functions w1, w2, , wN

ff nn

n

n

nn gfL )( n

nn gfL )( (One equation in N unknowns)

15 EE757, 2016, Dr. Mohamed Bakr

MoM (Cont’d)

N m gwfLw mn

nmn ,,2,1,,)(,

(N equations in N unknowns)

Taking the inner product of both sides with the mth weighting

function we obtain

In matrix form we can write gl mnmn

fLwfLwfLw

fLwfLwfLw

fLwfLwfLw

l

NNNN

N

N

mn

,,,

,,,

,,,

21

22212

12111

16 EE757, 2016, Dr. Mohamed Bakr

MoM (Cont’d)

,2

1

N

n

gw

gw

gw

g

N

m

,

,

,

2

1

The unknown coefficients are thus given by gl mmnn

1

The unknown function f can now be expressed in the

compact form

glff fffffmmn

-

nnn

N

nNnn

12

1

21

~~

17 EE757, 2016, Dr. Mohamed Bakr

MoM Example

Solve d2f/dx2=1+4x2, f(0)=f(1)=0 using MoM

We choose the basis functions as fn=xxn+1, n=1, 2, , N

f is thus approximated by

Also we choose wn=fn, n=1, 2, N (Galerkin’s approach)

our inner product is

We have Lfn=d2fn /dx2=n(n+1)xn-1

Show that lmn=<wm, Lfn>=mn/(m+n+1)

N

n

nn xxf

1

1)(

1

0

)()(, dxxgxfgf

))4)(2(2/()83(, mmmmgwg mm

18 EE757, 2016, Dr. Mohamed Bakr

MoM Example (Cont’d)

For N=1, we have l11=1/3, g1=11/30 1=11/10

For N=2, we have

For N=3, we have

Exact solution is obtained for N=3!

12/7

30/11

5/42/1

2/13/1

2

1

3/2

10/1

2

1

70/51

12/7

30/11

7/915/3

15/42/1

5/32/13/1

3

2

1

3/1

0

2/1

3

2

1

19 EE757, 2016, Dr. Mohamed Bakr

Types of Basis Functions

Entire domain basis functions fn are defined for the entire

domain of the function f

Subsectional basis functions are defined only over a

subsection of the domain of the function f

xn-1 xn-2 xn xn+1 xn+2

x

P(x-xn) 1

xn-1 xn-2 xn xn+1 xn+2

x

n n+1 n+2

n-1 n-2

Pulse functions in 1D

20 EE757, 2016, Dr. Mohamed Bakr

Types of Basis Functions (Cont’d)

xn-1 xn-2 xn xn+1 xn+2

x

T(x-xn) 1

xn-1 xn-2 xn xn+1 xn+2

x

Triangular functions in 1D

21 EE757, 2016, Dr. Mohamed Bakr

Types of Weighting Functions

N m gwfLw mn

nmn ,,2,1,,)(,

Recall that

If we choose wn=fn , n=1, 2, , N (Galerkin matching)

N m gffLf mn

nmn ,,2,1,,)(,

If we choose wn=(r-rn) , n=1, 2, , N (Point matching)

N m gfL mn

nmn ,,2,1,),()(),( rrrr

N m gfL mn

mnn ,,2,1),())(( rr

The two sides of the system equation are matched at a

number of space points