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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