EMLAB
1
2. Radiation integral
EMLAB
2
EM radiation
R
l
t
IlI
R
ej
Rt
jkR
/4)(
4)(rE
)(tI
Constant velocity
Constant accel-eration
Periodic motion
Accelerating charges radiate E and H proportional to 1/R.
+q
+q
+q
EMLAB
3Basic laws of EM theory
0
B
D
DJH
BE
t
t1) Maxwell’s equations
0
t
J2) Continuity equation (the relation between current density and
charge density in a space)
)(0
0
MHB
PED
3) Constitutive relation (explains the properties of materials)
4) Boundary conditions ( should be satisfied at the interface of two materials by E, H, D, B.)
EMLAB
4Continuity equation : Kirchhoff ’s current law
Q I
J
tSq nJ ˆ
Charges going out through dS.
nFor steady state, charges do not accumulate at any nodes, thus ρ become constant.
.currentSteady;0
t
J
t
dt
dddt
dd
dddtdQ
VVVC
VC
J
JaJ
aJ
dS
t
Jt
dQI
nn
differential form
integral form
Kirchhoff ’s current law
EMLAB
5
Boundary conditions
C unit vector tangential to the surface
tt EE 21
Snn DD 12
S n
Unit vector normal to the surface
S
h
Medium #1
Medium #2
τ
w
nn BB 21
KJhHH tt 21
0)( 21 SBBdd nnSVaBB
whJdwHHddSCS
aJsHaH )( tan2tan1
0)ˆˆ( 21
1
wwddC
SτEτEsEaE
ShdSdadS VV
)ˆˆ( 12 nDnDDD
h
EMLAB
6Two important vector identities
0)( A
0
0)()(11
CCSVdddd rArAaAA
1)
2)
0)()()()( 2112
1
2
2
1
1
2
2
1
rrrr
sssa
dsds
dds
ds
ddddd
CS
1C
1C
(ϕ : arbitrary scalar function)
(A: arbitrary vector function)
EMLAB
7
0)1( B
Potentials of time-varying EM theory
AB 0)( A
tt
tt
AE
AE
AE
AE
0)(
t
B
E)2(
t
D
JH)3(
ttt
AJ
EJA
1
AJA
Att
2
22
D)4(
t
t
)(2 A
A
EMLAB
8Lorentz condition
AB
To find unique value of vector potential A, the divergence and the curl of A should be known.
Only the curl of A is physically observed, divergence of A can be arbitrarily set.
0
t
A
For the above choice, (3), (4) become
Lorentz condition
JA
AJA
A
2
2
22
2
22 1
tct
2
2
22
2
22 1
tct
0
t
A Lorentz condition
EMLAB
9Fourier transform solution
JA
A
2
2
22 1
tc
2
2
22 1
tc
•boundary condition : free space
1. Because the number of variables are as many as four (x, y, z, t), we apply Fourier transform to the above equations.
dedec
detdet
tjtj
tjtj
~
2
1~~
2
1
),(~2
1),(,),(
~
2
1),(
2
22
rrrr
2. For a non-homogeneous differential equation, it is easier to substitute the source term with a delta function located at origin. (effectively it is an impulse response.)
~~~
2
22
c
EMLAB
10
Solution of Maxwell’s eqs for simple cases
EMLAB
11
,J ,J
Domain : infinite space
Domain : interior of a rectangular cavity
,J,J
Domain and boundary conditionsThe constraints on the behavior of electric and magnetic field near the interface of two media which have different electromagnetic properties. (e.g. PEC, PMC, impedance boundary, …)
Domain : interior of a circular cavity
waveguide
EMLAB
121-D example : Radiation due to Infinite current sheet
1. Using phasor concept in solving Helmholtz equation,
2
,~~~ 22
ckk JAA
xy
z
2. With an infinitely large surface current on xy-plane, variations of A with coordinates x and y become zero. Then the Laplacian is reduced to derivative with re-spect to z.
JAA ~~~
22
2
kz
)(ˆ 0 zJ sxJ
3. If the current sheet is located at z=0, it can be repre-sented by a delta function with an argument z. If the current flow is in the direction of x-axis, the only non-zero component of A is x-component.
)(~~
~
02
2
2
zJAkz
Asx
x
),(~
),(~
),(~
2
22 rJrArA
c
EMLAB
13
4. With a delta source, it is easier to consider first the re-gion of z≠0.
0~
~2
2
2
xx Ak
z
A
kz
kz
e
CA
jkz
x
sin
cos~
5. Four kinds of candidate solutions can satisfy the differ-ential equation only. Of those, exponential functions can be a propagating wave..
6. Solutions propagating in either direction are
)cos(}Re{
)cos(}Re{
)0(
)0(~)(
)(
2
1
kzte
kzte
zeC
zeCA
kztj
kztj
jkz
jkz
x
)0( z)0( z
7. The condition that A should be continuous at z=0 forces C1=C2
EMLAB
14
ssxx JdzzJdzAkdz
z
A00
22
2 ~)(
~~~
)0(2~~
2
2
jkCjkCejkCez
Adz
z
A jkjkxx
zjkx eCA
8. To find the value of C, integrate both sides of the orig-inal Helmholtz equation.
011~~~
0
0
jk
e
jk
eCdzAdzAdzA
jkjk
xxx
jk
JCJjkC s
s 2
~~
2 00
zjksx e
jk
JzA
2
~)(
~ 0
)0or 0(2
~~ˆ
~ 0
zzeJ
z
A zjksx yAB
)0()space free(377,2
~ˆ
~ 0 zjks e
Jj xAE
t
stjzjks dczJdee
kj
Jt
)/||(
2ˆ
2
)(~
2
1ˆ),( 0
||0 xxrA
2
)/||(ˆ),( 0 czJ
t s
xrB
2
)/||(ˆ),( 0 czJ
t s
xrE
EMLAB
15
)(ˆ 0 zJ sxJ
E
E
H
H
Propagating direction
An infinite current sheet generates uniform plane waves whose amplitude are uniform throughout space.
Plane wave 정의
E
H
Electric field : even symmetry
Magnetic field : odd symmetry
ttJ s cos)(0
Propagating direction
EMLAB
16
,JSource
JAA 22 k
Infinitesimally small current element in free space : 3D
JA
A
2
2
22 1
tc
EMLAB
17
Solution of wave equations in free space
JA
A
2
2
22 1
tc
2
2
22 1
tc
•Boundary condition: Infinite free space solution.
1. As the solutions of two vector potentials are identical, scalar potential is consid-ered first.
2. To decrease the number of independent variables (x, y, z, t), Fourier transform rep-resentation is used.
~~~
),(),(~,),(),(~
2
22
c
dtetdtet tjtj rrrr
3. For convenience, a point source at origin is considered.
EMLAB
18
)(22 r gkg
022 gkg
0)()(
0)(1 2
2
22
2
2
rgkr
rggk
r
rg
r
Green function of free space
ck
where
A suitable solution which is propagating outward from the origin is e-jkr.
kr
kr
e
r
Ag
jkr
cos
sinr
Aeg
jkr
1. The solution of the differential equation with the source function substituted by a delta function is called Green g, and is first sought.
2. With a delta source, consider first the region where delta function has zero value. Then, utilize delta function to find the value of integration constant.
3. With a point source in free space, the solution has a spherical symmetry. That is, g is independent of the variables , , and is a function of r only.
EMLAB
19
1)(22 VVVdgdkgd r
)0(01)1(4
4
sin
0
2
2
0 0 0
222
jk
jkr
jkr
V
ekA
drreAk
drddrr
eAkgdk
)0(4)1(4
sin)1(
0
22
0 2
2
AejkA
ddrr
ejkrA
dggd
jk
jk
SVa r
ergA
jkr
4)(,
4
1
.,4
),( rrrr
RR
eg
jkR
Green function of free space
4. To determine the value of A, apply a volume integral operation to both sides of the differential equations. The volume is a sphere with infinitesimally small radius and its center is at the origin.
5. With a source at r’, the solution is translated such that
EMLAB
20
~~~ 22 k
dV
)()'(~)(~ rrrr
)(),(),( 22 rrrrrr gkg rrrr
RR
eg
jkR
,4
),(
V
jkR
Vd
R
edg
4
)(~),(
)(~),(
~ rrr
rr
V
V
cRtj
tj
V
jk
tj
dR
cRt
ddeR
dede
det
4
)/,(
),(~2
1
4
1
4
),(~
2
1
),(~
2
1),(
)/(
r
rr
r
6. As the original source function can be represented by an integral of a weighted delta function, the solution to the scalar potential is also an integral of a weighted Green function.
7. Taking the inverse Fourier transform, the time domain solution is obtained as follows.
EMLAB
21
V
dR
cRtt
4
)/,(),(
rr
Retarded potential
(Retarded potential)
V
dR
cRtt
4
)/,(),(
rJrA
0
t
A
The distinct point from a static solution is that a time is retarded by R/c. This newly derived potential is called a retarded potential.
The vector potential also contains a retarded time variable.
Those A and are related to each other by Lorentz condition.
EMLAB
22Solution in time & freq. domain
' 4
)/,(),(
Vd
R
cRtt
r
r
V
dR
cRtt
4
)/,(),(
rJrA
V
jkR
dR
e
4
)(),(
rr
V
jkR
dR
e
4
)(),(
rJrA
V
jkR
V
jkR
V
jkR
deR
jkRd
R
e
dR
e
2
1ˆ4
1)(
4
4
)(),(
1),(
JRrJ
rJrArH
V
jkR
dR
kRjkR
R
kRjkR
R
e
k
)(
)(33)(1
4
1),(
4
2
2
2
2JRRJrE
EMLAB
23
V
jkR
dR
ej
4)ˆ(ˆ)( JRRJrE
V
jkR
dR
ejk
)ˆ(
4)( JRrH
Far field approximation
Electrostatic solution
V R
d2
ˆ
4
11),(
RJArH
Vd
R
24
ˆ)(),(
RrrE
Biot-Savart’s law
0k
Coulomb’s law
EMLAB
24Electric field in a phasor form
V
V
jkR
VV
dtR
dR
ej
dgj
gdj
)]ˆ(ˆ[4
1
4)]ˆ(ˆ[
)(1
'
JRRJ
JRRJ
rJJE
jj
t
)( AA
AE
''
)(1
)]()([1
VVdg
jdRg
jj
rJrJ
A
R
eRgdRg
jkR
V
4)(,)()(
'
rJA
EMLAB
25Radiation pattern of an infinitesimally small current
R
ezIjd
R
ej
jkR
V
jkR
4sinˆ
4)]ˆ(ˆ[
JRRJE
sinˆ)cosˆˆ()ˆ(ˆ
ˆ
00
0
JzJ
Jz
rJrrJ
J
ˆ
ˆˆ
0sincos
cossincossinsin
sincoscoscossin
ˆ
ˆ
ˆ r
z
y
x
z
r
sinz
I
EMLAB
26Example – wire antenna
coscos222 zrzrrrR rrz
o
r
C
zjkjkr
V
jkR
zdezJr
ej
dR
ej
cos)(4
sin
4)ˆ(ˆ)( JRRJrE2/l
2/l
02/)2/(sin
2/0)2/(sin)(
0
0
zlzlkI
lzzlkIzJ
2coscos
2cos
2sin 0 klkl
r
eIj
jkr
Current distributions along the length of a linear wire antenna.
EMLAB
27Poynting’s theorem and wave power
)()()( ttt HES
Electromagnetic wave power per unit area(Poynting vector)
}Re{2
1 *HES
Average wave power per unit area
EMLAB
28
r
z
o
r
R
cosz
cos)ˆˆ()ˆˆ(222 zrrrrrrrR rrrrrr
zrr ˆr
EMLAB
29
C
zjkjkr
C
rjkjkr
zdezJr
ejdezJ
r
ej
cos)ˆˆ( )(
4sin)(
4sin)( rrrE
N
n
njn
N
n
dnjkn eIeIAF
11
cos Array factor :
Array factor
1I
2I
3I
1z
2z
3z
4z
x
y
z
4I
r
z-directed arrayrr ˆr
rr ˆr
d
),cos( jnnn eIIkd
EMLAB
30
N
n
xjkn
jkrN
nn
neIer
klj
0
cos
0total sin
4ˆ
θEE
d
z
x0I2I
xl
1I
d
Array factor
0x 1x2x
x-directed array
Top view
1R 2R
1 2
1r 2r
r
Ox
r
EMLAB
31
Typical array configurations
EMLAB
32How to change currents on elementary antennas?
Magnitudes and phases of currents on elementary antennas can be changed by amplifiers and phase shifters.
N
n
dnjkneIAF
1
cos
EMLAB
33
Equi-phase surface
Equi-phase surface
,1,1,1,1,1 44
33
2210
jjjj eIeIeIeII
Pattern synthesis
EMLAB
34
1R 2R
1 2
1r 2r
r
Ox
r
(1) Two element array
2,1,1 10
dII
10
20
30
40
50
30
210
60
240
90
270
120
300
150
330
180 0
10
20
30
40
50
30
210
60
240
90
270
120
300
150
330
180 0
,12
,1,1
10
10
II
dII
(2) Two element array
Examples
EMLAB
35
10
20
30
40
50
30
210
60
240
90
270
120
300
150
330
180 0
10
20
30
40
50
30
210
60
240
90
270
120
300
150
330
180 0
(3) Five element array
,2
.1,1,1,1,1 44
33
2210
d
eIeIeIeII jjjj
(4) Five element array
0,2
.1,1,1,1,1 44
33
2210
d
eIeIeIeII jjjj
(5) Five element array
0,8.0
.1,1,1,1,1 44
33
2210
d
eIeIeIeII jjjj
10
20
30
40
50
30
210
60
240
90
270
120
300
150
330
180 0
Beam direction
EMLAB
36
phi=0:0.01:2*pi; %0<phi<2*pik=2*pi;d=0.5;% 0.5 lambda spacing.shi=k*d*cos(phi); alpha = pi*0.0;beta = exp(i*alpha);%Currents=[1,2*beta, 3*beta^2,2*beta^3,1*beta^4]; %Current excitationsCurrents=[1, 1*beta, 1*beta^2, 1*beta^3,1*beta^4]; %Current excitations E=freqz(Currents,1,shi); %E for different shi values E = DB(E)+30; % 최대값에서 30dB 범위까지 그림 .E = (E>0.).*E; polar(phi,E); %Generating the radiation pattern
Sample MATLAB codes
EMLAB
37
N-element linear array antenna
Uniform Array : Magnitudes of all currents are equal. Phases increase monotonically.
cosd
z
d
d
d
1r
cosd
2r3r
4r
Nr
1
2
3
4
N
y
1 cos1cos2cos
1
cos1
kdNjkdjkdj
N
n
kdnj
eee
eAF
)cos (
1
1
kd
eAFN
n
nj
EMLAB
38
121 Njjj eeeAF
)2/sin(
)2/sin(2/)1(2/2/
2/2/2/
2/
Ne
ee
ee
e
eAF Nj
jj
jNjNjN
j
jNNjjjj eeeeAFe 12
Difference :
j
jNjNj
e
eAFeAFe
1
1 1)1(
)2/sin(
)2/sin(
N
NAF
• Universal Pattern is symmetric about y = .p
• Width of main lobe decrease with N
• Number of sidelobes = (N-2)
• Widths of sidelobes = (2π/N)
• Side lobe levels decrease with increasing N.
EMLAB
39Visible and invisible regions
kdkd 0
Array Factor 의 특성
Array factor has a period of 2 p with re-
spect to ψ.
Of universal pattern, the range covered by
a circle with radius “kd” become visible
range.
The rest region become invisible range
)2()( AFAF
2
1
1
1coskd
kd
visible region
2nAF
Visible range of the lin-
ear array
EMLAB
40Grating Lobes Phenomenon
2
122 dkd
If the visible range includes more than
one peak levels of universal pattern,
unwanted peaks are called grating
lobes.
To avoid grating lobes, the following
condition should be met. 2
1)(f
1
1coskd
kd2
visible region
grating lobes
major lobe
They have the same strength !
2
Example :2
1/22 and For dkd , no grating lobe occurs
1/2 and 0For dkd , no grating lobe occurs
EMLAB
41Automotive radar antenna
EMLAB
42Analog beamforming : phase shifter
EMLAB
43
LPF A/D
DigitalSignal
Processing(Amplitude
&Phase)
~
Desired signal
direction
LPF A/D
LPF A/D
LPF A/D
Interference or
multipath
signal direction
Digital Beamformer (DBF)