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)