Prof. David R. JacksonDept. of ECE

Fall 2013

Notes 22

ECE 6340 Intermediate EM Waves

1

Radiation

Infinitesimal Dipole

Current model:

iz

I IlJ

x y x y z

1kl

y

x

l

z

I

y

x y

z

x

z l

2

Radiation (cont.)

Define

Then

1,

( ) 2 2

0

x xx

P x x

otherwise

( , , ) ( ) ( ) ( ) ( )izJ x y z Il P x P y P z

As 0, ( ) ( )x P x x

, , 0x y z Letting

( , , ) ( ) ( ) ( ) ( )izJ x y z Il x y z

( , , ) ( ) ( )izJ x y z Il ror

3

Maxwell’s Equations:

From (3):

(1)

(2)

0 (3)

/ (4)

ic

v

E j H

H J j E

H

E

1H A

(5)

Note: c accounts for conductivity.

Radiation (cont.)

Note: The Harrington book uses

H A

4

ˆ( , , ) ( ) ( )iJ x y z z Il r

From (1) and (5):

or

1( )E j A

( ) 0E j A

Hence E j A

E j A (6)

Radiation (cont.)

This is the “mixed potential” form for E.5

or

Substitute (5) and (6) into (2):

Use the vector Laplacian identity:

1( ) [ ]i

cA J j j A

2( ) icA k A J j

2 ( ) ( )A A A

The vector Laplacian has his nice property in rectangular coordinates:

2 2 2 2ˆ ˆ ˆx y zA x A y A z A

icH J j E

Radiation (cont.)

6

Hence

or

Then we have

2 2 ( )icA k A J A j

2 2( ) icA A k A J j

cA j

Lorenz Gauge:

2 2 iA k A J

Choose

Radiation (cont.)

2 2ck

Note: Lorenz, not Lorentz (as in Lorentz force law)!

7

Take the z component:

Hence we have

2 2 iz z zA k A J

2 2 ( ) ( )z zA k A Il r

Note:2 2 0x xA k A 2 2 0y yA k A

0xA 0yA so

Radiation (cont.)

8

Spherical shell model of 3D delta function:

2 2 ( ) ( )z zA k A Il r

Radiation (cont.)

y

x

a

z

Sa

The Az field should be symmetric (a function

of r only) since the source is.

2

1( )

4r r a

a

In spherical coordinates:

24 1a

V a

r dV r r dr

Note :

9

Assume

For

( ) ( )zA r R r

0r 2 2 0z zA k A

so that 2 22

10

d dRr k R

r dr dr

Next, let1

( ) ( )R r h rr

we have that

Radiation (cont.)

2 22

( ) 1 1( ) ( )

( ) ( )

dR rr r h r h r

dr r r

rh r h r

10

22

22

2

10

10

1( ) ( ) 0

d hrh h k

r dr rh

h rh h kr r

h r k h rr

( ) jkr jkrh r Ae Be

We choose exp (-jkr) for r > a to satisfy the radiation condition at infinity.

Solution:

Radiation (cont.)

We choose sin(kr) for r < a to ensure that the potential is finite at the origin.

( ) / .R r h r rRecall that

11

,

sin ,

jkrAe r ah r

B kr r a

Radiation (cont.)Hence

2 22 2

1 1( ) ( ) ( )

4

d dRr k R Il r Il r a

r dr dr a

22

1 1( ) ( ) ( )

4h r k h r Il r a

r a

The R function satisfies

2 1( ) ( ) ( )

4h r k h r Il r a

a

12

Also, from the differential equation we have

Radiation (cont.)

2 1( ) ( ) ( )

4

a a a

a a a

h r dr k h r dr Il r a dra

1( )

4h a h a Il

a

h a h a

We require

13

(BC #1)

(BC #2)

Hence

Radiation (cont.)

,

sin ,

jkrAe r ah r

B kr r a

sin

1cos ( )

4

jka

jka

Ae B ka

A jk e Bk ka Ila

Recall

14

(BC #2)

(BC #1)

Substituting Ae-jka from the first equation into the second, we have

Radiation (cont.)

1sin cos ( )

4jk B ka Bk ka Il

a

Hence we have

1 1

( )4 sin cos

B Ila k j ka ka

sin1( )

4 sin cos

jkae kaA Il

a k j ka ka

From the first equation we then have

15

Letting a 0, we have

Radiation (cont.)

sin1( )

4 sin cos

1( )

4

jkae kaA Il

a k j ka ka

Il

1( )

4A Il

Hence

1 1 1( )

4jkr jkr

zA r R r h r Ae Il e r ar r r

16

so

Final result:

ˆ ( )4

ˆˆ cos sin ( )4

jkr

jkr

eA z Il

r

er Il

r

Radiation (cont.)

yl

z

I

( , , ) ( ) ( )izJ x y z Il rx

Note: It is the moment Il that is important. 17

Extensions

(a) Dipole not at the origin

ˆ ( )4

jk r reA z Il

r r

y

x

r

z

r r

r

18

Extensions (cont.)

(b) Dipole not in the z direction

x

y

r

z

r r

rp̂

ˆ ( )4

jk r reA p Il

r r

19

(c) Volume current

y

x

( )iJ r

z

( )4

jk r ri

V

eA J r dV

r r

Extensions (cont.)

dl

ˆ

ˆ( )( )

i izJ dV z J dS dl

z I dl

dS

20

ˆ ( )4 4

jk r r jk r rie e

d A z Idl J dVr r r r

Hence

Consider the z component of the current:

The same result is obtained for the current components in the x and y directions, so this is a general result.

(d) Volume, surface, and filamental currents

( ) ( )4

( ) ( )4

ˆ( ) ( ) ( )4

jk r ri

V

jk r ris

S

jk r ri

C

eJ r dV

r r

eA r J r dS

r r

er I r d

r r

volume current density

surface current density

filament of current

( )

( )

ˆ ( )

i

i is

J dV

J dV J dS

I d

volume current density

surface current density

filamentary (wire) current

Extensions (cont.)

21Note: The current Ii is the current that flows in the direction of the unit vector.

ˆ ( ) ( )4

jk r ri

C

eA r r I r d

r r

Extensions (cont.)

22

Note: The current Ii is the current that flows in the direction of the contour.

Filament of current:

CA

BI

Fields

To find the fields:

1

1( 0).

c

H A

E H rj

valid for

1

c

E j A

j A Aj

Note: Ampere’s law (the equation above) for finding the electric field is an alternative to using the following equation:

23

cA j Recall :

Fields (cont.)

Results for infinitesimal, z-directed dipole at the origin:

2

2

1 11 cos

2

1 1 1( ) 1 sin

4 ( )

1 1( ) 1 sin

4

jkrr

jkr

jkr

IlE e

r jkr

IlE j e

r jkr jkr

IlH jk e

r jkr

c

with

24

Note on Dissipated PowerNote: If the medium is lossy, there will be an infinite amount of power dissipated by the infinitesimal dipole.

2

22 2

0 0 0

22 2 2

0 0 0

2 2 2

0

2 22 2

30

1

2

1sin

2

1sin

2

1 4 22

2 3 3

1 4 2 12 1 2

2 3 3 4

d

V

r

N Nr

c

P E dV

E r d d dr

E E r d d dr

E E r dr

Ilr dr

r

2

0

2

0

4sin sin

3

2cos sin

3

d

d

Note:

The N superscript denotes that the dependence is suppressed in the fields.

(power dissipation inside of a small spherical region V of radius )

25

Far Field

Far-field: ( 1)kr

2

10

~ ( ) sin4

~ ( ) sin4

r

jkr

jkr

E Or

jE Il e

r

jkH Il e

r

behaves as

E

H k

Note that The far-zone field acts like a homogeneous plane wave.

26

Radiated Power (lossless case)

*

2

22 2

0 0

2 2

0

1ˆP Re ( )

2

1Re

2

1Re

2

1sin

2

2sin

2

rad

S

S

S

E H r dS

E H dS

EdS

E r d d

E r d

The radius of the sphere is chosen as infinite in order to simplify the fields.

27

We assume lossless media here.

Radiated Power (cont.)

or2

2 2 2

0

22 3

0

P ( ) (sin ) sin4

( ) sin4

rad Il r dr

Il d

3

0

4sin

3d

Hence2

2 4P ( )

4 3rad Il

Integral result:

28

Use

k

22P ( )12rad

kIl

2k

Simplify using

22 4

P ( )4 3rad

kIl

22

2

4P ( )

12rad Il

Radiated Power (cont.)

Then

or

29

2

2P W3rad

lI

Note: The dipole radiation becomes significant when the dipole length is significant relative to a wavelength.

Radiated Power (cont.)

30

Far-Field Radiation From Arbitrary Current

( , , )4

jk r ri

V

eA J x y z dV

r r

y

x

( )iJ r

z

r r

r

r

r

r

31

12 2 2 2

12 2 2 2 2 2 2

2 2 2 2

2

2

2 2

ˆ1 2

r r x x y y z z

x y z x y z xx yy zz

r r r r r r r r

r r rr

r r

211 ~ 1 ...

2 8

xx x

Far- Field (cont.)

Use Taylor series expansion:

x

32

2 2

3

ˆ ˆ1 1 2 11

2 8

r r r r rr r r O

r r r r

2 2

2

1 1 1 1ˆ ˆ

2 2r r r r r r r r O

r r

As r

ˆr r r r r

Far- Field (cont.)

33

ˆr r r r r

Physical interpretation:

Far- Field (cont.)

Far-field point (at infinity)

y

x

z

r r

r

r

r

34

r̂

Hence

( )jkre

rr

Define:

( , ) ( )i jk r

V

a J r e dV

Far- Field (cont.)

ˆ

( ) ( , , )4

~ ( )4

jk r ri

V

jkri jkr r

V

eA r J x y z dV

r r

eJ r e dV

r

ˆ ˆ ˆ ˆsin cos sin sin cosk kr x k y k z k

(This is the k vector for a plane wave propagating towards the observation point.)

and

35

ˆr r r r r

The we have

( )jkre

rr

where

( , ) ( )i jk r

V

a J r e dV

Far- Field (cont.)

( ) ~ ,4

A r r a

sin cos sin sin cosk r x k y k z k

and

36

with

( , ) ( )i jk r

V

a J r e dV

Fourier Transform Interpretation

Then

sin cos

sin sin

cos

x

y

z

k k

k k

k k

Denote

( , ) ( , , ) x y zj k x k y k zi

V

a J x y z e dV

The array factor is the 3D Fourier transform of the current.

37

Magnetic Field

1H A

We next calculate the fields far away from the source, starting with the magnetic field.

38

In the far field the spherical wave acts as a plane wave, and hence

ˆjk jkr

1ˆH jk r A

~ ,4

A r a

1ˆ

4H jk r r a

Recall Hence

We then have

Note: Please see the Appendix for an alternative calculation of the far field.

ˆ~ ( , ) ( )4

jkH r a r

The far-zone electric field is then given by

1

1ˆ

1ˆ ˆ

4

ˆ ˆ4

c

c

c

E Hj

jkr Hj

jkjkr r a

j

j r r a

Electric FieldWe start with

39

ˆ ˆ ,4

,4

t

E j r r a

j a

Electric Field (cont.)We next simplify the expression for the far-zone electric field.

40

( ) ~ ,4

A r r a

tE j A

Recall that

Hence

Hence, we have

~ , ,tE j A r

Electric Field (cont.)

Note: The exact field is

E j A

41

The exact electric field has all three components (r, , ) in general, but in the far field there are only (, ) components.

Relation Between E and H

Hence

,EE

H H

ˆ4

ˆ4

ˆ4 / (4 )

t

jkH r a

jkr a

jk Er

j

or

1ˆH r E

Start with

~

4

t

t

E j A

j a

42

Recall:

1c ck

Also

~ , ,tE j A r

Summary of Far Field

( )jkre

rr

( , ) ( )i jk r

V

a J r e dV

( , , ) ~ ,4

A r r a

1ˆH r E

43

(keep only and components of A)

sin cos sin sin cosx y zk r k x k y k z k x k y k z

Poynting Vector

* *

* *

*

* *

1

21

~21 ˆ ˆ ˆ ˆ21

ˆ2

1ˆ

2

t t

S E H

E H

E E H H

r E H E H

E EE Er

44

Assuming a lossless media,

22 2

ˆ ˆ2 2 2

EE ES r r

22 2

* * *ˆ ˆ

2 2 2

EE ES r r

Poynting Vector (cont.)

In a lossless medium, the Poynting vector is purely real (no imaginary power flow in the far field).

45

Far-Field Criterion

Antenna(in free-space)

D : Maximum diameter

D

r r

How large does r have to be to ensure an accurate far-field approximation?z

The antenna is enclosed by a circumscribing sphere of diameter D.

/ 2D

y

x

rr

/ 2r D46

Far-Field Criterion (cont.)

Let Df = maximum phase error in the exponential term inside the integrand.

22

2

ˆ1 1ˆ

2 2

r rrr r r r r O

r r r

22

max

ˆ1

2 2

r rrk

r r

2 2

max

( / 2)

2 2

kr k D

r r

Hence

NeglectError

47

Set

Then

2

max 8

kD

r

omax rad (22.5 )

8

2

8 8

kD

r

Far-Field Criterion (cont.)

48

Hence, we have the restriction

2

0

2Dr

2 2 2

0 0

2 2kD D Dr

Far-Field Criterion (cont.)

max rad8

for

49

“Fraunhofer criterion”

Note on choice of origin:

2

0

2Dr

Far-Field Criterion (cont.)

The diameter D can be minimized by a judicious choice of the origin. This corresponds to selecting the best possible mounting point when mounting an antenna on a rotating measurement platform.

50

aD aD

aD D 2 aD D

Mounted at center Mounted at edge

D = diameter of circumscribing sphereDa = diameter of antenna

Far-Field Criterion (cont.)

51

It is best to mount the antenna at the center of the rotating platform.

Platform (ground plane)Antenna

Rotating pedestal

Wire Antenna

Assume

y

+h

z

I (z')

x-h

0( ) sinI z I k h z

+h-h

Wire antenna in free space

52

Wire Antenna (cont.)

Hence

( , ) ( )i jk r

V

a J r e dV

ˆ ( )iJ dV z I z dz

sin cos sin sin cos

cos

ˆ( , ) ( )

ˆ ( )

ˆ ( )

hjk r

h

hj x k y k z k

h

hjz k

h

a z I z e dz

z I z e dz

z I z e dz

53

cos0ˆ( , ) sin

hjz k

h

a z I k h z e dz

ˆ ˆ ˆ ˆˆ ˆ( , )

ˆ sin

t z z

z

a z a za

a

( ) ( , )4 ttE j A j r a

Wire Antenna (cont.)

Recall that

For the wire antenna we have

54

0 2

cos( cos ) cos( )ˆ( , ) (2 )

sin

kh kha z I

k

The result is

ˆ~ sin ( , )4

jkr

z

eE j a

r

Hence

or

0 2

ˆ~ sin ( , )4

cos( cos ) cos( )ˆ sin (2 )4 sin

jkr

z

jkr

eE j a

r

e kh khj I

r k

k

Simplify using

Wire Antenna (cont.)

55

We then have

0

cos( cos ) cos( )ˆ~2 sin

jkre kh khE j I

r

Wire Antenna (cont.)

Note: The pattern goes to zero at = 0, . (You can verify this by using L’Hôpital’s rule.)

56

Wire Antenna: Radiated Power

22 2

0 0

22

0

22

0

2 2 22 2 2

0

0

1sin

2

1(2 ) sin

2

sin

1 1 cos( cos ) cos( )sin

2 sin

radP E r d d

r E d

r E d

kh khr I d

r

57

Radiated Power (cont.)

or

2

20

0

cos( cos ) cos( )

4 sinrad

kh khP I d

58

Input Resistance

2

2

1(0)

22

(0)

rad in

radin

P R I

PR

I

2

20

0

cos( cos ) cos( )

4 sinrad

kh khP I d

59

Zin

Rin

jXin

I (0)

0

0

(0) sin

sin( )

I I k h z

I kh

2 2

0

cos( cos ) cos( )1

2 sin( ) sinin

kh khR d

kh

/2 Dipole: ,4 2

h kh

73inR

Input Resistance (cont.)

(resonant)

60

61

ExampleA small loop antenna

x

y

z

I

a0a

The current is approximately uniform)

0I I

0

2sin cos

0

0

cos jk aya a I e ad

Assume = 0: yE E

ˆˆ( , ) ya ya a

0 sinxk k

cosx a

ˆ ˆˆ ˆ cosl y y

By symmetry:

0yk

The x component of the array factor cancels for points and - .

62

Example (cont.)

2

cos1

0

cos 2jxe d j J x

0

0

0

00 1 0

4

2 sin4

jk r

jk r

eE j a

r

ej j aI J k a

r

0 1 02 sina j aI J k a

Hence

0 sinx k a

Identity:

We then have

63

020 0 0ˆ sin

4

jk rk a I eE

r

1 , 12

xJ x x

Approximation of Bessel function:

The result is

0

0 01 0

ˆ sin2

jk raI eE J k a

r

Hence

Example (cont.)

Appendix

64

In this appendix we derive the far field expression for the magnetic and electric field from a radiating current source by using the curl operations in spherical coordinates, instead of using the plane-wave approximation for the del operator.

Hence

1ˆ, sin

sin

1 1 1 1

sinr r

aa r a

r

ra raa a

r r r r r r

1,a O

r

2

1 1a O O

r r

Therefore

Magnetic Field (cont.)

Note that a is not a function of r, so these derivative terms are O(1).

65

Therefore

2

ˆ ˆ

1ˆ

ˆ~

ˆ

jkr

jkr

jkr

er r

r r r

jkrr e

r

jkr e

r

jkr

Magnetic Field (cont.)

ˆ~4

jkH a r

Hence

1~

4H a

ˆjkr 66

We then have

ˆ~ ( , ) ( )4

jkH r a r

Magnetic Field (cont.)

ˆ~ ( , ) ( )4 t

jkH r a r

or

Note: The t subscript can be added without changing the result.

The t subscript means transverse to r. That is, keep only

the and components.

67

ˆ~ ( , ) ( )4

jkH r a r

The far-zone electric field is then given by

1

1ˆ

4

c

c

E Hj

jkr a

j

Electric FieldWe start with

68

(This follows from the same reasoning as before.)

ˆ ˆ ˆ

ˆ~

r a r a r a

r a

Hence

ˆ ˆ ˆ~

ˆ ˆ

ˆ ˆ

t

r a r a jkr

jk r a r

jk r r a

jk a

Electric Field (cont.)

69

Electric Field (cont.)We next simplify the expression for the far-zone electric field.

70

( , , ) ~ ,4

A r r a

tE j A

Recall that Hence

1ˆ

4

1

4

4

c

t

c

t

jkE r a

j

jkjk a

j

j a