01 Electromagnetic Radiation

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Hon Tat Hui Electromagnetic Radiation

NUS/ECE EE4101

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Electromagnetic Radiation

1. Radiation Mechanism

When electric charges undergo acceleration or

deceleration, electromagnetic radiation will be

produced. Hence it is the motion of charges (i.e.,

currents) that is the source of radiation.

Yet not all current distributions will produce a strong

enough radiation for communication. We will firststudy some typical current distributions and the

radiation fields that they produce.


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2. Vector and Scalar PotentialsFrom Maxwells fourth equation:

0

0 A

For any vector function A,

AB

So we can write:


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ABE

j

jFrom Maxwells first equation:

Then,

0 AE j

For any scalar function ,

0

So we can write:j E A

That is, j E A


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From Maxwells second equation:

2 2

2 2

1

j

j j

j

k j

H J D

A J A

A A J A

A A J A

We can further specify the divergence of A according to

Lorentzs gauge as:

j A

2 2k


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Using Lorentzs gauge, we have:2 2k A A J

Now, the first, the second, and the last Maxwells

equations have been satisfied. To satisfy the third one,put into the third equation,j E A

2 2

j

k

D

E

A


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A and are called vector and scalar potentials and theysatisfy the following inhomogeneous Helmholtz equations:

2 2

2 2

2 2

k

k

k

A A J

Note that each component ofA is governed by the same

scalar equation as that for . Hence it suffices to solve

only one scalar equation, namely, the inhomogeneous

Helmholtz equation.


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3. Solutions to the Vector and Scalar PotentialsSolutions to the vector and scalar potentials are (see

Supplementary Notes):

'

'

1 '4

( ) '4

jkR

v

jkR

v

e' dvR

e

' dvR

R R

A R J R

R'R

R

R= field point

R = source point

R = |R-R|


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1

1

j

H A

E H

Once the potentials are known, the electric and magneticfields can be found from:


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4. Hertzian Dipole (length


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'''42

2

dzdydxR

eAI

x,y,z

d

d

jkR

A

zA

r

zyx

zzyyxx'R

222

222RR

Here x = y = 0 & z 0

(source at origin)

Very short dipoleIndependent of primed coordinate!

constant/ AIAzz zIJCurrent density:


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A has only thezcomponent. ConvertAz to spherical (Ar,

A,A) components first.

0

sin4

sin

cos4

cos

A

reIdAA

r

eIdAA

jkr

z

jkr

zr

Therefore:

re

Idzd

r

e

Ix,y,z

jkr

d

d

jkr

4

4

2

2

zzA


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r

e

krjkr

kIdjE

r

e

jkrr

IdE

r

e

jkr

jkIdH

EHH

jkr

jkr

r

jkr

r

2

2

1114

sin

11

2

cos

11

4

sin

0

E and H fields can now calculated and expressed inspherical coordinates as:

1

1

j

H A

E H


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Near fields: When kr


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Far fields:

(important case)

When kr>> 1, all terms vary with the

factors 1/r2 and 1/r3 vanish.

r

ekIdjH

r

ekIdjE

jkr

jkr

sin4

sin

4

Note that for far fields:

1. Er=Hr=E = 0.

2. E

H

and transverse to the rdirection, a TEM wave.

3. BothE andH are in phase..4. In free space, the wave impedance =E/H is equal

to 0, i.e., = 0 = 120 = 377.


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Electric field lines surrounding a Hertzian dipole at a given instant

See animation Far Field Electric 2D See animation Far Field Electric 3D


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5. Half-Wave Dipole (length = 0.5

)

R

Dipole

R= r -zcos

R

r

0forsin

0forsinsin)(

z'z'hkI

z'z'hkIz'hkIz'I

m

m

m

Assumption: zI )(z'I

h = /4

Far Field point R

(r,,)

y

x

z

z

See Supplementary Notes

for the exact derivation ofR.


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That is, the current distribution is a sinusoidal function asshown below:

I(z)

z

Im

z'hkIz'I m sin)(

0


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For a half-wave dipole, the exact field solutions are toocomplicated. Hence only the far fields will be

determined. The half-wave dipole can be considered as

an assembly of many Hertzian dipoles joined together.The far fields of the half-wave dipole are then the

summations of the far fields of the Hertizan dipoles.

r

edzIk

jdE

jkr

z sin4

)'(0'

Far-zone electric field of a Hertzian dipole at the origin:


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Far-zone electric field of a Hertzian dipole at an arbitraryposition R:

R

edzkI

jdE

jkR

sin4

)'(

Far-zone electric field of a half-wave dipole:

h

h

jkR

h

h

jkR

dR

ezI

kj

dR

ezkIjdEE

)'(4

sin

sin4

)'(


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Now put in the current expression for I(z) and use the

following substitutions forR (far-field approximation):

cos'

11

jkzjkrjkR eee

rR

Note:R = r -zcos

h

jkzjkr

m

h

h

jkzjkr

m

dzezhkr

ekIj

dzezhkr

ekIjE

0

cos'

cos'

')'(sin24

sin

')'(sin4

sin

We have,


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The integration can be performed to yield the following

result:

60jkr

m

eE j I Fr

EH

sin

cos2cos)( F

where


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6. Quarter-Wave Monopole

Equivalent to

(image theorem)

h = /4

h = /2

Large conducting plane


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6.1 Image Theorem

For antennas mounted over or near a ground plane (a

very large perfectly conducting plane), virtual

sources (images) can be place below the groundplane to account for reflections from the ground

plane. After introducing the image sources, the

electromagnetic field above the ground plane can beconsidered as a sum of the electromagnetic fields

due to the real sources (above the ground plane) and

the image sources (below the ground plane), with the

ground plane removed. This is the image theory.


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Note that the image theory can only be applied to find

the fields above the ground plane but not below the

ground plane. Below the ground plane, the

electromagnetic field is strictly zero.

6.2 Method to place the image currents

1. The image currents are at the same perpendicular

distances (for example along the z axis) from theground plane as the real currents.

2. The image currents have the same parallel

coordinates (for example the x and y coordinates)as the real currents.


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3. For vertical real currents, the image currents havethe same direction as the real currents. But for

horizontal real currents, the image currents have

the opposite directions as the real currents.

z

z

z

z

real current real current

image currrent image current


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Using the image theorem, a /4 monopole antenna fed by

a source at its base radiates the same far fields in the

region above the ground plane as a /2 dipole antenna.

But there is no radiation below the ground plane. This

situation applies to other vertical wire antennas placed

above a large conducting pane, such as a Hertzian dipole.


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Example 1

Find the maximum electric field intensity E of a half-wave

dipole at a distance of 10 km from the dipole. What is the

direction for maximum field intensity? Assume that thedipole carries a current whose maximum value is Im at the

middle point of the dipole and the current varies at a

frequency of 3 GHz.

Solution

For a half-wave dipole, the electric field intensity in the farfield region is:


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sin

cos2cos60

reIjE

jkr

m

It has only the component.

This field is maximum when = /2.

r

eIjE

jkr

m 602/

At 3 GHz, = 0.1 m, k= 2/= 20. Therefore at r= 10

km,

1000060

200000

2/

j

m

eIjE


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The variation ofE with time at r= 10 km is:

2000001032sin10000

60

10000

60Rekm10,

9

200000

2/

tI

ee

IjrtE

m

tjj

m


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References:

1. David K. Cheng, Field and Wave Electromagnetic, Addison-

Wesley Pub. Co., New York, 1989.

2. John D. Kraus,Antennas, McGraw-Hill, New York, 1988.

3. C. A. Balanis, Antenna Theory, Analysis and Design, John Wiley& Sons, Inc., New Jersey, 2005.

4. W. L. Stutzman and G. A. Thiele, Antenna Theory and Design,

Wiley, New York, 1998.

5. Fawwaz T. Ulaby, Applied Electromagnetics, Prentice-Hall, Inc.,

New Jersey, 2007.

6. Joseph A. Edminister, Schaums Outline of Theory and Problems

of Electromagnetics, McGraw-Hill, Singapore, 1993.7. Yung-kuo Lim (Editor), Problems and solutions on

electromagnetism, World Scientific, Singapore, 1993.

01 Electromagnetic Radiation

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