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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.