1. antenna is a structure designed for radiating and receiving em energy in a prescribed manner ...
TRANSCRIPT
Antenna is a structure designed for radiating and receiving EM energy in a prescribed manner
Far field region ( the distance where the receiving antenna is located far enough for the transmitter to appear as a point source)
The shape or pattern of the radiated field is independent of r in the far field.
Normalized power function or normalized radiation intensity
2
22Lr
max
( , , )( , )n
P rP
P
Directivity is the overall ability of an antenna to direct radiated power in a given direction.
An antenna’s pattern solid angle:
Total radiated power can be written as
Antenna efficiency e is measured as
3
( , )p nP d
maxmax max
( , ) 4( , )
( , )n
n ave p
PD D
P
2max .rad pP r P
.rad rad
rad diss rad diss
P Re
P P R R
If the current distribution of a radiating element is known, we can calculate radiated fields.
In general, the analysis of the radiation characteristics of an antenna follows the three steps below:
1. Determine the vector magnetic potential from known of assumed current on the antenna.
2. Find the magnetic field intensity from .
3. Find the electric field intensity from .
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4
From the point form of Gauss’s law for magnetic field,
0B ��������������
Define
therefore
we can express as
where Jd = current density at the point source (driving point) R = distance from the point source to the observation point (m)
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( ) 0A ��������������
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0
4dJA dVR
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From here we can determine , then find in free space.
We can then find the electric field from
The time-averaged radiated power is
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0 00 .rE a H
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The subscript “0” represents the observation point.
0 01
( , , ) Re( )2
P r E H
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W/m2.
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A short line of current that is short compared to the operating wavelength. This thin, conducting wire of a length dl carries a time-harmonic current
0( ) cos( )i t I t A
and in a phasor form 0jI I e A.
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The current density at the source seen by the observationpoint is
A differential volume of this current element is dV = Sdz.
.j Rd z
IJ e a
S
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Therefore
Then
Where at the observation point.
For short dipole, R r, thus we can write
Conversion into the spherical coordinate gives
.j Rd zJ dV Ie dza
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/ 20
0/ 24
j Rl
zl
eA I dza
R
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0A A����������������������������
00 .4
j r
ze
A Il ar
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cos sin .z ra a a 10
Therefore
We can then calculate for 0.B��������������
00 (cos sin ).4
j r
re
A Il a ar
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11
Multiply 2 to both nominator and denominator, so we have
We are interested in the fields at distances very far from the antenna, which is in the region where
1r
20
0 2
1sin .
4j r j
A Il e ar r
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2r
or
2R
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Under a far-field condition, we could neglect
and
Then
and
Finally, W/m2.
2
1
( )R
3
1.
( )R
0 sin4
j rIl eH j a
r
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00 00 sin .
4
j r
rIl e
E a H j ar
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2 2 2
202 2( , ) sin
32r
I lP r a
r
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Since the current along the short Hertzian dipole is uniform,we refer the power dissipated in the radial distance Rrad to I,
2maxrad pP r P
2
max
( , )( , ) sinn
P rP r
P
2 2 8sin sin sin
3p d d d
2 2 2
240 .I l
2
2rad
rad
I RP or 2 280 ( )rad
lR m.
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Assume a <<
A complicate derivation brings to
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0
0
sin4
j rIS eH a
r
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0 sin4
j rIS eE a
r
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If the loop contains N-loop coil then S = Na2
Longer than Hertizian dipole therefore they can generate higher radiation resistance and efficiency.
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Divide the dipole into small elements of Hertzian dipole. Then find and .
Figure of dipole
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The current on the two halves are Symmetrical and go to zero at the ends.
We can write Where
Assume = 0 for simplicity.
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( , ) ( ) cosi z t I z t
0
0
sin( ( )); 02 2( )
sin( ( )); 02 2
j
j
L LI e z z
I zL L
I e z z
From
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sin4
j rI dl edH j a
r
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0 / 20
/ 2 0
sin( ( )) sin ' sin( ( )) sin '4 2 2
j r j rL
L
I e L e LH j a z dz z dz
R R
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In far field but since small differences can be critical.
, 'r R ������������� �
j r j Re e
We can write
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cosR r z
( cos )j R j r ze e
0 / 2cos cos0
/ 2 0
sin( ( )) sin( ( ))4 2 2
sinL
j z j z
L
j rI L LH j a e z dz e z dz
e
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From
In our case
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2 2cos( ) sin( ) cos( )ax
ax ee c bx dx a c bx b c bx
a b
, cos , ,2l
x z a j c b
0cos( cos ) cos( )
2 22 sin
j rl l
I eH j a
r
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0 0rE a H Ha
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202
15( , ) ( ) r
IP r F a
r
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1. Find Pn(), calculate F() over the full range of for length L in terms of wavelength then find Fmax (this step requires Matlab)
2. Find p
3. Dmax (Directivity)
4. Rrad
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2
max
cos( cos ) cos( )2 2 2( ) sin
l l
dF
4
p
max
30( ) pF
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2L
2 2L
2
202 2
cos ( cos )15 2( , )sin
rI
P r ar
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20
max 2
15IP
r
2
2max
cos ( cos )( ) 2( )( ) sinn
FP
F