ene 429 antenna and transmission lines theory
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Lecture 9 Types of Antenna
1
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
( , , )( , )nP 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 DP
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 .
A
J
H
A
E
H
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)
B A
( ) 0A
A
0
4dJA dVR
5
From here we can determine , then find in free space.
We can then find the electric field from
The time-averaged radiated power is
B
H
0 00 .rE a H
The subscript “0” represents the observation point.
0 01( , , ) Re( )2P r E H
W/m2.
6
1. Hertzian dipole (electric dipole)
2. Small loop antenna (magnetic dipole)
3. Dipole antenna
7
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.
8
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 aS
9
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
/ 20
0/ 24
j Rl
zl
eA I dzaR
0A A
00 .4
j r
zeA Il ar
cos sin .z ra a a 10
Therefore
We can then calculate for 0.B
00 (cos sin ).4
j r
reA Il a ar
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
1 sin .4
j r jA Il e ar r
2r or 2R
12
Under a far-field condition, we could neglect
and
Then
and
Finally, W/m2.
2
1( )R
3
1 .( )R
0 sin
4
j rIl eH j ar
00 00 sin .4
j r
rIl eE a H j ar
2 2 2
202 2( , ) sin
32r
I lP r a
r
13
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
( , )( , ) sinnP rP rP
2 2 8sin sin sin 3p d d d
2 2 2
240 .I l
2
2rad
radI RP or 2 280 ( )rad
lR m.14
15
91cos(2 10 )i t
16
a) Pmax at r = 100 m
b) What is the time-averaged power density at P (100, /4, /2)?
c) Radiation resistance
17
Assume a <<
A complicate derivation brings to
18
0
0sin
4
j rIS eH ar
0 sin4
j rIS eE ar
If the loop contains N-loop coil then S = Na2
Longer than Hertizian dipole therefore they can generate higher radiation resistance and efficiency.
19
Divide the dipole into small elements of Hertzian dipole. Then find and .
Figure of dipole
H
E
20
H
E
The current on the two halves are Symmetrical and go to zero at the ends.
We can write Where
Assume = 0 for simplicity.
21
( , ) ( ) cosi z t I z t
0
0
sin( ( )); 02 2( )sin( ( )); 02 2
j
j
L LI e z zI z
L LI e z z
From
22
sin4
j rI dl edH j ar
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
In far field but since small differences can be critical.
, 'r R j r j Re e
We can write
23
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
From
In our case
24
2 2cos( ) sin( ) cos( )ax
ax ee c bx dx a c bx b c bxa b
, cos , ,2lx z a j c b
0cos( cos ) cos( )2 2
2 sin
j rl l
I eH j ar
0 0rE a H Ha
202
15( , ) ( ) rIP r F ar
where
25
2
cos( cos ) cos( )2 2( ) sin
l l
F
max
( )( ) ( )nFP F
20
max max2
15 ( )IP Fr
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
26
2
max
cos( cos ) cos( )2 2 2( ) sin
l l
dF
4p
max30 ( ) pF
Link to Matlab file
27
28
2L
2 2L
22
02 2
cos ( cos )15 2( , )sin
rIP r ar
20
max 2
15IPr
2
2max
cos ( cos )( ) 2( ) ( ) sinnFP F
Using Matlab, we get
29
p = 7.658
Dmax = 1.64
Rrad = 73.2
This is much higher than that of the Hertzian dipole.
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