ene 429 antenna and transmission lines theory
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ENE 429 Antenna and Transmission lines Theory. Lecture 14 Antenna problems and Radar. Review (1). Small loop antenna (magnetic dipole) Dipole antenna generates high radiation resistance and efficiency For far field region, where. Review (2). Half-wave dipole. - PowerPoint PPT PresentationTRANSCRIPT
Small loop antenna (magnetic dipole)
Dipole antenna generates high radiation resistance and efficiencyFor far field region,
where
0cos( cos ) cos( )
2 22 sin
j rl l
I eH j a
r
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202
15( , ) ( ) ,r
IP r F a
r
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2
cos( cos ) cos( )2 2( )
sin
l l
F
max
( )( )
( )n
FP
F
Half-wave dipole2
2max
cos ( cos )( ) 2( )( ) sinn
FP
F
p = 7.658, Dmax = 1.64, Rrad = 73.2
Image theory is employed to build a quarter-wave monopole antenna.
4
5
6
Monopole antenna is excited by a current source at its base.
Directivity is doubled and radiation resistance is half of that of dipole antenna.
7
The best operation: ground is highly conductive (or use counterpoise in case of remote antenna)
Shorter than /4 antenna arises highly capacitive input impedances, thus efficiency decreases.
Solution: inductive coil or top-hat capacitor
Inductive coil Top-hat capacitor
A group of several antenna elements in various configurations (straight lines, circles, triangles, etc.) with proper amplitude and phase relations, main beam direction can be controlled.
Improvement of the radiation characteristic can be done over a single-element antenna (broad beam, low directivity)
8
To simplify,1. All antennas are identical.2. Current amplitude is the same.3. The radiation pattern lies in x-y
plane
9
From
Consider ,
00 sin ,
4
j rIl eE j a
r
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2 0
0 .4
j rIl eE j a
r
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Let I1 = I0, I2 = I0ej,
since r1 and r2 >> d/2 for far field,
we can assume 1 2 and r1 r2 r.
10
1 21 0 2 0
1 24 4
j r j r
totI l I le e
E j a j ar r
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11
0 20 2 cos( cos )4 2 2
j r j
totI l e d
E j e ar
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But the exponential terms cannot be approximated, then
1 cos2d
r r
2 cos2d
r r
2 2 2
20 02 2( , ) 4cos ( cos )
2 232r
I l dP r a
r
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We can write this as
Funit = a unit factor or the maximum time-averaged power density for an individual element at
Farray = array factor =
whereThis depends only on distance d and relative current phase, .We can conclude that the pattern function of an array of identical elements is described by the product of the element factor and the array factor.
12
( , , )2
runit arrayP r F F a
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2
24cos ( )2
cos .d
We will simplify assumptions as follows:1. The array is linear, evenly spaced along the
line. 2. The array is uniform, driven by the same
magnitude current source with constant phase difference between adjacent elements.
13
2 3 ( 1)1 0 2 0 3 0 4 0 0, , , ,...j j j j N
NI I I I e I I e I I e I I e
2 ( 1)00 (1 ... )4
j rj j j N
totI l e
E j e e e ar
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2
2
sin ( )2
sin ( )2
array
N
F
(Farray)max = N2
Yagiuda (rooftop antenna)
14
Parasitic elements are indirectlydriven by current induced in themfrom the driven element.
Consider power transmission relation between transmitting and receiving antennas where particular antennas are aligned with same polarization.
15
Let Prad1 be Ptotal radiated by antenna 1 have a directivity Dmax1,
11 max12( , , )
4radPP r Dr
12 1 2 max1 22( , , ) .
4rad
rec
PP P r A D A
r
With reciprocal property,
Therefore, we have
21 2 1 max2 12( , , ) .
4rad
rec
PP P r A D A
r
max1 max2
1 2
.D DA A
Each variable is independent of one another, so each term has to be constant, we found that
16
max1 max22
1 2
4.
D DA A
Effective area (Ae) is much larger than the physical cross section.
More general expressions
We can also write
17
22
1( , , )rec
e
PA
P r
2 ( , ) ( , )4rad
rec t r
PP D A
r
2
( , ) ( , ) .4
rect r
rad
PD D
P r
Finally, consider Prad = etPin, Pout = erPrec, and Gt = etDt, Gr = erDr
18
2
( , ) ( , ) .4
outt r
in
PG G
P r
Friis transmission equation
Note: Assume - matched impedance condition between the transmitter circuitry/antenna and receiver
- antenna polarizations are the same.
Additional impedance matching network improves receiver performances
19
2 2
2( ) 4oc oc
recant in rad
V VP
Z Z R
in antZ Z
Since the receiver is matched, half the received power is dissipated in the load, therefore
Without the matching network,
20
212 2
Lrec
L
VP
Z
.LL oc
L ant
ZV V
Z Z
A monostatic radar system Some of energy is scattered by target so called
‘the echo signal’ received at the radar antenna.
Let Prad be the radiated power transmitted by the radar antenna, then the radiated power density P1(r, , ) at the target at the distance r away is
The power scattered by the target is then
s = radar cross section (m2)
11 2( , , ) ( , )
4radPP r Dr
2 1( , , )rad sP P r
This scattered power results in a radiated power density at the radar antenna of
Then
By manipulation of these equations, we have
or
2 12 2 2 2( , , ) ( , )
4 (4 )rad rad
s
P PP r D
r r
1 2 ( , , ) .rec eP P r A
221
3 41
( , )4
rec s
rad
PD
P r21
4 21
.4
rec se
rad
PA
P r
Radiation patterns for dipole antenna
http://www.amanogawa.com/archive/DipoleAnt/DipoleAnt-2.html
Ex1 Suppose a 0.5 dipole transmitting antenna’s power source is 12-V amplitude voltage in series with a 25 source resistance as shown. What is the total power radiated from the antenna with and without an insertion of a matching network?
0.5