Transcript 9 - web page for staff

Lecture 9 Types of Antenna
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
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)
2L2
r

The shape or pattern of the radiated field is independent
of r in the far field.
Normalized power function or normalized radiation
intensity


P( r ,  ,  )
Pn ( ,  ) 
Pmax
2

Directivity is the overall ability of an antenna to direct
radiated power in a given direction.
Pn ( ,  ) max 4
Dmax  D( ,  ) max 

Pn ( ,  ) ave  p

An antenna’s pattern solid angle:
 p   Pn ( ,  )d 

Total radiated power can be written as
Prad  r 2 Pmax  p .

Antenna efficiency e is measured as
Prad
Rrad
e

.
Prad  Pdiss Rrad  Rdiss
3


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 A from
known of assumed current J on the antenna.
2. Find the magnetic field intensity H from A.
3. Find the electric field intensity E from H .
4
From the point form of Gauss’s law for magnetic field,
 B  0
Define
therefore
we can express
B   A
  (  A)  0
A as
A
0 J d
dV

4 R
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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 From here we can determine B
, then find H in free space.
 We can then find the electric field from
E 0  0 a r  H 0 .
The subscript “0” represents the observation point.
 The time-averaged radiated power is

1
P(r , ,  )  Re( E 0  H 0 )
2
W/m2.
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1.
2.
3.
Hertzian dipole (electric dipole)
Small loop antenna (magnetic dipole)
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
i(t )  I 0 cos(t   )
and in a phasor form
A
I  I 0 e j A.
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The current density at the source seen by the observation
point is
I  j R
Jd  e
az .
S
A differential volume of this current element is dV = Sdz.
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Therefore
Then
J d dV  Ie j R dza z .
0 l / 2 e  j  R
A0 
I
dza z

4  l / 2
R
Where A0  A at the observation point.
For short dipole, R  r, thus we can write
0 e  j  r
A0 
Il
az.
4
r
Conversion into the spherical coordinate gives
a z  cos ar  sin  a .
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Therefore
0 e  j  r
A0 
Il
(cos  a r  sin  a ).
4
r
We can then calculate for
B0 .
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Multiply 2 to both nominator and denominator, so we have
 j
0
1 
2  j r
A0 
Il  e


 sin  a .
2
4
 r r 


We are interested in the fields at distances very far from the
antenna, which is in the region where
r
r

2
1
or R

2
12
Under a far-field condition, we could neglect
and
1
( R)2
1
.
3
( R)
Il  e  j  r
Then
H0  j
sin  a
4 r
Il0  e  j  r
E 0  0 a r  H 0  j
sin  a .
and
4
r
 0  2 I 2 l 2  2
2.
Finally,
W/m
P( r ,  )  
sin

a
r
2 2 
32

r 

13
P( r ,  )
Pn (r , ) 
 sin 2 
Pmax
8
 p   sin 2  d    sin 2  sin  d d 
3
Prad  r 2 Pmax  p
 2 I 2l 2
 40
.
2

Since the current along the short Hertzian dipole is uniform,
we refer the power dissipated in the radial distance Rrad to I,
I 2 Rrad
2 l 2
m.
or Rrad  80 ( )
Prad 

2
14
15
i  1cos(2 10 t )
9
a) Pmax at r = 100 m
16
b) What is the time-averaged power density at P (100, /4,
/2)?
c) Radiation resistance
17
Assume a << 
A complicate derivation brings to
0 IS  e  j  r
H
sin  a
40
r
0 IS  e  j  r
E
sin  a
4
r
If the loop contains N-loop coil then S = Na2
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
Longer than Hertizian dipole therefore they can generate
higher radiation resistance and efficiency.
Divide the dipole into small
elements of Hertzian dipole.
Then find H and E .
Figure of dipole
19
H
E
20
The current on the two halves are
Symmetrical and go to zero at the ends.
We can write i ( z , t )  I ( z ) cos t
Where
L
L

j
 I 0e sin(  ( 2  z )); 0  z  2
I ( z)  
 I e j sin(  ( L  z ));  L  z  0
2
2
 0
Assume  = 0 for simplicity.
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From
I  dl e  j  r
dH  j
sin  a
4
r
 j r
L/2 e
 I 0  0 e j r
L
L

H  j
a  
sin(  (  z )) sin  ' dz  
sin(  (  z )) sin  ' dz 
4
R
2
R
2
0
 L / 2

In far field r  R,    ' but e  j  r  e  j  R since small
differences can be critical.
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We can write
R  r  z cos 
 e j R  e j ( r  z cos )
 j r
 I0e
H  j
4
0
L/2
L
L


j  z cos 
j  z cos 
sin  a   e
sin(  (  z )) dz   e
sin(  (  z )) dz

2
2
0
L / 2
23
ax
e
ax
From  e cos(c  bx)dx  2 2  a sin(c  bx)  b cos(c  bx) 
a b
In our case
x  z, a  j  cos  , c 
l
2
, b  
l
l 

I 0 e  j  r  cos( 2 cos  )  cos( 2 ) 
H j
a


2 r
sin 




E  0 a r  H  0 H a
15I 02
P(r ,  ) 
F ( )a r
2
r
24
where
l
l 

cos(
cos

)

cos(
)

2
2
F ( )  

sin 




F ( )
Pn ( ) 
F ( ) max
 Pmax
2
15I 02

F ( ) max
2
r
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1.
2.
3.
Find Pn(), calculate F() over the full range of  for
length L in terms of wavelength then find Fmax (this step
requires Matlab)
2
l
l 
Find p

2
 cos( 2 cos  )  cos( 2 ) 

d



F ( ) max
sin 




Dmax (Directivity)

4. Rrad

4
p
30

F ( )max  p
26

Link to Matlab file
27
L
L
2


2

2


2 
15 I  cos ( 2 cos  ) 
P(r , ) 
ar
2


r 
sin 



15I 02
Pmax 
 r2
2
0
2

cos ( cos  )
F ( )
2
Pn ( ) 

F ( ) max
sin 2 
2
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Using Matlab, we get
p = 7.658
Dmax = 1.64
Rrad = 73.2 
This is much higher than that of the Hertzian dipole.
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