Dipole Antennas (Cont`d..)

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Transcript Dipole Antennas (Cont`d..)

Chapter 4
Antenna
Chapter Outlines
Chapter 4
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Antenna
Fundamental Parameters of Antennas
Electrically Short Antennas
Dipole Antennas
Monopole Antennas
Antenna Arrays
Helical Antennas
Yagi Uda Array Antennas
Antennas for Wireless Communications
2
Introduction
Wires passing an alternating current emit or radiate EM
energy. The shape and size of the current carrying structure
determine how much energy is radiated, direction of
propagation and how well the radiation is captured.
Henceforth, the structure to efficiently radiate in a preferred
direction is called transmitting antenna, while the other side
which is to capture radiation from preferable direction is
called receiving antenna.
In most cases, the efficiency and directional nature for an
antenna is the same whether its receiving or transmitting.
3
Introduction (Cont’d..)
Common types of antennas:
ure 8-1 (p. 389)
4
Introduction (Cont’d..)
Generic antenna network. The antenna acts as a
transducer between guided waves on the T-line and
waves propagating in space.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
5
4.1 Fundamental Parameters of
Antennas
To describe the performance of an antenna, definitions of
various parameters are discussed. The radiated power,
beam pattern, directivity, and efficiency are all
important parameters in characterizing antenna.
RADIATED POWER
Suppose transmitting antenna located at the origin of spherical
coordinate. From this coordinate system, there are three
components of radiated field, in r, θ and φ. But, the intensities
of these components vary with radial distance as 1/r, 1/r2 and
1/r3.
Fundamental Parameters of
Antennas (Cont’d..)
For
almost
all
practical
applications, a receiving antenna
located far enough away from the
transmitter (as a point source of
radiation)  far field region.
A distance r from the origin is
generally accepted as being in the
far field region if :
2L2
r Figure 8-3

(p. 392)
The spherical radiation pattern for an isotropic antenna.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
L is the length of the largest dimension on the antenna element,
and assumed L>λ. For smaller L, r should at least as large as λ
Fundamental Parameters of
Antennas (Cont’d..)
In the far field, the radiated waves resemble plane waves
propagating in ar direction where time harmonic fields related
by :
E S   0a r  H S
HS 
1
0
ar  ES
 0  120 
The time averaged power density vector of the wave is by
Poynting Theorem,

1
Pr , ,    Re E S  H*S
2
Where in the far field,
Pr , ,    Pr , ,   a r

Fundamental Parameters of
Antennas (Cont’d..)
So, the total power radiated by the antenna, Prad is found by
integrating it over a close spherical surface :
Prad   Pr , ,  .dS    Pr , ,  r sin dd
2
Example 1
In free space, suppose a wave propagating radially away
from an antenna at the origin has:
Hs 
Is
r
cos2  a ,
j
I

I
e
Where the driving current phasor S
0
Find:
•
ES
•
P(r,φ,θ)
•
Prad and Rrad
10
Solution to Example 1
To find ES, from time harmonic fields relation:
 Is

2
Es  a P  H s  oa r  
cos  a  ,
 r

o I s
cos 2  a
So, (a)E s 
r
Then, to find P(r,φ,θ)
j
 j



I
e

I
e
1
1
*
2
2
o o
o
P  Re E s  H s   Re 
cos  a 
cos  a 
2
2  r
r

2
1  Io 
(b) P  r ,  ,    o   cos 4  a r
2  r 
Solution to Example 1 (Cont’d..)
Then,
Prad  Pr ,  ,    dS 

1 2
I 0 Rrad
2
4
1
2 cos 
 0 I 0
a r r 2 sin dda r
2
r2


2
0
0

1
  0 I 0 2 cos 4  sin d
2

2
 0 I 0 2
5
2
2

120

I


o
Solving:
5
Rrad 
 96 2 
1 2
Io
2
(c) Rrad  950
 d
Fundamental Parameters of
Antennas (Cont’d..)
RADIATION PATTERNS
From Balanis book:
Fundamental Parameters of
Antennas (Cont’d..)
Radiation patterns usually indicate either electric field, E
intensity or power intensity. Magnetic field intensity, H has
the same radiation pattern as E related by η0.
The polarization or orientation of the E field vector is an
important consideration in an E field plot. A transmit receive
antenna pair must share same polarization for the most
efficient communication.
Fundamental Parameters of
Antennas (Cont’d..)
Coordinate System
Fundamental Parameters of
Antennas (Cont’d..)
Since the actual field intensity is not only depends on radial
distance, but also on how much power delivered to antenna, we
use and plot normalized function  divide the field or power
component with its maximum value.
E.g. the normalized power function or normalized radiation
intensity :
Pr , ,  
Pn  ,   
Pmax
Fundamental Parameters of
Antennas (Cont’d..)
If the antenna radiates EM waves equally in all directions, it is
termed as isotropic antenna, where the normalized power
function is equal to 1.
So,
Pn  ,  iso  1
In contrast with isotropic antenna, a directional antenna
radiates and receives preferentially in some direction.
Fundamental Parameters of
Antennas (Cont’d..)
The normalized radiation patterns for a generic antenna, called
polar plot. A 3D plot of radiation pattern can be difficult to
generate and work with, so take slices of the pattern and generate
2D plots (rectangular plots) for all θ at φ=π/2 and φ=3π/2
Polar
plot
Rectangular
plot (in dB)
Fundamental Parameters of
Antennas (Cont’d..)
Fundamental Parameters of
Antennas (Cont’d..)
The polar plot also can be in terms of dB. Where normalized E
field pattern,
E r , ,  
En  ,   
Emax
This will be identical to the power pattern in decibels if:
En  ,  dB  20 logEn  ,  
whereas
Pn  ,  dB  10 logPn  ,  
Fundamental Parameters of
Antennas (Cont’d..)
The are some zeros and nulls in radiation pattern, indicating no
radiations.
These lobes shows the direction of radiation, where main or
major lobe lies in the direction of maximum radiation. The other
lobes divert power away from the main beam, so that good
antenna design will seek to minimize the side and back lobes.
Beam’s directional nature is beamwidth, or half power
beamwidth or 3 dB beamwidth. It will shows the angular width
of the beam measured at the half power or -3 dB points.
Fundamental Parameters of
Antennas (Cont’d..)
For linearly polarized antenna, performance is often described in
terms of its principal E and H plane patterns.
E plane : the plane containing the E field vector & the direction
of max radiation
H plane : the plane containing the H field vector & the direction
of max radiation
For next figure,
•the x-z plane (elevation plane, φ=0) is the principal E-plane
•the x-y plane (azimuthal plane; θ=π/2) is the principal H-plane.
Fundamental Parameters of
Antennas (Cont’d..)
Fundamental Parameters of
Antennas (Cont’d..)
For this figure,
• Infinite number of
principal E-planes (elevation
plane, φ=φc).
• One principal H-plane
(azimuthal plane; θ=π/2).
Fundamental Parameters of
Antennas (Cont’d..)
DIRECTIVITY
A measure of how well an antenna radiate most of the power
fed into the main lobe. Before defining directivity, describe first
the antenna’s pattern solid angle or beam solid angle.
An arc with length equal to a
circle’s radius defines a radian.
Fundamental Parameters of
Antennas (Cont’d..)
An area equal to the square of a sphere’s
radius defines a steradian (sr).
A differential solid angle dΩ in sr is:
d  sin dd
For sphere, the solid angle is found by integrating dΩ :
2


 
sin dd  4 sr
 0  0
Fundamental Parameters of
Antennas (Cont’d..)
An antenna’s solid angle Ωp:
p 
 Pn  , d
To find the normalized power’s
average value taken over the entire
spherical solid angle :
Pn  ,  d  p

Pn  ,  ave 

4
d


Fundamental Parameters of
Antennas (Cont’d..)
The directive gain D(θ,φ) of an antenna is the ratio of the
normalized power in particular direction to the average
normalized power :
Pn  ,  
D ,   
Pn  ,  ave
The directivity Dmax is the maximum directive gain,
Dmax  D ,  max 
So,
Dmax
4

p
Pn  ,  max
Pn  ,  ave
where Pn  ,  max  1
Fundamental Parameters of
Antennas (Cont’d..)
Directivity in decibels as:
Dmax dB  10 logDmax 
Useful relation:
D ,    D ,  max Pn  ,  
Total radiated power as:
Prad  r 2 Pmax
 P , d
Or :
Prad  r 2 Pmax p
Fundamental Parameters of
Antennas (Cont’d..)
For (a), power gets radiated to the side and back lobes, so the
pattern solid angle is large and the directivity is small. For (b),
almost all the power gets radiated to the main beam, so pattern
solid angle is small and directivity is high.
Example 2
For this normalized radiation intensity,
Pn  ,   sin  sin  for 0     ,
2
3
0 otherwise.
Find the beamwidth, pattern solid angle and
the directivity.
31
Solution to Example 2
The beam is pointing in the +y direction.
WHY?!?!
Due to the beam having function in θ and φ, so that:
1
BW   BW  BW  .
2
From previous equation, Pn  ,  max  1 so that a 3dB
beamwidth is at half of total power = 0.5.
32
Solution to Example 2 (Cont’d..)
To find BWθ, we fix φ = π/2 to get:
Pn  ,    sin 2  1
and then set sin2θ equal to ½. Then,
 1
  sin 
  45 , so BW  180  45   45  90 .
 2
1
To find BWφ, we fix θ = π/2 to get:
Pn  ,    1sin 3 
33
Solution to Example 2 (Cont’d..)
and then set sin3θ equal to ½. Then,
  sin
So,
1

1 2 
1
3
  52.5 ,
so BW  180  52.5   52.5  75 .
1
BW   90  75   82.5 .
2
The pattern solid angle is:
 P   Pn d     sin 2  sin 3   sin  d d ,


0
0
 P   sin 3  d  sin 3 d , (note limits on  )
Solution to Example 2 (Cont’d..)
Where each integral is solved as follows:




0
0
0
0
y   sin 3 xdx   1  cos 2 x  sin xdx   sin xdx   cos 2 x sin xdx.
Please continue on your own!!
So,


0
0
 4  4 
 p  sin 3 d sin 3 d    
 3  3 
Finally,


 p  1.78sr
Dmax
and the directivity,
4
4


 7.1
 P 1.78
35
Fundamental Parameters of
Antennas (Cont’d..)
IMPEDANCE & EFFICIENCY
(a) A T-line terminated in a dipole antenna can be modeled
with an antenna impedance, Zant (b) consisting of resistive and
reactive components (c).
Fundamental Parameters of
Antennas (Cont’d..)
The antenna resistance consists of radiation resistance Rrad and
a dissipative resistance Rdiss that arises from ohmic losses in
the metal conductor. For antenna driven by phasor current,
Prad
1 2
 I 0 Rrad
2
and also
Pdiss
1 2
 I 0 Rdiss
2
For maximum radiated power, Rrad need to be as large as
possible but without being too large, for easily match with
the feedline.
Fundamental Parameters of
Antennas (Cont’d..)
So then the antenna efficiency, e
Prad
Rrad
e

Prad  Pdiss Rrad  Rdiss
The power gain G(θ,φ) is likely its directive gain plus efficiency,
where:
G ,    eD ,  
And the max power gain is when the directivity is max. It’s been
always expressed in dBi , indicating dB with respect to an
isotropic antenna.
4.2 Electrically Short Antennas
If the current distribution of a radiating element is known, it’s
possible to calculate the radiated fields by a direct integration
but, the integrals can be very complex.
For time harmonic fields, integration is performed to find a
phasor called retarded vector magnetic potential, which then
followed by simple differentiation to find the H field.
We will begin with a derivation of the retarded vector magnetic
potential, then find the radiated fields for Hertzian dipole
(infinitesimally short element with uniform current along its
length). From that, we can find the fields from longer
structures via integration e.g. small loop antenna.
Electrically Short Antennas
(Cont’d..)
Vector Magnetic Potential
In working with electric fields, E  V and analogous term
for H field is the vector magnetic potential A, often used for
antenna calculations.
Where in the point form of Gauss Law for magnetic fields,
B  0
With vector identity that states the div of the curl of any vector
A is zero. So,
B   A
Then we now seek a relation between vector A and a current
source.
Electrically Short Antennas
(Cont’d..)
From Biot Savart law,
Remember?!?
The vector magnetic
potential
at
the
observation point (o)
results from a current
density
distributed
about the volume vd.
0
a do
B0 
Jd 
dvd
4
Rdo2

Electrically Short Antennas
(Cont’d..)
Due to long derivation and by using vector identity, we could
get:
0
B0  0 
4
Then,
0
A0 
4
Jd
 Rdo2 dvd
Refer to text book!
Jd
 Rdo2 dvd
More properly,
0
A 0 x0 , y0 , z0  
4

J d  xd , y d , z d 
Rdo
2
dvd
Where
the
vector
potential at point o is a
function of the position
of current element
Electrically Short Antennas
(Cont’d..)
We have yet to consider time dependence, so:
0
A 0  x0 , y 0 , z 0 , t  
4



J d xd , y d , z d , t  Rdo / u p
dvd
Rdo
Where:


J d xd , yd , zd , t  Rdo / u p  J dse jt
Where Jds is the retarded phasor quantity:
J ds  J d xd , yd , z d e  jkRdo
k = β = 2π/λ
Electrically Short Antennas
(Cont’d..)
Using this current density, the phasor form is:
A 0S
0

4

J ds
dvd
Rdo
This is what we called as the time harmonic equation for the
retarded vector magnetic potential.
In phasor notation, the vector magnetic potential is related to
the magnetic flux density by :
B 0S    A 0S
Electrically Short Antennas
(Cont’d..)
From there, we could find H in free space :
H 0S  B 0S / 0
Because the radiation is propagating radially away from the
source, it is then a simple matter to find E, where:
E0S  0a r  H 0S
Finally, the time averaged power radiated is:

1
Pr , ,    Re E0 S  H*0 S
2

Electrically Short Antennas
(Cont’d..)
Hertzian Dipole
Suppose that a short line of current,
i(t )  I 0 cost   
It’s placed along the z axis as shown.
Electrically Short Antennas
(Cont’d..)
j
Here, the phasor current is: I s  I 0 e
To maintain constant current over its entire length, imagine a
pair of plates at the ends of the line that can store charge.
The stored charge at the ends resembles an electric dipole,
and the short line of oscillating current is then ‘Hertzian
Dipole’.
For Is in the +az direction through a cross sectional S, the
current density at the source seen by the observation point:
J ds
I s  jkR
 e
az
S
Electrically Short Antennas
(Cont’d..)
Geometrical arrangement of an
infinitesimal dipole and its
associated electric field
components on a spherical
surface.
Electrically Short Antennas
(Cont’d..)
A differential volume of this element,
dvd  Sdz
So,
J dsdvd  I s e  jkR dza z
Therefore,
A 0S
0

4
/2

 / 2
I s e  jkR
dza z
R
With assumption that the Hertzian dipole is very short,
integrate to get A0S :
A 0S
 0 I s e  jkR

az
4R
Electrically Short Antennas
(Cont’d..)
The unit vector az can be converted to its equivalent
spherical coordinates, so that:
A 0S
 0 I s e  jkR
cosa r  sin a 

4R
It is now a relatively straightforward matter to find B0S and
then,
H 0S
I s e  jkR

4R
1

 jk   sin a
R

Regroup to get:
H 0S
I s k 2e  jkR  j
1 


sin a
2


4
 kR kR 
Electrically Short Antennas
(Cont’d..)
The second terms drops off with increasing radius much faster
than the first term, where for far field condition:
1
1

kR kR2
Therefore,
I s ke jkR
H 0S  j
sin a
4R
Meanwhile, the far field value of E field :
I s ke jkR
E 0 S  j 0
sin a
4R
Electrically Short Antennas
(Cont’d..)
The time averaged power density at observation point is:
  I 2 2 k 2
Pr ,     0 0
 32 2 R 2

 2
 sin a
r


The term in brackets is
the max power density.
The sin2θ term is the
normalized radiation
intensity Pn(θ) plotted as :
Electrically Short Antennas
(Cont’d..)
The normalized radiation intensity can be used to find the
pattern solid angle,
p 

sin 2 d 
The directivity is then:

sin 2  sin dd  8 3
Dmax
4

 1.5
p
The total power radiated by a Hertzian dipole:
2 2 2


k
2
2  0 I0 
Prad  r Pmax p  r
 32 2 R 2


  40 2    I 2
0
 p

 

Electrically Short Antennas
(Cont’d..)
Also Rrad as:

Rrad  80 2  

2
For Hertzian dipole, where l<<λ, Rrad will be very small and the
antenna will not efficiently radiate power. Larger dipole
antennas, have much higher Rrad and thus more efficient.
Example 3
Suppose a Hertzian dipole antenna is 1 cm long
and is excited by a 10 mA amplitude current
source at 100 MHz. What is the maximum power
density radiated by this antenna at a 1 km
distance?
What is the antenna’s radiation
resistance?
55
Solution to Example 3
We should calculate the wavelength,
c
3x108 m s
cf,   
 3m.
6
f 100 x10 1 s
The max. power radiated is:
Pmax
o  I l
120  2 


32 r
32
32
2 2 2
o
2 2
2
 0.010   0.010 
2
2
10002
The antenna radiation resistance :
l 
2  0.01 
 80    80 
  8.8m

 3 
2
Rrad
2
2
2
pW
 0.052 2
m
Electrically Short Antennas
(Cont’d..)
Small Loop Antenna
A small loop of current located in the xy plane centered at the
origin  small loop antenna or magnetic dipole.
Electrically Short Antennas
(Cont’d..)
Substitute,
J dsdvd  I s e  jkRada
Giving us,
A 0S
 0 I s a e  jkR

da
4
R

Assume that a<<λ and A0s is in far field. It’s discovered that,
A 0S 
0 I s S
4R 2
1  jkRe  jkR sin a
Where, S  a 2 and thus,
H 0S
 0 I s Sk

sin e  jkRa
4 0 R
E0S
 0 I s Sk

sin e  jkRa
4R
Electrically Short Antennas
(Cont’d..)
The power density vector :
  2 2 I 2S 2k 2  2
0 0
 sin a
Pr ,   
r
 32  2 R 2 
0


Where,
Pmax 
 2 0 2 I 0 2 S 2 k 2
320 2 R 2
Since the normalized power function is the same as Hertzian
dipole, Ωp=8π/3, Dmax = 1.5
Electrically Short Antennas
(Cont’d..)
Try this!!
Calculation of Prad and Rrad yields
4 0 I 0  S 


Prad 
2
3
 
3
2
2
Rrad
4
S 
 320   
 2 
2
The fields for the small loop antenna is similar to Hertzian
dipole. It’s the dual of Hertzian (electric) dipole  magnetic
dipole.
The equations also valid for multi turn loop, as long as the loop
small compared to wavelength. For N circular loop, S=Nπa2 and
for square coil N loops, with each side length b, S=Nb2
4.3 Dipole Antennas
Dipole Antennas
A drawback to Hertzian dipole
as a practical antenna is its
small radiation resistance. A
longer will
have
higher
radiation resistance, becomes
more efficient. It as an L long
conductor conveniently placed
along the z axis with current
distribution i(z,t).
Dipole Antennas (Cont’d..)
Assume sinusoidal current distribution on each arm, where the
antenna is center-fed and the current vanishes at the end
points. The distribution:
iz, t   I s ( z ) cost
Where,

L

j
I
e
sin
k

z

 0  z  L/2
 0
2

I s ( z)  
I 0 e j sin k  L  z   L / 2  z  0

2

Division of the L long dipole into
a series of infinitesimal Hertzian
dipoles !!
Dipole Antennas (Cont’d..)
For simplicity, assume phase term =0, and make use of current
distribution term with magnetic field equation for a Hertzian
dipole to get :
0
L / 2  jkR


 jkR


 L
I 0k 
e
e
L


H 0S  j
a
sin k   z   sin  ' dz 
sin k   z   sin  ' dz


4
R
R


 2
 2
0
 L / 2



Dipole Antennas (Cont’d..)
For far field, the vectors r and R appear to be parallel, so that
θ’=θ and R=r, where:
e  jkR  e  jk r  z cos   e  jkr e jkz cos
After pulling the components
that don’t change with z, use
table of integrals and with
application to Euler’s identity,

 kL

 kL  

cos
cos


cos


 
 jkr
I e
 2

 2  a

H 0 S  H 0 S a  j 0
 
2 r 
sin 




Dipole Antennas (Cont’d..)
The vector E0S is then easily found from :
E0S  0a r  H 0S 0 H 0S a
The time averaged power radiated is:
Pr ,  
15 I 0 2
r 2
F   a r
Where, the pattern function is given by:
  kL

 kL  
 cos cos   cos  
2

 2 
F     

sin 




2
Dipole Antennas (Cont’d..)
It’s not generally equivalent to the normalized power function
since F(θ) can be greater than one.
Therefore, the normalized power function is:
Pn   
F  
F  max
And the max time averaged power density is then :
Pmax 
15 I 0 2
r
2
F  max
Dipole Antennas (Cont’d..)
Dipole Antennas (Cont’d..)
3D and 2D amplitude patterns for a thin dipole of l=1.25λ
Dipole Antennas (Cont’d..)
Half Wave Dipole Antennas
Because of its convenient radiation resistance, and because it’s
a smallest resonant dipole antenna, the half wavelength dipole
antenna merits special attention.
With kL/2 = π/2,


2 
2  cos  cos  
15 I 0 
2
a
Pr ,  
r

2 
2
r 
sin 



With the F(θ) is 1, the maximum
power density is:
Pr ,  
15 I 0 2
r 2
Dipole Antennas (Cont’d..)


cos 2  cos 
Therefore, the normalized power density is:
2

Pn   
cos 2 
The current distribution and normalized radiation pattern for a
half wave dipole antenna.
Dipole Antennas (Cont’d..)
Dipole Antennas (Cont’d..)
For half wave dipole…
 p  7.658
and with L = λ/2
Try this!!
We can then find the directivity as :
Dmax 
4
 1.640
p
Which is slightly higher than the directivity of Hertzian dipole,
the radiation resistance is given by:
Prad
1
 I 0 2 Rrad  r 2 Pmax p
2
Leads to: Rrad 
30

 p  73 .2
Dipole Antennas (Cont’d..)
This radiation resistance much higher than of Hertzian dipole,
where it radiates more efficiently  easier to construct an
impedance matching network for this antenna impedance.
This antenna impedance also contains a reactive components,
Xant, where for a λ/2 dipole antenna it is equal to 42.5Ω .
Therefore, total impedance by neglecting Rdiss.
Z ant  73.2  j 42.5
For impedance matching, need to make reactance zero (in
resonant condition). So, it can be achieved by making the
antenna slightly shorter (reduced in length until reactance
vanishes).
Example 4
Find the efficiency and maximum power gain of a λ/2
dipole antenna constructed with AWG#20 (0.406 mm
radius) copper wire operating at 1.0 GHz.
Compare your result with a 3 mm length dipole
antenna (Hertzian Dipole) if the center of this antenna
is driven with a 1.0 GHz sinusoidal current.
74
Solution to Example 4
We first find the skin depth of copper at 1.0 GHz,
1
1
 cu 

 2.09 10 6 m
f0
 1109 4 10 7 5.8 107




This is much smaller than the wire radius, so the wire area over
which current is conducted by:
S  2a cu  5.33 10 9 m 2
At 1 GHz, the wavelength is 0.3m and the λ/2 is 0.15m long. The
ohmic resistance is then:
Rdiss
1 

 0.485 
 S
Solution to Example 4 (Cont’d..)
Since the radiation resistance for half wave dipole is 73.2Ω, we
have :
73 .2
e
 0.99
73 .2  0.485 
A gain of:
Gmax  eDmax  0.991.640   1.63
Meanwhile for Hertzian Dipole, the ohmic resistance of the
small dipole is :
Rdiss
1 

 9.7 m
 S
Solution to Example 4 (Cont’d..)
To find the radiation resistance, with value of wavelength is
0.3m, thus :
Rrad
3 

3

10
m
2
 80
 79 m
 0.3m 


Therefore, its efficiency and gain:
79 m
e
 0.89
79 m  9.7m
Gmax  eDmax  0.891.5  1.34
Thus, the half wave dipole is clearly more efficient with a
higher gain than the short dipole.
4.4 Monopole Antennas
Consider the construction of half wave dipole for an AM radio
station broadcasting at 1 MHz. At this f, the wavelength is
300m long and the half wave dipole antenna must be 150m tall.
We can cut this in half, by employing image theory to build a
quarter wave monopole antenna that is only 75m tall!!
Monopole Antennas (Cont’d..)
Consider pair of charges, +Q and –Q (as electric dipole), where
the dashed line shows the location of zero potential surface. If
we slide a conductive plane over the zero potential surface, the
field lines in the upper half plane are unchanged.
Monopole Antennas (Cont’d..)
Note that the charge can be in any distribution (point charge,
line charge, surface or volume charge) and the image charge is
a mirror image of opposite polarity.
A monopole antenna is excited by a current source at its base.
By image theory, the current in the image will be the same with
the current in actual monopole. The pair of monopole resembles
a dipole antenna.
A monopole antenna placed over a conductive plane and half
the length of a corresponding dipole antenna will have identical
field patterns in the upper half plane.
Monopole Antennas (Cont’d..)
For the upper half plane (00< θ<900), the time averaged power,
max power density and normalized power density for the
quarter wave monopole is the same with half wave dipole. But
the pattern solid angle is different.
Monopole Antennas (Cont’d..)
Since the normalized power density is zero for (900< θ<1800),
the pattern solid angle:
p 
 Pn  d
Integrate over all space will be half value of Ωp for half wave
dipole. So, for quarter wave monopole,
 p  3.829
4
 Dmax 
 3.28
p
See that the radiation resistance is halved, and the antenna
impedance :
Rrad 
30

 p  36 .6
 Z ant  36.6  j 21.25
Summary of Key Antenna Parameters
4.5 Antenna Arrays
The antennas we have studied so far have all been
omnidirectional – no variation in φ. A properly spaced collection
of antennas, can have significant variation in φ leading to
dramatic improvements in directivity.
A Ka-Band Array Antenna
Antenna Arrays (Cont’d..)
An antenna array can be designed to give a particular shape of
radiating pattern. Control of the phase and current driving
each array element along with spacing of array elements can
provide beam steering capability.
For simplification:
 All
antenna elements are identical

The current amplitude is the same feeding each element.

The radiation pattern lies only in xy plane, θ=π/2
The radiation pattern then can be controlled by:

controlling the spacing between elements or

controlling the phase of current driving for each element
Antenna Arrays (Cont’d..)
For simple example, consider a pair of dipole antennas driven
in phase current source and separated by λ/2 on the x axis.
Assume each antenna radiates
independently, at far field point P,
the fields from 2 antennas will be
180 out-of-phase, owing to extra λ/2
distance travel by the wave from the
farthest antenna  fields cancel in
this direction. At point Q, the fields
in phase and adds. The E field is
then twice from single dipole,
fourfold increase in power 
broadside array  max radiation is
directed broadside to axis of
elements.
Antenna Arrays (Cont’d..)
Modify with driving the pair of dipoles
with current sources 180 out of phase.
Then along x axis will be in phase and
along y axis will be out of phase, as
shown by the resulting beam pattern 
endfire array  max radiation is
directed at the ends of axis containing
array elements.
Antenna Arrays (Cont’d..)
Pair of Hertzian Dipoles
Recall that the far field value of E field from Hertzian dipole at
origin,
I s ke jkR
E 0 S  j 0
sin a
4R
But confining our discussion to the xy plane where θ = π/2,
I s ke jkR
E 0 S  j 0
a
4R
Antenna Arrays (Cont’d..)
Consider a pair of z oriented
Hertzian dipole, with distance
d, where the total field is the
vector sum of the fields for both
dipoles and the magnitude of
currents the same but a phase
shift between them.
E0 S tot   E 0 S1  E 0 S 2
Where,
I s1  I 0
I s1ke jkR1
I s 2 ke jkR2
 j 0
a  j 0
a
4R1
4R2
I s 2  I 0 e j
Antenna Arrays (Cont’d..)
Assumption,
1    2
And,
R1  r  R2
Antenna Arrays (Cont’d..)
Where geometrically we could get,
d
R1  r  cos
2
d
R2  r  cos
2
Thus, the total E field becomes:
 jkR
I 0 ke
E0S tot   j 0
4R

 d
  j  k d cos   
j  k cos   
2
2 
e j 2 e  2
e  2
a


With Euler’s identity, the total E field at far field observation
point from two element Hertzian dipole array becomes :
 jkR
I 0ke
E0S tot   j 0
4R
 j

2e 2 cos k d cos   a

2 
 2


Antenna Arrays (Cont’d..)
To find radiated power,


1
   1
*
P r , ,    Re E S  H S   0 E0 S tot 2a r
2
 2  2
  I 2 2 k 2
 0 0
 32 2 R 2


 4 cos 2  k d cos   a
 r
 
2 
 2

It can be written as:
  
P r , ,    Funit Farray a r
 2 
Antenna Arrays (Cont’d..)
Unit factor, Funit is the max time averaged power density for an
individual antenna element at θ=π/2
  I 2 2 k 2
Funit   0 0
 32 2 R 2





An array factor, Farray is
Where,

Farray  4 cos 2  
2
  kd cos  
This is the pattern function resulting from an array of two
isotropic radiators.
Example 5
The λ/2 long antennas are
driven in phase and are λ/2
apart. Find:
the far field radiation pattern
for a pair of half wave dipole
shown.

the maximum power density 1
km away from the array if each
antenna is driven by a 1mA
amplitude current source at 100
MHz.

94
Solution to Example 5
At 100 MHz, λ = 3m, so that 1 km away is definitely in far field.
For a half wave dipole, we have :


2 
2  cos  cos  
15 I 0 
2
a
Pr ,   Pr , a r 
r

2 
2
r 
sin 



The unit factor can be found by evaluating above at θ = π/2 ,
Funit 
15 I 0 2
r 2
Solution to Example 5 (Cont’d..)
The array factor, with d = λ/2 and
driven at the same phase) :
=0 (due to antennas are
2 

Farray  4 cos  cos 
2

For the array, we now have :
60 I 0 2
 


2 
P r ,  ,    Funit Farray a r 
cos  cos 
2
2 

2

r
The normalized
power function is:
  
P r , ,  
2 
 


2 
Pn  ,   
 cos  cos 
 2  P r ,  ,  
2



 2  max
Solution to Example 5 (Cont’d..)
This can be plotted as :
But how to plot?!?!?!
Use MATLAB!!
The maximum radiated power density at 1000m is :
Pmax 
60 I 0
r
2
2

 
60 10
3 2
 1000 
2
 19
pW
m2
Antenna Arrays (Cont’d..)
N - Element Linear Arrays
The procedure of two-element array can be extended for an
arbitrary number of array elements, by simplifying assumptions :
The array is linear  antenna elements are evenly spaced, d
along a line.

The array is uniform  each antenna element driven by same
magnitude current source, constant phase difference, between
adjacent elements.

Antenna Arrays (Cont’d..)
I s1  I 0 , I s 2  I 0e j , I s3  I 0e 2 j ,.... I sN  I 0e j N 1
Antenna Arrays (Cont’d..)
The far field electric field intensity :


I 0 ke jkR
E 0 S tot   j 0
1  e j  e j 2  ...  e j  N 1 a r
4R
Where,
  kd cos  
Manipulate this series to get:
N 
sin 

2 

Farray 

sin 2  
2
2
With the max value as :
Farray max  N 2
Antenna Arrays (Cont’d..)
So then the normalized power pattern for these elements is :
N 
sin 

Farray
1
2 

Pn   

2
Farray

N
max
sin 2  
2
2


Example 6
Five antenna elements spaced λ/4 apart with
progressive phase steps 300. The antennas are
assumed to be linear array of z oriented dipoles
on the x axis. Find:

the normalized radiation pattern in xy plane

the plot of the radiation pattern.
102
Solution to Example 6
To find the array factor, first need to find psi, Ψ:
  kd cos   

 2   
    

  cos   30
    cos  
6
   4 
 180   2 
Inserting this ratio to array factor,
N 
5 
2  5
sin 
cos 
 sin 

2 
4
12 




 


sin 2  
sin 2  cos  
12 
2
4
2
Farray
Farray max  N 2  25
Solution to Example 6 (Cont’d..)
The normalized radiation pattern is :
5 
 5
sin 2 
cos 

1
12 
 4
Pn 
 
25
2 
sin  cos  
12 
4
The plot is :
But how to plot?!?!?!
Use MATLAB!!
Antenna Arrays (Cont’d..)
Parasitic Arrays
Not all the elements in array need be directly driven by current
source. A parasitic array typically one driven element and several
parasitic elements, the best known parasitic array is Yagi-Uda
antenna.
Antenna Arrays (Cont’d..)
On one side of the driven element is reflector  length and
spacing are chosen to cancel most of the radiation in that
direction, as well as to enhance the direction to forward or main
beam direction.

Several directors (four to six)  focus the main beam in the
forward direction  high gain and easy to construct.

Parasitic elements tend to pull down the Rrad of the driven
element. E.g. Rrad of dipole would drop from 73Ω to 20Ω when
used as the driven element in Yagi-Uda antenna.
But higher Rrad is more efficient, so use half wavelength folded
dipole antenna (four times Rrad of half-wave dipole!)
4.6 Helical Antennas
Diameter of ground
plane at least 3λ/4
Helical Antennas (Cont’d..)
Helical Antennas (Cont’d..)
Helical Antennas (Cont’d..)
Important Parameters
Helical Antennas (Cont’d..)
Helical Antennas (Cont’d..)
For this special case,
The radiated field is circularly polarized in all directions other
than θ = 00
Helical Antennas (Cont’d..)
Parameters for End Fire mode
Helical Antennas (Cont’d..)
Helical Antennas (Cont’d..)
Feed Design for Helical Antennas
Helical Antennas (Cont’d..)
Helical Antennas (Cont’d..)
4.7 Yagi Uda Array Antennas
Yagi Uda Array Antennas (Cont’d..)
Yagi Uda Array Antennas (Cont’d..)
Yagi Uda Array Antennas (Cont’d..)
Yagi Uda Array Antennas (Cont’d..)
Yagi Uda Array Antennas (Cont’d..)
Example 7
With above parameters, design a Yagi Uda
Array antenna by finding the element
spacing, lengths and total array length.
124
Solution to Example 7
from given directivity = 9.2 dB
Solution to Example 7(Cont’d..)
Solution to Example 7(Cont’d..)
Therefore,
Total array length: 0.8 λ
The spacing between directors: 0.2 λ
The reflector spacing: 0.2 λ
The actual elements length:
L3 = L5 : 0.447λ
L4 : 0.443λ
L1 : 0.490λ
Solution to Example 7(Cont’d..)
128
Solution to Example 7(Cont’d..)
Fig 10.25
129
Solution to Example 7(Cont’d..)
Fig 10.26
130
Yagi Uda Array Antennas (Cont’d..)
4.8 Antennas for Wireless
Communications
Parabolic Reflectors
Parabolic dish antenna
Parabolic reflector antenna
Antennas for Wireless
Communications (Cont’d..)
Cassegrain reflector antenna
All
of
these
parabolic
reflectors operates based on
the
geometric
optics
principle that a point source
of radiation placed at the
focal point of parabolic
reflector
will radiate the
energy incident on the dish
in a narrow and collimated
beam.
For high efficient, the dish must be significantly larger than
the radiation wavelength, and has a directive feed.
Antennas for Wireless
Communications (Cont’d..)
Patch Antennas
Other shapes such as circles, triangles and annular rings also
been used. It can be excited by an edge or probe fed, where its
location is chosen for impedance match between cable and
antenna.
Antennas for Wireless
Communications (Cont’d..)
Folded Dipole Antennas
A pair of half-wavelength dipole elements are
joined at the ends and fed from the center of
one of the pair. If the two sections are close
together (d on the order of λ/64), the
impedance will be four times greater than the
regular λ/2 dipole antenna. The directivity is
the same but the bandwidth is significantly
broader.
Antenna
End