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Antennas
Problem set 5
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Problem 1. An antenna is illuminated by a plane electromagnetic wave with linear
polarization that propagates along the dˆ i   xˆ direction. Suppose that the electric field makes
the angle of 60º with the y axis and has amplitude Einc  0.1V / m . The effective length of

the receiving antenna in the direction of arrival of the incoming wave is h er  zˆ  m  .
(a) Find the open-circuit voltage at the antenna terminals.
Solution
The voltage induced at the antenna terminals when they are in open-circuit is:
V oc  h er  Einc
0
is the incident electric, which must be perpendicular to dˆ i   xˆ because the
Here, Einc
0
incoming wave is transverse electromagnetic (TEM). This means that:
inc
inc
ˆ
ˆ
Einc
0  E1 y  E 2 z
inc
inc
Since the wave is linearly polarized, it is necessary that the components E1 , E 2 are either in
phase or in opposition of phase:
   arg  E 
inc
arg E1
inc
2
   arg  E   
inc
or arg E1
inc
2
[linear polarization]
This is equivalent to say that Einc
0 is real-valued apart from an overall phase factor:
j
inc
inc
inc
inc
ˆ
ˆ
Einc
real-valued.
0  e  E1 y  E2 z  with E1 , E2
Using the fact that the electric field makes an angle of 60º with the y-axis, it follows that:
j
º
º
ˆ
ˆ
Einc
0  e E0  cos 60 y  sin 60 z 
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where E0  E12  E22  Einc
 0.1[V / m] (there is not enough information in the wording of
0
the problem to decide which of the signs  is the correct one). Hence, using h er  zˆ one finds
that:
V oc   e j E0 sin 60º   e j 0.1sin 60º   e j 0.087 V 
In conclusion, the open-circuit voltage amplitude is V oc  87
 mV  .
(b) Represent the equivalent circuit of the antenna and identify all the parameters that are part
of the circuit. Suppose that the antenna impedance is Z a  70 .
Solution:
I 0
V
VL
Z L is the load impedance, connected at the antenna terminals.
(c) Find the power delivered to a load of 50 .
Solution:
From the equivalent circuit, it is possible to show that the power delived to the load is given
by the general formula (see the lecture notes)
Pr  Ci  Pr opt
Ci 
4 Ra RL
 Ra  RL    X a  X L 
2
2
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where  Pr opt 
2
V oc
is the power received by a matched load [maximum power that can be
8 Ra
extracted by the antenna from a given incident wave] (in the above, Z a  Ra  jX a and
Z L  RL  jX L ).
Substituting values:  Pr opt 
Ci 
2
V oc

8 Ra
0.087 2
 13.51W
8  70
4 Ra RL
 Ra  RL    X a  X L 
2
2

4  50  70
 50  70 
2
 0.972
Thus, the received power is Pr  0.972 13.51W  13.13W
Problem 2.
A short-dipole with length L  0.01m is illuminated by a plane wave with electric field
amplitude E  8 15 [V/m]. The oscillation frequency is 300/ [Hz].
(a) Find the power incident on the antenna per unit of area.
Solution:
The Poynting vector intensity (i.e., the power density) is:
S inc 
E
2
20
8

15

2  120
2

64  15
 4W / m2
2  120
(b) What is the directivity and maximum of the effective area of the antenna? Ignore losses.
Solution:
For an short dipole oriented along the z-direction:
D short-dipole  1.5
(same result as for an Hertz dipole)
The antenna effective area (parameter of the receiving antenna) can be related to the power
gain ( G ) (parameter of the transmitting antenna) through the universal formula:
Aef  ,   
02
2
G  ,    0 e g  ,   .
4
4
In the last identity, we used G  e g with e the antenna efficiency and g the directive gain.
From here, the maximum effective area is
Aef ,max 
02
2
2
max G  0 e max g  0 eD
4
4
4
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In conclusion,
Aef ,max 
02
eD
4
For a lossless antenna e=1 and we obtain:
2
Aef ,max
2
2
D  c  1.5  3  108  1.5 12
10  1.17 1012  m 2  !!!
 0 D

 
  
4
4  f 
4  300 /  
4
For small frequencies and lossless antennas the maximum effective area can be huge (many
orders of magnitude larger than the physical size of the antenna) because it varies with the
wavelength squared (note that D  1 )!
(c) Suppose that the antenna terminals are connected to a load adjusted to receive the
maximum power, and that the antenna axis is parallel to the incident electric field. What is the
received power? Ignore losses.
Solution:
The received power can be written in terms of the effective area using:
Pr  Ci C p Aef S inc
Here Ci is the impedance matching coefficient, which for a load adjusted to receive the
maximum (conjugated matched load) is Ci  1 . On the other hand, C p is the polarization
matching coefficient defined by:
Cp 
h er  Einc
0
2
h er Einc
0
2
2
Because of the reciprocity theorem, the effective height of the receiving antenna is identical to
that of the transmitting antenna:
h er  h e
For a linear antenna, (with axis oriented along z) the effective height is
h e  heθˆ .
(effective height of the transmitting antenna)
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This implies that for the short dipole C p 
θˆ  Einc
0
Einc
0
2
2
. The direction of maximum gain of the
short dipole is   90º . For   90º one has θˆ   zˆ and therefore C p
 90º

zˆ  Einc
0
Einc
0
2
2
 1,
where the last identity follows from the fact that the incident electric field is parallel to the
antenna axis (the z-axis).
In summary, in the conditions of the problem (incoming wave arrives along direction of
maximum gain   90º ), the received power is identical to the available power:
Pr  Aef ,max S inc
Substituting values Pr  1.17 1012  4  4.711012 Watt  !!! .
(d) Suppose now that the loss resistance of the antenna is 0.1 . Repeat the previous question.
Comment.
Solution:
Supposing that there is still impedance matching, the formula Pr  Aef ,max S inc still holds true.
As before, Aef ,max 
02
eD , but now the antenna efficiency is less than 1 because of the loss.
4
Clearly, we have:
Pr  4.71 1012  e Watt  (same result as before multiplied by the efficiency).
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To find the efficiency, we use:
Rrad
e
,
Rrad  RL
with Rrad
short-dipole
1 2
L2

0  2 the radiation resistance of the short
0
4 3
dipole (1/4 of the radiation resistance of an Hertz dipole).
We
have
Rrad
short-dipole
 20 2
L2
02
 20 2
2
0.012
2 0.01
20


 2 1015  .
(3 108 / (300  )) 2
( 106 ) 2
Since this value if much less than the loss resistance, we can write:
e
Rrad
 2  1014 . Thus, the received power is:
RL
Pr  0.094 Watt 
which is now a reasonable value. The lesson is: electrically small antennas are greatly
affected even by a small loss.
Using e 
Aef ,max 
Rrad
(which is possible when Rrad  RL ) one may generally find that (using
RL
02
eD ):
4
Aef ,max


0
 Rrad
1
1
1 0 2
D
D 6 L2 
D
L


RL
RL
4
4
24 RL
2
0
The disproportionaly large factor 02 disappears and in the end the effective area is
determined by the physical dimensions of the dipole ( L2 factor).
Problem 3. A half-wavelength dipole is oriented along the z-direction in free-space. The
antenna is illuminated by a incoming plane wave. The direction of incidence makes an angle
of 30º with respect to the antenna axis. The frequency of operation is 300MHz and the power
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available at the antenna terminals is 100W. Suppose that the antenna impedance is
Z a  73  j 42.5 .
(a) Find the amplitude of the incident electric field.
Solution:
The available power by definition is the power received by an antenna when there is both
impedance ( Ci  1 ) and polarization matching ( C p  1 ).
Pa  Aef S inc
The antenna effective area is given by:
Aef 
0 h er
4 Ra
2

0 h e
2
4 Ra
where the second identity used the reciprocity relation h er  h e . For a dipole antenna, we
know that (sinusoidal current approximation): h e  heθˆ with
he   L  /2


cos  cos  
2
2


k0
sin 
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Using   30º
and noting that

2  300 106
k0  
 2  m 1 
8
3 10
c
(the free-space
wavelength is 0  1m ), one finds that:
he   30º 
L   /2


cos  cos 30º 
2
2
  0.13 m

 
2
sin 30º
Hence, the effective area is:
Aef 
0 h e
4 Ra
2
2
120 0.13

 0.021m 2
4  73
The Poynting vector intensity is then:
S
inc
Pa 100 106


 4.65 103 W / m 2  .
0.021
Aef
The intensity of the incident electric field is therefore: E inc  20 S inc  1.87 V / m  .
(b) Suppose that the polarization of the incident wave is linear and the electric field is
contained in a plane   const. . Find the power delivered to a load of 50 .
Solution
We use Pr  Ci C p Aef S inc  Ci C p Pa  Ci C p  100 W . The impedance matching coefficient is:
Ci 
4 Ra RL
 Ra  RL    X a  X L 
2
2

4  50  73
 50  73
The polarization matching coefficient is C p 
2
 42.52
h er  Einc
0
h
r 2
e
E
 0.86
2
inc 2
0
. For a dipole antenna, h e  heθˆ and
hence:
Cp 
θˆ inc  Einc
0
Einc
0
2
2
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where θ̂inc is calculated for the direction determined by the incident wave. Since the incoming
ˆ inc  E φˆ inc . In order that the electric field is
wave is TEM, one can always write: Einc
2
0  E 1θ
contained in a plane   const. it is necessary that E 2  0 (incident field is contained in the
incident plane defined by the antenna axis and by the direction of incidence). Thus, in these
conditions:
Cp 
θˆ inc  E1θˆ inc
E1θˆ inc
2
2
1
This shows that Pr  0.86 1100W  86W .
(c) Repeat b) for the case in which the incident field is perpendicular to a   const. plane.
Solution
ˆ inc  E φˆ inc with E  0 . Then, C 
In this case, Einc
2
1
0  E 1θ
p
θˆ inc  E 2φˆ inc
E 2φˆ inc
2
2
 0 and the received
power is Pr  0.86  0  100W  0 . The incident electric field is horizontal (perpendicular to
the dipole axis), and due to this reason it does not interact with the antenna.
(d) Repeat b) for the case in which the incident field has right-circular polarization.
Solution
As previously mentioned, the incident field is generically of the form:
ˆ inc  E φˆ inc
Einc
2
0  E 1θ
The polarization of the incident field is
(i)
circular when E1  E 2 and arg E1  arg E 2  90º
(ii)
linear when arg E1  arg E 2
(iii)
elliptical all the other cases.
 
 
 
 
or arg E1  arg E 2  
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The condition (i) is equivalent to say that E1   jE 2 . One of the signs determines a rightcircular polarization (RCP) and the other a left-circular polarization (LCP), but we do not

ˆ inc  jφˆ inc
need to worry at the moment which is which. Substituting Einc
0  E1 θ
Cp 
θˆ inc  Einc
0
Einc
0
2

into
2
, we get:
Cp 

θˆ inc  θˆ inc  jφˆ inc
θˆ
inc
 jφˆ inc

2

2

1
2
2
1  j
2

1
2
The C p coefficient is independent of the sense of rotation of the electric field, i.e.
independent if the polarization is RCP or LCP. Thus, the received power is
Pr  0.86  0.5  100 W  43W ,
which is half the value obtained in question b). Indeed, the φ̂inc component of the incident
field does not interact with the antenna, and hence only half of the incident energy is available
to the antenna.
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