Exercise n. 1
Consider a horn antenna whose rectangular aperture has size A = 10 cm, B = 4 cm, operated at the frequency f =20GHz. Compute
1. the maximum gain G in natural units and in dB.
2. the full width of the main lobe between the zeros in the E and H plane 3. the direction and the level of the first sidelobe in the E and H plane
4. the magnitude of the radiated electric and magnetic field on axis, at the distance R = 5 km, when the horn has an input power Pin= 10 W.
Solution
1. The geometrical area of the aperture is Ag= AB = 40 cm2.
2. The aperture efficiency of a rectangular horn is νa = 0.8, hence the effective area is Aeq = νaAg= 0.8 · 40 = 32 cm2
3. The wavelength is λ = c/f = 3. · 108/20 = 1.5 cm
4. The maximum gain is G = (4π/λ2)Aeq = 179 = 22.5 dB. The directivity is the same:
antennas with size comparable to wavelength have ohmic efficiency very close to one.
5. Compute the far field distance rf f = 2D2/λ. The characteristic size D is the diagonal of the rectangle D = 10.77 cm and rf f = 1.54 m. Since the observation distance is R À rf f the concept of gain can be used. Hence, we compute the power density per unit surface in the observation point:
dPrad
dΣ = G Pin
4πR2 = 5.6977 · 10−6 From this we compute the electric field via
|E| = r
2Z0dPrad
dΣ = 65.5 mV/m and the magnetic field from the impedance relation
|H| = |E|
Z0 = 173 µA/m
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Exercise n. 2
Consider a radio link between a paraboloid and a half wavelength dipole antenna. The paraboloid has a diameter D = 1 m, is operated at the frequency f = 10 GHz, radiates a circularly polarized field and its input power is Pin= 0.1 W. The receiving dipole has ohmic efficiency η = 1, is located
α x
y z
x
y z
α
(a) (b)
x
Figure B.4. Radio link scheme.
on axis at the distance R = 20 km and perpendicular to the link direction. However it is not vertical, but forms the angle α with the vertical, as shown in Fig. B.4. Compute the received power as a function of the angle α.
Solution
• The geometrical area of the paraboloid aperture is Ag = πD2/4 = 0.7854 m2.
• The aperture efficiency of a paraboloid is νa= 0.6, hence the effective area is Aeq= νaAg= 0.6 · 0.7854 = 0.4712 m2
• The wavelength is λ = c/f = 3. · 108/10 = 3.0 cm
• The maximum gain is G = (4π/λ2)Aeq = 6580 = 38.2 dB. The directivity is the same:
antennas with size comparable to wavelength have ohmic efficiency very close to one.
• Compute the far field distance rf f = 2D2/λ. The characteristic size D is the diameter of the aperture and rf f = 66.67 m. Since the observation distance is R À rf f the Friis equation can be used.
• The maximum gain of the dipole (equal to the directivity since the ohmic efficiency is η = 1) is G = D = 1.643
• The polarization of the transmitting antenna is circular. The electric field lies in the x, y plane and the polarization vector can be written
ˆ
pT X = c(ˆx ± jˆy)
The double sign is necessary because the text does not specify whether the polarization is clockwise or counterclockwise. The constant c has the value that makes ˆpT X a unit vector:
|ˆpT X|2= ˆpT X· ˆp∗T X = c(ˆx ± jˆy) · c∗(ˆx ∓ jˆy) =
= |c|2(ˆx · ˆx ∓ jˆx · ˆy ± jˆy · ˆx + ˆy · ˆy) = |c|2(1 + 1) = 2|c|2 Alternatively,
|ˆpT X|2= |(ˆpT X)x|2+ |(ˆpT X)y|2= |c|2+ |c|2= 2|c|2
Hence c = 1/√ 2 and
ˆ
pT X = 1
√2(ˆx ± jˆy)
• The polarization of the receiving antenna is, by definition, the polarization of the E field radiated in the direction of the paraboloid if the dipole were used as transmitting antenna.
The link direction coincides with the one of maximum radiation and the E field is parallel to the antenna. Hence
ˆ
pRX= ±(cos αˆx − sin αˆy)
• Compute the polarization matching factor p = |ˆpT X· ˆpRX|2=
We see that p is independent of α and has the constant value of 1/2: a 3 dB loss, with respect to the case of polarization matching.
• The available power at the output terminals of the receiving antennas is given by Friis formula Pavail= PinGT XGRX
Consider a radio link between two identical paraboloids. The paraboloids have clockwise circular polarization. Compute the polarization matching factor.
Solution
Obviously, the result must be 1, since two identical antennas are polarization matched by definition.
The computation is, however, a little tricky.
Consider the transmitting antenna. Then ˆ
pT X = 1
√2(ˆxlocTX+ jˆylocTX)
With reference to Fig. B.5, we realize that the previous phasor describes really a clockwise polar-ization if observed from a point in the far field of the TX antenna. The axes have been identified with a T X subscript and a loc (as local) superscript because they are to be considered as attached to the transmitting antenna.
Now take this antenna, together with the reference system, and rotate it 180◦around the xlocT X axis in order to make it coincident with the receiving antenna. The reference system now is different from the original one and is indicated with the RX subscript. Clearly, the field radiated by the re-ceiving antenna if it were connected to a generator would have still clockwise circular polarization, as observed by a point in the far field of this RX antenna and its expression would be
ˆ
pRX= 1
√2(ˆxlocRX+ jˆylocRX)
loc
xTX
loc
yTX loc
zTX loc
xRX
loc
yRX
ETX
Figure B.5. Radio link scheme with two identical antennas.
Note that
ˆ
xlocRX= ˆxlocTX ˆ
ylocRX= −ˆylocTX ˆ
zlocRX= −ˆzlocTX Then we compute the polarization matching factor
p = |ˆpT X · ˆpRX|2=
¯¯
¯¯ 1
√2(ˆxlocT X + jˆylocT X) · 1
√2(ˆxlocRX+ jˆylocRX)
¯¯
¯¯
2
=1
2(1 + j · j · (−1)) = 1 as it should be.
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Exercise n. 4
A Hertzian dipole of length L = 2 m is fed by a transmission line with characteristic impedance Z∞ = 50 Ω. The dipole has an ohmic efficiency η = 0.8 and an input capacitance C = 30 pF.
Suppose that this dipole is required to radiate a total power Prad = 100 W at the frequency f = 10 MHz. Compute:
1. the voltage that must be applied at the dipole input
2. the reflection coefficient at the dipole input and the incident power on the line so that Prad = 100 W. Compute also the maximum and minimum voltage on the line and the VSWR.
Solution
1. The wavelength is λ = 3 · 108/107= 30 m The radiation resistance is
Rrad= Z0
2π 3
µL λ
¶2
= 3.5 Ω The input resistance is
Rin= Rrad
η = 4.4 Ω
The input impedance is
Zin= Rin− j 1
2πf C = 4.4 − j 530.5 Ω The input current Ia is such that the radiated power is
Prad= 1
2Prad|Ia|2 Hence
|Ia| =
r2Prad
Rrad =
r2 · 100
3.5 = 7.55 A The voltage Va to be applied is
|Va| = |Zin||Ia| = 4005 V 2. The input reflection coefficient is
Γa= Zin− Z∞
Zin+ Z∞ = 0.9985 e−j10.8◦ The input power is Prad/η = 125 W. The necessary incident power is
P+= Pin
1 − |Γa|2 = 40.3 kW On the other hand P+=12|VZ+|2
∞ , hence
|V+| =p
2Z∞P+= 2008 V The maximum voltage is then
Vmax= |V+|(1 + |Γa|) = 4013 V and the minimum
Vmin= |V+|(1 − |Γa|) = 3.1 V The VSWR is
V SW R = 1 + |Γa|
1 − |Γa| = 1291
Obviously, using this antenna in these conditions is absolutely ridiculous and a matching device is to be used.
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Exercise n. 5
Find the sizes A, B of a rectangular horn with gain G = 19 dB at the frequency f = 12 GHz and
symmetrical main lobe, as measured between the zeros, i.e. with equal beamwidth in the E and H planes. Compute also the full beamwidth.
Solution
The first zero in the E plane is located at ηy = π, from which sin ϑEz1= λ
B
Similarly, the first zero in the H plane is located at ηx= 3π/2, from which sin ϑHz1= 3λ
2A If ϑHz1= ϑEz1, then B = 2A/3.
The geometrical area is
Ag= AB =2A2 3 The maximum effective area is
Aeq= νaAg= νa2A2 3 The maximum gain is
G =4π
λ2Aeq =8πνa 3
µA λ
¶2
from which we get
A λ =
r 3
8πνaG The required gain is GdB = 19 dB, i.e.
G = 10GdB/10= 79.43 We find then A = 3.44λ = 8.600 cm, since the wavelength is
λ = c
f = 3 · 108
12 · 109 = 2.5 · 10−2m Moreover B = 5.733 cm and
ϑEz1= arcsin µλ
B
¶
= arcsin(0.436) = 25.852◦ The full beamwidth (between the zeros) is
BW = 2ϑEz1= 2ϑHz1= 51.704◦
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Exercise n. 6
Consider a paraboloid antenna with the following radiation pattern at the frequency f = 5 GHz:
g(ϑ,ϕ) = G0cosαϑ for ϑ ≤ π 2
= 0 for ϑ > π 2
The full half-power beam-width is FHPBW=10◦. Compute G0 and estimate the antenna diameter.
Solution Next recall the normalization condition
1
Compute the integral, noting that g(ϑ,ϕ) does not depend on ϕ, so that the ϕ integral equals 2π:
1 In conclusion, the normalization condition yields
G0= 2(α + 1) = 364.36
that is (G0)dB = 25.61 dB. To find the diameter of the antenna, compute the maximum effective area and the required antenna diameter is
D = r4Ag
π = 47.06 cm
Fig. B.6 shows the radiation pattern in cartesian coordinates, Fig. B.7 shows the same pattern in polar coordinates.
−1000 −80 −60 −40 −20 0 20 40 60 80 100 50
100 150 200 250 300 350 400
ϑ
Figure B.6. Radiation pattern in cartesian coordinates
100 200
300 400
30
210
60
240
90
270 120
300 150
330
180 0
Figure B.7. Polar radiation pattern