Mustansiriyah University Subject: Antennas

Faculty of Engineering Class: 3rd-year

Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

Antenna Effective Length and Effective Areas

An antenna in the receiving mode, whether it is in the form of a wire, horn,
aperture, array, dielectric rod, etc., is used to capture (collect) electromagnetic
waves and to extract power from them, as shown in Figures 6.1 (a) and (b). For
each antenna, an equivalent length and a number of equivalent areas can then be
defined. These equivalent quantities are used to describe the receiving
characteristics of an antenna, whether it be a linear or an aperture type, when a
wave is incident upon the antenna.

Figure 6.1: Uniform plane wave incident upon dipole and aperture antennas.

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Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

1- Effective Length
The effective length of an antenna is a quantity that is used to determine the
voltage induced on the open-circuit terminals of the antenna when a wave
impinges upon it. The effective length le for an antenna is usually a complex
vector quantity represented by
𝑙𝑒 (𝜃, 𝜑) = 𝑙𝜃 (𝜃, 𝜑)𝑎̂𝜃 + 𝑙𝜑 (𝜃, 𝜑)𝑎̂𝜑
The effective height is a far-field quantity and it is related to the far-zone
field Ea radiated by the antenna, with current Iin in its terminals
𝐸𝑎 = 𝐸𝜃 𝑎̂𝜃 + 𝐸𝜑 𝑎̂𝜑

The effective length represents the antenna in its transmitting and receiving
modes, and it is particularly useful in relating the open-circuit voltage Voc of
receiving antennas. This relation can be expressed as
𝑉𝑜𝑐 = 𝐸 𝑖 𝑙𝑒
where,
𝑉𝑜𝑐 : open-circuit voltage at antenna terminals
𝐸 𝑖 : incident electric field
𝑙𝑒 : vector effective length

Voc can be thought of as the voltage induced in a linear antenna of length 𝑙𝑒

when 𝑙𝑒 and 𝐸 𝑖 are linearly polarized. the effective length of a linearly polarized
antenna receiving a plane wave in a given direction is defined as “the ratio of the
magnitude of the open-circuit voltage developed at the terminals of the antenna
to the magnitude of the electric-field strength in the direction of the antenna
polarization.

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Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

2- Effective Area (Aperture) Ae

The effective antenna aperture is the ratio of the available power at the
terminals of the antenna to the power flux density of a plane wave incident upon
the antenna, which is matched to the antenna in terms of polarization. If there is
no specific direction chosen, the direction of maximum radiation intensity is
implied.
𝑃𝐿
𝐴𝑒 = ,
𝑊𝑖
where
Ae is the effective aperture, m2,
PL is the power delivered from the antenna to the load, W,
Wi is the power flux density (Poynting vector magnitude) of the incident
wave, W/m2.
Using the Thevenin equivalent of a receiving antenna, we can show that
equation relates the antenna impedance and its effective aperture as
|𝐼𝐴 |2 𝑅𝐿 ⁄2 |𝑉𝐴 |2 𝑅𝐿
𝐴𝑒 = = ∙
𝑊𝑖 2𝑊𝑖 [(𝑅𝑟 +𝑅𝑙 +𝑅𝐿 )2 + (𝑋𝐴 + 𝑋𝐿 )2 ]
Under condition of conjugate matching, 𝑅𝐴 = 𝑅𝑟 +𝑅𝑙 = 𝑅𝐿 , 𝑋𝐴 = −𝑋𝐿
|𝑉𝐴 |2 1
𝐴𝑒 = ∙
8𝑊𝑖 (𝑅𝑟 +𝑅𝑙 )
For aperture type antennas, the effective area is smaller than the physical
aperture area. Aperture antennas with constant amplitude and phase distribution
across the aperture have the maximum effective area, which is practically equal
to the geometrical area. The effective aperture of wire antennas is much larger
than the surface of the wire itself. Sometimes, the aperture efficiency of an
antenna is estimated as the ratio of the effective antenna aperture and its
physical area:
𝜀𝑎𝑝 = 𝐴𝑒 ⁄𝐴𝑃

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Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

Example:
A uniform plane wave is incident upon a very short dipole. Find the effective
area Ae and the aperture efficiency, assuming that the radiation resistance is
Rr= 80 (πl/ λ)2, and that the field is linearly polarized along the axis of the dipole.
Compare Ae with the physical surface of the wire if l = λ /50 and d = λ / 300, where
d is the wire’s diameter.
Solution: -
Since the dipole is very short, we can neglect the conduction losses. Wire
antennas do not have dielectric losses. Therefore, we assume that Rl = 0. Under
conjugate matching (which is implied unless specified otherwise),
|𝑉𝐴 |2 1
𝐴𝑒 = ∙
8𝑊𝑖 (𝑅𝑟 )
The dipole is very short and we can assume that the E-field intensity is the
same along the whole wire. Then, the voltage created by the induced
electromotive force of the incident wave is
𝑉𝐴 = |𝐸| ∙ 𝑙
The Poynting vector has a magnitude of
|𝐸|2
𝑃 = 𝑊𝑖 =
2𝜂
Then,
|𝐸|2 ∙ 𝑙 2 2𝜂 (λ /50)2 2𝜂
𝐴𝑒 = ∙ = ∙
8|𝐸|2 (𝑅𝑟 ) 8 80 (π𝑙/ λ)2
3𝜆2
𝐴𝑒 = = 0.119 ∙ λ2
8𝜋
The physical surface of the dipole is
λ λ
𝐴𝑃 = 𝜋𝑑𝑙 = 𝜋 ∙ ∙ = 2.1 × 10−4 ∙ λ2
300 50
The aperture efficiency of this dipole is then
0.119 ∙ λ2
𝜀𝑎𝑝 = 𝐴𝑒 ⁄𝐴𝑃 = = 568.2
2.1 × 10−4 ∙ λ2

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Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

Relation Between the Directivity D0 and the Effective Aperture Ae

The simplest derivation of this relation goes through two stages.
Stage 1:
Prove that the ratio D0/Ae is the same for any antenna. Consider two antennas:
A1 and A2. Let A1 be the transmitting antenna, and A2 be the receiving one. Let
the distance between the two antennas be R as shown in Figure 6.2.

Figure 6.2: Two antennas separated by a distance R.

The power density generated by A1 at A2 is
𝐷1 Π1
𝑊1 =
4𝜋𝑅 2
Here, Π1 is the total power radiated by A1 and D1 is the directivity of A1.
The power received by A2 and delivered to its load is
𝐷1 Π1
𝑃𝐿1→2 = 𝐴𝑒2 ∙ 𝑊1 = ∙𝐴
4𝜋𝑅2 𝑒2
where 𝐴𝑒2 is the effective area of A2.
𝑃𝐿1→2
𝐷1 ∙ 𝐴𝑒2 = 4𝜋𝑅 2
Π1

Now, let A1 be the receiving antenna and A2 be the transmitting one. We

can derive the following:
𝑃𝐿2→1
𝐷2 ∙ 𝐴𝑒1 = 4𝜋𝑅 2
Π2

69
Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

If Π1 = Π2 , then, according to the reciprocity principle in electromagnetics

(The reciprocity in antenna theory states that if antenna #1 is a transmitting
antenna and antenna #2 is a receiving antenna, then the ratio of transmitted to
received power Ptra /Prec, will not change if antenna #1 becomes the receiving
antenna and antenna #2 becomes the transmitting one).

𝑃𝐿1→2 = 𝑃𝐿2→1 . therefore,

𝐷1 𝐷2
𝐷1 ∙ 𝐴𝑒2 = 𝐷2 ∙ 𝐴𝑒1 ⟹ = =𝛶
𝐴𝑒1 𝐴𝑒2
We thus proved that 𝛶 is the same for every antenna.
Stage 2:
Find the ratio 𝛶 = 𝐷𝑚𝑎𝑥 ⁄𝐴𝑒 for an infinitesimal dipole. The maximum directivity
𝑖𝑑
of a very short dipole (infinitesimal dipole) is 𝐷𝑚𝑎𝑥 = 1.5. the effective aperture
3𝜆2
of infinitesimal dipole is 𝐴𝑖𝑑
𝑒 = as previously mentioned. Then,
8𝜋

𝐷𝑚𝑎𝑥 1.5
𝛶= = 2 ∙ 8𝜋
𝐴𝑒 3𝜆
𝐷𝑚𝑎𝑥 4𝜋
𝛶= = 2
𝐴𝑒 𝜆
The above equation is true if there are no dissipation, polarization mismatch, and
impedance mismatch in the antenna system. If those factors are present, then
𝜆2
𝐴𝑒 = (1 − |Γ|2
) ∙ |𝛒
̂𝑤 ∙ 𝛒 |2
̂𝑎 ∙ ( ) 𝑒 ∙ 𝐷𝑚𝑎𝑥
4𝜋
2 2
𝜆2
𝐴𝑒 = (1 − |Γ| ) ∙ |𝛒 ̂𝑎 | ∙ ( ) 𝐺𝑚𝑎𝑥
̂𝑤 ∙ 𝛒
4𝜋
From these equations, we can obtain a simple relation between the antenna
beam solid angle ΩA and effective area Ae
𝜆2 𝜆2
𝐴𝑒 = 𝐷 =
4𝜋 𝑚𝑎𝑥 Ω𝐴

70
Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

Other Antenna Equivalent Areas

Before, we have defined the antenna effective area (or effective aperture) as
the area, which when multiplied by the incident wave power density, produces the
power delivered to the load (the terminals of the antenna) PL. In a similar manner,
we define the antenna scattering area As. It is the area, which when multiplied
with the incident wave power density, produces the re-radiated (scattered) power:
𝑃𝑟𝑒 |𝐼𝐴 |2 ∙ 𝑅𝑟 2
𝐴𝑠 = = ,𝑚
𝑊𝑖 2𝑊𝑖
In the case of conjugate matching,
|𝑉𝐴 |2 𝑅𝑟 |𝑉𝐴 |2 𝑅𝑟
𝐴𝑠 = ∙ = ∙ , 𝑚2
8𝑊𝑖 (𝑅𝑟 +𝑅𝑙 )2 8𝑊𝑖 (𝑅𝐴 )2
The loss area Al is the area, which when multiplied by the incident wave
power density, produces the dissipated (as heat) power of the antenna.
𝑃𝑙 |𝐼𝐴 |2 ∙ 𝑅𝑙 2
𝐴𝑙 = = ,𝑚
𝑊𝑖 2𝑊𝑖
In the case of conjugate matching,
|𝑉𝐴 |2 𝑅𝑙 |𝑉𝐴 |2 𝑅𝑙
𝐴𝑙 = ∙ 2
= ∙ 2
, 𝑚2
8𝑊𝑖 (𝑅𝑟 +𝑅𝑙 ) 8𝑊𝑖 (𝑅𝐴 )
The capture area Ac is the area, which when multiplied with the incident
wave power density, produces the total power intercepted by the antenna:
𝑃𝑐 |𝐼𝐴 |2 ∙ (𝑅𝑟 +𝑅𝑙 + 𝑅𝐿 ) 2
𝐴𝑐 = = ,𝑚
𝑊𝑖 2𝑊𝑖
In the case of conjugate matching,
|𝑉𝐴 |2 (𝑅𝑟 +𝑅𝑙 + 𝑅𝐿 ) |𝑉𝐴 |2 (𝑅𝐴 + 𝑅𝐿 ) |𝑉𝐴 |2 1
𝐴𝑐 = ∙ = ∙ = ∙ , 𝑚2
8𝑊𝑖 (𝑅𝑟 +𝑅𝑙 )2 8𝑊𝑖 (𝑅𝐴 )2 4𝑊𝑖 𝑅𝐴

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Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

The capture area 𝐴𝑐 is the sum of the effective area 𝐴𝑒 , the loss area 𝐴𝑙 and the
scattering area 𝐴𝑠 :
𝐴𝑐 = 𝐴𝑒 + 𝐴𝑙 + 𝐴𝑠
When conjugate matching is achieved,
𝐴𝑒 = 𝐴𝑙 + 𝐴𝑠 = 0.5𝐴𝑐
If conjugate matching is achieved for a loss-free antenna, then
𝐴𝑒 = 𝐴𝑠 = 0.5𝐴𝑐

72
Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

Example:
A lossless resonant half-wavelength dipole antenna, with input impedance
of 73 ohms, is connected to a transmission line whose characteristic impedance is
50 ohms. Assuming that the pattern of the antenna is given approximately by
𝑼 = 𝑩𝟎 𝒔𝒊𝒏𝟑 𝜽
Find the maximum absolute gain of this antenna.
Solution:
4𝜋 𝑈(𝜃, 𝜑)
𝐷(𝜃, 𝜑) =
∏

∏ = ∯ 𝑈(𝜃, 𝜑) 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑
𝛺𝐴

2𝜋 𝜋
= ∯ 𝑈(𝜃, 𝜑) 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 = ∫ ∫ 𝑈(𝜃, 𝜑) 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑
0 0
4𝜋
2𝜋 𝜋 2𝜋 𝜋
3 4
3𝜋 2
∏=∫ ∫ 𝐵0 𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 = ∫ ∫ 𝐵0 𝑠𝑖𝑛 𝜃 𝑑𝜃 𝑑𝜑 = 𝐵0 ( )
0 0 0 0 4
𝜋 3𝜋
Note:- ∫0 𝑠𝑖𝑛4 𝜃 𝑑𝜃 =
8

4𝜋 𝑈(𝜃, 𝜑) 4𝜋 𝐵0 𝑠𝑖𝑛3 𝜃 16
𝐷(𝜃, 𝜑) = = 2 = 𝑠𝑖𝑛3 𝜃
∏ 3𝜋 3𝜋
𝐵0 ( )
4
4𝜋 𝑈𝑚𝑎𝑥 16
𝐷𝑚𝑎𝑥 = = = 1.697 𝑤ℎ𝑒𝑛 𝜃 = 𝜋/2
∏ 3𝜋
Since the antenna is lossless, then the antenna radiation efficiency 𝑒 = 1.
4𝜋 𝑈(𝜃, 𝜑)
𝐺(𝜃, 𝜑) = = 𝑒. 𝐷(𝜃, 𝜑)
𝑃𝑖𝑛
4𝜋 𝑈𝑚𝑎𝑥
𝐺𝑚𝑎𝑥 = = 𝑒. 𝐷𝑚𝑎𝑥
𝑃𝑖𝑛
𝐺𝑎𝑏𝑠 (𝜃, 𝜑) = 𝑒𝑟 𝑒. 𝐷(𝜃, 𝜑) = (1 − |Γ|2 ) 𝑒. 𝐷(𝜃, 𝜑)

73
Mustansiriyah University Subject: Antennas
Faculty of Engineering Class: 3rd-year
Electrical Engineering Department Sixth Lecture
Teachers: Asst. Prof. Dr. Eng. Malik Al-Khalidi
Prof. Eng. Zaid Assad

𝐺𝑎𝑏𝑠,𝑚𝑎𝑥 = 𝑒𝑟 𝑒. 𝐷𝑚𝑎𝑥 = (1 − |Γ|2 ) 𝑒. 𝐷𝑚𝑎𝑥

𝑍1𝑛 − 𝑍𝑐
Γ=
𝑍𝑖𝑛 + 𝑍𝑐
73 − 50 2
𝑒𝑟 = (1 − | | ) = 0.965
73 + 50
𝐺𝑎𝑏𝑠,𝑚𝑎𝑥 = 0.965(1.697) = 1.6376
𝐺𝑎𝑏𝑠,𝑚𝑎𝑥 (𝑑𝐵) = 10 𝑙𝑜𝑔10 (1.6376) = 2.142

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