The effective aperture (also known as the effective area) of an antenna is the area over which the antenna collects energy from the incident wave and deliv-ers it to the receiver load. If the power density in the wave incident from the (θ, φ) direction is S at the antenna and Pr(θ, φ) is the power delivered to the load connected to the antenna, then the effective aperture, Ae, is defined as
Ae(θ, φ) = Pr(θ, φ)
S m2 (2.80)
Referring to the equivalent circuit of the receiving antenna and the load shown in Fig. 2.17, the power delivered to the load, ZL, connected to the antenna terminals is
Pr = 1
2|IR|2RL (2.81)
where, RL is the real part of the load impedance, ZL= RL+ jXL. Let Za= Ra+ jXabe the antenna impedance and Va be Thevenin’s equivalent source corresponding to the incident plane wave. The real part of the antenna impedance can be further divided into two parts, i.e., Ra= Rrad+ Rloss, where Rrad is the radiation resistance and Rloss is the loss resistance corre-sponding to the power lost in the ohmic and dielectric losses in the antenna.
If the antenna is conjugate-matched to the load so that maximum power can be transferred to the load, we have ZL= Za∗ or RL= Ra and XL=−Xa. It is seen from the equivalent circuit that the power collected from the plane wave is dissipated in the three resistances, the receiver load, RL, the radiation resistance, Rrad, and the loss resistance, Rloss. For a
ZL
IR z
ZL
Za
Va
~
Fig. 2.17 Receiving antenna and its equivalent circuit
conjugate-match, the current through all three resistances is
I = Va
RL+ Rrad+ Rloss = Va
2RL (2.82)
and the three powers are computed using the formulae
Pr= 1
2|I|2RL= 1 2
|Va|2
(2RL)2RL= |Va|2
8RL (2.83)
Pscat= 1
2|I|2Rrad = |Va|2Rrad
8R2L (2.84)
Ploss= 1
2|I|2Rloss = |Va|2Rloss
8R2L (2.85)
where Pr is the power delivered to the receiver load, Ploss is the power dissipated in the antenna, and Pscat is the power scattered, since there is no physical resistance corresponding to the radiation resistance. The total power collected by the antenna is the sum of the three powers
Pc = Pr+ Pscat+ Ploss (2.86) If the power density in the incident wave is S, then the effective collecting aperture, Ac, of the antenna is the equivalent area from which the power is collected
Ac(θ, φ) = Pc(θ, φ)
S m2 (2.87)
This area is split into three parts—Ae: the effective aperture corresponding to the power delivered to the receiver load, Aloss: the loss aperture corre-sponding to the power loss in the antenna, and As: the scattering aperture corresponding to the power re-radiated or scattered by the antenna. These are given by
Ae(θ, φ) = Pr(θ, φ)
S = |Va(θ, φ)|2
8RLS m2 (2.88)
Aloss(θ, φ) = Ploss(θ, φ)
S = |Va(θ, φ)|2Rloss
8RL2S m2 (2.89) As(θ, φ) = Pscat(θ, φ)
S = |Va(θ, φ)|2Rrad
8R2LS m2 (2.90) All these are functions of the incident direction (θ, φ).
Consider an antenna radiating into free space. Let Pt1be the total power input to the antenna. If all the power is radiated using an isotropic radiator, the power density at a distance R from the antenna is
S0 = Pt1
4πR2 W/m2 (2.91)
If the gain of the antenna is G1(θ, φ), the power density will be larger by that amount in the (θ, φ) direction, i.e.
S = Pt1G1(θ, φ)
4πR2 (2.92)
Let us label the transmitting antenna as antenna 1.
Now consider a receiving antenna (labelled as antenna 2), kept at a dis-tance R from the transmitting antenna. Let the effective receiving aperture of this antenna be Ae2. Thus, the power delivered to the matched load con-nected to the receiving antenna is
Pr2 = Pt1G1Ae2
4πR2 (2.93)
or the power-transfer ratio is Pr2
Pt1 = G1Ae2
4πR2 (2.94)
The input impedance of the receiver is the load to the receiving antenna.
Therefore, the power delivered to the matched load is the power measured by the receiver. We assume that the transmitting antenna is also matched to the source.
If we interchange the positions of the transmitter and the receiver and maintain the conjugate-match at both the antenna ports, the power-transfer ratio will be
Pr1
Pt2 = G2Ae1
4πR2 (2.95)
where G2 is the gain of antenna 2 and Ae1 is the effective aperture of an-tenna 1. Since the ports are assumed to be matched, the power-transfer ratios are the same from the third result of the reciprocity theorem. Hence we can write
G1Ae2
4πR2 = G2Ae1
4πR2 (2.96)
or
G1
Ae1 = G2
Ae2 (2.97)
Since the antennas are arbitrarily excited, this result shows that the gain-to-effective aperture ratio is a constant for any antenna.
Now, to find the constant we need to relate the fields to the antenna parameters in the transmit and receive modes. Since the above result is independent of the antenna, we select the simplest of the antennas, the Hertzian dipole, to obtain the constant. For simplicity, we also assume that the radiation efficiency of the Hertzian dipole is unity, thus, the directivity and gain are equal, i.e., D(θ, φ) = G(θ, φ).
From the previous analysis [Eqn (2.54)], the radiation resistance of the Hertzian dipole is
Rrad = 2 3πη
dl λ
2
= πη D
dl λ
2
(2.98) D = 3/2 of a Hertzian dipole is used to get the last part of Eqn (2.98). If E is the electric field strength, the power density at the antenna is
S = 1 2
E2
η (2.99)
The open-circuit voltage induced in the dipole, which is also Thevenin’s equivalent voltage shown in the circuit, is
Va= Edl (2.100)
assuming E and dl are oriented in the same direction (Fig. 2.17). Power delivered to a matched load Rrad is
Pr = |Va|2
8Rrad (2.101)
Dividing the power in the matched load by the power density, we get the effective aperture area of the Hertzian dipole as
Ae= |Va|2 8Rrad
2η
|E|2 (2.102)
Substituting the expressions for Va and Rrad and simplifying, we get D
Ae = 4π
λ2 (2.103)
If the radiation efficiency is not unity, we may replace D by G and write G
Ae = 4π
λ2 (2.104)
Thus, the ratio of the gain and the effective aperture is equal to 4π/λ2 for any antenna.
EXAMPLE2.16
For a Hertzian dipole with radiation efficiency less than 1, show that the gain and the effective aperture are related by
G Ae
= 4π λ2
Solution: Consider a Hertzian dipole which is terminated in a matched load ZL= RL+ jXLand receiving electromagnetic energy. Let Ra= Rrad+ Rloss be the antenna resistance and, under matched condition, RL= Ra. If Va is the open-circuit voltage, the power delivered to the matched load is
Pr= |Va|2
8RL = |Va|2 8Ra
Substituting Ra= Rrad+ Rloss into Eqn (2.52) and rearranging Ra= Rrad
κ
Therefore, the power delivered to the load is Pr= |Va|2κ
8Rrad
If E is the electric field strength of the incident wave, the average power density associated with the wave is given by
S = 1