## OPEN-CIRCUIT VOLTAGE

It should be noted that the vector effective height h is defined for the situation where the antenna is used for transmission. Let us consider how h is related to receiving antenna operation.

Figure 3 shows two antenna systems in which antenna 1 with vector effective height h is used as a transmitted antenna in (a) and as a receiving antenna in (b). 101 is the terminal current and V01 is the open terminal voltage (open-circuit

voltage).

Antenna 2 is an infinitesimal dipole antenna with vector effective height hd = lz, which is used as receiving and transmitting antennas in (a) and (b), respectively.

The radiation field E1 from antenna 1 induces an open – circuit voltage at antenna 2

V02 — E1 ‘ ^Z

-jkR

= (-J 30 k—— IQ1h) • hd

Using Eq. (1), the radiation field from antenna 2 with terminal current 102 is written as

E2 — — j 30 k — /02hd

The open-circuit voltage V01 at antenna 1 induced by the radiation field from antenna 2 satisfies the relationship according to the [reciprocity theorem (1)]

^01I01 — V02I02

Substituting Eq. (4) into Eq. (6) yields

Пі = – j 30 k-

Using Eq. (5) and replacing V01 and E2 with V0 and Einc, respectively, Eq. (7) is written as

V0 — Einc • h

(a) |

L Г |

02 |

Ant. 2, h |

d |

Figure 3. Determination of open-circuit voltage at antenna 1. (a) Antenna 1 for transmission. Antenna 2 is an infinitesimal dipole for reception. (b) Antenna 1 for reception. Antenna 2 is an infinitesimal dipole antenna for transmission. |

^ Ic: |

Ant. 2, h |

d |

(b) |

(9) |

Equation (8) means that the open-circuit voltage is given by the inner product of the incident wave field and the antenna vector effective height. Example. Let us obtain the maximum open-circuit voltage for a half-wavelength dipole antenna (L = A/2) located on the z axis. The maximum open circuit voltage is obtained when the polarization of an incident wave Einc is parallel to the dipole; that is, the incident wave illuminates the dipole from the в = 90° direction. When the half-wavelength dipole has a current distribution of I(z’) = I0 cos kz’ over the antenna conductor from z’ = —A/4 to z’ = +A/4, Eq. (3) yields S = I0 A/wz using r • R = 0. Hence, h = A/wz. From Eq. (8), the maximum open-circuit voltage is |V0| = |Einc|A/w. EQUIVALENT CIRCUIT OF A RECEIVING ANTENNA The original receiving antenna problem shown in Fig. 4(a) can be handled by superimposing two cases, shown in Fig. 4(b) and (c). In Fig. 4(b) an EM plane wave illuminating the antenna induces an open-circuit voltage of V0 = Einc • h. In Fig. 4(c) the antenna acts as a transmitting antenna with terminal current 1. Superimposing these two cases leads to a relationship of |

Vrec — V0 Vtrans |

where Vrec = ZiI and Vtrans = ZAI, with ZA defined as the antenna input impedance. From Eq. (9), the terminal current 1 |

(15) (16) |

I |

I |

Vre ^inc |

+ |

-V Einc |

Z |

+ |

L |

+ |

Vn |

Vtrans |

|E. • h|2 w = L = w L 4 r L max |E. • h|2 |E. • h|2 ‘ inc ^ * inc ^ |

E 4(-^а+га) 4Rl |

(a) |

I = |

(11) (12) |

Figure 4. A receiving antenna. (a) Original receiving antenna problem. (b) Antenna with open terminals. (c) Transmitting antenna.

is given as

V0

(10)

Therefore, the equivalent circuit for Eq. (10) is as shown in Fig. 5, where the open-circuit voltage V0 is used as a voltage generator with an internal impedance ZA.

Let us express the load impedance ZL and the antenna input impedance ZA as

ZL – RL + jXL

ZA – (RA + rA) + jXA

It follows that half of the power provided by the generator is delivered to the antenna load. Note that WLmax in Eq. (15) is the maximum power that can be delivered to the load ZL, because the antenna is perfectly matched to the load. Also, note that WR is recognized as the scattered (reradiated) power from the receiving antenna (1,4,5).

Example. The relationship WR = WL obtained when the impedance is matched can be checked by a numerical technique called the method of moments (3), where the incident power Winc and the received power WL are calculated. The scattered power is obtained from Winc — WL. For a center-load dipole antenna (4), the relationship WR = WL is obtained when the antenna length L is less than 0.8A. Limits on the validity of the equivalent circuit shown in Fig. 5 are discussed elsewhere (1,4,5).

RECEIVING CROSS-SECTION AND APERTURE EFFICIENCY

The receiving cross-section Ar of an antenna is defined as the ratio of the power WL [Eq. (13)] received by the antenna to the Poynting power (power density of the incident wave, Pn = |EiJ2/Zo, where Zo = 120Ш):

_|Emc. h|2RL Z0 r lzA +ZLI2 |Emc|2 |

where Ra and rA are the radiation resistance and the loss resistance (not contributing to the radiation), respectively. Then, the power delivered to the antenna load is given as |

Ar = |

(17) |

Using an impedance mismatch factor Mimp defined as WL WLmaxMimp, Eq. (17) is rewritten as |

V0|2 Rl |

(13) |

|ZA + ZL|2 and the power consumed in the generator is given as |V0|2(RA+rA) A Iza+zlI2 |

l^nc-bl2 M 4 (RA+rA) |

Zn |

Ar – |

|Eind2 zo |Eincl2 |

(18) |

.|Einc • h|2 |

Mm |

4R« |

(14) |

where Rl = Ra + rA = Ra/^ (Appendix Eq. (a1)) is used. Using the absolute gain Ga [see Eq. (a5)], Eq. (18) is written as |

When an impedance matching condition is satisfied [i. e., when Zl is complex conjugate of Za(Rl = RA + rA and XL = |

= AlG lE-nc-hl2 r 4я a |Emc|2|h|2 |

Mim |

(19) |

Because the received power is proportional to the square of the open-circuit voltage, that is, |Einc • h|2, the third factor |Einc • h|2/|Einc|2|h|2 in Eq. (19), is the reduction factor of the received power from polarization mismatch. Let the third factor be denoted as Mpoi, which has a maximum value of 1 for the case in which h is a real constant multiplied by the complex conjugate of Einc. Eq. (19) now can be written as |

Ar = t— GMM |

(20) |

a pol imp |

4n |

The maximum receiving cross-section, which is called the effective area Aeff, is obtained when Mpol = Mimp = 1. |

^eff – 4jr Ga |

(21) |

ation power density from a reference antenna that has the same input power as the test antenna:

G(e^) — |

(a4) |

W |

Z |

W |

Ga — nD |

Let the receiving antenna gain be defined as the ratio of the effective area of the antenna, Aeff, to the effective area of an isotropic antenna, Aeff iso = A2/4w. Then, from Eq. (21), the receiving antenna gain is equal to the absolute gain when the same antenna is used as a transmitting antenna. Example. An isotropic antenna has a gain of Ga = 1 (=0 dB) by definition. Using Eq. (21), the effective area is calculated to be Aeff = 0.0796A2. An infinitesimal dipole, whose vector effective height is |h| = l|sin в|, has a gain of Ga = 1.5 ( = 1.76 dB) at в = 90° and an effective area of Aeff = 0.119A2. A halfwavelength dipole antenna with |h| = A/w yields an effective area of 0.130A2, because Ga = 1.64 (=2.15 dB). Note that all Gas are calculated using Appendix (a6) and (a7) with ^ = 1. When a receiving antenna has an aperture Aap which is much larger than the wavelength A, the performance of the receiving antenna is evaluated by how efficiently the aperture is utilized for reception. Since the receiving antenna of Aap has the potential to collect an EM wave power of Wap = PincAap, the ratio of the received power PincAeff to Wap is defined as the aperture efficiency ^ap, where |

where E0 is the far-field radiated from the reference antenna, and Win, 0 is the power input to the reference antenna. Note that the maximum value of the gain is conventionally used for the antenna gain if the coordinates (в, ф) for the direction of interest are not specified. When an isotropic antenna (hypothetical antenna radiating with uniform radiation power density in all directions) is chosen as the reference antenna, the gain is called the absolute gain and denoted as Ga. Equation (a4) becomes |

|E(R, 9, 0)|2/Win |E0I2/Wln>0 |

4nR2 |E(R, Є, ф)^ |

Ga = |

(a5) |

(a6) |

— n |

Zn |

Using Eq. (1) and Eq. (a2), |

4nR2 |E(R, Є, ф)^ |

rad |

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