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The Equivalent Circuit of the Receiving AntennaСтр 1 из 2Следующая ⇒ Abstract The Norton equivalent circuit of a receiving antenna consists of a current generator feeding two admittances in parallel, Ya and YL. Admittance YL represents the load, and the power dissipated in YL, evaluated by “circuit” methods, correctly represents the actual electromagnetic power delivered to the load. The symbol Ya represents the input admittance of the antenna radiating in its transmitting mode. Whether the power dissipated in Ya, again evaluated by circuit methods, is equal to the power scattered by the antenna has been a matter of controversy for quite a few years. The present note seeks to review thet recurring problem.
Introduction We shall derive the equivalent circuit of the typical antenna system shown in Figure 1, but the analysis can be extended to other types of antennas (see Section 6). The configuration consists of a load (a receiver), a pickup antenna (for example, a coaxial line). The main ideas of the analysis can be developed by making a few simplifying assumptions: -The antenna radiates in free space -The boundary walls of the antenna system are perfectly conducting, and -The wave guide carries a single mode The extension to multimode lines and to exterior regions Ve containing linear media (possibly anisotropic and/or nonreciprocal) is fairly straightforward, and is not germane to the present discussion. If the waveguide carries only one mode, the transverse fields in the reference cross section Sg – chosen sufficiently far away from discontinuities for all non-propagating modes to be attemuated will be of the form
Where α is normalized according to the condition
In Figure 1, space is divided into two regions& -The volume VL of the load, bounded by Sg and the perfectly conducting walls, Sw and -The exterior volume, Ve, bounded by Sg, Sw, and a spherical surface, S∞, of very large radius. The firlds in Ve, result from two contributions^ -The fields ES, HS, generated by the sources О in the presence of a metalized (short-circuited) surface, Sg, and
Хрень
- The fields Ea, Ha, due to the existence of a tangential electric field V α in the radiating aperture Sg (the transmitting mode) When the antenna is fed from the left the ratio (-l)/V has a given value Ya, namely, the admittance of the transmitting antenna. The radiated field is of the form E=Cea, In the far field, in particular
Similarly, on the load side, the ratio (I/V) has a given value YL It is to be noted that the field problem embodied in Figure 1 is a particular case of the more general problem shown in Figure 2. Here, (unxE) is given on Se, and J in Ve. The formal solution for E is
In this equation, Ve is the volume exterior to Se, r0 is an exterior point, and the (Green’s) dyadic is formed from the column vectors Gx, Gy, Gz. Thus
Vector -curlcurl Формула
Where Is a Huygens equivalent magnetic current In the application of Equation to Figure 1 JmS is vanishes on Sw Thus,
The first term on the right-hand side is the field, Es, generated by О in the presence of a short-circuited Sg. The second term is the contribution from the antenna. It has been previously denoted by Vea, hence we write
In the far field
The fields ES, HS, can further be split into the incident fields, Ei, Hi generated by the currents О radiating in free space, and the scantered fields, ESC, HSC, generated by the currebts induced on Sw and the metalized Sg. In Ve we write
Or
Power Budget for Two Sources Consider first a single source of current, J, immersed in a volume, V, bounded by a surface, S. By inserting Maxwell’s equations into the relationship
One obtains
Хрень The surface integral may be written in more detail as Формула The first two terms are the powers radiated individually by J1, and J2, and the third term is a power interaction term, which must be carefully kept in the analysis. It is clear from the presence of that “combined” term that radiated powers do not add up, except under special conditions. The point is illustrated by the examples discussed in Section 4. On the source side of Equation, we may similarly write. Формула Here again the combined effects за the sources generate an interaction term, which may either increase or decrease the total power provided by the sources with respect to the sun of the individual powers. Формула In these expressions, К is the distance to a common phase-refence point, O, F1 and F2 are functions of 0 and φ, and Ro is the characteristic impedance, of free space. The radiated power, Equation, now becomes Формула Формула We note it Формула Формула Формула Формула In the presence of two equal sources (i.e., with I1=I2=I) the total far fields is Формула Giving a radiated power Формула The factor For anti-parallel currents (i.e., for I1=-I2=I) Формула At small (kol), the two currents form a dipole line, and the two sources together radiate a power Формула That power approaches zero with kol, fundamentally because (+I) and (-I) interfere more and more destructively as their mutual distance decreases. The same kind of behavior holds in three dimensions. For example, consider the two equal electric dipoles shown in Figure 66. The fields stemming from Pel are Формула Хрень Формула Формула The leading factor is the sum of the individual radiated powers. The double integral term represents the interaction. For small (kol), Prad becomes Формула Assume, as a final exercise, that Pel in Figure 6b is left untouched, but that Pe2 is replaced by a similarly located and oriented magnetic dipole Pm. The radiation fields of Pm are Формула Формула Формула On the basis of Equation (28), the radiated power consists of three terms: Формула Or Формула The first term is due to the source radiating in the presence of a short-circuited Sg. The second and third terms represent the contribution of the magnetic current (un x E). We note that from Equation(19), Prad2 is precisely the power, Equation (23), evaluated by the rules of circuit theory. The proof rests on the fact that in the transmitting mode, the flux of power through Sg is equal to the flux through S. The forst flux is Формула The second flux, through S, is precisely Prad Since all terms in Equation (50) depend on the presence of the antenna, it is difficult to pinpoint the “scattered power”, since it does not exist as a separate entity, and the observable is actually the total radiated power. In some cases (see the example mentioned in Section 4) the interaction term vanishes, and one may be inclined, in that case, to call Prad the scattered power (more precisely, the power due to loading with YL). A Simple Application Consider again the two dipoles shown in Figure 6b, and assume that Pel is a given, active element (the source), and the Pel is a dipole moment induced in the passive, parasitic element 2. That element is in the form ofa short linear antenna, with terminals AB (Figure 7a). We shall evaluate the “s” and “a” fields for that simple configuration. The “s” fields are obtained by short-circuiting AB. At sufficiently large distances l, the induced Pe2 is proportional to Pel, and we may write Формула Where Keja is an casily determined factor. The far field of dipole Pel is a given in Equation(36), and similar relationships can be invoked for Pe2. A few simple steps lead to a far field Формула If we open the terminals AB, and load them by an impedance Zl, we create a configuration similar to that a Figure 1, where cross section Sg is now replaced by the terminals AB. The field Es in Equation (8) in now Equation (54), and the radiated field contributed by loading the antenna with ZL can be determined by means of either the Thevenin or the Norton equivalent circuits of the receiving antenna (Figure 7b). The resulting dipole Pe2 can be written as CejyPel. It radiates a field Формула The total field E=ES+Ea is equation (54), with Keja replaced by (Keja+Ceeb). It isow a simple matter to derive the radiated power, an to identify the interaction terms. For example, the term Prad in Equation (52) takes the form Формула The interacting sources are Pel and the “shorted” Pe2 given in Equation (53). The integration over can be performed, if needed by means of the relationship Формула Abstract The Norton equivalent circuit of a receiving antenna consists of a current generator feeding two admittances in parallel, Ya and YL. Admittance YL represents the load, and the power dissipated in YL, evaluated by “circuit” methods, correctly represents the actual electromagnetic power delivered to the load. The symbol Ya represents the input admittance of the antenna radiating in its transmitting mode. Whether the power dissipated in Ya, again evaluated by circuit methods, is equal to the power scattered by the antenna has been a matter of controversy for quite a few years. The present note seeks to review thet recurring problem.
Introduction We shall derive the equivalent circuit of the typical antenna system shown in Figure 1, but the analysis can be extended to other types of antennas (see Section 6). The configuration consists of a load (a receiver), a pickup antenna (for example, a coaxial line). The main ideas of the analysis can be developed by making a few simplifying assumptions: -The antenna radiates in free space -The boundary walls of the antenna system are perfectly conducting, and -The wave guide carries a single mode The extension to multimode lines and to exterior regions Ve containing linear media (possibly anisotropic and/or nonreciprocal) is fairly straightforward, and is not germane to the present discussion. If the waveguide carries only one mode, the transverse fields in the reference cross section Sg – chosen sufficiently far away from discontinuities for all non-propagating modes to be attemuated will be of the form
Where α is normalized according to the condition
In Figure 1, space is divided into two regions& -The volume VL of the load, bounded by Sg and the perfectly conducting walls, Sw and -The exterior volume, Ve, bounded by Sg, Sw, and a spherical surface, S∞, of very large radius. The firlds in Ve, result from two contributions^ -The fields ES, HS, generated by the sources О in the presence of a metalized (short-circuited) surface, Sg, and
Хрень
- The fields Ea, Ha, due to the existence of a tangential electric field V α in the radiating aperture Sg (the transmitting mode) When the antenna is fed from the left the ratio (-l)/V has a given value Ya, namely, the admittance of the transmitting antenna. The radiated field is of the form E=Cea, In the far field, in particular
Similarly, on the load side, the ratio (I/V) has a given value YL It is to be noted that the field problem embodied in Figure 1 is a particular case of the more general problem shown in Figure 2. Here, (unxE) is given on Se, and J in Ve. The formal solution for E is
In this equation, Ve is the volume exterior to Se, r0 is an exterior point, and the (Green’s) dyadic is formed from the column vectors Gx, Gy, Gz. Thus
Vector -curlcurl Формула
Where Is a Huygens equivalent magnetic current In the application of Equation to Figure 1 JmS is vanishes on Sw Thus,
The first term on the right-hand side is the field, Es, generated by О in the presence of a short-circuited Sg. The second term is the contribution from the antenna. It has been previously denoted by Vea, hence we write
In the far field
The fields ES, HS, can further be split into the incident fields, Ei, Hi generated by the currents О radiating in free space, and the scantered fields, ESC, HSC, generated by the currebts induced on Sw and the metalized Sg. In Ve we write
Or
The Equivalent Circuit of the Receiving Antenna
On the basis of Maxwell’s equations, satisfied by both the “s” and “a” fields
Integrating over all space (but excluding the volume VL of the load) yields
This equation is obtained by involving the following properties:
1. Ea and Es are perpendicular to the perfectly conducting walls properties, Sw 2. Es is perpendicular to Sg 3. The “a” and “s” fields satisfy the radiation conditions, hence the bracketed term in Equation is O(l/R3) at large distances R. The contribution from a large spherical surface of radius К there-fore vanishes in the limit In the “s” pattern, the transverse magnetic field on Sg is
Where lg is the short-circuit current induced under illumination by the exterior sources J. Alternately, the the short-circuit current density on Sg is
In Equation, we may set Ea-Vaα. On the basis of Equation, this gives
In the configuration of Figure 1, an equivalent voltage Va, appears in cross section Sg, to the left of Sg
To the right of Sg,
Expressing continuity of Ht across Sg gives
The Norton equivalent circuit of Figure 3 expresses these relationships. It can be invoked to predict the fields. It can also be used to evaluate the power flowing to the load. This is done by applying Poynting's the orem (discussed in Section 31) to YL. Thus, using peak values for field quantities,
where gL=ReYL. This is the power budget dor the enclosed volume VL of the load, assumed to contain only passive materials. Wу note that Poynting's vector is tangent to the metallic walls, and hence that is flux throught the boundary of VL reduces to the flux throught Sg. The electromagnetically derived power, Equation, is also the value predicted by circuit theory. For the power dissipated in Ya, circuit theory would also give
Central to the discussion in whether Pa has any electromagnetic meaning, for example, in terms of the power scattered by the antenna. Under matched conditions, with Ya = YL, Pa=PL, one could (erroneously) conclude that only half the available power goes to the matched load, while the other half is reradiated back to the space from whence it came. Under these circumstances, the antenna absorption efficiency could not exceed 50℅, an apparent paradox. There are consequently reasons to wonder whether the equivalent circuit, which correctly to evaluate radiated power from the receiving antenna. Such doubts were already voiced more than half a century ago by Silver, who wrote that the radiated energy was modified by the interaction between the fields. This remark forms the basis of the arguments presented in the next section.   Что делать, если нет взаимности? А теперь спустимся с небес на землю. Приземлились? Продолжаем разговор...  Живите по правилу: МАЛО ЛИ ЧТО НА СВЕТЕ СУЩЕСТВУЕТ? Я неслучайно подчеркиваю, что место в голове ограничено, а информации вокруг много, и что ваше право...  Что делает отдел по эксплуатации и сопровождению ИС? Отвечает за сохранность данных (расписания копирования, копирование и пр.)...  ЧТО ТАКОЕ УВЕРЕННОЕ ПОВЕДЕНИЕ В МЕЖЛИЧНОСТНЫХ ОТНОШЕНИЯХ? Исторически существует три основных модели различий, существующих между... Не нашли то, что искали? Воспользуйтесь поиском гугл на сайте:
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, (Fa dimensionless)
for example, is the solution of
satisfies the radiated conditions. In Equation 
on Sg, and it
is the sun of the individually radiated powers. The integral in the term between brackets therefore represents the influence of the interaction term. At small distances, i.e., for kol<1, this term approaches one, hence Prad becomes four times the power radiated by I, or, equivalently, the power radiated by 2l.
