
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, Y_{a }and Y_{L. }Admittance Y_{L} represents the load, and the power dissipated in Y_{L}, evaluated by “circuit” methods, correctly represents the actual electromagnetic power delivered to the load. The symbol Y_{a }represents the input admittance of the antenna radiating in its transmitting mode. Whether the power dissipated in Y_{a}, 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 V_{e} 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 S_{g} – chosen sufficiently far away from discontinuities for all nonpropagating 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 V_{L} of the load, bounded by S_{g }and the perfectly conducting walls, S_{w} and The exterior volume, V_{e}, bounded by S_{g}, S_{w}, and a spherical surface, S_{∞, }of very large radius. The firlds in V_{e}, result from two contributions^ The fields E_{S}, H_{S}, generated by the sources О in the presence of a metalized (shortcircuited) surface, S_{g}, and
Хрень
 The fields E_{a}, H_{a}, due to the existence of a tangential electric field V α in the radiating aperture S_{g }(the transmitting mode) When the antenna is fed from the left the ratio (l)/V has a given value Y_{a}, namely, the admittance of the transmitting antenna. The radiated field is of the form E=Ce_{a}, In the far field, in particular , (F_{a }dimensionless) Similarly, on the load side, the ratio (I/V) has a given value Y_{L} 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, (u_{n}xE) is given on S_{e}, and J in V_{e}. The formal solution for E is In this equation, V_{e} is the volume exterior to S_{e}, r_{0} is an exterior point, and the (Green’s) dyadic is formed from the column vectors G_{x}, G_{y}, G_{z}. Thus
Vector for example, is the solution of curlcurl Формула
Where satisfies the radiated conditions. In Equation Is a Huygens equivalent magnetic current In the application of Equation to Figure 1 J_{mS} is on S_{g}, and it vanishes on S_{w} Thus,
The first term on the righthand side is the field, E_{s}, generated by О in the presence of a shortcircuited S_{g}. The second term is the contribution from the antenna. It has been previously denoted by Ve_{a}, hence we write In the far field
The fields E_{S}, H_{S,} can further be split into the incident fields, E_{i}, H_{i} generated by the currents О radiating in free space, and the scantered fields, E_{SC}, H_{SC}, generated by the currebts induced on S_{w} and the metalized S_{g}. In V_{e} 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 J_{1}, and J_{2}, 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 phaserefence point, O, F_{1 }and F_{2 }are functions of 0 and φ, and R_{o} 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 I_{1}=I_{2}=I) the total far fields is Формула Giving a radiated power Формула The factor 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 k_{o}l<1, this term approaches one, hence P^{rad} becomes four times the power radiated by I, or, equivalently, the power radiated by 2l. For antiparallel currents (i.e., for I_{1}=I_{2}=I) Формула At small (k_{o}l), the two currents form a dipole line, and the two sources together radiate a power Формула That power approaches zero with k_{o}l, 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 P_{el} 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 (k_{o}l), P^{rad} becomes Формула Assume, as a final exercise, that P_{el} in Figure 6b is left untouched, but that P_{e2} is replaced by a similarly located and oriented magnetic dipole P_{m}. The radiation fields of P_{m} 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 shortcircuited S_{g}. The second and third terms represent the contribution of the magnetic current (u_{n} x E). We note that from Equation(19), P^{rad}_{2 }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 S_{g} is equal to the flux through S. The forst flux is Формула The second flux, through S, is precisely P^{rad} 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 P^{rad} the scattered power (more precisely, the power due to loading with Y_{L}). A Simple Application Consider again the two dipoles shown in Figure 6b, and assume that P_{el} is a given, active element (the source), and the P_{el} 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 shortcircuiting AB. At sufficiently large distances l, the induced P_{e2} is proportional to P_{el}, and we may write Формула Where Ke^{ja} is an casily determined factor. The far field of dipole P_{el} is a given in Equation(36), and similar relationships can be invoked for P_{e2}. A few simple steps lead to a far field Формула If we open the terminals AB, and load them by an impedance Z_{l}, we create a configuration similar to that a Figure 1, where cross section S_{g} is now replaced by the terminals AB. The field E_{s} in Equation (8) in now Equation (54), and the radiated field contributed by loading the antenna with Z_{L} can be determined by means of either the Thevenin or the Norton equivalent circuits of the receiving antenna (Figure 7b). The resulting dipole P_{e2} can be written as Ce^{jy}P_{el}. It radiates a field Формула The total field E=E_{S}+E_{a} is equation (54), with Ke^{ja} replaced by (Ke^{ja}+Ce^{eb}). It isow a simple matter to derive the radiated power, an to identify the interaction terms. For example, the term P^{rad} in Equation (52) takes the form Формула The interacting sources are P_{el} and the “shorted” P_{e2} 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, Y_{a }and Y_{L. }Admittance Y_{L} represents the load, and the power dissipated in Y_{L}, evaluated by “circuit” methods, correctly represents the actual electromagnetic power delivered to the load. The symbol Y_{a }represents the input admittance of the antenna radiating in its transmitting mode. Whether the power dissipated in Y_{a}, 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 V_{e} 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 S_{g} – chosen sufficiently far away from discontinuities for all nonpropagating 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 V_{L} of the load, bounded by S_{g }and the perfectly conducting walls, S_{w} and The exterior volume, V_{e}, bounded by S_{g}, S_{w}, and a spherical surface, S_{∞, }of very large radius. The firlds in V_{e}, result from two contributions^ The fields E_{S}, H_{S}, generated by the sources О in the presence of a metalized (shortcircuited) surface, S_{g}, and
Хрень
 The fields E_{a}, H_{a}, due to the existence of a tangential electric field V α in the radiating aperture S_{g }(the transmitting mode) When the antenna is fed from the left the ratio (l)/V has a given value Y_{a}, namely, the admittance of the transmitting antenna. The radiated field is of the form E=Ce_{a}, In the far field, in particular , (F_{a }dimensionless) Similarly, on the load side, the ratio (I/V) has a given value Y_{L} 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, (u_{n}xE) is given on S_{e}, and J in V_{e}. The formal solution for E is In this equation, V_{e} is the volume exterior to S_{e}, r_{0} is an exterior point, and the (Green’s) dyadic is formed from the column vectors G_{x}, G_{y}, G_{z}. Thus
Vector for example, is the solution of curlcurl Формула
Where satisfies the radiated conditions. In Equation Is a Huygens equivalent magnetic current In the application of Equation to Figure 1 J_{mS} is on S_{g}, and it vanishes on S_{w} Thus,
The first term on the righthand side is the field, E_{s}, generated by О in the presence of a shortcircuited S_{g}. The second term is the contribution from the antenna. It has been previously denoted by Ve_{a}, hence we write In the far field
The fields E_{S}, H_{S,} can further be split into the incident fields, E_{i}, H_{i} generated by the currents О radiating in free space, and the scantered fields, E_{SC}, H_{SC}, generated by the currebts induced on S_{w} and the metalized S_{g}. In V_{e} 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 V_{L} of the load) yields
This equation is obtained by involving the following properties:
1. E_{a} and E_{s} are perpendicular to the perfectly conducting walls properties, S_{w} 2. E_{s} is perpendicular to S_{g} 3. The “a” and “s” fields satisfy the radiation conditions, hence the bracketed term in Equation is O(l/R^{3}) at large distances R. The contribution from a large spherical surface of radius К therefore vanishes in the limit In the “s” pattern, the transverse magnetic field on S_{g} is Where l_{g} is the shortcircuit current induced under illumination by the exterior sources J. Alternately, the the shortcircuit current density on S_{g} is In Equation, we may set E_{a}V_{a}α. On the basis of Equation, this gives
In the configuration of Figure 1, an equivalent voltage V_{a}, appears in cross section S_{g}, to the left of S_{g}
To the right of S_{g},
Expressing continuity of H_{t }across S_{g }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 Y_{L}. Thus, using peak values for field quantities, where g_{L}=ReY_{L}. This is the power budget dor the enclosed volume V_{L} 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 V_{L }reduces to the flux throught S_{g}. The electromagnetically derived power, Equation, is also the value predicted by circuit theory. For the power dissipated in Y_{a}, circuit theory would also give
Central to the discussion in whether P_{a} has any electromagnetic meaning, for example, in terms of the power scattered by the antenna. Under matched conditions, with Y_{a} = Y_{L}, P_{a}=P_{L}, 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 вызывает тренды на фондовых и товарных рынках Объяснение теории грузового поезда Первые 17 лет моих рыночных исследований сводились к попыткам вычислить, когда этот... Система охраняемых территорий в США Изучение особо охраняемых природных территорий(ООПТ) США представляет особый интерес по многим причинам... ЧТО ПРОИСХОДИТ, КОГДА МЫ ССОРИМСЯ Не понимая различий, существующих между мужчинами и женщинами, очень легко довести дело до ссоры... Что делать, если нет взаимности? А теперь спустимся с небес на землю. Приземлились? Продолжаем разговор... Не нашли то, что искали? Воспользуйтесь поиском гугл на сайте:
