Antenna Handbook VOLUME I FUNDAMENTALS AND MATHEMATICAL TECHNIQUES Edited by Y. T. La Electromagnetics Laboratory Department of Electrical and Computer Engineering University of Illinois- Urbana s. W. Lee Electromagnetics Laboratory Department of Electrical and Computer Engineering University of Illinois- Urbana -,'.,' r ;-,-,:;_, CT: :DI B r-' -" .. " .. -' ""1 m t,.. Basics 1-25 into the space. If the antenna is lossy, the radiated power P3 is only a portion of P2 The ratio (57) is called the radiation efficiency. Three Gains in State (k, U) As indicated in (39) the antenna radiates in two orthogonal polarizations (u: v). Alternatively, its polarization can be represented by (within a phase factor) a single unitary vector U in the manner described in (46) and (47). Then the radiation intensity in watts per steradian in state (k, U) is J(k) = IA(kW = IA(k, uW + IA(k, vW (58) which includes radiation in all (both) polarizations. The radiated power P3 by the antenna in space is P3 = Jll dO J211 dcp [J(k) sin 0] o 0 (59) The general definition of a gain (dimensionless) in the direction k and for all polarizations is G(k) = 4n J(k) = 4n IA(k)12 = 4n intensity of the antenna in direction k (60) Pn Pn a reference power where Pn is a reference power. Depending on Pm there are three commonly used gains: (a) realized gain GI(k) if Pn = PI = power incident at the antenna, (b) gain G2(k) if Pn = P2 = power accepted by the antenna, (c) directivity D(k) if Pn = P3 = power radiated by the antenna. The relations among the three gains are The three gains are graphically illustrated in the lower half of Fig. 14. Three Gains in State (k, u) (61) The antenna has a polarization U as defined in (47). In defining gains we may specify a preferred polarization u. Then the radiation intensity in state (k, u) is J(k, u) = IA(k, uW = IA(k),u*12 (62) in watts per steradian. The ratio 1-26 Fundamentals and Mathematical Techniques

..",.PJ ---. P 2 ... ' RADIATION INTENSITY IN (k, u) IA(k,uW-IA(k)ou'12 P MISMATCH. PL EFFICIENCY. P3 POWERS: INCIdENT ACCEPTED RADIATED G(k) - g(k,u) + g(k, v) IA(k)12-IA(k,u)12+ IA(k,vW RADIATION INTENSITY IN (k, U) POLARIZATION EFFICIENCY IUou'I2 Fig, 14, Six definitions of gain (a b means at = b). (Courtesy G. A. Deschamps) = J(k, u) = lu. *12 P J(k) u (63) is called the polarization efficiency in state (k, u) for an antenna that has polarization U. With respect to state (k, u) we define a gain 4.rr 4.rr I 12 g(k, u) = P J(k, u) = P A(k)'u* = pG(k) n n. = 4.rr intensity of the antenna in state (k, u) a reference power (64) Again, depending on the reference power Pm there are three gains g .. g2, and d, which are analogous to the cases associated with (60). These three gains are illustrated in the upper half of Fig. 14. The gain defined in (64) is called the partial gain for a specific polarization u. The (total) gain G(k) in (60) is the sum of partial gains for any two orthogonal polarizations: which follows from (58). Peak Gains G(k) = g(k, u) + g(k, v) All of the six gains depend on the observation direction k = (J, f/J). In applications we often use the maximum of a gain as a function of k. Thus, corresponding to (60), we define a peak gain for all polarizations by G = max G(k) = IA(kW n (65) Basics 1-27 In a similar manner we define a peak gain for a specific polarization u. When a gain is given without a specified observation direction, it is customarily assumed to be the peak gain. Gain-Related Terms When defining antenna terms related to gain, we can use anyone of the six definitions of gain. If the radiation in all polarizations is of interest, we use G(k) in (60). If the radiation in a specific polarization is of interest, we use g(k, u) in (62). With this understanding we use G(k) to represent a typical gain in the discusston . below. An isotropic radiator is a hypothetical lossless antenna having equal radiation intensity in all directions, i.e., J(k) == a constant. If an input power Pn were fed to an isotropic radiator, its radiation intensity in watts per steradian would be (66) Here Pn can be anyone of three powers explained at the beginning of this section. In terms ot- (66) the gain definition in (60) has the following interpretation: intensity of the antenna in direction k = ~ - - ~ - - ~ ~ ~ - - ~ - - ~ - - ~ ~ ~ ~ - - - - - - - - - -intensity of an isotropic radiator fed by the same power (67) We often express the dimensionless G by its decibel (dB) value: 1OloglOG. Sometimes we write dB as dBi, where the letter "i" emphasizes that the gain is over an isotropic radiator. Consider an antenna with gain G(k) fed by an input power Pn. Its radiation intensity J(k) would be the same as that for an isotropic radiator if the latter were fed with an input power given in watts by EIRP = PnG(k) (68) where EIRP stands for equivalent (effective) isotropically radiated power. We often express EIRP by lOloglOPnG in decibels referred to 1 W (dBW). When an antenna is used for receiving, a figure of merit of the antenna is defined by peak gain of the antenna - = - - ~ - - - - - = = ~ - - - - - - - - - - - - - - -Ta noise temperature of the antenna G (69) Ta, which is usually given in kelvins (K), is discussed in Chapter 2. 1-28 Fundamentals and Mathematical Techniques 11. Pattern Approximation by (cos 6)9 The far-field patterns of many aperture-type antennas have two characteristics: (a) A single major lobe exists in the forward half-space z > 0 and the lobe maximum is in the +z direction. (b) The radiation in the backward half-space z < 0 i.s negligible. For these types of antennas their far fields can be approximated by simple analytical functions described in this section. For an x-polarized antenna, such as the flanged waveguide shown in Fig. 15, we have (for 0 O. Note that F, is not a good approximation of F for z < o. (b) Electric current alone (Fig. lOc). The equivalent source in (51) consists of both electric and magnetic surface currents. An alternative equivalent source is (2Js> 0), which doubles the electric current in (51) and contains no magnetic current. The field produced by (2Js,0) in the absence of ~ , is F2, which is the second approximation of F for z > o. (c) Magnetic current alone (Fig. lOd). Another equivalent source is (0, 2Ks) , which doubles the magnetic current in (51) and contains no electric current. The field produced by it in the absence of ~ l is F3, which is the third approximation of F for z > O. Version (a) above is sometimes known as the Stratton-Chu formula. Field Fl is the average of F2 and F3 Kirchhoff's approximation has been applied to transmission through aperture problems in which the aperture is not necessarily in a planar screen, such as radiation from an open-ended waveguide. 8. Scattering by an Obstacle Consider the scattering problem sketched in Fig. 11. The obstacle (scatterer) is illuminated by an incident plane wave Fi given by (52) where kl = wave vector with magnitude k = wv',UE and pointing in the direction of propagation of Fi Z = y-l = VflJt: Ul = a unitary vector which describes the polarization of Fi and is orthogonal to kl C = amplitude of Fi in (watt)1I2 (meter)-' We say that Fi is in state (k" ud. In the presence of the obstacle the total field 2-22 p \ Y. SOURCE'l.k \ 1 Fundamentals and Mathematical Techniques . N Fig. 11. We are interested in the scattered far field at r with state (k2' U2) when an obstacle is illuminated by an incident plane wave Fi with state (kb u,). everywhere is the sum of Fi and a scattered field F. At a far-field observation point r, field F is represented by a spherical wave {E(r)} ....., } e-ikr, H(r) y'Yk2 x A(k2) r (53) The distance r is measured from a reference point 0 in the vicinity of the obstacle. Vector k2 has a magnitude equal to k and is in the direction of Or. The amplitude vector A(k2) given in (watt) 112 may be decomposed into two orthogonal components, as discussed in Section 7, Chapter 1. Let us concentrate on a particular component of A(k2) with polarization U2, namely, AuI. We introduce the notation (after G. A. Deschamps) A(2, 1) = A(k2)uI (54) which represents the scattering amplitude in state 2 (direction k2 and polarization U2) due to an incident plane wave in state 1. Using this notation, we will introduce some definitions and theorems for the scattering problem in Fig. 11. Bistatic Cross Section The bistatic cross section (BCS) in square meters from state 1 to state 2 is defined by (55 a) Theorems and Formulas 2-23 431 (intensity of scattered field in state 2) = - - - - ~ ~ ~ ~ ~ ~ ~ - - - - - - - - - - ~ - - ~ - -power density of incident plane wave in state 1 (56a) 431 (intensity of scattered field in state 1') = - - - - ~ ~ ~ ~ ~ ~ ~ - - ~ - - - - - - ~ - - ~ ~ power density of incident plane wave in state 1 (56b) We will consider two examples of ReS. (a) For a smooth conductor whose dimension is large in terms of wavelength and is illuminated by a linearly polarized incident field, its ReS is approximately independent of polarization and is given by (57a) where R land R2 are two principal radii of curvature of the conducting body at the specular point P (Fig. 12). The point P is determined by the relation in which the surface normal N is in the opposite direction ofkl. We assume that there is only one such specular point. If the conductor is a sphere of radius a, the use of (57a) leads to ReS ::: 7ra2, a well-known result. (b) For a rectangular conducting plate of dimension a x b in the z = 0 plane, the ReS is again approximately independent of polarization and is given by Res - 1 (k b 0 sin a Sinf3)2 = - a cos 0----3l a f3 where (00, d, we have e(x y) = y (Z 1-) 112 cos(.!!. x) , {3 cd c (70a) h(x,y) = (Z k/{3)-1 Z x e (70b) where Z = y-I = ()11f:)"2 {3 = k[l - (n/kc)21'/2 (71) As for the characteristic impedance, there are at least four commonly used definitions, namely, ZcI = Zk/{3 Zc2 = 2(d/c)ZcI Zc3 = Zc4 = (n/2)(d/c)ZcI (72a) (72b) (72c) (72d) Thus there are at least four different ways to define the (modal) voltage and current in the rectangular waveguide, corresponding to the four choices of Zc in (72). Once a choice of Zc is made, we find the (total) voltage and current at the reference plane z = 0 are given by Theorems ana Formulas v = V+ + V- = VZc(a + b) 1= r + r = ffc(a - b) The input impedance of the antenna is defined by 1 V Zin = Yin = I The relations among r, Ziri' and Yin are Zin = Yc = 1 + r Zc Yin 1 - r r = (Zin/Zc) - 1 = 1 - (YinIYc) (Zin/ZJ + 1 1 + (YinIYc) 2-29 (73) (74) (75) (76) (77) It is clear from (76) that antenna impedance Zin also depends on the choice of Zc. The power transmitted (radiated) from the source into the free space via the antenna is given by p, = Re{ J J (E x H*).ZdxdY} = lal2 - Ibl2 = la12(1 - Ir12) = Re{VI*} (78a) (78b) (79) The reflected power back toward the source (due to the mismatch of Zc and Zin) is Ib12. 10. Three Ideal Sources for Transmitting Antennas As discussed in the previous section, an antenna as a one-port device can be described by either wave amplitudes (a, b), or by circuit parameters (V, I). When the antenna is used for transmitting, a source is connected to the transmission line (or feed waveguide). Following Deschamps [6], we will introduce three principal ideal sources: (a) unit amplitude source (defined by a = 1 (watt)1/2) (b) unit voltage source (V = 1 volt) ( c) unit current source (I = 1 ampere) Their graphical representation is shown in Fig. 16. Note that the dot in each source indicates the propagation direction of the incident wave leaving the dot, the positive voltage terminal at the dot, or the current out of the dot. The term "ideal" is used because those sources have special internal impedance Z.n viz., 2-30 Fundamentals and Mathematical Techniques a a-I 8 ~ --V-I 1-1 b-r b c Fig. 16. Three ideal sources for transmitting antennas. (a) Zs = Zc for the amplitude source, in which Zc is the characteristic impedance of the transmission line. Thus the source is matched to the transmission line, and there is no reflection at the source-transmission-line junction. (However, the mismatch at the antenna-transmission-line junc-tion may still exist; Zc mayor may not be equal to Zin') (b) Zs = 0 for the voltage source so that there is no internal voltage drop within the source. (c) Zs -+ 00 for the current source so that the total source current enters the transmission line. In Fig. 16 we use the zigzag line, straight line, and gap inside the circles to indicate the source impedances. For a transmitting antenna anyone of the three ideal sources may be used as the excitation. For each case, we list in Table 1 the internal feed waveguide quantities (V, 1, a, b) and the external radiation quantities (radiated field F and its power PI)' Here F stands for vector fields (E, H). The subscripts (1,2,3) are used to identify excitations due to a unit (amplitude, voltage, current) source. For example, at is the value of a when the excitation is a unit amplitude source, and F3 is the value of F when the excitation is a unit current source. The three fields (Ft, F2, F,,) in Table 1 are related by (80) F = nc 13 F = ~ V Z . (1 + Zin) 13 F 3 1 - rat t 2 ( Zc at t (81) where at = 1 (watt)t/2, V2 = 1 volt, and 13 = 1 ampere. 11. Three Ideal Meters for Receiving Antennas When an antenna is used for receiving, the source is replaced by a receiver or, for our present purpose, a meter. Again, following Deschamps we introduce three principal ideal meters: (a) Amplitude meter, which measures the incoming traveling-wave amplitude b at a reference plane z = 0, and has internal source impedance Zs = Zc such that the meter is matched to the feed waveguide. (The antenna impedance Zin mayor may not be matched to Zc-) Note that b is a complex number, including both magnitude and phase information. Theorems and-Formulas 2-31 Table 1. Relationships for Transmitting Antennas ~ Amplitude Voltage Current Quantity V (volt) v'2,.( 1 + r)a I V2 = 1 Zin/3 1 (ampere) VY;(1 - r)al YinV2 13 = 1 a (watt) 1/2 al = 1 1 vv:. ( Yin) T Yc 1 + Yc V2 1 (Zin ) . Tv'2,. ZC + 1 h b (watt)112 fal ~ VY; (1 - i;) V2 ~ vz:: (Zin - 1) 1 2 c Z 3 c Pt (watt) (1 - If12)a12 V/Re Yin I/ReZin Field FI F2 F3 Source Zc (matched) o (short) 00 (open) impedance (b) Voltmeter, which measures the voltage Vat a reference plane z = 0, and has an infinite internal source impedance (Zs -+ 00) such that V is the open-circuit voltage. (c) Ammeter, which measures the current 1 at a reference plane z = 0, and has a zero internal source impedance (Zs = 0) such that 1 is the short-circuit current. The graphical representation of meters is given in Fig. 17. Note our convention that the circles represent sources, while squares represent meters (compare Figs. 16 and 17). When the receiving antenna is illuminated by an incident field we may connect any of the three ideal meters in Fig. 17 to its feed waveguide, corresponding to matched-load, open-circuit, and short-circuit situations. Table 2 lists the internal feed waveguide quantities (V, I, a, b). The subscripts (4, 5,6) are used to identify the use of an (amplitude meter, voltmeter, ammeter). For the same incident field the three situations are related by (82) a-O 8-.. -.. ----v 1=0 v-O a b b c Fig. 17. Three ideal meters for receiving antennas. (a) Amplitude. (b) Voltage. (c) Current. 2-32 Fundamentals and Mathematical Techniques Table 2. Relationships for Receiving Antennas ~ Amplitude Voltmeter Ammeter Quantity V (volt) VZcb4 Vs 0 1 (ampere) -ffcb4 0 16 a (wattt2 0 1 TffcVs 1 ' TVZch b (watt) 112 b4 1 Tffcvs 1 -Tffc/6 Source impedance Zc (matched) 00 (open) o (short) (83) Commonly, Vs is known as the open-circuit voltage, and 16 as the short-circuit current of a receiving antenna. 12. Reciprocity between Antenna Transmitting and Receiving The reciprocity theorem in a circuit is well known. Its application to antennas, however, is not simple for the reason explained below. If viewed from the transmission line, the antenna looks like a circuit element whose transmitting and receiving properties are describable by two (complex) numbers: (a, b) or (V, 1), as discussed in Section 9. Outside the antenna in the free-space region, either the radiated field of the transmitting antenna or the incident field on the receiving antenna is more complex. They are vector fields characterized by polarization and spatial variation, which are not describable by circuit quantities. Hence the reciprocity for an antenna cannot be simply stated by the usual exchange of sources and meters. In this section we will give two reciprocity relations for antennas. Reciprocity Involving General Incident Fields Consider the transmitting situation in Fig. 18a, where the antenna is excited by a unit amplitude source with al = 1 (watt)1/2. The radiated field is FI = (EJ, "d. In the receiving situation (Fig. 18b) the same antenna is connected to an amplitude meter with matched impedance (so that a4 = 0), and is illuminated by an incident field F4 = (E4' "4). Then a reciprocity states [6] (84) Theorem. and-Formulaa a z-o I I I 0)-1 I 81 I b) I 2-33 b . Fig. 18. An antenna in transmitting excited with a unit-amplitude source, and in receiving connected with an amplitude meter. (a) Transmitting. (b) Receiving. Here S is an arbitrary closed surface which encloses the antenna but excludes the source of F4. Its outward unit normal is N. It is emphasized that F4 is the incident field that would exist in the absence of the antenna in Fig. 18b, and does not include the scattered field F;' from the antenna. It can be shown, however, that (84) remains valid if F4 is replaced by F4 + F;', because the cross flux (F;'SF.) = O. Reciprocity Involving Plane Waves In many applications we are interested in a special case of Fig. 18, namely, the radiated field F. in the transmitting situation is known in the far-field zone (Fig. 19a) , and the incident field F4 in the receiving situation is a plane wave (Fig. 19b). We express the radiated field F. by 1:-0 I I I I 0)-1 I e: I b) I I I a I F = {E1(r)} _ { V Z ~ ( k ) } e-ikr, H.(r) y'Vk x A(k) r (85) b Fig. 19. Same antenna as Fig. 18 except incident field F4 is a plane wave. (a) Transmitting. (b) Receiving. 2-34 Fundamentals and Mathematical Techniques where Z = y-l = (pJe) 1/2. We represent the incident field F4 by (86) whose amplitude is C in (watt)I12/meter, propagation vector (-k), and polarization vector u*. Then a reciprocity relation states [6] b4 = -j). ( ~ ) [A(k)u*] (87) where)' = 'brlk is the free-space wavelengtn. If the open-circuit voltage and short-circuit current in the receiving situation are of interest, we may use (87) or (84) in conjunction with (82) and (83). The interpretation of (87) is as follows: The received amplitude b4 of an antenna under the matched condition due to the incidence of a plane wave in state (-k, u*) is proportional to the antenna radiation far-field amplitude in state (k, u). [Remember that (-k, u*) and (k, u) have the same polarization.] Antenna Effective Length Consider the receiving situation in Fig. 19b, where the antenna is connected to a matched meter and is illuminated by an incident plane wave F4 defined in (86). The effective length h is defined by the relation (88) where V 5 is the open-circuit voltage of the transmission line at a reference plane z = 0, and E4(r = 0) is the electric field at a reference point 0 of the incident plane wave given in (86). From (88), (87), and (82) we conclude that h = vf( 1 + i:)( -j :J A(k) (89) where Z = (pJe) 1/2. Clearly h is a complex vector in meters. The relation in (89) relates the effective length h and the far-field amplitude A(k) of an antenna. Transmitting Field in Terms of Effective Length Excited by a unit amplitude source (a = 1), the transmitted far field El is given in (85). Now if the same antenna is excited by a unit current source (I = 1), the corresponding transmitted far field E3 is then given by 1 ( Z. ) e-jkr E3(r) - 2 VZcZ 1 + i: A(k) -,-, (90) where we have used the formula (81) in relating El and E3. Replacing A by h in accordance with (89), we rewrite (90) as Theorems andbrmulas which may be used as an alternative definition of the effective length h. Receiving Cross Section 2-35 (91) Consider the receiving situation in Fig. 19b, where the antenna is connected to a matched amplitude meter. The incident plane wave is given in (86), with amplitude C and state (-k, u*). The received wave amplitude is b4, given in (81) . . Then we define the receiving cross section (effective area) a in state (k, u) of the , antenna by a(k, u) = \bC4\2 received power of the antenna under matched condition = (92) power density of incident plane wave in state (-k, u*) in square meters. Making use of (87), we have (93) A partial gain of the antenna in state (k, u) is defined by (Section 10, Chapter 1) g.(k, u) = 4n \ d. A(k).u* \2 (94) From (93) and (94) we have 4n g.(k, u) = ;e a(k, u) (95) where a;.. = l2/4n is sometimes known as the receiving cross section of the fictitious isotropic radiator. As discussed in Section 10, Chapter 1, the partial gain g, in a pre-ferred polarization u is related to the (total) gain G, in all (both) polarizations by (96) where U describes the polarization of the antenna in direction k. Substituting (96) in (95) gives (97) which relates the receiving cross section and the gain of an antenna. The factor 2-36 Fundamentals and Mathematical Techniques (98) is called the polarization efficiency (or polarization mismatch/actor). We emphasize that unitary vector u* (not u) describes the polarization of the incoming plane wave in direction (-k), while U describes the polarization of the receiving antenna in the outgoing direction (+k). For example, let k = z and U = (i - jY)/V2 for a right-hand circularly polarized antenna. If the incoming plane wave is also right-hand circularly polarized, we have u* = (i + jY)/V2 and p = 1. 13. The Radar Equation and Friis Transmission Formula The configuration of a bistatic radar is sketched in Fig. 20. The obstacle (target) in the vicinity of point 0 is illuminated by an incident wave from the transmitting antenna at point 1. A part of the scattered energy is received by a receiving antenna at point 2. Both distances R I and R2 are large in terms of wavelength so that the scatterer is in the far-field zones of the antennas. The problem at hand is to determine the power ratio P 2/ PI, where PI = power incident from the generator to the transmitting antenna, P2 = power received by the receiver via the receiving antenna. To this end we need to define the antennas and the obstacle more precisely, as below. (a) In the vicinity of point 0, the radiated field of the transmitting antenna is in R 1 - DISTANCE ill R2 ~ DISTANCE 20 TRANSMITTING ANTENNA Fig. 20. A bistatic radar. RECEIVING ANTENNA RECEIVER -Theorems 2-37 the direction kl and has a polarization described by a unitary vector VI' In other words, the state of the transmitting antenna at point 0 is (kl' VI)' The gain of the transmitting antenna in the direction kl is GI(kl), which is related to the directivity DI(kl) by (99) Here 1] I is the transmitting antenna efficiency accounting for the conductor and dielectric losses, and r I is the reflection coefficient accounting for the impedance , mismatch between the transmitting antenna and its feed (Section 10, Chapter 1). # (b) If the receiving antenna were used for transmitting (Fig. 21), the state of its radiated field at point 0 would be (k2' V2), and its gain would be G2(k2). A relation similar to (99) holds for G2 and D2. (c) Due to the illumination from the transmitting antenna the obstacle produces a scattered field at point 2, with state (ko = -k2, Vo). Its bistatic cross section from state (kb VI) to state (ko, Vo) is denoted by the BCS. With the above description we may now calculate various power quantities. The power density Wo of the radiated field at point 0 from the transmitting an-tenna is (100) in watts per square meter. The power density W2 of the scattered field at point 2 from the obstacle is ""-2 Fig. 21. The receiving antenna of Fig. 20 is now used for transmitting. 2-38 Fundamentals and Mathematical Techniques (101) The received power P2 by the receiver at point 2 via the receiving antenna may be calculated from (92) and (97) with the result A.2 P2 = W2 4.7lIUoU212G2(k2) (102) Combining (100) through (102), we have the power ratio for the bistatic radar in Fig. 20, namely, (103) which is known as the radar (range) equation. Next, we will consider a special case of (103). Let us remove the obstacle in Fig. 21, and study the direct power transmission from point 1 to point 2 (Fig. 22). Making use of the following relations: Un = UI, BCS = 4.7lRl then (103) becomes the Friis transmission formula for the far-field transmission between two antennas, namely, RECEIVER Fig. 22. Power transmission between two antennas. Theorems anclFormulas 2-39 (104) The factor (A./4.71R)2 is called the free-space loss factor, and it accounts for the loss due to the spherical spread of the transmitted field. The polarization efficiency (105) accounts for the loss due to polarization mismatch between the transmitting and the receiving antennas. If both antennas have the same polarization, UI = Ul and , p = 1. (The conjugate on Ul is due to the fact that U2 refers to a propagation direction opposite that of UI.) Note the symmetry between 1 and 2 in the right-hand side of (104), implying that the same formula applies if the roles of the receiving and transmitting antennas are interchanged in Fig. 22. 14. Noise Temperature or an Antenna For a high-resolution antenna a crucial factor that limits its ability to detect a weak signal is the antenna noise, which is the subject of the present section. Antenna Noise Temperature The receiving antenna sketched in Fig. 23 receives the desired signal as well as noise. We denote the available noise power at terminall: by Pa in watts. It is a common practice to express Pa in terms of an (effective) antenna noise temperature Ta in kelvins via the relation E' RECEIVER TRANSMISSION LINE E I I I I I ANTENNA I I Fig. 23. A receiving antenna receives thermal noises from external noise emitters. 2-40 Fundamentals and Mathematical Techniques (106) where k' = Boltzmann's constant = 1.38 x 10-23 11K !l.f = bandwidth of the antenna receiving system in hertz The interpretation of (106) is that antenna noise power Pa is equal to that of a matched resistor whose physical temperature is Ta. Hereafter we will use Pa and 'Fa interchangeably. Clearly, 'Fa depends on the antenna receiving characteristics and external noise emitter. Brightness Temperature of an Emitter Every object with its physical temperature above the absolute zero (0 K) is an emitter of thermal energy in the form of electromagnetic waves. Within a narrow frequency band the amount of energy radiated in direction k and polarization u or simply in state (k, u) is proportional to a parameter called the brightness temperature Tb(k, u). The latter is related to the physical temperature T;, of the emitter via the relation (107) where the dimensionless parameter e(k, u) is called the emissivity in state (k, u) of the emitter. To determine e(k, u) we illuminate the emitter by an incident plane wave in state (-k, u*). See Fig. 24. Then incident power absorbed by emitter e(k, u) = incident power intercepted by emitter For a large emitter an approximate formula for calculating its emissivity is e(k, u) == 1 - If(k, u)i2 (108) Here f(k, u) is the reflection coefficient at the surface of the emitter. Since there are two independent polarizations we can define two independent (partial) brightness temperatures Tb(k, u) and Tb(k, v) in a given direction k, for each choice of orthogonal, unitary, polarization vectors (u, v). A common choice of (u, v) is u = horizontal polarization, v = vertical polarization (109) The sum of the two (partial) brightness temperatures gives the total brightness temperature (110) which is proportional to the total energy radiated by the emitter in direction k in Theorems and Formulas 2-41 ~ ~ PLANE / WAVE Fig. 24. To determine the emissivity (k, u) of a noise emitter we illuminate the emitter by an incident plane wave of state (-k, u*). both polarizations. A blackbody absorbs all the incoming energy impinging on it (a perfect absorber). Its emissivity is unity (a perfect emitter), and 'I2Tb(k) = Tb(k, u) = Tb(k, v) = 1'p for any direction k. Calculation of AntenlUl Noise Temperature The power received by a receiving antenna can be traced to three sources: the desired signal, interference from other coherent radiators, and incoherent noise from noise emitters. Fig. 25 shows some important noise emitters in the free-space environment. Let us concentrate on a typical noise emitter (No.1 in Fig. 23). Its contribution to the antenna noise temperature can be calculated from the following formula: where 1 Jlf J21f Ta = 4Jl 0 sinOdO 0 dlj>[Tb(-k,u*)d(k,u) + Tb(-k,v*)d(k,v)] (111) (u, v) = unitary vectors describing two orthogonal polarizations d(k, u) = (partial) directivity of the antenna in state (k, u) Tb( -k, u*) = (partial) brightness temperature of the emitter in state (-k, u*). Note that states (k, u) and (-k, u*) have the opposite duections but the same polarization If there is more than one noise emitter in space, the superposition principle applies to the calculation of Ta. This is so because noise emitters are incoherent and the superposition of powers (temperatures) is permissible. For an idealized omni-directional antenna, which radiates equally in both polarizations, we have d(k, u) = d(k, v) = 112, and 2-42 Fundamentals and Mathematical Techniques AMPLIAER NOISE GROUND THERMAL NOISE I -tl RADIO STARS TRANMISSION-LINE NOISE DISCHARGE LIGHTS IGNITION NOISE TV STATION Fig. 25. Important noise emitters for a receiving antenna. (Courtesy Y. T. Lo) 1 Jll J2.:rr To = TaO = 8.1l 0 sin 0 dO 0 d [Tb( -k, u*) + Tb( -k, v*)] (112) Fig_ 26 presents some typical values of TaO of an omnidirectional antenna, together with the noise temperature of a typical receiver. Noise Power at the Receiver's Terminal Corresponding to the antenna noise temperature Ta in (111), the available noise power at terminal in Fig. 23 is given by Pa according to (106). This incoming power propagates through the transmission line, and arrives at the receiver's terminal with its value in watts equal to P'a = k' + (1 -1])Tol where 1] = power transmission efficiency between terminals and To = physical temperature of the transmission line (113) Theorems anclf.ormulas 60 50 :.: 40 ! 30 , 20 Q I ... ' 10 0 -10 10 I ATMOSPHERIC SUMMER: 2200-2400 HOURS. NYC Z WINTER: 0800-1200 HOURS. NYC t;/ i'. ..... ,/ r--V ""- URBAN '-.HUMAN-MADE '-* >-D 1.5 D 1.5 R, Z; Rr . -*kL 1 so that sin (kL cos lJ)/kL cos lJ -+ I. 3-16 Fundamentals and Mathematical Techniques analysis. In Chart 6, results are presented for an arbitrary-length dipole and for a half-wavelength dipole (L = lI2). The Traveling-Wave Antenna The traveling-wave antenna [8,10,13,15,16] is defined as an antenna whose conductors support a current distribution described by a simple traveling wave. In comparison to the dipole antenna, which supports two counter-directed traveling waves giving rise to the sinusoidal (standing-wave) distribution, the traveling-wave antenna supports a single wave. A simple example of a traveling-wave antenna and its attendant characteristics are presented in Chart 7. For this antenna the current distribution is given by I(z) = I(.,e-jkz, where 10 is its amplitude and k the free-space wave number. The important feature of the traveling-wave antenna is that the maximum radiation is neither end-fire (8 = 0) nor broadside (8 = n/2). Rather it occurs at an intermediate angle and creates a single major conical lobe independent of antenna length and tilted in the direction of wave travel with minor lobes determined by length. An approximation for the angle of the main lobe from the end-fire direction [13,16] is given by 8m = cos-'[1- 0.371/(Ul)]. The Small Loop Antenna The small loop antenna [8,9,10,11,13,14,15] of radius a, carrying a uniform current I, can be analyzed using the techniques presented previously. The characteristics for the small loop are presented in Chart 8. The striking similarity between the expressions in Chart 8 for the small loop and those for the short dipole in Chart 4 is immediately evident. Clearly, the electric current I on a dipole of length L generates fields which are obviously related to those of a loop of radius a carrying an electric current I. Since the fields radiated by such a loop are identical with those radiated by a magnetic dipole [1] if where m is the magnetic current, we can obtain the fields due to a small current loop from those of a short electric dipole using duality. The procedure is as follows: (a) The original problem is a linear electric current I of length L in a medium with constitutive parameters E and Jl. Obtain the solution to this problem. (b) By the duality of Chart 3, construct the solution of the problem of a linear magnetic current m and length L in a medium with constitutive param-eters E and Jl. This is obtained from the solution in (a) above by setting I, = VEfJlm = jkl(na2)fL, E,(r) = VJlfEH2(r), and H,(r) = Note that the magnetic current is spatially orthogonal to the plane of the loop. (c) Replace mL by jWJl/(na2) to provide the solution for the small loop radiating element. Techniques for"1.ow-Frequency Problems Chart 6. Characteristics for a Dipole with Sinusoidal Current Distribution I(z) 10 sin ko(Ll2 - Izl), -Ll2 z Ll2 Arbitrary-Length Dipole Az(r) I( e-jkr cos [(kLl2) cos 0] - cos (kLl2) , 21fr ksin20 Er(r) Hr(r) Pr[1l,12] 8. e-jkr cos [(kLl2) 0] - cos (kLl2) ly E 2:rrr sm 0 .i...l e-jkr cos [(kLl2) cos 0] - cos (kLl2) 'I" 1 0 2:rrr sin 0 (if/(? in {sin kL [Si(kt.) - Y2Si(2kL)] where + (1 + coskL)[ln(kLy) - Ci(kL)] - COS2kL [In (2kLy) - Ci(2kL)]} far zone 3-17 Si(x) = JX sinx dx, o x Iny = 0.5772 Er(r) Pr Rr D Half-Wavelength Dipole (L = A12) l e-jkr cos [(:rr/2) cos 0] o 2:rcr k sin20 8. e-jkr cos [(:rr:2) cos 0] 1 Y E J 2:rrr sm 0 fifM - Ci(2:rc)] = 0.194Y,u/d(,z [In (2:rcy) - Ci(2:rc)] = 0.194Y,u/E(Rr = 73 g in free space) 4cos2[(:rr/2) cos 0] = 1644 cos2[(:rr/2) cos 0] sin20[ln(2:rry)-Ci(2:rr)] sin20 4 In (2:rry) - Ci (2:rr) = 1.644 3-18 Fundamentals and Mathematical Techniques Chart 7. Characteristics for the Traveling-Wave Antenna I(z) loe-;kz, 0 z L (l L) e-;kr e-;k(U2)(; -0056) sin [(kLl2)(1 - cos 8)] o 43fT. (kLI2)(1 - cos 8) 0" (l L) . 8 e-;kr -;k(LI2)(1-OO56) sin [(kLI2)(1 - cos 8)] IWP. 0 sm 43fT e (kL/2)(1 _ cos 8) .i. 'k (l L) . 8 e-;kr -;k(U2)(1-OO56) sin [(kLl2)(1 - cos 8)] T I 0 sm 43fT e (kLI2)(1 - cos 8) {In(2Jly) -1 + In (2L1,t) - Ci(2kL) + {In(2Jly) -1 + In (2L1,t) - Ci(2kL) + The Perfect Ground Plane The perfect ground plane, i.e., the planar, perfectly conducting surface with conductivity a 00, is often found in the environment of antenna structures. Since the previous characterizations have been for antennas in homogeneous media with real E and 11, the introduction of conducting bodies such as infinite planes will require a modification. The boundary conditions at any perfectly conducting surface require that the tangential electric field satisfy, with n(r) the outward normal to the surface and r on the surface, n(r) x E(r) = 0 (33) and that the magnetic field satisfy n(r) H(r) = 0 (34) The exclusion of fields from the interior of the surface leads to the generation of surface charge and current densities given by n(r) E(r) = es(r)/E n(r) x H(r) = Jir) (35) Techniques foi Low-Frequency Problema Chart 8. Characteristics of the Small Loop Antenna /(cp) A(r) E(r) H(r) Er(r) Hr(r) Pr Rr GD(), cp) D r = a, () = n/2, -jkr ( k 1 ) r (lna2) cos () ear + ? A e 2 J -jkr ( k 1 ) -8(/na2)sin()- k - - - "2 r,;- 4/2na4 y"Ek -6-4nr r r 1.5 sin2() 1.5 3-19 An arbitrarily oriented antenna located above a perfectly conducting ground plane must generate a vanishing electric field at the surface in order to satisfy the boundary condition on the electric field. Such a condition can be satisfied by creating an equivalent problem where the ground plane is removed and an image source is introduced to produce, when combined with the original source, a vanishing tangential electric field at the location of the plane. The appropriate images for electric and magnetic currents are shown in Fig. 2. A close inspection will show that the tangential electric and normal magnetic fields vanish at the plane surface and that the fields are identical in the upper half-space. 3-20 a b Fundamentals and Mathematical Techniques IMAGE SOURCE Fig. 2. Equivalent source distributions for sources above perfectly conducting ground planes. (a) Original problem. (b) Equivalent problem. The theory of images can be applied to configurations other than infinite planes. In fact, a source in the presence of any surface composed of intersecting planes can be represented in terms of the source and multiple images [1, 8]. This concept applies to both exterior and interior (e.g., waveguide) problems. An example of the use of images is provided by the monopole antenna driven against a perfectly conducting ground plane. The fields radiated by the monopole into the upper half-space are identical with those of the dipole. The radiated power is contained only in the upper half-space and thus is only half of that for the dipole antenna. Hence we can immediately infer that the input voltage to the monopole need only be one-half that of the dipole to produce the same fields in the upper half-space. Thus Pr = /flo2 4 ~ ([sin (kL)][Si(kL) - 1/ZSi(2kL)] + [1 - cos (kL)][ln (kLy) - Ci(kL)] - [cos (kL)l[ln (2kLy) - Ci(2kL)]} The input resistance is given by Rr = Prllo2, and for a quarter-wavelength monopole Rr = / f 8 ~ {In(2Jly) - Ci(2Jl)} = 36.5 n Techniques forLow-Frequency Problems 3-21 The Rectangular-Aperture Antenna The rectangular-aperture antenna [1,3,8,9,10,11,13,15], shown in Fig. 3, has many features in common with the electric dipole. In the following we will illus-trate the use of equivalence and images to provide a convenient mechanism for evaluating its fields. In each case the electric-field distribution in the aperture will be assumed to be known. Then the fields produced by the aperture can be obtained using equivalence, images, and the field representations presented earlier. For an assumed field distribution in the aperture in a perfectly conducting plane, the pictorial representation of Fig. 4a is appropriate. Using equivalence, the , situation in Fig. 4b applies and the fields in the half-space z > 0 are identical with those in the original problem but zero for z < O. In this figure Ja = i x Hand K" = -i x E. As a result of the vanishing fields for z < 0, the perfectly conducting plane is completed through the aperture as shown in Fig. 4c. Using images, the plane can be removed and the situation in Fig. 4d holds. Note that the fields are the correct fields for the half-space z > 0 and not for z < O. The fields radiated by the aperture distribution can be evaluated using equations in Charts 1 and 2 with only a magnetic current source or an equivalent electric vector potential. In the following we tabulate the expressions necessary to construct the radiation fields and certain characteristics for some commonly encountered aperture distributions defined in general as z / / / / r / / / / / v /'" / _-----...1 17- OD x Fig. 3. The rectangular aperture or slot antenna in a perfectly conducting plane. 3-22 Fundamentals and Mathematical Techniques (1-00 (1-00 (1-00 I J J J I ZERO ZERO I AELDS AELDS I I I I tE J. J. J ~ K. K. I I I ----.z I a b c d I I Fig. 4. Equivalent source distributions and their environments for fields in the right half-space. (a) Aperture field distribution in a perfectly conducting plane. (b) Using equivalence with (a). (c) Perfectly conducting plane is completed through aperture. (d) Using images to (c). In the far zone the electric vector potential is given by F = x __ e - dz' o(z') . dy' eiky' sin 8sintp dx' EQ(X',y')ei/ex'sin8costp 1 -ikrJoo JLJ2 JL/2 2.1l' r -00 -L 12 -L 12 y (37) and the radiated fields by Er(r) = jkr x F = jk !F! e(O, cp) Hr(r) = VE/p, r x Er(r) = jWE!F! 11(0, cp) where e( 0, cp) = q, cos 0 cos cp + 8 sin cp b( 0, qJ) = r x e( 0, cp) = q, sin cp - 8 cos 0 cos cp (38) (39) Since the far-zone fields are simply related to the far-zone electric vector potential, only the electric vector potential will be presented in the following charts. In Chart 9 the far-zcrre electric vector potential is presented for various lengths and widths of rectangular apertures with uniform excitation. Also presented are some radiation characteristics for the short, thin-slot antenna excited by a uniform transverse electric field. The results presented for the short, thin slot are consistent with the relationship between the impedances of complementary structures [13], I.e. , Techniques for Low-Frequency Problems Chart 9. Far-Zone Characteristics for Apertures with Uniform Aperture Distribution Uniform Aperture Distribution E"(x,y,z) = yEod(z), Aribtrary Length: Lx;, Ly " e-ikr sin [(kLx;12) sin o cos . R y y z Fig. 6, continued. Techniques for Low-Frequency Problems 3-31 Impedance: -3-dB Gain: Resistive at Bandwidth dB above Pattern Type Conllguratlon Jr, R(O) Percent Isotropic Dipole Polarization Number II y Slot In large -@tt ground plane L x 350 70 2.14 0 h E L-A/2 ...... I-!/d-29 Vertical y x full-wave -N-. loop 45 13 3.14 1 h F D-A/',; D/d-36 Helical over II y reflector screen, ~ . tube 6A long 130 200 10.14 8 eire. G coded Into 6 turns A/4 apart II y ~ Rhombic ~ 1 6000 R ,., x 600 100 16.74 14.5 h H L-9A 1-9A/2 Parabolic II y with folded ~ - . dipole feed 300 30 14.74 12.5 h H (A/2) D-5A/2 y Hom, #-. coaxial feed 50 15.14 h H 35 13 L-3A R L f 1-3A II G y I ~ ....... -f-II y II .. , . Y Y II H ~ - -Fig. 6, continued. 3-32 -/ / a / / Fundamentals and Mathematical Techniques Fig. 7. Images for imperfect ground analysis. / / / / / b z z Fig. 8. Plane-wave reflection at an interface. Techniques for Low-Frequency Problems 3-33 radiated field El(1, cp) due to the given antenna is computed for the case when it is located in an infinite homogeneous medium at its original location. Then the reflected field F R ( (1, cp) is calculated from (54) where Er( (1, cp) is the field due to the image of the original source in a perfect ground plane and p is the unit vector normal to the plane of incidence (p = z x .at). Finally, the total field E( (1, cp) is the sum of the two contributions: (55) Chart 11 illustrates the steps and the results of a procedure for evaluating the fields of a vertical and horizontal short dipole carrying a constant current 10 The procedure for analyzing an arbitrary antenna, though more involved, can proceed in a similar manner. The Fresnel reflection coefficients are strictly valid only for plane waves and are therefore not rigorously valid for antennas near ground planes [13,23,24,25]. An-ays The subject of antenna arrays has been extensively documented and will be discussed in Chapters 13, 14, and 17. Previous work is described in the writings of Bach and Hansen [26], Kraus [14], Jasik [10], Stutzman and Thiele [13], Weeks [9], and Ma [27]. Here we will merely show the relationship of some aspects of array analysis to previously discussed subjects in this chapter. In the following we will present expressions which are useful for expressing the fields due to an array of identical antenna elements, each having an identical current distribution differing by, at most, a complex scaling constant. The similarity of current distribution implies identical orientation of the elements. The field due to an electric current source J(r) can be evaluated using the equations in Charts 1 and 2. Some elements in an antenna array are shown in Fig. 9. Using element number 1 as the reference element in an N-element array we establish our coordinate system definitions so that a linear shift of the nth element will cause it to be co-incident with element 1. In this case such a shift between coincident points is r n - rl or r' n - r' I. The relationship between the element current distributions is given by (56) The magnetic vector potential in the far zone for the nth element is . . 1 e-jkr f .. , An(r) = aneJkr'(r" - rl) 4- -- Jt(r')eJkr.r dv' :IT r v (57) or equivalently, Chart 11. Evaluation of Field Components for a Short Dipole of Constant Current in Presence of a f4'initely Conducting Ground z z I I to' P.o .y tl' P.o' (] 10 J( Vertical HOrizontal Step I. The field, due to a primary source in free space, is evaluated: -jk(r-h cos 8) Er( 0, 91) = 0 jWlJ.o/oL sin 0 e 4.7lr Ay = loL Er(O,91) = jWlJ.rl x t x A . e-jk(r-h cos 8) = }wlJ.o/oL A (-coS91cP - cosOsm916) .7lr Step 2. The field, due to the image source, is evaluated: . . e-jk(r+"oos8) E/(O, 91) = 6}wlJ.o/oLsmO e-jk(r+h cos 8) E/(O,91) = jWlJ.o/oL A..... (cos 914> + cosOsin916) Step 3. Evaluate the ground-reflected field using (54): ER(O,91) = RyE/(O,91) ER(O,91) = RyE/(O,91) + (Rh - Ry)(E/(O,91)'44> e-jk(r+h cos 8) = jWlJ.oloL A..... (-Rh'COS 91 4> + RycosOsin 91 6) Step 4. Evaluate the total field using (55): -jkr . E( 0, 91) = 6 jWlJ.o/oL sin 0 _e - (elkh cos 8 + R e -jkh cos 8) 4.7lr v -jkr E(O, 91) = jWIJ.I..JoL e4.7lr [-4> cos 0 (ejkhOOS8 + Rhe-jkhCOs8) - OcosOsin91(ejkhCOS8 - Rve-jkhCoS8)] Techniques for Low-Frequency Problems Fig. 9. Antenna array geometry. (58) The far-zone magnetic vector potential for the array is then N A(r) = At(r) L aneikr'(r" - rl) (59) n=t where the summation, denoted by [(8, cp), is referred to as the array [actor: N [(8,cp) = L aneju'(r" - rl) (60) n=t This factorization separates the element contribution or element pattern At(r) from the array pattern [(8, cp), which depends only on the relative source strengths and 3-36 Fundamentals and Mathematical Techniques locations. This factoring is involved in a procedure referred to as pattern multiplication. The electric and magnetic fields in the far zone can then be calculated using (16) and (17) in Chart 2: E(r) = jwp, 1(0, cp) r x r x A(r) H(r) = -jk 1(0, cp) r x A(r) (61) Chart 12 presents the radiation fields for N-element arrays of uniform current (Io) and equally phased dipoles of spacing d. The computations liave been performed using the preceding equations for two cases, namely, arrays of collinear elements and parallel elements. The procedure used in deriving the directive gain for short elements (kL ~ 1) follows that in [26]. Results obtained using the expressions for directive gain at broadside (0 = nl2, cp = n/2) in decibels relative to a single isotropic source are presented versus element spacing d in Fig. 10 for a collinear array and in Fig. 11 for the parallel array. The number of elements N is a parameter. Some reference data for practical arrays is also presented in Fig. 12 [21]. In the previous discussion the current distribution on each element was assumed to be known and unperturbed by the presence of the other elements in the array. For adequate interelement spacing this is a valid assumption, but for close spacing the interaction of the elements can cause perturbations of the elemental current distributions and, of course, the input impedance for the elements. The rigorous treatment of these effects requires a consistent treatment of the boundary-value problems, as discussed later in this chapter. 3. Integral Equations in Antenna Analysis In previous discussions the current distributions on radiating elements were assumed to be known. This, however, is generally not the case since the precise description of the current distribution on a metallic structure such as an antenna in the presence of an exciting source, such as a voltage generator at its terminals, involves the solution of a complicated boundary-value problem. In the following we will describe the integral equations which can be solved for the unknown source distributions induced by specified excitations, and in Part 2 we will consider the numerical solution of these equations and some associated issues. Perfectly Conducting Wires and Bodies Here we will focus our attention on radiating structures composed of perfect electric conductors over which the boundary conditions given by (33), (34), and (35) must hold. Furthermore, we will devote our attention mainly to wire struc-tures, with conducting bodies touched on briefly. In order to facilitate ensuing discussions concerning integral equations, questions concerning validity of each specific equation, the existence or uniqueness of solutions, and various features of the limiting process which reduce the integral representations for radiated fields to integral equations for unknown source dis-Chart 12. Radiation Fields for Vertical Collinear and Parallel Arrays of Short Dipoles Supporting Equal Amplitude and Equal Phase Currents z z JIC. ~ II .1(1 I I II ~ l I til til x J( Collinear Parallel Elemental Vector Potential e-ikr sin [(kLl2) cos 0] AI(r) = Z 4Jlr (/nL) (kLl2) cos 0 N f(O.rp) = 2: eik(II-I)tlcosO = eilk(N-I)dI2ICososin[(kNdI2)cosO] 11=1 sin [(kdI2) cosO] Er(r) = 6jw,u(/nL)sinOe-ikrsin[(kLl2)cosO] 4Jlr (kLl2) COS 0 X eilk(N-I)dI2Icoso sin [(kNdI2) cos 0] sin [(kdI2) cos 0] Go(O.rp) = sin20 sin2[ (kNdI2) cos 0]/sin2[ (kdI2) cos (J] N-I e-ikr sin [(kLl2) cosO] A1(r) = Z 4Jlr (/nL) (kLl2) cos 0 Array Factor (all = 1) f(O. rp) = eilk(N-I)dI2IsinOsinq> sin [(kNdI2) sin Osinrp] sin [(kdI2) sin 0 sin rp] Far-Zone Field Er(r) = 6 jw,u(/nL) sin 0 e-ikr sin [(kLl2) cos 0] 4Jlr (kLl2) cos 0 x eik(N-I)(dI2)sinOsinq> sin [(kNdI2) sin Osin rp] sin [(kdI2) sin 0 sin rp] Directive Gain Go(O. rp) = sin20 sin2[(kNdI2) sin 0 sin rp]/sin2[(kdl2) sin 0 sin rp] N-I I I 2N/3 + 4 2: (N - m)[(sinmkd)/(mkd)3 - (cosmkd)/(mkdf] 2N/3 + 2 2: (N - m)({[(mkdf - 1]sinmkd}/(mkd)3 + (cosmkd)/(mkd)2) m=! m=1 3-38-14 12 / II ~ '/ !/iI '/ IV/, I / Ii 2 IlL ~ o o If /L ~ ::...-0.2 Fundamentals and Mathematical Techniques ./ ........... V --~ V ""' ...... L """ N-;7 /' V -......... ....... ./ V ............ """ V ~ V --...... ./ " ........ L V ~ V ~ ...... L ~ V )/ ~ -~ --... V V V --./ V V V ~ -............. ./ ~ .J' / V ~ V/ V V ~ 2 ~ V lL "" V 0.4 0.6 0.8 1.0 1.2 1.4 1.6 ELEMENT SPACING (d/A)-WAVELENGTHS Fig. 10. Gain of a collinear array of short dipoles relative to isotropic source. tributions are necessarily glossed over. For further information the reader is referred to Stratton [2], Silver [3], and Poggio and Miller [7]. The field equations in Chart 1 are a convenient point to begin the construction of the integral equations. The electric-field representations will be used since, for antennas, the driving source is most easily specified in terms of voltage or electric field. Later, the magnetic-field representation will be discussed in relation to large conducting surfaces. The representations in Chart 1 are for the fields due to volumetric distributions of sources. For electric current sources constrained to a surface S (which may be considered to be the boundary of V in Chart 1) the integral representations are simply modified in that the volume densities become surface densities and the volume integral becomes a surface integral over S. The integral representations for the electric field due to electric sources over S are shown in Chart 13. The general boundary-value problem of determining the current distribution on a perfect electric conducting surface is approached using an integral equation. The boundary condition on S is stated as n(r) x Et(r) = 0, rES (62) Techniques fOF- Low-Frequency Problems 16 -~ ~ \ / /" / ~ -~ " ,r ",. ~ \' N-Y ./ .JII / 14 / [!Y / ",. ~ \1 V/ lY /' ~ ' V j / .-L 1/1 ~ / y / \ ~ V' ,/ ./ "'"'-./ I II / ~ '/ ' , \-V / ~ ./ II Vi / Y ~ \ 1\ ~ / ..,.. VII 1/ / ' ..... ~ ~ VII J ~ , ~ V ./ ............ VI 1/ 2/ / "" ~ il J .;V ........ ~ ..... r// ~ / 2 j ~ 1/ '/ ./ . ~ o o 0.2 0.4 0.6 0.8 1.0 1.2 ELEMENT SPACING (d/}..)-WAVELENGTHS .JII " / ./ ..,.. ." V ~ 1.4 Fig. ll. Gain of equiphased short-dipole array relative to isotropic source. 3-39 . with n(r) the outwardly pointing normal to Sand E,(r), the total electric field at the surface. The total field is composed of an incident or driven portion Ei(r) and a portion generated by the induced surface sources J.\.(r') and pAr') referred to as E(r). The boundary condition then requires n(r) x E(r) = -n(r) x Ei(r) , (63) The field component E(r) is given by the integral representations in Chart 5 but with the observation point r on the surface. The procedure of taking the observation point to the surface must be performed delicately due to the singularity in