Chapter 6 Integral solutions of Maxwell’s equations 6.1 Vector Kircho? solution: method of Stratton and Chu One of the most powerful tools for the analysis of electromagnetics problems is the integral solution to Maxwell’s equations formulated by Stratton and Chu [187, 188]. These authors used the vector Green’s theorem to solve for ? E and ? H in much the same way as is done in static ?elds with the scalar Green’s theorem. An alternative approach is to use the Lorentz reciprocity theorem of § 4.10.2, as done by Fradin [74]. The reciprocity approach allows the identi?cation of terms arising from surface discontinuities, which must be added to the result obtained from the other approach [187]. 6.1.1 The Stratton–Chu formula Consider an isotropic, homogeneous medium occupying a bounded region V in space. Themediumisdescribedbypermeability ?μ(ω), permittivity ?epsilon1(ω), andconductivity ?σ(ω). The region V is bounded by a surface S, which can be multiply-connected so that S is the union of several surfaces S 1 ,...,S N asshowninFigure6.1;theseareusedtoexclude unknown sources and to formulate the vector Huygens principle. Impressed electric and magnetic sources may thus reside both inside and outside V. Wewishtosolvefortheelectricandmagnetic?eldsatapoint r within V. Todothiswe employ the Lorentz reciprocity theorem (4.173), written here using the frequency-domain ?elds as an integral over primed coordinates: ? contintegraldisplay S bracketleftbig ? E a (r prime ,ω)× ? H b (r prime ,ω)? ? E b (r prime ,ω)× ? H a (r prime ,ω) bracketrightbig · ?n prime dS prime = integraldisplay V bracketleftbig ? E b (r prime ,ω)· ? J a (r prime ,ω)? ? E a (r prime ,ω)· ? J b (r prime ,ω)? (6.1) ? H b (r prime ,ω)· ? J ma (r prime ,ω)+ ? H a (r prime ,ω)· ? J mb (r prime ,ω) bracketrightbig dV prime . (6.2) Notethatthenegativesignontheleftarisesfromthede?nitionof ?n astheinward normal to V asshowninFigure6.1.WeplaceanelectricHertziandipoleatthepoint r = r p where we wish to compute the ?eld, and set ? E b = ? E p and ? H b = ? H p in the reciprocity theorem, where ? E p and ? H p are the ?elds produced by the dipole (5.88)–(5.89): ? H p (r,ω)= jω?×[?pG(r|r p ;ω)], (6.3) ? E p (r,ω)= 1 ?epsilon1 c ?× parenleftbig ?×[?pG(r|r p ;ω)] parenrightbig . (6.4) Figure 6.1: Geometry used to derive the Stratton–Chu formula. We also let ? E a = ? E and ? H a = ? H, where ? E and ? H are the ?elds produced by the impressed sources ? J a = ? J i and ? J ma = ? J i m within V that we wish to ?nd at r = r p . Since the dipole ?elds are singular at r = r p , we must exclude the point r p with a small spherical surface S δ surroundingthevolume V δ asshowninFigure6.1.Substitutingthese?eldsinto(6.2) we obtain ? contintegraldisplay S+S δ bracketleftbig ? E × ? H p ? ? E p × ? H bracketrightbig · ?n prime dS prime = integraldisplay V?V δ bracketleftbig ? E p · ? J i ? ? H p · ? J i m bracketrightbig dV prime . (6.5) A useful identity involves the spatially-constant vector ?p and the Green’s function G(r prime |r p ): ? prime × bracketleftbig ? prime ×(G ?p) bracketrightbig =? prime [? prime ·(G ?p)] ?? prime2 (G ?p) =? prime [? prime ·(G ?p)] ? ?p? prime2 G =? prime (?p ·? prime G)+ ?pk 2 G, (6.6) where we have used ? prime2 G =?k 2 G for r prime negationslash= r p . We begin by computing the terms on the left side of (6.5). We suppress the r prime de- pendence of the ?elds and also the dependencies of G(r prime |r p ). Substituting from (6.3) we have contintegraldisplay S+S δ [ ? E × ? H p ] · ?n prime dS prime = jω contintegraldisplay S+S δ bracketleftbig ? E ×? prime ×(G ?p) bracketrightbig · ?n prime dS prime . Using ?n prime · [ ? E ×? prime ×(G ?p)] = ?n prime · [ ? E ×(? prime G × ?p)] = (?n prime × ? E)·(? prime G × ?p) we can write contintegraldisplay S+S δ [ ? E × ? H p ] · ?n prime dS prime = jω?p · contintegraldisplay S+S δ [?n prime × ? E] ×? prime GdS prime . Figure 6.2: Decomposition of surface S n to isolate surface ?eld discontinuity. Next we examine contintegraldisplay S+S δ [ ? E p × ? H] · ?n prime dS prime =? 1 ?epsilon1 c contintegraldisplay S+S δ bracketleftbig ? H ×? prime ×? prime ×(G ?p) bracketrightbig · ?n prime dS prime . Use of (6.6) along with the identity (B.43) gives contintegraldisplay S+S δ [ ? E p × ? H] · ?n prime dS prime =? 1 ?epsilon1 c contintegraldisplay S+S δ braceleftbig ( ? H × ?p)k 2 G ? ?? prime × bracketleftbig (?p ·? prime G) ? H bracketrightbig +(?p ·? prime G)(? prime × ? H) bracerightbig · ?n prime dS prime . We would like to use Stokes’s theorem on the second term of the right-hand side. Since the theorem is not valid for surfaces on which ? H has discontinuities, we break the closed surfacesinFigure6.1intoopensurfaceswhoseboundarycontoursisolatethedisconti- nuitiesasshowninFigure6.2.Thenwemaywrite contintegraldisplay S n =S na +S nb ?n prime ·? prime × bracketleftbig (?p ·? prime G) ? H bracketrightbig dS prime = contintegraldisplay Gamma1 na +Gamma1 nb dl prime · ? H(?p ·? prime G). For surfaces not containing discontinuities of ? H the two contour integrals provide equal and opposite contributions and this term vanishes. Thus the left-hand side of (6.5) is ? contintegraldisplay S+S δ bracketleftbig ? E × ? H p ? ? E p × ? H bracketrightbig · ?n prime dS prime = ? 1 ?epsilon1 c ?p · braceleftbiggcontintegraldisplay S+S δ bracketleftbig jω?epsilon1 c (?n prime × ? E)×? prime G + k 2 (?n prime × ? H)G + ?n prime ·( ? J i + jω?epsilon1 c ? E)? prime G bracketrightbig dS prime where we have substituted ? J i + jω?epsilon1 c ? E for ? prime × ? H and used ( ? H × ?p)· ?n prime = ?p ·(?n prime × ? H). Now consider the right-hand side of (6.5). Substituting from (6.4) we have integraldisplay V?V δ ? E p · ? J i dV prime = 1 ?epsilon1 c integraldisplay V?V δ ? J i · bracketleftbig ? prime ×? prime ×(?pG) bracketrightbig dV prime . Using (6.6) and (B.42), we have integraldisplay V?V δ ? E p · ? J i dV prime = 1 ?epsilon1 c integraldisplay V?V δ braceleftbig k 2 (?p · ? J i )G +? prime · [ ? J i (?p ·? prime G)] ?(?p ·? prime G)? prime · ? J i bracerightbig dV prime . Figure 6.3: Geometry of surface integral used to extract E at r p . Replacing ? prime · ? J i with ?jω?ρ i from the continuity equation and using the divergence theorem on the second term on the right-hand side, we then have integraldisplay V?V δ ? E p · ? J i dV prime = 1 ?epsilon1 c ?p · bracketleftbiggintegraldisplay V?V δ (k 2 ? J i G + jω?ρ i ? prime G)dV prime ? contintegraldisplay S+S δ (?n prime · ? J i )? prime GdS prime bracketrightbigg . Lastly we examine integraldisplay V?V δ ? H p · ? J i m dV prime = jω integraldisplay V?V δ ? J i m ·? prime ×(G ?p)dV prime . Use of ? J i m ·? prime ×(G ?p) = ? J i m ·(? prime G × ?p) = ?p ·( ? J i m ×? prime G) gives integraldisplay V?V δ ? H p · ? J i m dV prime = jω?p · integraldisplay V?V δ ? J i m ×? prime GdV prime . We now substitute all terms into (6.5) and note that each term involves a dot product with ?p. Since ?p is arbitrary we have ? contintegraldisplay S+S δ bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime + + 1 jω?epsilon1 c contintegraldisplay Gamma1 a +Gamma1 b (dl prime · ? H)? prime G = integraldisplay V?V δ bracketleftbigg ? ? J i m ×? prime G + ?ρ i ?epsilon1 c ? prime G ? jω ?μ ? J i G bracketrightbigg dV prime . The electric ?eld may be extracted from the above expression by letting the radius of the excluding volume V δ recede to zero. We ?rst consider the surface integral over S δ . ExaminingFigure6.3weseethat R =|r p ? r prime |=δ, ?n prime =? ? R, and ? prime G(r prime |r p ) = d dR parenleftbigg e ?jkR 4π R parenrightbigg ? prime R = ? R parenleftbigg 1 + jkδ 4πδ 2 parenrightbigg e ?jkδ ≈ ? R δ 2 as δ → 0. Assuming ? E is continuous at r prime = r p we can write ? lim δ→0 contintegraldisplay S δ bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime = lim δ→0 integraldisplay Omega1 1 4π bracketleftBigg ( ? R × ? E)× ? R δ 2 +( ? R · ? E) ? R δ 2 ? jω ?μ( ? R × ? H) 1 δ bracketrightBigg δ 2 dOmega1 = lim δ→0 integraldisplay Omega1 1 4π bracketleftbig ?( ? R · ? E) ? R +( ? R · ? R) ? E +( ? R · ? E) ? R bracketrightbig dOmega1 = ? E(r p ). Here we have used integraltext Omega1 dOmega1 = 4π for the total solid angle subtending the sphere S δ . Finally, assuming that the volume sources are continuous, the volume integral over V δ vanishes and we have ? E(r,ω)= integraldisplay V parenleftbigg ? ? J i m ×? prime G + ?ρ i ?epsilon1 c ? prime G ? jω ?μ ? J i G parenrightbigg dV prime + + N summationdisplay n=1 integraldisplay S n bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime ? ? N summationdisplay n=1 1 jω?epsilon1 c contintegraldisplay Gamma1 na +Gamma1 nb (dl prime · ? H)? prime G. (6.7) A similar formula for ? H can be derived by placing a magnetic dipole of moment ?p m at r = r p and proceeding as above. This leads to ? H(r,ω)= integraldisplay V parenleftbigg ? J i ×? prime G + ?ρ i m ?μ ? prime G ? jω?epsilon1 c ? J i m G parenrightbigg dV prime + + N summationdisplay n=1 integraldisplay S n bracketleftbig (?n prime × ? H)×? prime G +(?n prime · ? H)? prime G + jω?epsilon1 c (?n prime × ? E)G bracketrightbig dS prime + + N summationdisplay n=1 1 jω ?μ contintegraldisplay Gamma1 na +Gamma1 nb (dl prime · ? E)? prime G. (6.8) We can also obtain this expression by substituting (6.7) into Faraday’s law. 6.1.2 The Sommerfeld radiation condition In § 5.2.2 we found that if the potentials are not to be in?uenced by e?ects that are in?nitely removed, then they must obey a radiation condition. We can make the same argumentaboutthe?eldsfrom(6.7)and(6.8). Letusallowoneoftheexcludingsurfaces, say S N , to recede to in?nity (enclosing all of the sources as it expands). As S N →∞any contributions from the ?elds on this surface to the ?elds at r should vanish. Letting S N be a sphere centered at the origin, we note that ?n prime =??r prime and that as r prime →∞ G(r|r prime ;ω) = e ?jk|r?r prime | 4π|r ? r prime | ≈ e ?jkr prime 4πr prime , ? prime G(r|r prime ;ω) = ? R parenleftbigg 1 + jkR 4π R 2 parenrightbigg e ?jkR ≈??r prime parenleftbigg 1 + jkr prime r prime parenrightbigg e ?jkr prime 4πr prime . Substituting these expressions into (6.7) we ?nd that lim S N →S ∞ contintegraldisplay S N bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime ≈ lim r prime →∞ integraldisplay 2π 0 integraldisplay π 0 braceleftbigg bracketleftbig (?r prime × ? E)× ?r prime +(?r prime · ? E)?r prime bracketrightbig parenleftbigg 1 + jkr prime r prime parenrightbigg + jω ?μ(?r prime × ? H) bracerightbig e ?jkr prime 4πr prime r prime 2 sinθ prime dθ prime dφ prime ≈ lim r prime →∞ integraldisplay 2π 0 integraldisplay π 0 braceleftbig r prime bracketleftbig jk ? E + jω ?μ(?r prime × ? H) bracketrightbig + ? E bracerightbig e ?jkr prime 4π sinθ prime dθ prime dφ prime . Since this gives the contribution to the ?eld in V from the ?elds on the surface receding to in?nity, we expect that this term should be zero. If the medium has loss, then the exponential term decays and drives the contribution to zero. For a lossless medium the contributions are zero if lim r→∞ r ? E(r,ω)<∞, (6.9) lim r→∞ r bracketleftbig η?r × ? H(r,ω)+ ? E(r,ω) bracketrightbig = 0. (6.10) To accompany (6.8) we also have lim r→∞ r ? H(r,ω)<∞, (6.11) lim r→∞ r bracketleftbig η ? H(r,ω)? ?r × ? E(r,ω) bracketrightbig = 0. (6.12) We refer to (6.9) and (6.11) as the ?niteness conditions, and to (6.10) and (6.12) as the Sommerfeld radiation condition, for the electromagnetic ?eld. They show that far from the sources the ?elds must behave as a wave TEM to the r-direction. We shall see in § 6.2 that the waves are in fact spherical TEM waves. 6.1.3 Fields in the excluded region: the extinction theorem TheStratton–Chuformulaprovidesasolutionforthe?eldwithintheregion V,external to the excluded regions. An interesting consequence of this formula, and one that helps us identify the equivalence principle, is that it gives the null result ? H = ? E = 0 when evaluated at points within the excluded regions. We can show this by considering two cases. In the ?rst case we do not exclude the particular region V m , but do exclude the remaining regions V n , n negationslash= m. Then the electric ?eld everywhere outside the remaining excluded regions (including at points within V m ) is, by (6.7), ? E(r,ω)= integraldisplay V+V m parenleftbigg ? ? J i m ×? prime G + ?ρ i ?epsilon1 c ? prime G ? jω ?μ ? J i G parenrightbigg dV prime + + summationdisplay nnegationslash=m integraldisplay S n bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime ? ? summationdisplay nnegationslash=m 1 jω?epsilon1 c contintegraldisplay Gamma1 na +Gamma1 nb (dl prime · ? H)? prime G, r ∈ V + V m . In the second case we apply the Stratton–Chu formula only to V m , and exclude all other regions. We incur a sign change on the surface and line integrals compared to the ?rst case because the normal is now directed oppositely. By (6.7) we have ? E(r,ω)= integraldisplay V m parenleftbigg ? ? J i m ×? prime G + ?ρ i ?epsilon1 c ? prime G ? jω ?μ ? J i G parenrightbigg dV prime ? ? integraldisplay S m bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime + + 1 jω?epsilon1 c contintegraldisplay Gamma1 na +Gamma1 nb (dl prime · ? H)? prime G, r ∈ V m . Each of the expressions for ? E is equally valid for points within V m . Upon subtraction we get 0 = integraldisplay V parenleftbigg ? ? J i m ×? prime G + ?ρ i ?epsilon1 c ? prime G ? jω ?μ ? J i G parenrightbigg dV prime + + N summationdisplay n=1 integraldisplay S n bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime ? ? N summationdisplay n=1 1 jω?epsilon1 c contintegraldisplay Gamma1 na +Gamma1 nb (dl prime · ? H)? prime G, r ∈ V m . This expression is exactly the Stratton–Chu formula (6.7) evaluated at points within the excluded region V m . The treatment of ? H is analogous and is left as an exercise. Since we may repeat this for any excluded region, we ?nd that the Stratton–Chu formula returns the null ?eld when evaluated at points outside V. This is sometimes referred to as the vector Ewald–Oseen extinction theorem [90]. We must emphasize that the ?elds within the excluded regions are not generally equal to zero; the Stratton–Chu formula merely returns this result when evaluated there. 6.2 Fields in an unbounded medium Two special cases of the Stratton–Chu formula are important because of their applica- tion to antenna theory. The ?rst is that of sources radiating into an unbounded region. The second involves a bounded region with all sources excluded. We shall consider the former here and the latter in § 6.3. Assuming that there are no bounding surfaces in (6.7) and (6.8), except for one surface that has been allowed to recede to in?nity and therefore provides no surface contribution, we ?nd that the electromagnetic ?elds in unbounded space are given by ? E = integraldisplay V parenleftbigg ? ? J i m ×? prime G + ?ρ i ?epsilon1 c ? prime G ? jω ?μ ? J i G parenrightbigg dV prime , ? H = integraldisplay V parenleftbigg ? J i ×? prime G + ?ρ i m ?μ ? prime G ? jω?epsilon1 c ? J i m G parenrightbigg dV prime . We can view the right-hand sides as superpositions of the ?elds present in the cases where (1) electric sources are present exclusively, and (2) magnetic sources are present exclusively. With ?ρ i m = 0 and ? J i m = 0 we ?nd that ? E = integraldisplay V parenleftbigg ?ρ i ?epsilon1 c ? prime G ? jω ?μ ? J i G parenrightbigg dV prime , (6.13) ? H = integraldisplay V ? J i ×? prime GdV prime . (6.14) Using ? prime G =??G we can write ? E(r,ω)=?? integraldisplay V ?ρ i (r prime ,ω) ?epsilon1 c (ω) G(r|r prime ;ω)dV prime ? jω integraldisplay V ?μ(ω) ? J i (r prime ,ω)G(r|r prime ;ω)dV prime =?? ? φ e (r,ω)? jω ? A e (r,ω), where ? φ e (r,ω)= integraldisplay V ?ρ i (r prime ,ω) ?epsilon1 c (ω) G(r|r prime ;ω)dV prime , ? A e (r,ω)= integraldisplay V ?μ(ω) ? J i (r prime ,ω)G(r|r prime ;ω)dV prime , (6.15) aretheelectricscalarandvectorpotentialfunctionsintroducedin § 5.2. Using ? J i ×? prime G = ? ? J i ×?G =?×( ? J i G) we have ? H(r,ω)= 1 ?μ(ω) ?× integraldisplay V ?μ(ω) ? J i (r prime ,ω)G(r|r prime ;ω)dV prime = 1 ?μ(ω) ?× ? A e (r,ω). (6.16) These expressions for the ?elds are identical to those of (5.56) and (5.57), and thus the integral formula for the electromagnetic ?elds produces a result identical to that obtained using potential relations. Similarly, with ?ρ i = 0, ? J i = 0 we have ? E =? integraldisplay V ? J i m ×? prime GdV prime , ? H = integraldisplay V parenleftbigg ?ρ i m ?μ ? prime G ? jω?epsilon1 c ? J i m G parenrightbigg dV prime , or ? E(r,ω)=? 1 ?epsilon1 c (ω) ?× ? A h (r,ω), ? H(r,ω)=?? ? φ h (r,ω)? jω ? A h (r,ω), where ? φ h (r,ω)= integraldisplay V ?ρ i m (r prime ,ω) ?μ(ω) G(r|r prime ;ω)dV prime , ? A h (r,ω)= integraldisplay V ?epsilon1 c (ω) ? J i m (r prime ,ω)G(r|r prime ;ω)dV prime , are the magnetic scalar and vector potentials introduced in § 5.2. 6.2.1 The far-zone ?elds produced by sources in unbounded space Many antennas may be analyzed in terms of electric currents and charges radiating in unbounded space. Since antennas are used to transmit information over great distances, the ?elds far from the sources are often of most interest. Assume that the sources are contained within a sphere of radius r s centered at the origin. We de?ne the far zone of the sources to consist of all observation points satisfying both r greatermuch r s (and thus r greatermuch r prime ) and kr greatermuch 1. For points in the far zone we may approximate the unit vector ? R directed from the sources to the observation point by the unit vector ?r directed from the origin to the observation point. We may also approximate ? prime G = d dR parenleftbigg e ?jkR 4π R parenrightbigg ? prime R = ? R parenleftbigg 1 + jkR R parenrightbigg e ?jkR 4π R ≈ ?r jk e ?jkR 4π R = ?r jkG. (6.17) Using this we can obtain expressions for ? E and ? H in the far zone of the sources. The approximation (6.17) leads directly to ?ρ i ? prime G ≈ bracketleftbigg j ? prime · ? J i ω bracketrightbigg (?r jkG) =? k ω ?r bracketleftbig ? prime ·(G ? J i )? ? J i ·? prime G bracketrightbig . Substituting this into (6.13), again using (6.17) and also using the divergence theorem, we have ? E(r,ω)≈? integraldisplay V jω ?μ bracketleftbig ? J i ? ?r(?r · ? J i ) bracketrightbig GdV prime + ?r k ω?epsilon1 c contintegraldisplay S (?n prime · ? J i )GdS prime , where the surface S surrounds the volume V that contains the impressed sources. If we let this volume slightly exceed that needed to contain the sources, then we do not change the value of the volume integral above; however, the surface integral vanishes since ?n prime · ? J i = 0 everywhere on the surface. Using ?r × (?r × ? J i ) = ?r(?r · ? J i ) ? ? J i we then obtain the far-zone expression ? E(r,ω)≈ jω?r × bracketleftbigg ?r × integraldisplay V ?μ(ω) ? J i (r prime ,ω)G(r|r prime ;ω)dV prime bracketrightbigg = jω?r × bracketleftbig ?r × ? A e (r,ω) bracketrightbig , where ? A e is the electric vector potential. The far-zone electric ?eld has no r-component, and it is often convenient to write ? E(r,ω)≈?jω ? A eT (r,ω) (6.18) where ? A eT is the vector component of ? A e transverse to the r-direction: ? A eT =??r × bracketleftbig ?r × ? A e bracketrightbig = ? A e ? ?r(?r · ? A e ) = ? θ ? A eθ + ? φ ? A eφ . We can approximate the magnetic ?eld in a similar fashion. Noting that ? J i ×? prime G = ? J i ×(jk?rG) we have ? H(r,ω)≈?j k ?μ(ω) ?r × integraldisplay V ?μ(ω) ? J i (r prime ,ω)G(r|r prime ,ω)dV prime ≈? 1 η jω?r × ? A e (r,ω). With this we have ? E(r,ω)=?η?r × ? H(r,ω), ? H(r,ω)= ?r × ? E(r,ω) η , in the far zone. To simplify the computations involved, we often choose to approximate the vector potential in the far zone. Noting that R = radicalbig (r ? r prime )·(r ? r prime ) = radicalBig r 2 + r prime 2 ? 2(r · r prime ) and remembering that r greatermuch r prime for r in the far zone, we can use the leading terms of a binomial expansion of the square root to get R = r radicalBigg 1 ? 2(?r · r prime ) r + parenleftbigg r prime r parenrightbigg 2 ≈ r radicalbigg 1 ? 2(?r · r prime ) r ≈ r bracketleftbigg 1 ? ?r · r prime r bracketrightbigg ≈ r ? ?r · r prime . (6.19) Thus the Green’s function may be approximated as G(r|r prime ;ω) ≈ e ?jkr 4πr e jk?r·r prime . (6.20) Here we have kept the approximation (6.19) intact in the phase of G but have used 1/R ≈ 1/r in the amplitude of G. We must keep a more accurate approximation for the phase since k(?r · r prime ) may be an appreciable fraction of a radian. We thus have the far-zone approximation for the vector potential ? A e (r,ω)≈ ?μ(ω) e ?jkr 4πr integraldisplay V ? J i (r prime ,ω)e jk?r·r prime dV prime , which we may use in computing (6.18). Let us summarize the expressions for computing the far-zone ?elds: ? E(r,ω)=?jω bracketleftBig ? θ ? A eθ (r,ω)+ ? φ ? A eφ (r,ω) bracketrightBig , (6.21) ? H(r,ω)= ?r × ? E(r,ω) η , (6.22) ? A e (r,ω)= e ?jkr 4πr ?μ(ω)?a e (θ,φ,ω), (6.23) ?a e (θ,φ,ω) = integraldisplay V ? J i (r prime ,ω)e jk?r·r prime dV prime . (6.24) Here ?a e is called the directional weighting function. This function is independent of r and describes the angular variation, or pattern, of the ?elds. In the far zone ? E, ? H, ?r are mutually orthogonal. Because of this, and because the ?elds vary as e ?jkr /r, the electromagnetic ?eld in the far zone takes the form of a spherical TEM wave, which is consistent with the Sommerfeld radiation condition. Power radiated by time-harmonic sources in unbounded space. In § 5.2.1 we de?ned the power radiated by a time-harmonic source in unbounded space as the total time-average power passing through a sphere of very large radius. We found that for a Hertzian dipole the radiated power could be computed from the far-zone ?elds through P av = lim r→∞ integraldisplay 2π 0 integraldisplay π 0 S av · ?rr 2 sinθ dθ dφ where S av = 1 2 Re braceleftbig ˇ E × ˇ H ? bracerightbig is the time-average Poynting vector. By superposition this holds for any localized source. Assuming a lossless medium and using phasor notation to describe the time-harmonic Figure 6.4: Dipole antenna in a lossless unbounded medium. ?elds we have, by (6.22), S av = 1 2 Re braceleftBigg ˇ E ×(?r × ˇ E ? ) η bracerightBigg = ?r ˇ E · ˇ E ? 2η . Substituting from (6.21), we can also write S av in terms of the directional weighting function as S av = ?r ˇω 2 2η parenleftbig ˇ A eθ ˇ A ? eθ + ˇ A eφ ˇ A ? eφ parenrightbig = ?r k 2 η (4πr) 2 parenleftbigg 1 2 ˇa eθ ˇa ? eθ + 1 2 ˇa eφ ˇa ? eφ parenrightbigg . (6.25) We note that S av describes the variation of the power density with θ,φ, and is thus sometimes used as a descriptor of the power pattern of the sources. Exampleofacurrentsourceradiatingintoanunboundedmedium: thedipole antenna. A common type of antenna consists of a thin wire of length 2l and radius a, fedatthecenterbyavoltagegeneratorasshowninFigure6.4.Thegeneratorinduces an impressed current on the surface of the wire which in turn radiates an electromagnetic wave. For very thin wires (a lessmuch λ, a lessmuch l) embedded in a lossless medium, the current may be accurately approximated using a standing-wave distribution: ? J i (r,ω)= ?z ? I(ω)sin [k(l ?|z|)]δ(x)δ(y). (6.26) We may compute the ?eld produced by the dipole antenna by ?rst ?nding the vector potential from (6.15) and then calculating the magnetic ?eld from (6.16). The electric ?eld may then be found by the use of Ampere’s law. Weassumealosslessmediumwithparametersμ,epsilon1. Substitutingthecurrentexpression into (6.15) and integrating over x and y we ?nd that ? A e (r,ω)= ?z μ ? I 4π integraldisplay l ?l sin k(l ?|z prime |) e ?jkR R dz prime (6.27) where R = radicalbig (z ? z prime ) 2 +ρ 2 and ρ 2 = x 2 + y 2 . Using (6.16) we have ? H =?× 1 μ ? A e =? ? φ 1 μ ? ? A ez ?ρ . Writing the sine function in (6.27) in terms of exponentials, we then have ? H φ = j ? I 8π bracketleftBigg e jkl integraldisplay 0 ?l ? ?ρ e ?jk(R?z prime ) R dz prime ? e ?jkl integraldisplay 0 ?l ? ?ρ e ?jk(R+z prime ) R dz prime + + e jkl integraldisplay l 0 ? ?ρ e ?jk(R+z prime ) R dz prime ? e ?jkl integraldisplay l 0 ? ?ρ e ?jk(R?z prime ) R dz prime bracketrightBigg . Noting that ? ?ρ e ?jk(R±z prime ) R =±ρ ? ?z prime e ?jk(R±z prime ) R [R ?(z ? z prime )] =?ρ 1 + jkR R 3 e ?jk(R±z prime ) we can write ? H φ = j ? Iρ 8π bracketleftBigg ?e jkl e ?jk(R?z prime ) R [R +(z ? z prime )] vextendsingle vextendsingle vextendsingle vextendsingle 0 ?l ? e ?jkl e ?jk(R+z prime ) R [R ?(z ? z prime )] vextendsingle vextendsingle vextendsingle vextendsingle 0 ?l + + e jkl e ?jk(R+z prime ) R [R ?(z ? z prime )] vextendsingle vextendsingle vextendsingle vextendsingle l 0 + e ?jkl e ?jk(R?z prime ) R [R +(z ? z prime )] vextendsingle vextendsingle vextendsingle vextendsingle l 0 bracketrightBigg . Collecting terms and simplifying we get ? H φ (r,ω)= j ? I(ω) 4πρ bracketleftbig e ?jkR 1 + e ?jkR 2 ?(2 cos kl)e ?jkr bracketrightbig (6.28) where R 1 = radicalbig ρ 2 +(z ? l) 2 and R 2 = radicalbig ρ 2 +(z + l) 2 . For points external to the dipole the source current is zero and thus ? E(r,ω)= 1 jωepsilon1 ?× ? H(r,ω)= 1 jωepsilon1 braceleftbigg ??ρ ? ?z ? H φ (r,ω)+ ?z 1 ρ ? ?ρ [ρ ? H φ (r,ω)] bracerightbigg . Performing the derivatives we have ? E ρ (r,ω)= j η ? I(ω) 4π bracketleftbigg z ? l ρ e ?jkR 1 R 1 + z + l ρ e ?jkR 2 R 2 ? z ρ (2 cos kl) e ?jkr r bracketrightbigg , (6.29) ? E z (r,ω)=?j η ? I(ω) 4π bracketleftbigg e ?jkR 1 R 1 + e ?jkR 2 R 2 ?(2 cos kl) e ?jkr r bracketrightbigg . (6.30) Theworkofspecializingtheseexpressionsforpointsinthefarzoneisleftasanexercise. Instead, we shall use the general far-zone expressions (6.21)–(6.24). Substituting (6.26) into (6.24) and carrying out the x and y integrals we have the directional weighting function ?a e (θ,φ,ω) = integraldisplay l ?l ?z ? I(ω)sin k(l ?|z prime |)e jkz prime cosθ dz prime . Writing the sine functions in terms of exponentials we have ?a e (θ,φ,ω) = ?z ? I(ω) 2 j bracketleftbigg e jkl integraldisplay l 0 e jkz prime (cosθ?1) dz prime ? e ?jkl integraldisplay l 0 e jkz prime (cosθ+1) dz prime + + e jkl integraldisplay 0 ?l e jkz prime (cosθ+1) ? e ?jkl integraldisplay 0 ?l e jkz prime (cosθ?1) bracketrightbigg . Carrying out the integrals and simplifying, we obtain ?a e (θ,φ,ω)= ?z 2 ? I(ω) k F(θ,kl) sinθ where F(θ,kl) = cos(kl cosθ)? cos kl sinθ is called the radiation function. Using ?z = ?r cosθ ? ? θ sinθ we ?nd that ?a eθ (θ,φ,ω)=? 2 ? I(ω) k F(θ,kl), ?a eφ (θ,φ,ω) = 0. Thus we have from (6.23) and (6.21) the electric ?eld ? E(r,ω)= ? θ jη ? I(ω) 2π e ?jkr r F(θ,kl) (6.31) and from (6.22) the magnetic ?eld ? H(r,ω)= ? φ j ? I(ω) 2π e ?jkr r F(θ,kl). (6.32) We see that the radiation function contains all of the angular dependence of the ?eld and thus describes the pattern of the dipole. When the dipole is short compared to a wavelength we may approximate the radiation function as F(θ,kl lessmuch 1) ≈ 1 ? 1 2 (kl cosθ) 2 ? 1 + 1 2 (kl) 2 sinθ = 1 2 (kl) 2 sinθ. (6.33) So a short dipole antenna has the same pattern as a Hertzian dipole, whose far-zone electric ?eld is (5.93). We may also calculate the radiated power for time-harmonic ?elds. The time-average Poynting vector for the far-zone ?elds is, from (6.25), S av = ?rη | ˇ I| 2 8π 2 r 2 F 2 (θ,kl), and thus the radiated power is P av = η | ˇ I| 2 4π integraldisplay π 0 F 2 (θ,kl)sinθ dθ. This expression cannot be computed in closed form. For a short dipole we may use (6.33) to approximate the power, but the result is somewhat misleading since the current on a short dipole is much smaller than ? I. A better measure of the strength of the current is its value at the center, or feedpoint, of the dipole. This input current is by (6.26) merely ? I 0 (ω) = ? I(ω)sin(kl). Using this we ?nd P av ≈ η | ˇ I 0 | 2 4π 1 4 (kl) 2 integraldisplay π 0 sin 3 θ dθ = η π 3 | ˇ I 0 | 2 parenleftbigg l λ parenrightbigg 2 . This is exactly 1/4 of the power radiated by a Hertzian dipole of the same length and currentamplitude(5.95). Thefactorof 1/4 comesfromthedi?erencebetweenthecurrent of the dipole antenna, which is zero at each end, and the current on the Hertzian dipole, which is constant across the length of the antenna. It is more common to use a dipole antenna that is a half wavelength long (2l = λ/2), since it is then nearly resonant. With this we have through numerical integration the free-space radiated power P av = η 0 | ˇ I 0 | 2 4π integraldisplay π 0 cos 2 parenleftbig π 2 cosθ parenrightbig sinθ dθ = 36.6| ˇ I 0 | 2 and the radiation resistance R r = 2P av | ˇ I(z = 0)| 2 = 2P av | ˇ I 0 | 2 = 73.2 Omega1. 6.3 Fields in a bounded, source-free region In § 6.2 we considered the ?rst important special case of the Stratton–Chu formula: sources in an unbounded medium. We now consider the second important special case of a bounded, source-free region. This case has important applications to the study of microwave antennas and, in its scalar form, to the study of the di?raction of light. 6.3.1 The vector Huygens principle We may derive the formula for a bounded, source-free region of space by specializing the general Stratton–Chu formulas. We assume that all sources of the ?elds are within the excluded regions and thus set the sources to zero within V. From (6.7)–(6.8) we have ? E(r,ω)= N summationdisplay n=1 integraldisplay S n bracketleftbig (?n prime × ? E)×? prime G +(?n prime · ? E)? prime G ? jω ?μ(?n prime × ? H)G bracketrightbig dS prime ? ? N summationdisplay n=1 1 jω?epsilon1 c contintegraldisplay Gamma1 na +Gamma1 nb (dl prime · ? H)? prime G, (6.34) and ? H(r,ω)= N summationdisplay n=1 integraldisplay S n bracketleftbig (?n prime × ? H)×? prime G +(?n prime · ? H)? prime G + jω?epsilon1 c (?n prime × ? E)G bracketrightbig dS prime + + N summationdisplay n=1 1 jω ?μ contintegraldisplay Gamma1 na +Gamma1 nb (dl prime · ? E)? prime G. (6.35) This is known as the vector Huygens principle after the Dutch physicist C. Huygens, who formulated his “secondary source concept” to explain the propagation of light. According to his idea, published in Trait′e de la lumi`ere in 1690, points on a propagating wavefront aresecondarysourcesofsphericalwavesthataddtogetherinjusttherightwaytoproduce the ?eld on any successive wavefront. We can interpret (6.34) and (6.35) in much the same way. The ?eld at each point within V, where there are no sources, can be imagined to arise from spherical waves emanated from every point on the surface bounding V. The amplitudes of these waves are determined by the values of the ?elds on the boundaries. Thus, we may consider the boundary ?elds to be equivalent to secondary sources of the ?elds within V. We will expand on this concept below by introducing the concept of equivalence and identifying the speci?c form of the secondary sources. 6.3.2 The Franz formula The vector Huygens principle as derived above requires secondary sources for the ?elds within V that involve both the tangential and normal components of the ?elds on the boundingsurface. Sinceonlytangentialcomponentsarerequiredtoguaranteeuniqueness within V, we seek an expression involving only ?n × ? H and ?n × ? E. Physically, the normal component of the ?eld is equivalent to a secondary charge source on the surface while the tangential component is equivalent to a secondary current source. Since charge and current are related by the continuity equation, speci?cation of the normal component is super?uous. To derive a version of the vector Huygens principle that omits the normal ?elds we take the curl of (6.35) to get ?× ? H(r,ω)= N summationdisplay n=1 ?× contintegraldisplay S n (?n prime × ? H)×? prime GdS prime + N summationdisplay n=1 contintegraldisplay S n ?× bracketleftbig (?n prime · ? H)? prime G bracketrightbig dS prime + + N summationdisplay n=1 ?× contintegraldisplay S n jω?epsilon1 c (?n prime × ? E)GdS prime + N summationdisplay n=1 1 jω ?μ contintegraldisplay Gamma1 na +Gamma1 nb ?× bracketleftbig (dl prime · ? E)? prime G bracketrightbig dS prime . (6.36) Now, using ? prime G =??G and employing the vector identity (B.43) we can show that ?× bracketleftbig f (r prime )? prime G(r|r prime ) bracketrightbig =?f (r prime ) braceleftbig ?× bracketleftbig ?G(r|r prime ) bracketrightbigbracerightbig + bracketleftbig ?G(r|r prime ) bracketrightbig ×?f (r prime ) = 0, since ?×?G = 0 and ? f (r prime ) = 0. This implies that the second and fourth terms of (6.36) are zero. The ?rst term can be modi?ed using ?× braceleftbigbracketleftbig ?n prime × ? H(r prime ) bracketrightbig G(r|r prime ) bracerightbig = G(r|r prime )?× bracketleftbig ?n prime × ? H(r prime ) bracketrightbig ? bracketleftbig ?n prime × ? H(r prime ) bracketrightbig ×?G(r|r prime ) = bracketleftbig ?n prime × ? H(r prime ) bracketrightbig ×? prime G(r|r prime ), giving ?× ? H(r,ω)= N summationdisplay n=1 ?× contintegraldisplay S n ?× bracketleftbig (?n prime × ? H)G bracketrightbig dS prime + N summationdisplay n=1 ?× contintegraldisplay S n jω?epsilon1 c (?n prime × ? E)GdS prime . Finally, using Ampere’s law ?× ? H = jω?epsilon1 c ? E in the source free region V, and taking the curl in the ?rst term outside the integral, we have ? E(r,ω)= N summationdisplay n=1 ?×?× contintegraldisplay S n 1 jω?epsilon1 c (?n prime × ? H)GdS prime + N summationdisplay n=1 ?× contintegraldisplay S n (?n prime × ? E)GdS prime . (6.37) Similarly ? H(r,ω)=? N summationdisplay n=1 ?×?× contintegraldisplay S n 1 jω ?μ (?n prime × ? E)GdS prime + N summationdisplay n=1 ?× contintegraldisplay S n (?n prime × ? H)GdS prime . (6.38) These expressions together constitute the Franz formula for the vector Huygens principle [192]. 6.3.3 Love’s equivalence principle Love’sequivalenceprincipleallowsustoidentifytheequivalentHuygenssourcesforthe ?elds within a bounded, source-free region V. It then allows us to replace a problem in the bounded region with an “equivalent” problem in unbounded space where the source- excluding surfaces are replaced by equivalent sources. The ?eld produced by both the real and the equivalent sources gives a ?eld in V identical to that of the original problem. This is particularly useful since we know how to compute the ?elds within an unbounded region by employing potential functions. We identify the equivalent sources by considering the electric and magnetic Hertzian potentials produced by electric and magnetic current sources. Consider an impressed electric surface current ? J eq s and a magnetic surface current ? J eq ms ?owing on the closed surface S in a homogeneous, isotropic medium with permeability ?μ(ω) and complex permittivity ?epsilon1 c (ω). These sources produce ? Π e (r,ω)= contintegraldisplay S ? J eq s (r prime ,ω) jω?epsilon1 c (ω) G(r|r prime ;ω)dS prime , (6.39) ? Π h (r,ω)= contintegraldisplay S ? J eq ms (r prime ,ω) jω ?μ(ω) G(r|r prime ;ω)dS prime , (6.40) which in turn can be used to ?nd ? E =?×(?× ? Π e )? jω ?μ?× ? Π h , ? H = jω?epsilon1 c ?× ? Π e +?×(?× ? Π h ). Upon substitution we ?nd that ? E(r,ω)=?×?× contintegraldisplay S 1 jω?epsilon1 c bracketleftbig ? J eq s G bracketrightbig dS prime +?× contintegraldisplay S [? ? J eq ms ]GdS prime , ? H(r,ω)=??×?× contintegraldisplay S 1 jω ?μ bracketleftbig ? ? J eq ms G bracketrightbig dS prime +?× contintegraldisplay S ? J eq s GdS prime . These are identical to the Franz equations (6.37) and (6.38) if we identify ? J eq s = ?n × ? H, ? J eq ms =??n × ? E. (6.41) These are the equivalent source densities for the Huygens principle. We now state Love’s equivalence principle [39]. Consider the ?elds within a homoge- neous, source-free region V with parameters (?epsilon1 c , ?μ) bounded by a surface S. We know how to compute the ?elds using the Franz formula and the surface ?elds. Now consider a second problem in which the same surface S exists in an unbounded medium with identical parameters. If the surface carries the equivalent sources (6.41) then the electro- magnetic ?elds within V calculated using the Hertzian potentials (6.39) and (6.40) are identical to those of the ?rst problem, while the ?elds calculated outside V are zero. We see that this must be true since the Franz formulas and the ?eld/potential formulas are identical, and the Franz formula (since it was derived from the Stratton–Chu formula) gives the null ?eld outside V. The two problems are equivalent in the sense that they produce identical ?elds within V. The ?elds produced by the equivalent sources obey the appropriate boundary condi- tions across S. From (2.194) and (2.195) we have the boundary conditions ?n ×( ? H 1 ? ? H 2 ) = ? J s , ?n ×( ? E 1 ? ? E 2 ) =? ? J ms . Here ?n points inward to V, ( ? E 1 , ? H 1 ) are the ?elds within V, and ( ? E 2 , ? H 2 ) are the ?elds within the excluded region. If the ?elds produced by the equivalent sources within the excluded region are zero, then the ?elds must obey ?n × ? H 1 = ? J eq s , ?n × ? E 1 =? ? J eq ms , which is true by the de?nition of ( ? J eq s , ? J eq sm ). Note that we can extend the equivalence principle to the case where the media are di?erent internal to V than external to V. See Chen [29]. With the equivalent sources identi?ed we may compute the electromagnetic ?eld in V using standard techniques. Speci?cally, we may use the Hertzian potentials as shown above or, since the Hertzian potentials are a simple remapping of the vector potentials, we may use (5.60) and (5.61) to write ? E =?j ω k 2 bracketleftbig ?(?· ? A e )+ k 2 ? A e bracketrightbig ? 1 ?epsilon1 c ?× ? A h , ? H =?j ω k 2 bracketleftbig ?(?· ? A h )+ k 2 ? A h bracketrightbig + 1 ?μ ?× ? A e , where ? A e (r,ω)= contintegraldisplay S ?μ(ω) ? J eq s (r prime ,ω)G(r|r prime ;ω)dS prime (6.42) = contintegraldisplay S ?μ(ω)[?n prime × ? H(r prime ,ω)]G(r|r prime ;ω)dS prime , (6.43) ? A h (r,ω)= contintegraldisplay S ?epsilon1 c (ω) ? J eq ms (r prime ,ω)G(r|r prime ;ω)dS prime (6.44) = contintegraldisplay S ?epsilon1 c (ω)[??n prime × ? E(r prime ,ω)]G(r|r prime ;ω)dS prime . (6.45) At points where the source is zero we can write the ?elds in the alternative form ? E =?j ω k 2 ?×?× ? A e ? 1 ?epsilon1 c ?× ? A h , (6.46) ? H =?j ω k 2 ?×?× ? A h + 1 ?μ ?× ? A. (6.47) Figure 6.5: Geometry for problem of an aperture in a perfectly conducting ground screen illuminated by an impressed source. By superposition, if there are volume sources within V we merely add the ?elds due to these sources as computed from the potential functions. 6.3.4 The Schelkuno? equivalence principle With Love’s equivalence principle we create an equivalent problem by replacing an excluded region by equivalent electric and magnetic sources. These require knowledge of both the tangential electric and magnetic ?elds over the bounding surface. However, the uniqueness theorem says that only one of either the tangential electric or the tangential magnetic?eldsneedbespeci?edtomakethe?eldswithin V unique. Thuswemaywonder whether it is possible to formulate an equivalent problem that involves only tangential ? E or tangential ? H. It is indeed possible, as shown by Schelkuno? [39, 169]. When we use the equivalent sources to form the equivalent problem, we know that they produce a null ?eld within the excluded region. Thus we may form a di?erent equivalent problem by ?lling the excluded region with a perfect conductor, and keeping the same equivalent sources. The boundary conditions across S are not changed, and thus by the uniqueness theorem the ?elds within V are not altered. However, the manner in which we must compute the ?elds within V is changed. We can no longer use formulas for the ?elds produced by sources in free space, but must use formulas for ?elds produced by sources in the vicinity of a conducting body. In general this can be di?cult since it requires the formation of a new Green’s function that satis?es the boundary condition over the conducting body (which could possess a peculiar shape). Fortunately, we showed in § 4.10.2 that an electric source adjacent and tangential to a perfect electric conductor produces no ?eld, hence we need not consider the equivalent electric sources (?n × ? H) when computing the ?elds in V. Thus, in our new equivalent problem we need the single tangential ?eld ??n × ? E. This is the Schelkuno? equivalence principle. There is one situation in which it is relatively easy to use the Schelkuno? equivalence. Consider a perfectly conducting ground screen with an aperture in it, as shown in Figure 6.5.Weassumethattheaperturehasbeenilluminatedinsomewaybyanelectromagnetic wave produced by sources in region 1 so that there are both ?elds within the aperture and electric current ?owing on the region-2 side of the screen due to di?raction from the edges of the aperture. We wish to compute the ?elds in region 2. We can create an equivalent problem by placing a planar surface S 0 adjacent to the screen, but slightly o?set into region 2, and then closing the surface at in?nity so that all of the screen plus region 1 is excluded. Then we replace region 1 with homogeneous space and place on S 0 the equivalent currents ? J eq s = ?n× ? H, ? J eq ms =??n× ? E, where ? H and ? E are the ?elds on S 0 in the original problem. We note that over the portion of S 0 adjacent to the screen ? J eq ms = 0 since ?n × ? E = 0, but that ? J eq s negationslash= 0. From the equivalent currents we can compute the ?elds in region 2 using the potential functions. However, it is often di?cult to determine ? J eq s over the conducting surface. If we apply Schelkuno?’s equivalence, we can formulate a second equivalent problem in which we place into region 1 a perfect conductor. Then we have the equivalent source currents ? J eq s and ? J eq ms adjacent and tangential to a perfect conductor. By the image theorem of § 5.1.1 we can replace this problem by yet another equivalent problem in which the conductor is replaced by the images of ? J eq s and ? J eq ms in homogeneous space. Since the image of the tangential electric current ? J eq s is oppositely directed, the ?elds of the electric current and its image cancel. Since the image of the magnetic current is in the same direction as ? J eq ms , the ?elds produced by the magnetic current and its image add. We also note that ? J eq ms is nonzero only over the aperture (since ?n × ? E = 0 on the screen), and thus the ?eld in region 1 can be found from ? E(r,ω)=? 1 ?epsilon1 c (ω) ?× ? A h (r,ω), where ? A h (r,ω)= integraldisplay S 0 ?epsilon1 c (ω)[?2?n prime × ? E ap (r prime ,ω)]G(r|r prime ;ω)dS prime and ? E ap is the electric ?eld in the aperture in the original problem. We shall present an example in the next section. 6.3.5 Far-zone ?elds produced by equivalent sources The equivalence principle is useful for analyzing antennas with complicated source distributions. The sources may be excluded using a surface S, and then a knowledge of the ?elds over S (found, for example, by estimation or measurement) can be used to compute the ?elds external to the antenna. Here we describe how to compute these ?elds in the far zone. Given that ? J eq s = ?n × ? H and ? J eq ms =??n × ? E are the equivalent sources on S,wemay compute the ?elds using the potentials (6.43) and (6.45). Using (6.20) these can be approximated in the far zone (r greatermuch r prime , kr greatermuch 1) as ? A e (r,ω)= ?μ(ω) e ?jkr 4πr ?a e (θ,φ,ω), ? A h (r,ω)= ?epsilon1 c (ω) e ?jkr 4πr ?a h (θ,φ,ω), (6.48) where ?a e (θ,φ,ω) = contintegraldisplay S ? J eq s (r prime ,ω)e jk?r·r prime dS prime , ?a h (θ,φ,ω) = contintegraldisplay S ? J eq sm (r prime ,ω)e jk?r·r prime dS prime , (6.49) are the directional weighting functions. To compute the ?elds from the potentials we must apply the curl operator. So we must evaluate ?× bracketleftbigg e ?jkr r V(θ,φ) bracketrightbigg = e ?jkr r ?×V(θ,φ)+? parenleftbigg e ?jkr r parenrightbigg × V(θ,φ). The curl of V is proportional to 1/r in spherical coordinates, hence the ?rst term on the right is proportional to 1/r 2 . Since we are interested in the far-zone ?elds, this term can be discarded in favor of 1/r-type terms. Using ? parenleftbigg e ?jkr r parenrightbigg =??r parenleftbigg 1 + jkr r parenrightbigg e ?jkr r ≈??r jk e ?jkr r , kr greatermuch 1, we have ?× bracketleftbigg e ?jkr r V(θ,φ) bracketrightbigg ≈?jk?r × bracketleftbigg e ?jkr r V(θ,φ) bracketrightbigg . Using this approximation we also establish ?×?× bracketleftbigg e ?jkr r V(θ,φ) bracketrightbigg ≈?k 2 ?r × ?r × bracketleftbigg e ?jkr r V(θ,φ) bracketrightbigg = k 2 e ?jkr r V T (θ,φ) where V T = V ? ?r(?r · V) is the vector component of V transverse to the r-direction. With these formulas we can approximate (6.46) and (6.47) as ? E(r,ω)=?jω ? A eT (r,ω)+ jk ?epsilon1 c (ω) ?r × ? A h (r,ω), (6.50) ? H(r,ω)=?jω ? A hT (r,ω)? jk ?μ(ω) ?r × ? A e (r,ω). Note that ?r × ? E =?jω?r × ? A eT + jk ?epsilon1 c ?r × ?r × ? A h . Since ?r × ? A eT = ?r × ? A e and ?r × ?r × ? A h =? ? A hT ,wehave ?r × ? E = η bracketleftbigg ?jω ? A hT ? jk ?μ ?r × ? A e bracketrightbigg = η ? H. Thus ? H = ?r × ? E η and the electromagnetic ?eld in the far zone is a TEM spherical wave, as expected. Example of ?elds produced by equivalent sources: an aperture antenna. As an example of calculating the ?elds in a bounded region from equivalent sources, let us ?nd the far-zone ?eld in free space produced by a rectangular waveguide opening intoaperfectly-conductinggroundscreenofin?niteextentasshowninFigure6.6.For simplicity assume the waveguide propagates a pure TE 10 mode, and that all higher-order Figure 6.6: Aperture antenna consisting of a rectangular waveguide opening into a con- ducting ground screen of in?nite extent. modes excited when the guided wave is re?ected at the aperture may be ignored. Thus the electric ?eld in the aperture S 0 is ? E a (x, y) = ?yE 0 cos parenleftBig π a x parenrightBig . We may compute the far-zone ?eld using the Schelkuno? equivalence principle of § 6.3.4. We exclude the region z < 0 + using a planar surface S which we close at in?nity. We then ?ll the region z < 0 with a perfect conductor. By the image theory the equivalent electric sources on S cancel while the equivalent magnetic sources double. Since the only nonzero magnetic sources are on S 0 (since ?n × ? E = 0 on the screen), we have the equivalent problem of the source ? J eq ms =?2?n × ? E a = 2?xE 0 cos parenleftBig π a x parenrightBig on S 0 in free space, where the equivalence holds for z > 0. We may ?nd the far-zone ?eld created by this equivalent current by ?rst computing the directional weighting function (6.49). Since ?r · r prime = ?r ·(x prime ?x + y prime ?y) = x prime sinθ cosφ + y prime sinθ sinφ, we ?nd that ?a h (θ,φ,ω) = integraldisplay b/2 ?b/2 integraldisplay a/2 ?a/2 ?x2E 0 cos parenleftBig π a x prime parenrightBig e jkx prime sinθ cosφ e jky prime sinθ sinφ dx prime dy prime = ?x4π E 0 ab cosπ X π 2 ? 4(π X) 2 sinπY πY where X = a λ sinθ cosφ, Y = b λ sinθ sinφ. Here λ is the free-space wavelength. By (6.50) the electric ?eld is ? E = jk 0 epsilon1 0 ?r × ? A h where ? A h is given in (6.48). Using ?r × ?x = ? φcosθ cosφ + ? θ sinφ we ?nd that ? E = jk 0 abE 0 e ?jkr r parenleftBig ? θ sinφ + ? φcosθ cosφ parenrightBig cos(π X) π 2 ? 4(π X) 2 sin(πY) πY . The magnetic ?eld is merely ? H = (?r × ? E)/η. 6.4 Problems 6.1 Beginning with the Lorentz reciprocity theorem, derive (6.8). 6.2 Obtain (6.8) by substitution of (6.7) into Faraday’s law. 6.3 Show that (6.8) returns the null result when evaluated within the excluded regions. 6.4 Show that under the condition kr greatermuch 1 the formula for the magnetic ?eld of a dipole antenna (6.28) reduces to (6.32), while the formulas for the electric ?elds (6.29) and (6.30) reduce to (6.31). 6.5ConsiderthedipoleantennashowninFigure6.4.Insteadofastanding-wave current distribution, assume the antenna carries a traveling-wave current distribution ? J i (r,ω)= ?z ? I(ω)e ?jk|z| δ(x)δ(y), ?l ≤ z ≤ l. Find the electric and magnetic ?elds at all points away from the current distribution. Specialize the result for kr greatermuch 1. 6.6 A circular loop of thin wire has radius a and lies in the z = 0 plane in free space. A current is induced on the wire with the density ? J(r,ω)= ? φ ? I(ω)cos [k 0 a(π ?|φ|)]δ(r ? a) δ(θ ?π/2) r , |φ|≤π. Compute the far-zone ?elds produced by this loop antenna. Specialize your results for the electrically-small case of k 0 a lessmuch 1. Compute the time-average power radiated by, and the radiation resistance of, the electrically-small loop. 6.7 Consider a plane wave with the ?elds ? E = ? E 0 ?xe ?jkz , ? H = ? E 0 η ?ye ?jkz , normally incident from z < 0 on a square aperture of side a in a PEC ground screen at z = 0. Assume that the ?eld in the aperture is identical to the ?eld of the plane wave with the screen absent (this is called the Kirchho? approximation). Compute the far-zone electromagnetic ?elds for z > 0. 6.8 Consider a coaxial cable of inner radius a and outer radius b, opening into a PEC ground plane at z = 0. Assume that only the TEM wave exists in the line and that no higher-order modes are created when the wave re?ects from the aperture. Compute the far-zone electric and magnetic ?elds of this aperture antenna.