—This work derives exact expressions for the voltage and current inducedinto a two conductors non isolated trans mis- sion lines by an incident plane wave. The methodology is to us e the transmission line radiating properties to derive scatt ering matrices and make use of reciprocity to derive the response to the incident wave. The analysis is in the frequency domain and it considers transmission lines of any small electric cr oss section, incident by a plane wave from any incident directio n and any polarisation. The analytic results are validated by suc cessful comparison with ANSYS commercial software simulation resu lts, and compatible with other published results. Index Terms —electromagnetic theory, guided waves, electro- magnetic interference I. INTRODUCTION This work calculates the voltage and differential current developed on an ideal two-conductors TEM transmission line (TL) of any small electric cross section, connected to passi ve (lumped) loads and hit by a monochromatic plane wave, as shown in Figure 1. We derive both amplitude and phase for Fig. 1. Conﬁguration of a two ideal conductors transmission line (TL), connected at both sides to passive loads: ZL(left) and ZR(right), hit by a monochromatic plane wave propagating toward the centre of coordinates. The cross section is electrically small and may be of any shap e. The loads are located at the terminations of the TL, and are shown farther a way, because of technical drawing limitations. the voltage and current developedalong the TL, and hence the powers delivered to the passive loads. The conﬁguration as deﬁned here is a scattering problem, requiring a full wave solution to set the tangential compone nt of the electric ﬁeld to 0 on the surface of the TL. However,we shall formulate here an analytic solution to this problem , which is compared with a full wave HFSS solution. As shown in [1], the characteristic impedance Z0of the TL and an equivalent separation distance dcan be found by an electrostatic cross section analysis, leading to a twin l ead equivalent (examples for determining Z0anddare shown in Appendix B of [1]). The twin lead equivalent is shown in Figure 2. When the TL is excited at its port(s), this equivale nt Fig. 2. Twin lead equivalent of the analysed transmission li ne, deﬁned by the separation distance dbetween the conductors, and the characteristic impedance Z0, both computed by the cross section analysis described in Appendix B of [1]. The TL is of length 2L. twin lead radiates the same far ﬁelds as the actual TL. To emphasise, we do notsolve for the twin lead geometry, but for an arbitrary cross section in Figure 1, the twin lead is on ly used as an intermediate tool for the calculations. The incident plane wave shown in Figure 1 propagates toward the coordinates origin with phase e−jk·r, so that the wavenumber vector k=−/hatwiderkpoints toward the origin. Expanding /hatwiderin Cartesianunitvectors,thephasecanbewritten as ejk[xsinθcosϕ+ysinθsinϕ+zcosθ], (1) where θandϕare the spherical angles which represent the direction of the plane wave arrival. The polarisation of the incident plane wave is shown in Figure 3. It travels toward the centre of coordinates, perpe n- dicular to the θ, ϕplane, with a polarisation angle αfrom the θaxis, so that the E ﬁeld at the origin, where its phase is 0 (see Eq. (1)) is given by E=E0(/hatwideθcosα+/hatwideϕsinα) (2) or its components Eθ=E0cosα;Eϕ=E0sinα. (3) 2 Fig. 3. The incident plane wave propagates toward the centre of coordinates in the −brdirection. In spherical coordinates the local equiphase su rface is θ, ϕand the polarisation is at angle αfrom the θaxis, so that at the origin E=E0(bθcosα+bϕsinα)(see Eq. (2)). Such problem has been handled in [2]–[6], but not for an arbitrary cross section (as in Figure 1), and not for a genera l incident plane wave (as in Figure 3). The methodology used in these works is based on a generalisation of the telegraph equations for transmission lines, for the twin lead geometr y indicated in Figure 2. Given C′andL′the capacitance and inductance per unit length, using Z0=/radicalbig L′/C′and the identity√ L′C′= 1/c, we write the (inhomogeneous) telegraph equations in the following compact form dV dz+jk(Z0I) =Vs (4) Z0dI dz+jkV=Z0Is (5) where the source terms VsandIsare induced by HyandEx ﬁelds of the incident plane wave, integrated along the xaxis of the twin lead (see Figure 2), as follows: Vs=jkη0/integraldisplayd 0Hy(x, z)dx≃jkη0Hy(z)d(6) Z0Is=−jk/integraldisplayd 0Ex(x, z)dx≃ −jkE x(z)d,(7) andη0= 377Ωis the free space impedance.The last ( ≃) parts of Eqs. (6) and (7) are consistent with our framework which is restricted to small electric cross section ( kd≪1). The coupled non-homogeneous differential equations (4) and (5) can be decoupled into two second order equations: d2V dz2+k2V=dVs dz−jk(Z0Is) (8) Z0d2I dz2+k2Z0I=Z0dIs dz−jkVs (9)In [2] the authors wrote the general solution for those secon d order differential equations, subject to the termination c ondi- tions: V(−L) =−ZLI(−L) ; V(L) =ZRI(L),(10) however they used in the expression for the sources the implicit assumption that the incident ﬁeld has only a Hy component. In [3] this solution has been simpliﬁed, and in [4]–[6] this formalism has been extended to multiconductor TL. The results derived in this work satisfy Eqs. (4)-(10), but we use a different technique to derive them. We shall use our knowledge on the radiation properties of TL [1] to derive the receiving properties, i.e. the response of a TL to an incident monochromatic plane wave described in Figure 3. To determine the voltage along the TL, we segment it into M parallel ports as shown in Figure 4. The ports 1 and Mare Fig. 4. Mparallel ports along the TL, adjacent ports are at distance ∆z. Ports 1 and Mare deﬁned for the TL impedance Z0, and the middle ports 2 ..M−1are deﬁned for a high impedance ZH. Two additional ports representing far antennas matched for the bθand bϕpolarizations are deﬁned. Those ports are named θandϕ, and are deﬁned for the TL impedance Z0. deﬁned for the TL impedance Z0, and the middle ports 2 .. M−1are deﬁned for a high impedance ZH→ ∞. We deﬁne two additional ports representing far antennas matched for the /hatwideθand/hatwideϕpolarizations of the ﬁeld radiated by the TL, deﬁning a system of M+ 2ports. The additional ports, M+ 1and M+ 2, named θandϕ, may be deﬁned for any impedance, and for convenience we choose to deﬁne them for the TL impedance Z0. The scattering matrix of this system is shown schematically in Figure 5. Using the formalism developed in [1], we know the /hatwideθand/hatwideϕcomponents of the radiated electric ﬁeld (summarised in Appendix A). Those ﬁeld components are translated into voltages using a normalisation explain ed in Appendix A, enabling us to calculate the generalised S matri x (see Appendix B) describing this system. Here we excite this system with an incident voltage, scaled to the incident electric ﬁeld in Eq. (2) by V+=E0d, (11) which deﬁnes the voltage components (equivalent to Eq. (3)) V+ θ=V+cosα;V+ ϕ=V+sinα. (12) 3 Fig. 5. Schematic diagram of a (M+ 2)×(M+ 2)scattering matrix describing the systems deﬁned in Figures 4 and 6. The ﬁrst por ts 1.. M are the TL ports in the ﬁgures, and the two additional ports are th e farθorϕ polarised antennas, named θ,ϕ. In both cases Sθϕ=Sϕθ= 0, because they are orthogonally polarised, and S11=SMM=Sθθ=Sϕϕ= 0. As will be shown, the elements Siifor1< i < M are∓1for the matrices deﬁned by the conﬁgurations in Figures 4 and 6, respectively. We note that ports 1... Mare matched with the Z0impedance at ports 1 and MandZH(i.e. inﬁnite) on the middle ports, which is exactly the conﬁguration of a TL matched at both terminations, on which we want to obtain the voltages. Those are obtained by multiplying the S matrix by the column vector  0 . . 0 V+ θ V+ ϕ (13) of adequate excitation elements according to Eq. (12), and given∆zis arbitrarilysmall (or Marbitrarilylarge)we obtain thevoltageasfunctionof zalongthematchedTLforthegiven incident plane wave. The solution for matched TL is after tha t generalised for any terminations in Figure 1. Similarly,the analytic solutionfor the currentis obtaine dby segmenting the TL into Mserial ports as shown in Figure 6. Ports 1 and Mare deﬁned for the TL impedance Z0, and the middle ports 2 .. M−1are deﬁned for a tiny impedance ZT→0. Here we deﬁne too the additional ports representing far antennas matched for the /hatwideθand/hatwideϕpolarizationsof the ﬁeld radiatedbytheTL,obtaininga differentsystemof M+2ports. For this system we also calculate a generalized S matrix, but deﬁned for current waves (see Appendix B). Here the methodology is similar to the one explained for the voltages , the matched at both terminations TL result is obtained using Fig. 6. Mserial ports along the TL, adjacent ports are at distance ∆z. Ports 1 and Mare deﬁned for the TL impedance Z0, and the middle ports 2 .. M−1are deﬁned for a tiny impedance ZT. The “+” is deﬁned at the right side of the port, so that the port current is deﬁned in the +zdirection. Note also that port 1 is physically identical to the one deﬁned Fig ure 4 and so is portM, up to the sign. Here we deﬁne as well the two additional ports named θandϕ, deﬁned for the TL impedance Z0. Z0impedance at ports 1 and MandZT(i.e. 0) on the middle ports, so we ﬁrst obtain the current on the matched TL and then generalise the result for any loads. The advantages and novelties of our derivation are: •The method of using the S matrix reciprocity to derive receiving characteristics from transmitting characteris tics is original and may be extended to additional conﬁgura- tions. •We do not limit ourselves to the twin lead cross section, presenting a general algorithm for TL of any small electric cross section. •Our formulation is for a general incident plane wave, i.e. froman arbitrarydirection θ,ϕ, andarbitrarypolarisation α. •The solution of Eqs. (8) and (9) in [2] subject to general terminationconditions(10)isextremelycomplicated,and it is not obvious in this method how one ﬁrst gets a simple solution for a matched at both terminations TL, and how to further generalise it. In our solution based on S parameters, this emerges naturally. The work is organized as follows. In Section II we carry out all the analytic derivations for the voltage and differe ntial current along a TL hit by a monochromatic plane wave. We start with the voltage on a TL with matched loads and generalise the result for a TL terminated in any loads ( ZL andZR, as in Figure 1). This derivation is technical but quite lengthy, therefore after deriving the voltage we give the ﬁn al results for the current, which is calculated in Appendix C. Those calculations use the far E ﬁeld radiated from segments of the TL (based on [1] and summarised in Appendix A) and the properties of the generalised scattering matrix summar ised in Appendix B. In Section III we describe the full wave HFSS simulations 4 performed, and explain some delicate issues regarding the measurements of voltage and current. To be mentioned that a TL hit by a monochromaticplane wave may develop common mode currents. However, our derivation of the current is bas ed on the reciprocity of the system deﬁned in Figure 6. The cur- rents used to deﬁne the scattering matrix are differential, and for those differential currents we use the far ﬁelds develop ed in [1]. Hence, when we use the reciprocity to express the current developed on the TL as function of the plane wave excitation, we obtain only the differential part of the curr ent, and this drawback applies also to the method used in [2]–[6]. If the purpose is to calculate the power on the loads, this is not a limitation, because the common mode does not affect this power, but this has to be taken into account in the curren t measurement in Section III. In Section IV-A we validate the analytic results with full wave HFSS simulations for a cross section which is not a twin lead, showing that this formalism works for a general electrically small cross section. We compare the theoretic al results obtained in Sections II for the voltage and differen tial current with full wave HFSS solutions for matched and non matched transmission lines. In Section IV-B we prove the analytical results obtained in Section II are fully compati ble with the results obtained in [2]–[6] and therefore satisfy Eqs. (4)-(10). InSectionVwecalculategeneralexpressionsforthepowers transferred to the loads, and interpret them in terms of the transmission radiation patterns in [1] and incident plane w ave polarisation. We show that the power in the “left” load is closely related to the transmitting properties of a source a t the left termination and viceversa. In Section VI we show a full analysisoftheconnectionbetweentheradiationcharacter istics of TL calculated in [1] and the receiving characteristics calculated in this work. The work is ended with some concluding remarks. Note: through this work, the phasor amplitudes are RMS values, hence there is no 1/2 in the expressions for power. Also, it is worthwhile to mention that the results of this wor k depend on physical sizes relative to the wavelength, and hen ce are valid for all frequencies satisfying the condition of sm all electric cross section. II. DERIVATION OF THE VOLTAGE AND DIFFERENTIAL CURRENT ALONG THE TL We start with the voltages, analysing the system described in Figure 4. As explained in the introduction, to ﬁnd the voltages on a matched TL, i.e. for ZL=ZR=Z0in Figure 2, one needs only the cross elements between the group of ports 1... Mand the group θ,ϕof the S matrix. However, to generalise the solution for the voltage along a TL loaded by and ZLandZR, one needs additional elements of the S matrix, we therefore calculate here all the elements of the matrix for the system deﬁned in Figure 4, schematically shown in Figure 5. This matrix satisﬁes V−=SV+, (14)whereSis an(M+2)×(M+2)matrix and V+,V−are col- umn vectors of incoming and outgoing voltages, respectivel y. We start with the submatrix with indices 1 to M. Feeding amiddleport1< n < M , located at zn=−L+ (n−1)∆z, (15) where ∆z=2L M−1(16) withaforward(entering)voltage V+ nandterminatingtheother TLports 1≤i≤Mbytheimpedancesdeﬁnedforthoseports (i.e.Z0forports1and M,andZHforthemiddleports)results in a forwardwave from z=zntoz=Landa backwardwave fromz=zntoz=−L, because the waves encounter at the intermediate ports a very high impedance ZH→ ∞. The impedance seen at port nisZ0/2, so the reﬂection coefﬁcient is Sn,n=Z0/2−ZH Z0/2+ZH≃ −1+Z0 ZH, therefore the port voltage is Vn=V+ n(1 +Sn,n) =V+ nZ0/ZH. The outgoing voltages at ports i/ne}ationslash=nare V− i/negationslash=n=Vne−jk∆z|i−n|=V+ nZ0 ZHe−jk∆z|i−n|.(17) and for i=n V− n=V+ nSn,n=V+ n/parenleftbigg −1 +Z0 ZH/parenrightbigg (18) This results in the following (partial) ncolumn of the S matrix S1≤i≤M ,1<n<M =/braceleftigg Z0 ZHe−jk∆z|i−n|i/ne}ationslash=n −1 +Z0 ZH≃ −1i=n,(19) see uppersignin Figure5.Forcolumn n= 1orM, theresults are similar, only replaceZ0 ZHby 1, hence S1≤i≤M , n=1,M=/braceleftbigg e−jk∆z|i−n|i/ne}ationslash=n 0 i=n(20) Eqs. (19) and (20) deﬁne the upper left square of the S matrix, i.e. all the elements connecting the TL ports 1 to M- see Figure 5, on which the diagonal elements are shown, the upper sign relates to this system. Now we add the columns Sθ,nandSϕ,n(last columns in Figure 5), representing the far /hatwideθand/hatwideϕpolarisation antennas, respectively. Feeding a middleport1< n < M with the voltage V+ n and all other ports by their matched load, we calculate the fa r ﬁelds given in Eqs. (A.1)-(A.3),scaled to the voltages V− θand V− ϕaccording to Eq. (A.10). We start with the contribution of the xcurrents given in Eqs. (A.1) and (A.2), see Figure 7. It is convenient to expres s the results using the distances from the feeding point to the terminations: l2=L−zn;l1=zn+L, (21) shown in Figure 7, where znis deﬁned in Eq. (15). Hence in all following expressions, the value of nis “hidden” in l1 5 Fig. 7. Currents in the TL while feeding port nwithV+ n. Deﬁning I≡ V+ n/ZH, the current in the feeding line at z=znis2Iin the positive x direction, and the currents at z=L(portM) and z=−L(port 1) are Ie−jkl2andIe−jkl1, respectively, both in the negative xdirection. The z directed current waves are the forward wave in the region [zn, L]and the backward current wave in the region [−L, zn], to be used in Eq. (A.3). For the forward wave we use the current at the middle point zm2= (zn+ L)/2, which is Ie−jkl2/2, while for the backward wave we use the current atzm1= (zn−L)/2:−Ie−jkl1/2(minus because it ﬂows in the −z direction). andl2. To shorten the notations in the forthcoming results we introduce two deﬁnitions f1≡e−jk(l1+l2cosθ)/2sin(kl1cos2(θ/2)) f2≡e−jk(l2−l1cosθ)/2sin(kl2sin2(θ/2)).(22) Using Eqs.(A.1) for each of the three currents at zw=zn,L and−L, scaling with (A.10), after some algebra we obtain: V− θ(x)=−jV+ n(Z0/ZH)[f1+f2] cosθcosϕ(23) As evident from Eqs.(A.1) and (A.2), V− ϕ(x)is identical to the above, up to replacing cosθcosϕby−sinϕ, we therefore have V− ϕ(x)=jV+ n(Z0/ZH)[f1 +f2] sinϕ. (24) Next we calculate the contributions of the forward current wave in the region z= [zn, L]and the backward current wave in the region z= [−L, zn](see Figure 7), using the upper and lower signs in Eq. (A.3), respectively. For the forward wave we use the current in the middle point zm2= (zn+L)/2, having the phase e−jkl2/2for a TL length h=l2, while for the backward wave we use the current at zm1= (zn−L)/2, having the phase e−jkl1/2, for a TL length h=l1. Also the backward current is deﬁned in the −zdirection, it therefore is taken with the minus sign. The middle points can be written more compactly as zm2=l1/2 ; zm1=−l2/2, (25) and after scaling with (A.10), we obtain: V− θ(z)= 2jV+ n(Z0/ZH)cosϕ[f2cos2(θ/2)−f1sin2(θ/2)] (26) Now we add the xandzdirected currents contributions to V− θ=V− θ(x)+V− θ(z), using 1−cos(θ) = 2 sin2(θ/2)and 1 + cos( θ) = 2 cos2(θ/2), after some algebra we obtain: V− θ=jV+ n(Z0/ZH)[f2−f1] cosϕ (27) V− ϕhas only the contribution of the xdirected currents, given in Eq (24), hence V− ϕ=V− ϕ(x). (28)Results(27)and(28)deﬁnethe Sθ,nandSϕ,nmatrixelements respectively, for the columns 1< n < M : Sθ,1<n<M =j(Z0/ZH)[f2−f1] cosϕ(29) Sϕ,1<n<M =j(Z0/ZH)[f1+f2] sinϕ.(30) For column n= 1orM, the results are similar, only replace Z0 ZHby 1, hence Sθ, n=1,M=j[f2−f1] cosϕ (31) Sϕ, n=1,M=j[f1+f2] sinϕ. (32) The transpose elements are found by the reciprocity conditi on Si,jZj=Sj,iZi(see Appendix B, Eq. B.8), where ZiandZj are the impedances for which ports iandjhave been deﬁned, respectively.Giventheports1, M,θandϕaredeﬁnedfor Z0of the TL and ports2 .. M−1are deﬁned for ZH, the transposed relations Sn,θandSn,ϕare given by Eqs. (31) and (32), for alln: Sn,θ=j[f2−f1] cosϕ (33) Sn,ϕ=j[f1+f2] sinϕ. (34) When issuing and excitation V+ θfrom the θport and matching all other ports, the voltages at ports non the TL are V− n= V+ θSn,θ. Given the number Mis not limited, one can get the continuous voltage on the TL for the excitation of the θport: Vθ(z) =jV+ θ[f2(z)−f1(z)] cosϕ (35) where the zdependence in f1,2(see Eq. (22)) is in l1andl2, deﬁned according to Eq. (21), using zforzn. This voltage develops on the TL due to a /hatwideθpolarised plane wave coming from coordinates (θ, ϕ)at inﬁnity. Similarly for an excitation V+ ϕfrom the ϕport, while matching all other ports, one obtains the continuous voltag e on the TL: Vϕ(z) =jV+ ϕ[f1(z) +f2(z)] sinϕ. (36) which develops due to a /hatwideϕpolarised plane wave coming from coordinates (θ, ϕ)at inﬁnity. The total voltage developed on the matched TL is Vθ(z) + Vϕ(z), where the intensities of V+ θandV+ ϕare according to Eq. (12). It comes out V(z) =jV+[f2cos(ϕ−α)−f1cos(ϕ+α)](37) Eq. (37) is the ﬁnal result for the voltage on a matched TL, i.e. terminated at both ends with resistors Z0. We now generalize the above result for a TL terminated with any load s at ports 1 and M:ZL(left) and ZR(right), respectively (see Figure 2). We express the generalized results in terms of the reﬂection coefﬁcients: ΓL=ZL−Z0 ZL+Z0andΓR=ZR−Z0 ZR+Z0.(38) For the matched case, the only incoming voltage waves are V+ θandV+ ϕ, while all other V+ 1≤i≤M= 0. Here we have two additional incoming voltages at the mismatched ports 1 and M: V+ 1= ΓLV− 1andV+ M= ΓRV− M, (39) 6 while at the matched (with ZH) ports n= 2.. M−1the incoming voltages are 0: V+ 2≤n≤M−1= 0. (40) We use the general connection for 1≤n≤M(i.e. on TL) but1≤i≤M+ 2, i.e. all excitations including θandϕ: V− n=M+2/summationdisplay i=1Sn,iV+ i=Sn,1V+ 1+Sn,MV+ M+V(z) = Sn,1ΓLV− 1+Sn,MΓRV− M+V(z) (41) where the only non zero terms are i= 1, M, θandϕ(see Eq(40)),andthelast two terms i=θ, ϕrepresentthematched voltage at z=zn, given in (35), (36), summed in (37). In the last form of Eq. (41), we used (39). Forn= 1, Eq. (41) becomes V− 1= ΓRV− Me−jk2L+V(−L). (42) because S1,1= 0,S1,M=SM,1=e−jk2Landzn=−L. Forn=M, Eq. (41) reads V− M= ΓLV− 1e−jk2L+V(L). (43) because SM,M= 0andzn=L. Using Sn,1=e−jkl1,Sn,M=e−jkl2, for2≤n≤M−1, Eq. (41) becomes V− 2≤n≤M−1=e−jkl1ΓLV− 1+e−jkl2ΓRV− M+V(z),(44) where in principle V(z)here excludes the terminations, but as we shall see this exclusion is not necessary. We solve now Eqs. (42) and (43) for V− 1andV− Mand obtain: V− 1=ΓRe−j2kLV(L) +V(−L) 1−ΓLΓRe−j4kL(45) V− M=ΓLe−j2kLV(−L) +V(L) 1−ΓLΓRe−j4kL. (46) To avoid confusions the generalised (non-matched) results are subindexed “NM”. The total (non matched) voltage VNM(zn) =V+ n+V− n. For ports 1 or M, this voltage is VNM(−L) =V− 1(1 + Γ L) (47) VNM(L) =V− M(1 + Γ R). (48) For the ports 2≤n≤M−1,V+ n= 0(40),therefore Eq. (44) describes the total voltage on those ports. Taking the limit z→ −Lof Eq. (44), using (42) we ﬁnd it reduces to (47) and similarly the limit z→Lof Eq. (44) reduces to (48), therefore Eq. (44) describes the non matched voltage on the TL at all ports (in the continuum, for all z) VNM(z) =V(z) + ∆V(z), (49) where ∆V(z)is the correctionterm due to non matching(ﬁrst two terms in (44)): ∆V(z) =e−jkl1ΓLV− 1+e−jkl2ΓRV− M,(50) andV− 1andV− Mare given in Eqs. (45) and (46). This concludes the voltage developed on the TL due to a monochromatic plane wave, for a TL matched at both ends,the solution is Eq. (37) and for general terminations we add the correction term in Eq. (50). A similar calculation is carried out to ﬁnd the differential current developed on the TL due to a monochromatic plane wave.ThisisdonebyanalysingthesystemdeﬁnedinFigure6. Because this calculation is lengthy, and similar to this car ried outonthesystemdeﬁnedinFigure4,itisdoneinAppendixC, and the results are given below. The currents on a TL matched on both terminations due to θorϕpolarisations are: Iθ(z) =−j(V+ θ/Z0)[f1(z) +f2(z)] cosϕ(51) Iϕ(z) =j(V+ ϕ/Z0)[f1(z)−f2(z)] sinϕ,(52) and the total current for matched TL is their sum: I(z) =−j(V+/Z0)[f1cos(ϕ+α) +f2cos(ϕ−α)](53) It is easy to check that Eqs. (37) and (53) satisfy the termina - tion conditions (10) with ZL=ZR=Z0. For a TL terminated in any impedances ZL,ZR, the differential current along the TL, INM(i.e. non matched) is INM(z) =I(z) + ∆I(z), (54) where ∆I(z)is the correction term due to non matching: ∆I(z) =−e−jkl1ΓLI− 1−e−jkl2ΓRI− M,(55) andI− 1andI− Mare given explicitly in Eqs. (C.26) and (C.27), and also satisfy V− 1=−Z0I− 1;V− M=Z0I− M(56) The solutions (49) and (54) satisfy the termination condi- tions (10). III. FULL WAVE HFSSSIMULATIONS We describe in this Section the HFSS simulations done for the scattering problem deﬁned in Figure 1. The results of thi s simulations are compared in the next section with the analyt ic results. We used a non twin lead cross section (used also in [1]), shown in Figure 8. For this cross section one knows analytically the distance din the twin lead representation (shown as red points in Figure 8) by image theory, yielding d=/radicalbig s2−(2a)2= 2.537cm, (57) and also the characteristic impedance Z0=η0 πln/parenleftbiggd+s 2a/parenrightbigg = 105 .6 Ω (58) while for other cross sections those can be determined by an electrostatic ANSYS 2D Maxwell simulation, as shown in Appendix B of [1]. The electric ﬁeld of a plane wave is by default E0= 1V/m in the HFSS simulation, so to normalise the results for E0d= 1V we divide the measured results by the value of din Eq. (57). For convenience, we shall use a ﬁxed TL length of l≡ 2L= 125cm, and test for different frequencies. We measure the voltage and current along the TL from z=−61.25cm to 7 Fig. 8. Cross section of two parallel cylinders: the distanc e between the centres is s= 3.59cm, and the diameters are 2a= 2.54cm. The red points show the current images which deﬁne the twin lead representa tion, and the distance between them d= 2.537cm is calculated in Eq. (57). z= 61.25cm at intervals of 6.125cm, in total at 21 points. At the TL terminations z=−Landz=L, we use inactive lumped ports deﬁned for the impedance we need at those terminations. Two-conductors transmission lines excited onlyat termina- tions, develop the TEM mode, so that both EzandHzare 0. In such case one can measure the voltage by/integraltext E·dlfrom the “+” to the “-” conductor and the current using/contintegraltext H·dl around the “+” conductor, both on anyintegration path. In the case analyzed here, the TL is excited by an external planewave,therefore,dependingonthe incidenceof thiswa ve Ezand/or Hzare not necessarly 0, we therefore need more careful deﬁnitions for the voltage and current measurement s, as described in the following subsections. A. Voltage measurement The voltage measured by the integral/integraltextE·dlin the cross section depends on the chosen integration path if Hz/ne}ationslash= 0, as shown in Figure 9. To deﬁne the correct path we look at the deﬁnitions of the parallel ports in Figure 4. Those have been deﬁnedonthe x−zplane,sothat onlyxdirectedcurrentsﬂow through the port, and this fact has been used in the calculati on of the S matrix in Section II, see Figure 7. Therefore,to be consistent with the parallel ports deﬁniti on, the correct path to measure the voltage is path A(on the x axis) shown in Figure 9, and this path is used in all the voltag e measurements in Section IV. B. Current measurement For the case Ez= 0, the current can be measured by/contintegraltext H· dlaround the “positive” conductor, on any integration path, like path Cin Figure 10. In the opposite case consisting in an incident plane wave for which Ez/ne}ationslash= 0(as shown in the ﬁgure), not only the integration path has to be tight around t he conductor, but also, due to common mode current, one has to Fig. 9. Cross section voltage measurement on two possible pa thesAand B. In case Hz= 0, all pathes lead to the same result, namelyR AE·dl=R BE·dl. However, if Hz/negationslash= 0, as for the incident plane wave shown here, the result of the path integral depends on the paths used, and the correct voltage measurement isR AE·dl, i.e. along the xaxis consistent with the parallel ports deﬁnition in Figure 4. Fig. 10. Cross section differential current measurement fo r the case Hz/negationslash= 0 requires integration on tight loops around each conductor f rom which one obtains thecurrents IAandIB(Eq.(59)).Thedifferential current iscalculated in Eq. (60). measure around both conductors, on paths AandB, as shown in Figure 10, to obtain the currents IA=/contintegraldisplay AH·dl;IB=/contintegraldisplay BH·dl.(59) AsonenotesfromthederivationinSectionC, theSmatrixhas been derived from differential currents, so the analytic re sults (51) and (52) represent differential currents. To compare t he HFSS simulation results to the analytic results we calculat e the simulation differential current: I= (IA−IB)/2; (60) IfHz= 0,IBtrivially reduces to −IA, so that the differential current is the current on the “positive” conductor, which ma y be calculated by integrating on any path around it, like path s AorCin Figure 10. IV. VALIDATION OF THE ANALYTIC RESULTS A. Comparisons with HFSS full wave solution 1) Matched transmission line: We validate in this section the analytic results for matched TL in Eqs (37) and (53) 8 by comparison with full wave solution of ANSYS HFSS simulation results, described in the previous section. We analyse three examples, each from a main incidence direction, by plane waves traveling along the x,yorzaxes. In all examples the incident electric ﬁeld intensity satisﬁ es Ed=η0Hd= 1V. The half lenght of the TL is L= 0.625m (or the length 2L= 1.25m). Inthe ﬁrst examplewe examinea planewavetravelingfrom θ=π, alongthe zaxis, colinearwith the TL,havingthephase e−jkz, as shown in Figure 11. For θ=π, Eqs. (37) and (53) Fig. 11. Matched TL illuminated by a −bxpolarised plane wave from θ=π. (normalized by Z0) reduce to V(z) =V+je−jkLsin[k(L−z)] cos( ϕ−α)(61) Z0I(z) =−V+je−jkLsin[k(L−z)] cos( ϕ−α),(62) Clearly, at θ=π, the angle ϕis ill deﬁned and so is the polarisation angle α(Figure 3), but the difference ϕ−αis meaningful.Themaximumvoltageandcurrentoccurat α=ϕ orπ−ϕ, representing two opposite polarisations. Choosing α=ϕthe polarisation in Figure 3 becomes /hatwideθcosϕ+/hatwideϕsinϕ, which equals −/hatwidexas shown in Figure 11 and using V+= 1V Eqs. (61) and (62) become V(z) =je−jkLsin[k(L−z)] (63) Z0I(z) =−je−jkLsin[k(L−z)], (64) which behave oscillatory according to the frequency, with 0 value at right termination at z=L. We remark that for this plane wave Ez=Hz= 0therefore the voltage and current can be measured on any paths (see Figures 9 and 10). Figures 12-14 show the voltage and normalized current for the−/hatwidexpolarised plane wave from θ=πin Figure 11 for frequencies 30, 60 and 120MHz, or L/λ= 1/16, 1/8 and 1/4, respectively. The voltage at the left termination at z=−L isV(−L) =je−jkLsin(2kL), resulting in 0.27 + j0.65, (1 +j)/√ 2and 0 for the cases L/λ= 1/16, 1/8 and 1/4, respectively. The normalized currents Z0I(−L)gets the minusof the abovevalues,satisfyingthe terminationcondi tion V(−L) =−Z0I(−L). In the next example we use a plane wave hitting from θ= ϕ=π/2with phase ejkyas shown in Figure 15. For this direction Eqs. (37) and (53) (normalized by Z0) reduce to V(z) =V+sinα[1−e−jkLcos(kz)] (65) Z0I(z) =jV+sinα e−jkLsin(kz), (66) The maximum is for α=±π/2, so using α=π/2, results according to Figure 3 in a /hatwideϕpolarisation which is −/hatwidexat θ=ϕ=π/2, as shown in Figure 15. We note that this is the-1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Real(V) z/λ-1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Imag(V) z/λ -1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Z0Imag(I) z/λ Fig. 12. Real and imaginary parts of the voltage V(z)and normalized current Z0I(z)forthe plane waveincidence shown inFigure 11,atfrequency 30MHz orL/λ= 1/16. The continuous line is the analytic solution in Eqs. (63) an d (64) and the stars are the ANSYS simulation results. -1.5-1-0.500.511.5 -0.1-0.0500.050.1Real(V) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Imag(V) z/λ -1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Imag(I) z/λ Fig. 13. Same as Figure 12, for frequency 60MHz or L/λ= 1/8. -1.5-1-0.500.511.5 -0.2-0.100.10.2Real(V) z/λ-1.5-1-0.500.511.5 -0.2-0.100.10.2Imag(V) z/λ -1.5-1-0.500.511.5 -0.2-0.100.10.2Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.2-0.100.10.2Z0Imag(I) z/λ Fig. 14. Same as Figure 12, for frequency 120MHz or L/λ= 1/4. same plane wave shown in Figure 9, so that here the path on 9 Fig. 15. Matched TL illuminated by a −bxpolarised plane wave from (θ= π/2, ϕ=π/2)(as also shown in Figure 9). The view is from the positive x axis direction, so that only the “upper positive” conductor is seen. which one measures the voltage matters and must be path A in Figure 9. Using V+= 1V and α=π/2in Eqs. (65) and (66) result in V(z) = 1−e−jkLcos(kz) (67) Z0I(z) =je−jkLsin(kz), (68) Figures 16-18 show the voltage and normalized current for the−/hatwidexpolarised plane wave from θ=π/2andϕ=π/2in Figure11forfrequencies30,60and120MHz,or L/λ= 1/16, 1/8 and 1/4, respectively. -1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Real(V) z/λ-1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Imag(V) z/λ -1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Z0Imag(I) z/λ Fig. 16. Real and imaginary parts of the voltage V(z)and normalized current Z0I(z)for theplane waveincidence shown in Figure15,at frequency 30MHz orL/λ= 1/16. The continuous line is the analytic solution in Eqs. (67) an d (68) and the stars are the ANSYS simulation results. As expected, the voltage is an even function of zand the current an odd function of z. Speciﬁcally, the voltages at the terminations at z=−LandLare0.146+ j0.354,0.5 +j0.5 and 1 for the cases L/λ= 1/16,1/8 and1/4, respectively.The normalized currents have the same values at z=L, satisfying the termination condition V(L) =Z0I(L), and minus the above values at z=−L, satisfying V(−L) =−Z0I(−L). The voltage in the middle of the TL at z= 0are 0.076+j0.383, 1−1/√ 2 +j/√ 2and 1+j for the cases L/λ= 1/16, 1/8 and 1/4, respectively, and the current at z= 0is of course 0. In the next example we use a plane wave incident from θ=π/2andϕ= 0, with phase ejkxas shown in Figure 19.-1.5-1-0.500.511.5 -0.1-0.0500.050.1Real(V) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Imag(V) z/λ -1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Imag(I) z/λ Fig. 17. Same as Figure 16, for frequency 60MHz or L/λ= 1/8. -1.5-1-0.500.511.5 -0.2-0.100.10.2Real(V) z/λ-1.5-1-0.500.511.5 -0.2-0.100.10.2Imag(V) z/λ -1.5-1-0.500.511.5 -0.2-0.100.10.2Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.2-0.100.10.2Z0Imag(I) z/λ Fig. 18. Same as Figure 16, for frequency 120MHz or L/λ= 1/4. ForthisdirectionEqs.(37)and(53)(normalizedby Z0)reduce to V(z) =−jV+cosα e−jkLsin(kz) (69) Z0I(z) =−V+cosα[1−e−jkLcos(kz)].(70) Here the maximum is obtained for α= 0orπ. Using α=π deﬁnes according to Figure 3 a −/hatwideθpolarisation which equals /hatwidezatθ=π/2andϕ= 0, as shown in Figure 19. This is the same plane wave mentioned in Figure 10, which requires current measurement by integrating on the tight trajectori esA andBinFigure10,usingEq.(60)todeterminethedifferential current. Using V+= 1V and α=πin Eqs. (65) and (66) result in V(z) =je−jkLsin(kz) (71) Z0I(z) = 1−e−jkLcos(kz), (72) Figures 20-22 show the voltage and normalized current for the/hatwidezpolarised plane wave from θ=π/2andϕ= 0in Figure19forfrequencies30,60and120MHz,or L/λ= 1/16, 1/8 and 1/4, respectively. 10 Fig. 19. Matched TL illuminated by a bzpolarised plane wave from (θ= π/2, ϕ= 0)(as also shown in Figure 10). -1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Real(V) z/λ-1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Imag(V) z/λ -1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.06-0.04-0.0200.020.040.06Z0Imag(I) z/λ Fig. 20. Real and imaginary parts of the voltage V(z)and normalized current Z0I(z)for theplane waveincidence shown in Figure19,at frequency 30MHz orL/λ= 1/16. The continuous line is the analytic solution in Eqs. (71) an d (72) and the stars are the ANSYS simulation results. As expected, the voltage is an odd function of zand the current an even function of z. The voltage developed for the incident ﬁeld in Figure 19 (Eq. 71) has the value of the normalized current developed for the incident ﬁeld in Figure 15 (Eq. 68) and viceversa. -1.5-1-0.500.511.5 -0.1-0.0500.050.1Real(V) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Imag(V) z/λ -1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Imag(I) z/λ Fig. 21. Same as Figure 20, for frequency 60MHz or L/λ= 1/8.-1.5-1-0.500.511.5 -0.2-0.100.10.2Real(V) z/λ-1.5-1-0.500.511.5 -0.2-0.100.10.2Imag(V) z/λ -1.5-1-0.500.511.5 -0.2-0.100.10.2Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.2-0.100.10.2Z0Imag(I) z/λ Fig. 22. Same as Figure 20, for frequency 120MHz or L/λ= 1/4. 2) Non matched transmission line: We compare here sev- eral unmatched cases for the −/hatwidexpolarised plane wave from θ=π/2, and ϕ=π/2, shown in Figure 15 for frequency 60 MHz. The matched solutions are given in Eqs. (67) and (68), and for this frequency ( kL=π/4) are: V(z) = 1−e−jπ/4cos(kz) (73) Z0I(z) =je−jπ/4sin(kz), (74) as shown in Figure 17. We generalize them for non matched cases using Eqs (49) and (54), obtaining the correction terms due to non matching (Eqs. (50) and (55)) ∆V(z) =ΓLe−jkz+ ΓRejkz−2jΓLΓRcos(kz)√ 2(1 + Γ RΓL)(75) Z0∆I(z) =ΓLe−jkz−ΓRejkz−2ΓLΓRsin(kz)√ 2(1 + Γ RΓL)(76) Figures 23-25 show the voltage and normalized current for the cases: ZL=Z0/2andZR= 2Z0,ZL=ZR=Z0/2and ZL=ZR= 2Z0, respectively. B. Compatibility with previous works We show in this subsection that our analytic results are compatible with the results obtained by other authors [2]–[ 6], hence satisfy Eqs. (4) and (5). Using Eqs. (50), (55) and (56), it is easy to check that d∆V dz+jk(Z0∆I) = 0 ; Z0d∆I dz+jk∆V= 0,(77) which is clear, because the induced sources depend only on the incident ﬁeld and not on the loads. Therefore, it is enough to show that Eqs. (4) and (5) are satisﬁed by the matched solution V(z),I(z)in Eqs. (37), (53), for a general incident plane wave. First we need the phase of the incident plane wave (Eq. (1)) on the TL at x≃0,y= 0and−L≤z≤L, which comes out ejkzcosθ, (78) 11 -1.5-1-0.500.511.5 -0.1-0.0500.050.1Real(V) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Imag(V) z/λ -1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Imag(I) z/λ Fig. 23. Real and imaginary parts of the voltage V(z)and normalized current Z0I(z)for theplane waveincidence shown in Figure15,at frequency 60MHz (orL/λ= 1/16), but for loads ZL=Z0/2andZR= 2Z0, orΓL=−1/3 andΓR= 1/3. The continuous line is the analytic solution, i.e. Eq. (73) plus correction (75) and Eq. (74) plus correction (76) and the sta rs are the ANSYS simulation results. -1.5-1-0.500.511.5 -0.1-0.0500.050.1Real(V) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Imag(V) z/λ -1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Imag(I) z/λ Fig. 24. Same as Figure 23 for ZL=ZR=Z0/2orΓL= ΓR=−1/3. -1.5-1-0.500.511.5 -0.1-0.0500.050.1Real(V) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Imag(V) z/λ -1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Real(I) z/λ-1.5-1-0.500.511.5 -0.1-0.0500.050.1Z0Imag(I) z/λ Fig. 25. Same as Figure 23 for ZL=ZR= 2Z0orΓL= ΓR= 1/3.so the incident ExandHyﬁelds along the TL are Ex(z) =Exejkzcosθ;Hy(z) =Hyejkzcosθ,(79) where the values ExandHyare understood as the values at the origin. Using the unit vectors identity /hatwidex=/hatwidersinθcosϕ+ /hatwideθcosθcosϕ−/hatwideϕsinϕ, given the incident E ﬁeld has only θ andϕcomponents, the xcomponent of the electric ﬁeld (at origin) is Ex=Eθcosθcosϕ−Eϕsinϕ= E0(cosαcosθcosϕ−sinαsinϕ)(80) where for the second expression we used Eq. (3). Using Eqs. (11), (79) and (80), Z0Isin Eq. (7) comes out Z0Is=jkV+(sinαsinϕ−cosαcosθcosϕ)ejkzcosθ,(81) To derive the Hycomponent of the incident ﬁeld we note that the H components of the incident plane wave are: Hθ=Eϕ η0;Hϕ=−Eθ η0(82) Using the unit vectors identity /hatwidey=/hatwidersinθsinϕ+ /hatwideθcosθsinϕ+/hatwideϕcosϕ, given the incident H ﬁeld has only θandϕcomponents, the ycomponent of H (at origin) is Hy=Hθcosθsinϕ+Hϕcosϕ= 1 η0[Eϕcosθsinϕ−Eθcosϕ], (83) where in the second expression we used Eq. (82). Using Eqs. (3) and (11), this can be rewritten as η0Hy=V+[sinαcosθsinϕ−cosαcosϕ],(84) Using Eqs. (79), Vsin Eq. (6) comes out Vs=jkV+[sinαcosθsinϕ−cosαcosϕ]ejkzcosθ,(85) Here it is left to show that Eqs. (81) and (85) equal to the LHS of Eqs. (5) and (4) respectively, applied on the matched solution V(z),I(z)given in Eqs. (37), (53).We start with two intermediate results for the functions f1(z)andf2(z)deﬁned in Eq. (22): d f1/dz+jkf1=kcos2(θ/2)ejkzcosθ d f2/dz−jkf2=−ksin2(θ/2)ejkzcosθ,(86) and obtain dV dz+jk(Z0I) =−jkV+[cos2(θ/2)cos( ϕ+α)+ sin2(θ/2)cos( ϕ−α)]ejkzcosθ(87) Z0dI dz+jkV=jkV+[sin2(θ/2)cos( ϕ−α)− cos2(θ/2)cos( ϕ+α)]ejkzcosθ(88) Usingsimpletrigonometricidentitiesoneﬁndsthatresult s(87) and (88) are identical to Eqs. (85) and (81), respectively. 12 V. THE POWER TRANSFERRED TO THE LOADS We calculate here the power transferred to the loads for the general case of loads ZLandZRas shown in Figures 1 and 2. We need ﬁrst the voltage and current for the matched at both teminations case at the terminations( z=±L).FromEqs. (37) and (53) we obtain V(−L) =jV+e−jkLsin[2kLsin2(θ/2)] cos( ϕ−α)(89) V(L) =−jV+e−jkLsin[2kLcos2(θ/2)] cos( ϕ+α)(90) and Z0I(−L) =−V(−L) ; Z0I(L) =V(L)(91) satisfy the boundary conditions. From Eqs. (89)-(91), the powers on the “left” and “right” loads for the matched at both teminations case, are P(−L)≡ −V(−L)I∗(−L) = |V+|2 Z0sin2[2kLsin2(θ/2)] cos2(ϕ−α)(92) P(L)≡V(L)I∗(L) = |V+|2 Z0sin2[2kLcos2(θ/2)] cos2(ϕ+α)(93) We now generalise the powers for a TL with any loads. We set z=±Lin Eqs. (49), (54), or simpler, use directly Eqs. (47), (48), (C.28) and (C.29), obtaining VNM(−L) = (1 + Γ L)ΓRe−j2kLV(L) +V(−L) 1−ΓLΓRe−j4kL(94) VNM(L) = (1 + Γ R)ΓLe−j2kLV(−L) +V(L) 1−ΓLΓRe−j4kL(95) and ZLINM(−L) =−VNM(−L) ; ZRINM(L) =VNM(L) (96) satisfy the boundary conditions. Now we express the power on the “left” load ZLby PNM(−L) =Re{−VNM(−L)I∗ NM(−L)}and on the “right” loadZRbyPNM(L) =Re{VNM(L)I∗ NM(L)}. Using Eqs. (94)- (96) one obtains PNM(−L) =1− |ΓL|2 |1−ΓLΓRe−j4kL|2[P(−L) +|ΓR|2P(L)− 2PmixRe{ΓRe−j2kL}] (97) PNM(L) =1− |ΓR|2 |1−ΓLΓRe−j4kL|2[P(L) +|ΓL|2P(−L)− 2PmixRe{ΓLe−j2kL}]. (98) The generalised results for the powers are expressed in terms of the matched-TL powers P(−L)andP(L)(given in Eqs. (92)-(93)) plus an additional “mixed” term Pmixgiven by Pmix≡V(L)I∗(−L)≡ −V(−L)I∗(L) = |V+|2 Z0sin[2kLsin2(θ/2)] sin[2 kLcos2(θ/2)] cos(ϕ−α)cos(ϕ+α) (99)Those results may be understood from the radiation prop- erties of the TL in [1]. We remark P(−L)is proportional tosin2[2kLsin2(θ/2)], according to the radiation pattern of a TL carrying a forward wave only and maximised for the polarisation radiated in this case: /hatwidep+(θ, ϕ) =/hatwideθcosϕ+/hatwideϕsinϕ⇒α=ϕ,(100) see Section 2.1 in [1]. Also, we see P(−L) = 0for a polarisation orthogonal to /hatwidep+. This suggests that P(−L), i.e. the power into the left load for the matched on both sides TL is closely related to transmitting properties of a source at the left termination, issuing a forward wave. Similarly, P(L)is proportional to sin2[2kLcos2(θ/2)], ac- cording to the radiation pattern of a TL carrying a backward wave only and maximised for the polarisation radiated in thi s case: /hatwidep−(θ, ϕ) =/hatwideθcosϕ−/hatwideϕsinϕ⇒α=−ϕ,(101) see Section 2.2 in [1], and P(L) = 0for a polarisation orthogonal to /hatwidep−. This suggests that P(L), i.e. the power into the right load for the matched on both sides TL is closely related to transmitting properties of a source at the right termination, issuing a backward wave. The general non matched case given in Eqs. (97)-(98) is affected by both radiation patterns and polarisations, a nd their combination found in the mixed term in Eq. (99). We remark that for a polarisation orthogonal to either /hatwidep+or/hatwidep−, Pmix= 0and either P(−L)orP(L)are 0, in which case the received powers are either according to sin2[2kLcos2(θ/2)] orsin2[2kLsin2(θ/2)]. The formal connection between the transmitting and receiv- ing properties of TL is analysed the next section. VI. CONSISTENCY BETWEEN TRANSMITTING -RECEIVING RADIATION PROPERTIES To analyse the connection between the results of receiving electromagnetic radiation shown in this work and the result s of transmitting (radiating) electromagnetic radiation in [1], we have to summarise below some results from [1] and rewrite them in a convenient form. The far electric ﬁeld radiated by a TL carrying a forward wave only as in the upper panel of Figure 26, is given in Fig. 26. Upper panel: radiating TL fed by a current source Iatz=−L, matched at z=L. Lower panel: the same TL in receive mode, with a voltmeter at z=−Lmeasuring the open circuit voltage Voc. Section 2.1 of [1] in terms of the value of I+in the middle 13 of the TL at z= 0. We rewrite it here in terms of the current source Iin the standard form for a radiating antenna fed by a current E+=jkη0G(r)Il+ eff, (102) wherel+ effis the effective antenna length for the radiation of a TL carrying a forward wave only, is given by l+ eff= 2jde−jkLsin[2kLsin2(θ/2)]/hatwidep+(103) where the polarisation vector /hatwidep+is deﬁned in Eq. (100). The radiation resistance for this case (given in Eq.(36) of [1]) is rrad=η0 2π(kd)2[1−sinc(4 kL)], (104) which does not need a “+” superscript, being valid also for a backward wave only. The radiation pattern (directivity) fo r a forward wave only is D+(θ) =η0k2 4πrrad|l+ eff|2= 2sin2[2kLsin2(θ/2)] [1−sinc(4 kL)],(105) as given in Section 2.1, Eq. (13) in [1]. When dealing with radiationfromTL, which is usually small relativeto the pow er carried by the TL, we may use the term “radiation losses” for the power lost to radiation. So in the upper panel of Figure 26 , the radiated power is |I|2rrad, and the power transmitted by the TL is |I|2Z0, and clearly rrad≪Z0. But in the context of the radiating properties we have to consider the radiatio n efﬁciency, deﬁned as the radiated power, divided by the tota l power into the antenna. In our case this is er=rrad rrad+Z0≃rrad Z0. (106) which is small, meaning that the (matched) TL is not an efﬁcient antenna. Eq. (106) is valid for both forward only or backward only wave, hence does not need a superscript. The antenna gain is the directivity multiplied by the efﬁciency : G+(θ) =D+(θ)er=(kd)2η0sin2[2kLsin2(θ/2)] πZ0,(107) Now, in receive mode, calculating the scalar product be- tweentheincidentplanewave(Eq.(2))andtheeffectivelen gth in Eq. (103) yields the open circuit voltage measured in the lower panel of Figure 26 Voc=E·l+ eff= 2jE0de−jkLsin[2kLsin2(θ/2)] cos( ϕ−α), (108) which is compared with the open end voltage obtained from the formalism developed in this work. So using Eq. (94) with ΓR= 0andΓL= 1(open) yield exactly the result (108) where we identify V+=E0daccordingto Eq. (11). Also, the absolute value of Vocis maximal for a matched polarisation, i.e. for α=ϕ. We replace now the voltmeter in the lower panel of Fig- ure 26 by a load ZL(left). Eq. (97), reduces for ΓR= 0 toPNM(−L) = (1 − |ΓL|2)P(−L), so maximum power is obtained for ΓL= 0(ZL=Z0), yielding Eq. (92). Using the matched polarisation ( α=ϕ) in (92) results in the maximum received power: Prec=|V+|2 Z0sin2[2kLsin2(θ/2)].(109)Dividingthisby the Poyntingvector S=E2 0/η0, resultsin the effective receiving cross section area A+(for receiving into the left termination - the equivalent of transmitting a forw ard wave) A+=Prec S=d2η0 Z0sin2[2kLsin2(θ/2)].(110) which equals exactly toλ2 4πG+. A similar analysis for the backward wave, using the current source at the z=Ltermination (pointing upward) and the matched load at z=−Lin Figure 26 yields the ﬁeld radiated by a backward wave current: E−=jkη0G(r)Il− eff, (111) wherel− effis the effective antenna length for the radiation of a TL carrying a backward wave only, is given by l− eff=−2jde−jkLsin[2kLcos2(θ/2)]/hatwidep−,(112) and the polarisation vector /hatwidep−is deﬁned in Eq. (101).For this case, the gain is the directivity function D−given in Eq. (17) of [1] multiplied by the efﬁciency in Eq. (106), yields G−(θ) =(kd)2η0sin2[2kLcos2(θ/2)] πZ0,(113) The effective receiving cross section area A−comes out A−=d2η0 Z0cos2[2kLcos2(θ/2)], (114) satisfying the relation A−=λ2 4πG−. VII. C ONCLUSIONS We derived in this work the voltage and differential current developed on an ideal two-conductors TEM transmission line (TL) of any small electric cross section, connected to passi ve loads and hit by a monochromatic plane wave, as shown in Figure 1. For this derivation we used our knowledge on the radiation properties of TL [1] to build S matrices which describe the radiating system and used the reciprocity to derive the curr ent and voltage induced on the TL. This methodology allowed us to derive ﬁrst the voltage and current on a matched at bothteminationsTL,yieldingtherelativelysimpleexpres sions given in Eqs. (37) and (53). The generalisation to any loads is then obtained, using the S matrix. We validated our analytic results in Section IV-A for both the matched on two sides TL case and the non matched case for different plane wave incidences. We also showed in Section IV-B that our analytic results are compatible with t he methodology used in previous works [2]–[6]. The simplicity of this proof, requiring the application of Eqs. (4) and (5) onlyon the matched at both teminations solution (see Eq. 77), emphasises the added value of this work. In Section V, we derived the powers on the loads and showedthe connectionbetweenthose powersandthe radiatin g properties of the TL. For the matched at both teminations TL, the power on the “left” load has the same spatial depencence on the incident plane wave direction as the radiation patter n 14 of a forward wave, and is maximised for the polarisation of a radiating forward wave /hatwidep+(see Eq. (100)). Similarly, for the matched at both teminationsTL, the power on the “right” load has the same spatial depencence on the incident plane wave direction as the radiation pattern of a backward wave, and is maximisedforthepolarisation /hatwidep−(seeEq.(101)).Thegeneral non matched case is affected by both radiation patterns and polarisations, and their combination found in a mixed term. In Section VI we showed that the formal relations between receiving cross section area and antenna gain are satisﬁed f or the TL. APPENDIX A FAR RADIATED EFIELD AND ITS NORMALIZATION ToderivetheSmatricesweneedthecontributionofcurrents shown in Figure A.1 to the far ﬁeld. Based on [1], a current element Idin the xdirection (where Iis the current and dthe separation distance in the twin lead representation), a t z=zw(w=wire) contributes the following E ﬁeld in θandϕ polarisations (see left panel of Figure A.1). Fig. A.1. Currents contributing to radiation. The left pane l shows a xdirected current Iat location zwon the TL (which may represent a parallel port or a load), contributes to the far E ﬁeld according to Eqs. Eqs.( A.1) and (A.2). The right panel shows a TL segment of width hcarrying a forward and/or backward current wave, I±evaluated at the middle location zm, contributing to the far E ﬁeld according to Eq. (A.3). Eθ(x)=−jkG(r)η0dIejkzwcosθcosθcosϕ(A.1) Eϕ(x)=jkG(r)η0dIejkzwcosθsinϕ,(A.2) where the subscript (x)shows that this is a xdirected current contribution. A TL section of lenght haround z=zm(m=middle) carryingaforwardand/orbackwardcurrentcontributesthe fol- lowing far θpolarised E ﬁelds (see right panel of Figure A.1). Eθ(z)=−2kG(r)η0dI±(zm)ejkzmcosθ(1±cosθ)cosϕ sin[kh(1∓cosθ)/2] (A.3) where the upperand lower signs are for forwardand backward waves respectively, and the values used for the forward or backward currents are in the middle of the line at zm, both currents deﬁned in the +zdirection. The subscript (z)shows that this is a zdirected current contribution. In the ﬁnal result, Eθ=Eθ(x)+Eθ(z)andEϕ=Eϕ(x). To work out the S matrices (Section II and Appendix C), one needs to scale the radiated EθandEϕﬁelds to outgoingvoltages V− θandV− ϕat the ports θandϕ(Section II), or to outgoing currents I− θandI− ϕ(Appendix C). Those ports have been deﬁned for the impedance Z0, which is the charateristic impedance of the analysed TL. To make those portsmatched,we use theantennamodelshowninFigureA.2. The model assumes a real antenna impedance ZAmatched Fig. A.2. The antenna model assumes a real antenna impedance ZAmatched by an ideal transformer to the θorϕport of impedance Z0. via an ideal transformer to the θorϕport of impedance Z0. Using the effective length of the antenna leff(for the given incidence direction), one expresses the open circuit volta ge on the antenna [9] Voc=E leff, (A.4) where EisEθorEϕ. The θorϕport being matched, the antenna “sees” a matched load, so that the actual voltage on the antenna terminals is Voc/2, and the outgoing voltage at the port is V−=1 2E leff/radicalbig Z0/ZA. (A.5) Now to activate the antenna in transmit mode, we feed the θ orϕport by V+, obtaining the current feeding the antenna IA=VA/ZA=V+//radicalbig ZAZ0 (A.6) Using the well known formula for the far E ﬁeld radiated by a dipole [9] E0=jkη0G(r)IAleff (A.7) where leffis the effective dipole length discussed before, into the same direction [9]. The ﬁeld radiated from the antenna(s ) has been named E0to distinguish it from the previously discussed ﬁeld travelling toward the antenna(s). Expressi ng it as function of V+, we obtain E0=jkη0G(r)leffV+//radicalbig ZAZ0 (A.8) This ﬁeld E0radiated from the far antenna(s), is the ﬁeld incident on the TL (or one of its components), mentioned in Figure 3. We want to normalise it so that E0d=V+(see Eq. (11)), considering the ﬁeld incident on the TL as a plane wave. Imposing condition (11) on Eq. (A.8), results in leff=√ZAZ0 jkη0G(r)d(A.9) Given the far antennas are only tools to build the S matrices and from them infer the voltage and current along the TL, the “actual” value of leffis of no interest, so to express the ﬁelds 15 radiated by the TL EθandEϕas voltages, we set lefffrom (A.9) into Eq. (A.5), obtaining V− θorϕ=Z0Eθorϕ 2jkη0G(r)d(A.10) For the case we need to express the ﬁelds radiated by the TLEθandEϕas outgoing currents(Section C), we use I−= −V−/Z0, obtaining: I− θorϕ=−Eθorϕ 2jkη0G(r)d(A.11) APPENDIX B GENERALISED SCATTERING MATRIX In this appendix we deﬁne the generalized scattering ma- trix and explain its reciprocity properties. The generaliz ed scattering matrix [7] for an arbitrary network (as shown in Figure B.1) is deﬁned by the following matrix equation Fig. B.1. An Nports arbitrary network interfaced to transmission lines o f characteristic impedances Z1,Z2, ...ZN. V−=SV+, (B.1) whereV±are column vectors for the incoming and outgoing voltage waves at the ports of the network, where each port has its own characteristic impedance, as shown in Figure B.1 . The voltage and current waves at the ports satisfy V+=ZI+;V−=−ZI−, (B.2) whereZis a diagonal matrix of the characteristic impedances in Figure B.1: Zij=Ziδij, (B.3) so one can easily apply functions on them, as follows. Normalisingthevoltagesandcurrentsateachportaccordin g to a±≡√ Z−1V±;b±≡√ ZI±,(B.4) and setting into Eq. (B.1), result in a−=√ Z−1S√ Za+, (B.5)which deﬁne the (ordinary) scattering matrix, for which we use a lower case “s” s≡√ Z−1S√ Z. (B.6) It is known [7]–[12] that the reciprocity condition for the ordinary scattering matrix is sij=sji. (B.7) Using this in Eq. (B.6) results in the reciprocity condition for the generalized scattering matrix SijZj=SjiZi (B.8) Scattering matrices are usually deﬁned for voltage waves, b ut in this work we need to deﬁne a generalized scattering matrix for current waves as follows I−=SII+, (B.9) and we name it SIto distinguish it from the regular scattering matrix in (B.1). Using Eq. (B.2), one obtains V−=−ZSIZ−1V+. (B.10) Comparing it with (B.1) we get the relation S=−ZSIZ−1. (B.11) Setting it in (B.6) results in s=−√ ZS√ Z−1, (B.12) and using (B.7) we obtain SI ijZi=SI jiZj (B.13) APPENDIX C DERIVATION OF THE CURRENT ALONG THE TL We deﬁne a system of M+ 2ports, the ﬁrst Mports are serial ports on the TL (shown in Figure 6) and 2 additional ports are far antennas at polarisations /hatwideθand/hatwideϕ. Our purpose is to determine the TL current for an incidence of a plane wave, we therefore calculate the matrix Sdeﬁned by I−=SI+, (C.1) althoughthisis not the usual deﬁnitionof the scatteringma trix (seeAppendixB).Forsimplicity,westillcallit S,buttakeinto consideration that any reﬂection coefﬁcient has an opposit e sign. Feeding port 1< n < M , located at znwith a forward (entering) current I+ nand terminating the other TL ports 1≤ i≤Mby the impedances deﬁned for those ports (i.e. Z0 for ports 1 and M, andZTfor the middle ports) results in a forward wave from z=zntoz=Land a backward wave fromz=zntoz=−L, because the waves encounter at the intermediate ports a very small serial impedance ZT→0. The reﬂection coefﬁcient at port nisSn,n=ZT−2Z0 ZT+2Z0≃ −1 +ZT Z0, so the port current In=I+ n(1 +Snn)≃I+ nZT Z0. The currentwaves at ports i/ne}ationslash=ngive rise to outgoingcurrents at the ports given by: I− i/negationslash=n=Ine−jk∆z|i−n|=I+ nZT Z0e−jk∆z|i−n|.(C.2) 16 and for i=n I− n=I+ nSn,n=I+ n/parenleftbigg −1 +ZT Z0/parenrightbigg (C.3) This results in the following (partial) ncolumn of the S matrix S1≤i≤M ,1<n<M =/braceleftigg ZT Z0e−jk∆z|i−n|i/ne}ationslash=n −1 +ZT Z0≃ −1i=n,(C.4) see lowersign in Figure5.Forcolumn n= 1orM, theresults are similar, only replaceZT Z0by 1, hence S1≤i≤M , n=1,M=/braceleftbigg e−jk∆z|i−n|i/ne}ationslash=n 0 i=n(C.5) Eqs. (C.4) and (C.5) deﬁne the upper left square of the S matrix, i.e. all the elements connecting the TL ports 1 to M, andnowwe calculatethe two additionalelementsbelongingt o portsθandϕ(last two columnsin Figure 5),using Eqs. (A.1)- (A.3) and scaling the ﬁelds with Eq. (A.11). Feeding a middleport1< n < M with the current I+ nand terminating all other ports with matched loads, we calculate the far ﬁelds given in Eqs. (A.1)-(A.3), scaled to the currents I− θandI− ϕaccording to Eq. (A.11). As in the voltage calculation section, the value of nis hidden in the parameters l1andl2(Eq. (21)), see Figure C.1. We start with the contribution of the xcurrents given in Eqs. (A.1) and (A.2). The xdirected currents are shown in Figure C.1. Using Eqs.(A.1) for the two currents at Land Fig. C.1. Currents in the TL while feeding port nwithI+ n. Deﬁning I≡ I+ nZT/Z0, thexdirected currents are Ie−jkl1in port 1 at z=−Land −Ie−jkl2in port Matz=L. Thezdirected current waves are the forward wave in the region [zn, L]and the backward current wave in the region [−L, zn], to be used in Eq. (A.3). For the forward wave we use the curren t at the middle point zm2= (zn+L)/2, which is Ie−jkl2/2, while for the backward waveweusethecurrent at zm1= (zn−L)/2which us Ie−jkl1/2. −L, scaling with (A.11), we obtain after some algebra: I− θ(x)=jI+ n(ZT/Z0)[f2−f1] cosθcosϕ, (C.6) where f1,2are deﬁned in Eq. (22). As evident from Eqs.(A.1) and (A.2), I− ϕ(x)is identical to the above, up to replacing cosθcosϕby−sinϕ, we therefore have I− ϕ(x)=jI+ n(ZT/Z0)[f1−f2] sinϕ (C.7) Next we calculate the contributions of the forward current wave in the region z= [zn, L]and the backward current wave in the region z= [−L, zn](see Figure C.1), using the upper and lower signs in Eq. (A.3), respectively. For the forwardwave we use the current in the middle point zm2, having the phase e−jkl2/2for a TL length h=l2, while for the backward wave we use the current at zm1, having the phase e−jkl1/2, for a TL length h=l1, both currents are deﬁned in the +z direction, hence taken with plus sign. Using zm1,2in Eq. (25), after scaling with (A.11), we obtain: I− θ(z)=−2jI+ n(ZT/Z0)[f2cos2(θ/2) +f1sin2(θ/2)] cos ϕ (C.8) Now we add the xandzdirected currents contributions to I− θ=I− θ(x)+I− θ(z), using 1−cos(θ) = 2 sin2(θ/2)and1 + cos(θ) = 2 cos2(θ/2), after some algebra we obtain: I− θ=−jI+ n(ZT/Z0)[f1+f2] cosϕ (C.9) I− ϕhas only the contribution of the xdirected currents, given in Eq (C.7), so that I− ϕ=I− ϕ(x)(C.10) Results (C.9) and (C.10) deﬁne the Sθ,nandSϕ,nmatrix elements respectively, for the columns 1< n < M : Sθ,1<n<M =−j(ZT/Z0)[f1+f2] cosϕ(C.11) Sϕ,1<n<M =j(ZT/Z0)[f1−f2] sinϕ (C.12) For column n= 1orM, the results are similar, only replace ZT Z0by 1, hence Sθ, n=1,M=−j[f1+f2] cosϕ (C.13) Sϕ, n=1,M=j[f1−f2] sinϕ (C.14) The transpose elements are found by the reciprocity conditi on Si,jZi=Sj,iZj(see AppendixB, Eq.B.13),where ZiandZj are the impedances for which ports iandjhave been deﬁned, respectively. Note that this reciprocity condition is diff erent from the one used for the voltage calculation in Section II (see Appendix B). Given the ports 1, M,θandϕare deﬁned for Z0of the TL and ports 2 .. M−1are deﬁned for ZT, the transposed relations Sn,θandSn,ϕare given by Eqs. (C.13) and (C.14), for all n: Sn,θ=−j[f1+f2] cosϕ (C.15) Sn,ϕ=j[f1−f2] sinϕ (C.16) When issuing an excitation I+ θfrom the θport and matching all other ports, the currents at ports non the TL are I− n= I+ θSn,θ. Given the number Mis not limited, one can get the continuous current on the TL for the excitation of the θport: Iθ(z) =−jI+ θ[f1(z) +f2(z)] cosϕ(C.17) and similarly for the ϕport: Iϕ(z) =jI+ ϕ[f1(z)−f2(z)] sinϕ (C.18) where the zdependence in f1,2(see Eq. (22)) is in l1andl2, deﬁned according to Eq. (21), using zforzn. Expressingthose in terms of V+ θ=Z0I+ θ, andV+ ϕ=Z0I+ ϕ results in Eqs. (51) and (52) in the main text. The total current developed on the TL is I(z) =Iθ(z) +Iϕ(z) (C.19) 17 This concludes the results on a TL matched on both ends. We generalize the above results for TL terminated with any loads at ports 1 and M:ZL(left) and ZR(right), respectively. We express the generalized results in terms of the reﬂection coefﬁcients ΓLandΓR, given in Eqs. (38). For currents they are used with the minus sign. At ports 1 and Mwe have: I+ 1=−ΓLI− 1andI+ M=−ΓRI− M,(C.20) and at ports n= 2.. M(matched with ZT), I+ 2≤n≤M−1= 0. (C.21) We use the general connection for 1≤n≤M(i.e. on TL) but1≤i≤M+ 2, i.e. all excitations including θandϕ: I− n=M+2/summationdisplay i=1Sn,iI+ i=Sn,1I+ 1+Sn,MI+ M+I(z) = −Sn,1ΓLI− 1−Sn,MΓRI− M+I(z) (C.22) where the only non zero terms are i= 1, M, θandϕ(see Eq (C.21)), and the last two terms i=θ, ϕrepresent the matched currents at z=zn, given in (C.17), (C.18), summed in (C.19). In the last form of Eq. (C.22), we used (C.20). Forn= 1, Eq. (C.22) becomes I− 1=−ΓRI− Me−jk2L+I(−L). (C.23) because S1,1= 0,S1,M=SM,1=e−jk2Landzn=−L. Forn=M, Eq. (C.22) reads I− M=−ΓLI− 1e−jk2L+I(L). (C.24) because SM,M= 0andzn=L. Using Sn,1=e−jkl1,Sn,M=e−jkl2, for2≤n≤M−1, Eq. (C.22) becomes I− 2≤n≤M−1=−e−jkl1ΓLI− 1−e−jkl2ΓRI− M+I(z),(C.25) where in principle I(z)here excludes the terminations, but as we shall see this exclusion is not necessary. We solve now Eqs. (C.23) and (C.24) for I− 1andI− Mand obtain: I− 1=−ΓRe−j2kLI(L) +I(−L) 1−ΓLΓRe−j4kL(C.26) I− M=−ΓLe−j2kLI(−L) +I(L) 1−ΓLΓRe−j4kL.(C.27)Weneedthetotal(nonmatched)current INM(zn) =I+ n+I− n for ports 1 to Mon the TL, i.e. the outgoing I−plus the incoming I+. For ports 1 or M, this current is INM(−L) =I− 1(1−ΓL) (C.28) INM(L) =I− M(1−ΓR). (C.29) For the ports 2≤n≤M−1,I+ n= 0(C.21), therefore Eq.(C.25)describesthetotalcurrentonthoseports.Takin gthe limitz→ −Lof Eq. (C.25),using (C.23)we ﬁnd it reducesto (C.28) and similarly the limit z→Lof Eq. (C.25) reduces to (C.29),thereforeEq. (C.25)describesthe nonmatchedvolt age on the TL at all ports (in the continuum, for all z). Hence, INM(z) =I(z) + ∆I(z), where ∆I(z)is the correction term due to non matching: ∆I(z) =−e−jkl1ΓLI− 1−e−jkl2ΓRI− M,(C.30) andI− 1andI− Mare given in Eqs. (C.26) and (C.27).