Free space transmission lines in receiving antenna operation
aa r X i v : . [ phy s i c s . c l a ss - ph ] S e p Free space transmission lines in receiving antennaoperation
Reuven Ianconescu ∗ ,Vladimir Vulfin †† Shenkar College of Engineering and Design, Ramat Gan, Israel, [email protected] ∗ Ben-Gurion University of the Negev, Beer Sheva 84105, Israel, [email protected]
Abstract —This work derives exact expressions for the voltageand current induced into a two conductors non isolated transmis-sion lines by an incident plane wave. The methodology is to usethe transmission line radiating properties to derive scatteringmatrices and make use of reciprocity to derive the responseto the incident wave. The analysis is in the frequency domainand it considers transmission lines of any small electric crosssection, incident by a plane wave from any incident direction andany polarisation. The analytic results are validated by successfulcomparison with ANSYS commercial software simulation results,and compatible with other published results.
Index Terms —electromagnetic theory, guided waves, electro-magnetic interference
I. I
NTRODUCTION
This work calculates the voltage and differential currentdeveloped on an ideal two-conductors TEM transmission line(TL) of any small electric cross section, connected to passive(lumped) loads and hit by a monochromatic plane wave, asshown 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: Z L (left) and Z R (right), hit bya monochromatic plane wave propagating toward the centre of coordinates.The cross section is electrically small and may be of any shape. The loads arelocated at the terminations of the TL, and are shown farther away, becauseof technical drawing limitations. the voltage and current developed along the TL, and hence thepowers delivered to the passive loads.The configuration as defined here is a scattering problem,requiring a full wave solution to set the tangential componentof 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 Z of theTL and an equivalent separation distance d can be found byan electrostatic cross section analysis, leading to a twin leadequivalent (examples for determining Z and d are shown inAppendix B of [1]). The twin lead equivalent is shown inFigure 2. When the TL is excited at its port(s), this equivalent Fig. 2. Twin lead equivalent of the analysed transmission line, definedby the separation distance d between the conductors, and the characteristicimpedance Z , both computed by the cross section analysis described inAppendix B of [1]. The TL is of length L . twin lead radiates the same far fields as the actual TL. Toemphasise, we do not solve for the twin lead geometry, butfor an arbitrary cross section in Figure 1, the twin lead is onlyused as an intermediate tool for the calculations.The incident plane wave shown in Figure 1 propagatestoward the coordinates origin with phase e − j k · r , so thatthe wavenumber vector k = − b r k points toward the origin.Expanding b r in Cartesian unit vectors, the phase can be writtenas e jk [ x sin θ cos ϕ + y sin θ sin ϕ + z cos θ ] , (1)where θ and ϕ are the spherical angles which represent thedirection of the plane wave arrival.The polarisation of the incident plane wave is shown inFigure 3. It travels toward the centre of coordinates, perpen-dicular to the θ, ϕ plane, with a polarisation angle α from the θ axis, so that the E field at the origin, where its phase is 0(see Eq. (1)) is given by E = E ( b θ cos α + b ϕ sin α ) (2)or its components E θ = E cos α ; E ϕ = E sin α. (3) Fig. 3. The incident plane wave propagates toward the centre of coordinatesin the − b r direction. In spherical coordinates the local equiphase surface is θ, ϕ and the polarisation is at angle α from the θ axis, so that at the origin E = E ( b θ cos α + b ϕ sin α ) (see Eq. (2)). Such problem has been handled in [2]–[6], but not for anarbitrary cross section (as in Figure 1), and not for a generalincident plane wave (as in Figure 3). The methodology usedin these works is based on a generalisation of the telegraphequations for transmission lines, for the twin lead geometryindicated in Figure 2. Given C ′ and L ′ the capacitanceand inductance per unit length, using Z = p L ′ /C ′ andthe identity √ L ′ C ′ = 1 /c , we write the (inhomogeneous)telegraph equations in the following compact form dVdz + jk ( Z I ) = V s (4) Z dIdz + jkV = Z I s (5)where the source terms V s and I s are induced by H y and E x fields of the incident plane wave, integrated along the x axisof the twin lead (see Figure 2), as follows: V s = jkη Z d H y ( x, z ) dx ≃ jkη H y ( z ) d (6) Z I s = − jk Z d E x ( x, z ) dx ≃ − jkE x ( z ) d, (7)and η = 377Ω is the free space impedance. The last ( ≃ ) partsof Eqs. (6) and (7) are consistent with our framework whichis restricted to small electric cross section ( kd ≪ ).The coupled non-homogeneous differential equations (4)and (5) can be decoupled into two second order equations: d Vdz + k V = dV s dz − jk ( Z I s ) (8) Z d Idz + k Z I = Z dI s dz − jkV s (9) In [2] the authors wrote the general solution for those secondorder differential equations, subject to the termination condi-tions: V ( − L ) = − Z L I ( − L ) ; V ( L ) = Z R I ( L ) , (10)however they used in the expression for the sources theimplicit assumption that the incident field has only a H y component. In [3] this solution has been simplified, and in[4]–[6] this formalism has been extended to multiconductorTL.The results derived in this work satisfy Eqs. (4)-(10), butwe use a different technique to derive them. We shall useour knowledge on the radiation properties of TL [1] to derivethe receiving properties, i.e. the response of a TL to anincident monochromatic plane wave described in Figure 3. Todetermine the voltage along the TL, we segment it into M parallel ports as shown in Figure 4. The ports 1 and M are Fig. 4. M parallel ports along the TL, adjacent ports are at distance ∆ z .Ports 1 and M are defined for the TL impedance Z , and the middle ports2 .. M − are defined for a high impedance Z H . Two additional portsrepresenting far antennas matched for the b θ and b ϕ polarizations are defined.Those ports are named θ and ϕ , and are defined for the TL impedance Z . defined for the TL impedance Z , and the middle ports 2 .. M − are defined for a high impedance Z H → ∞ . We definetwo additional ports representing far antennas matched for the b θ and b ϕ polarizations of the field radiated by the TL, defininga system of M + 2 ports. The additional ports, M + 1 and M + 2 , named θ and ϕ , may be defined for any impedance,and for convenience we choose to define them for the TLimpedance Z . The scattering matrix of this system is shownschematically in Figure 5. Using the formalism developed in[1], we know the b θ and b ϕ components of the radiated electricfield (summarised in Appendix A). Those field componentsare translated into voltages using a normalisation explained inAppendix A, enabling us to calculate the generalised S matrix(see Appendix B) describing this system.Here we excite this system with an incident voltage, scaledto the incident electric field in Eq. (2) by V + = E d, (11)which defines the voltage components (equivalent to Eq. (3)) V + θ = V + cos α ; V + ϕ = V + sin α. (12) Fig. 5. Schematic diagram of a ( M + 2) × ( M + 2) scattering matrixdescribing the systems defined in Figures 4 and 6. The first ports 1.. M arethe TL ports in the figures, and the two additional ports are the far θ or ϕ polarised antennas, named θ , ϕ . In both cases S θϕ = S ϕθ = 0 , because theyare orthogonally polarised, and S = S MM = S θθ = S ϕϕ = 0 . As willbe shown, the elements S ii for < i < M are ∓ for the matrices definedby the configurations in Figures 4 and 6, respectively. We note that ports 1... M are matched with the Z impedanceat ports 1 and M and Z H (i.e. infinite) on the middle ports,which is exactly the configuration of a TL matched at bothterminations, on which we want to obtain the voltages.Those are obtained by multiplying the S matrix by thecolumn vector .. V + θ V + ϕ (13)of adequate excitation elements according to Eq. (12), andgiven ∆ z is arbitrarily small (or M arbitrarily large) we obtainthe voltage as function of z along the matched TL for the givenincident plane wave. The solution for matched TL is after thatgeneralised for any terminations in Figure 1.Similarly, the analytic solution for the current is obtained bysegmenting the TL into M serial ports as shown in Figure 6.Ports 1 and M are defined for the TL impedance Z , andthe middle ports 2 .. M − are defined for a tiny impedance Z T → . Here we define too the additional ports representingfar antennas matched for the b θ and b ϕ polarizations of the fieldradiated by the TL, obtaining a different system of M +2 ports.For this system we also calculate a generalized S matrix,but defined for current waves (see Appendix B). Here themethodology is similar to the one explained for the voltages,the matched at both terminations TL result is obtained using Fig. 6. M serial ports along the TL, adjacent ports are at distance ∆ z . Ports1 and M are defined for the TL impedance Z , and the middle ports 2 .. M − are defined for a tiny impedance Z T . The “+” is defined at the rightside of the port, so that the port current is defined in the + z direction. Notealso that port 1 is physically identical to the one defined Figure 4 and so isport M , up to the sign. Here we define as well the two additional ports named θ and ϕ , defined for the TL impedance Z . Z impedance at ports 1 and M and Z T (i.e. 0) on the middleports, so we first obtain the current on the matched TL andthen generalise the result for any loads.The advantages and novelties of our derivation are: • The method of using the S matrix reciprocity to derivereceiving characteristics from transmitting characteristicsis 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 smallelectric cross section. • Our formulation is for a general incident plane wave, i.e.from an arbitrary direction θ , ϕ , and arbitrary polarisation α . • The solution of Eqs. (8) and (9) in [2] subject to generaltermination conditions (10) is extremely complicated, andit is not obvious in this method how one first gets asimple solution for a matched at both terminations TL,and how to further generalise it. In our solution based onS parameters, this emerges naturally.The work is organized as follows. In Section II we carryout all the analytic derivations for the voltage and differentialcurrent along a TL hit by a monochromatic plane wave.We start with the voltage on a TL with matched loads andgeneralise the result for a TL terminated in any loads ( Z L and Z R , as in Figure 1). This derivation is technical but quitelengthy, therefore after deriving the voltage we give the finalresults for the current, which is calculated in Appendix C.Those calculations use the far E field radiated from segmentsof the TL (based on [1] and summarised in Appendix A) andthe properties of the generalised scattering matrix summarisedin Appendix B.In Section III we describe the full wave HFSS simulations performed, and explain some delicate issues regarding themeasurements of voltage and current. To be mentioned that aTL hit by a monochromatic plane wave may develop commonmode currents. However, our derivation of the current is basedon the reciprocity of the system defined in Figure 6. The cur-rents used to define the scattering matrix are differential, andfor those differential currents we use the far fields developedin [1]. Hence, when we use the reciprocity to express thecurrent developed on the TL as function of the plane waveexcitation, we obtain only the differential part of the current,and this drawback applies also to the method used in [2]–[6].If the purpose is to calculate the power on the loads, this isnot a limitation, because the common mode does not affectthis power, but this has to be taken into account in the currentmeasurement in Section III.In Section IV-A we validate the analytic results with fullwave HFSS simulations for a cross section which is not atwin lead, showing that this formalism works for a generalelectrically small cross section. We compare the theoreticalresults obtained in Sections II for the voltage and differentialcurrent with full wave HFSS solutions for matched and nonmatched transmission lines. In Section IV-B we prove theanalytical results obtained in Section II are fully compatiblewith the results obtained in [2]–[6] and therefore satisfyEqs. (4)-(10).In Section V we calculate general expressions for the powerstransferred to the loads, and interpret them in terms of thetransmission radiation patterns in [1] and incident plane wavepolarisation. We show that the power in the “left” load isclosely related to the transmitting properties of a source at theleft termination and viceversa. In Section VI we show a fullanalysis of the connection between the radiation characteristicsof TL calculated in [1] and the receiving characteristicscalculated in this work.The work is ended with some concluding remarks.Note: through this work, the phasor amplitudes are RMSvalues, hence there is no 1/2 in the expressions for power.Also, it is worthwhile to mention that the results of this workdepend on physical sizes relative to the wavelength, and henceare valid for all frequencies satisfying the condition of smallelectric cross section.II. D ERIVATION OF THE VOLTAGE AND DIFFERENTIALCURRENT ALONG THE
TLWe start with the voltages, analysing the system describedin Figure 4.As explained in the introduction, to find the voltages on amatched TL, i.e. for Z L = Z R = Z in Figure 2, one needsonly the cross elements between the group of ports 1... M andthe group θ , ϕ of the S matrix.However, to generalise the solution for the voltage along aTL loaded by and Z L and Z R , one needs additional elementsof the S matrix, we therefore calculate here all the elements ofthe matrix for the system defined in Figure 4, schematicallyshown in Figure 5. This matrix satisfies V − = SV + , (14) where S is an ( M +2) × ( M +2) matrix and V + , V − are col-umn vectors of incoming and outgoing voltages, respectively.We start with the submatrix with indices 1 to M . Feedinga middle port < n < M , located at z n = − L + ( n − z, (15)where ∆ z = 2 LM − (16)with a forward (entering) voltage V + n and terminating the otherTL ports ≤ i ≤ M by the impedances defined for those ports(i.e. Z for ports 1 and M , and Z H for the middle ports) resultsin a forward wave from z = z n to z = L and a backward wavefrom z = z n to z = − L , because the waves encounter at theintermediate ports a very high impedance Z H → ∞ .The impedance seen at port n is Z / , so the reflectioncoefficient is S n,n = Z / − Z H Z / Z H ≃ − Z Z H , therefore the portvoltage is V n = V + n (1 + S n,n ) = V + n Z /Z H . The outgoingvoltages at ports i = n are V − i = n = V n e − jk ∆ z | i − n | = V + n Z Z H e − jk ∆ z | i − n | . (17)and for i = nV − n = V + n S n,n = V + n (cid:18) − Z Z H (cid:19) (18)This results in the following (partial) n column of the Smatrix S ≤ i ≤ M , 1] cos ϕ (29) S ϕ, HFSS SIMULATIONS We describe in this Section the HFSS simulations done forthe scattering problem defined in Figure 1. The results of thissimulations are compared in the next section with the analyticresults. We used a non twin lead cross section (used also in[1]), shown in Figure 8. For this cross section one knowsanalytically the distance d in the twin lead representation(shown as red points in Figure 8) by image theory, yielding d = p s − (2 a ) = 2 . cm , (57)and also the characteristic impedance Z = η π ln (cid:18) d + s a (cid:19) = 105 . (58)while for other cross sections those can be determined byan electrostatic ANSYS 2D Maxwell simulation, as shown inAppendix B of [1].The electric field of a plane wave is by default E = 1 V/min the HFSS simulation, so to normalise the results for E d = 1 V we divide the measured results by the value of d in Eq. (57).For convenience, we shall use a fixed TL length of l ≡ L = 125 cm, and test for different frequencies. We measurethe voltage and current along the TL from z = − . cm to Fig. 8. Cross section of two parallel cylinders: the distance between thecentres is s = 3 . cm, and the diameters are a = 2 . cm. The red pointsshow the current images which define the twin lead representation, and thedistance between them d = 2 . cm is calculated in Eq. (57). z = 61 . cm at intervals of 6.125cm, in total at 21 points. Atthe TL terminations z = − L and z = L , we use inactivelumped ports defined for the impedance we need at thoseterminations.Two-conductors transmission lines excited only at termina-tions, develop the TEM mode, so that both E z and H z are 0.In such case one can measure the voltage by R E · dl fromthe “+” to the “-” conductor and the current using H H · dl around the “+” conductor, both on any integration path.In the case analyzed here, the TL is excited by an externalplane wave, therefore, depending on the incidence of this wave E z and/or H z are not necessarly 0, we therefore need morecareful definitions for the voltage and current measurements,as described in the following subsections. A. Voltage measurement The voltage measured by the integral R E · dl in the crosssection depends on the chosen integration path if H z = 0 , asshown in Figure 9. To define the correct path we look at thedefinitions of the parallel ports in Figure 4. Those have beendefined on the x − z plane, so that only x directed currents flowthrough the port, and this fact has been used in the calculationof the S matrix in Section II, see Figure 7.Therefore, to be consistent with the parallel ports definition,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 voltagemeasurements in Section IV. B. Current measurement For the case E z = 0 , the current can be measured by H H · dl around the “positive” conductor, on any integration path,like path C in Figure 10. In the opposite case consisting inan incident plane wave for which E z = 0 (as shown in thefigure), not only the integration path has to be tight around theconductor, but also, due to common mode current, one has to Fig. 9. Cross section voltage measurement on two possible pathes A and B . In case H z = 0 , all pathes lead to the same result, namely R A E · dl = R B E · dl . However, if H z = 0 , as for the incident plane wave shown here, theresult of the path integral depends on the paths used, and the correct voltagemeasurement is R A E · dl , i.e. along the x axis consistent with the parallelports definition in Figure 4.Fig. 10. Cross section differential current measurement for the case H z = 0 requires integration on tight loops around each conductor from which oneobtains the currents I A and I B (Eq. (59)). The differential current is calculatedin Eq. (60). measure around both conductors, on paths A and B , as shownin Figure 10, to obtain the currents I A = I A H · dl ; I B = I B H · dl . (59)As one notes from the derivation in Section C, the S matrix hasbeen derived from differential currents, so the analytic results(51) and (52) represent differential currents. To compare theHFSS simulation results to the analytic results we calculatethe simulation differential current: I = ( I A − I B ) / (60)If H z = 0 , I B trivially reduces to − I A , so that the differentialcurrent is the current on the “positive” conductor, which maybe calculated by integrating on any path around it, like paths A or C in Figure 10.IV. V ALIDATION OF THE ANALYTIC RESULTS A. Comparisons with HFSS full wave solution1) Matched transmission line: We validate in this sectionthe analytic results for matched TL in Eqs (37) and (53) by comparison with full wave solution of ANSYS HFSSsimulation results, described in the previous section.We analyse three examples, each from a main incidencedirection, by plane waves traveling along the x , y or z axes.In all examples the incident electric field intensity satisfies Ed = η Hd = 1 V. The half lenght of the TL is L = 0 . m(or the length L = 1 . m).In the first example we examine a plane wave traveling from θ = π , along the z axis, colinear with the TL, having the phase e − jkz , as shown in Figure 11. For θ = π , Eqs. (37) and (53) Fig. 11. Matched TL illuminated by a − b x polarised plane wave from θ = π . (normalized by Z ) reduce to V ( z ) = V + je − jkL sin[ k ( L − z )] cos( ϕ − α ) (61) Z I ( z ) = − V + je − jkL sin[ k ( L − z )] cos( ϕ − α ) , (62)Clearly, at θ = π , the angle ϕ is ill defined and so is thepolarisation angle α (Figure 3), but the difference ϕ − α ismeaningful. The maximum voltage and current occur at α = ϕ or π − ϕ , representing two opposite polarisations. Choosing α = ϕ the polarisation in Figure 3 becomes b θ cos ϕ + b ϕ sin ϕ ,which equals − b x as shown in Figure 11 and using V + = 1 VEqs. (61) and (62) become V ( z ) = je − jkL sin[ k ( L − z )] (63) Z I ( z ) = − je − jkL sin[ k ( L − z )] , (64)which behave oscillatory according to the frequency, with 0value at right termination at z = L . We remark that for thisplane wave E z = H z = 0 therefore the voltage and currentcan be measured on any paths (see Figures 9 and 10).Figures 12-14 show the voltage and normalized current forthe − b x polarised plane wave from θ = π in Figure 11 forfrequencies 30, 60 and 120MHz, or L/λ = 1 / , 1/8 and 1/4,respectively. The voltage at the left termination at z = − L is V ( − L ) = je − jkL sin(2 kL ) , resulting in . 27 + j . , (1 + j ) / √ and 0 for the cases L/λ = 1 / , 1/8 and1/4, respectively. The normalized currents Z I ( − L ) gets theminus of the above values, satisfying the termination condition V ( − L ) = − Z I ( − L ) .In the next example we use a plane wave hitting from θ = ϕ = π/ with phase e jky as shown in Figure 15. For thisdirection Eqs. (37) and (53) (normalized by Z ) reduce to V ( z ) = V + sin α [1 − e − jkL cos( kz )] (65) Z I ( z ) = jV + sin α e − jkL sin( kz ) , (66)The maximum is for α = ± π/ , so using α = π/ , resultsaccording to Figure 3 in a b ϕ polarisation which is − b x at θ = ϕ = π/ , as shown in Figure 15. We note that this is the -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 R ea l ( V ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 I m ag ( V ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 Z I m ag ( I ) z/ λ Fig. 12. Real and imaginary parts of the voltage V ( z ) and normalized current Z I ( z ) for the plane wave incidence shown in Figure 11, at frequency 30MHzor L/λ = 1 / . The continuous line is the analytic solution in Eqs. (63) and(64) and the stars are the ANSYS simulation results. -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z I m ag ( I ) z/ λ Fig. 13. Same as Figure 12, for frequency 60MHz or L/λ = 1 / . -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 Z I m ag ( I ) z/ λ Fig. 14. Same as Figure 12, for frequency 120MHz or L/λ = 1 / . same plane wave shown in Figure 9, so that here the path on Fig. 15. Matched TL illuminated by a − b x polarised plane wave from ( θ = π/ , ϕ = π/ (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 + = 1 V and α = π/ in Eqs. (65) and (66) resultin V ( z ) = 1 − e − jkL cos( kz ) (67) Z I ( z ) = je − jkL sin( kz ) , (68)Figures 16-18 show the voltage and normalized current forthe − b x polarised plane wave from θ = π/ and ϕ = π/ inFigure 11 for frequencies 30, 60 and 120MHz, or L/λ = 1 / ,1/8 and 1/4, respectively. -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 R ea l ( V ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 I m ag ( V ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 Z I m ag ( I ) z/ λ Fig. 16. Real and imaginary parts of the voltage V ( z ) and normalized current Z I ( z ) for the plane wave incidence shown in Figure 15, at frequency 30MHzor L/λ = 1 / . The continuous line is the analytic solution in Eqs. (67) and(68) and the stars are the ANSYS simulation results. As expected, the voltage is an even function of z and thecurrent an odd function of z . Specifically, the voltages at theterminations at z = − L and L are . 146 + j . , . j . and 1 for the cases L/λ = 1 / , 1/8 and 1/4, respectively. Thenormalized currents have the same values at z = L , satisfyingthe termination condition V ( L ) = Z I ( L ) , and minus theabove values at z = − L , satisfying V ( − L ) = − Z I ( − L ) .The voltage in the middle of the TL at z = 0 are0.076+j0.383, − / √ j/ √ and 1+j for the cases L/λ = 1 / , 1/8 and 1/4, respectively, and the current at z = 0 is of course 0.In the next example we use a plane wave incident from θ = π/ and ϕ = 0 , with phase e jkx as shown in Figure 19. -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z I m ag ( I ) z/ λ Fig. 17. Same as Figure 16, for frequency 60MHz or L/λ = 1 / . -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 Z I m ag ( I ) z/ λ Fig. 18. Same as Figure 16, for frequency 120MHz or L/λ = 1 / . For this direction Eqs. (37) and (53) (normalized by Z ) reduceto V ( z ) = − jV + cos α e − jkL sin( kz ) (69) Z I ( z ) = − V + cos α [1 − e − jkL cos( kz )] . (70)Here the maximum is obtained for α = 0 or π . Using α = π defines according to Figure 3 a − b θ polarisation which equals b z at θ = π/ and ϕ = 0 , as shown in Figure 19. This isthe same plane wave mentioned in Figure 10, which requirescurrent measurement by integrating on the tight trajectories A and B in Figure 10, using Eq. (60) to determine the differentialcurrent.Using V + = 1 V and α = π in Eqs. (65) and (66) result in V ( z ) = je − jkL sin( kz ) (71) Z I ( z ) = 1 − e − jkL cos( kz ) , (72)Figures 20-22 show the voltage and normalized current forthe b z polarised plane wave from θ = π/ and ϕ = 0 inFigure 19 for frequencies 30, 60 and 120MHz, or L/λ = 1 / ,1/8 and 1/4, respectively. Fig. 19. Matched TL illuminated by a b z polarised plane wave from ( θ = π/ , ϕ = 0) (as also shown in Figure 10). -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 R ea l ( V ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 I m ag ( V ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5-0.06 -0.04 -0.02 0 0.02 0.04 0.06 Z I m ag ( I ) z/ λ Fig. 20. Real and imaginary parts of the voltage V ( z ) and normalized current Z I ( z ) for the plane wave incidence shown in Figure 19, at frequency 30MHzor L/λ = 1 / . The continuous line is the analytic solution in Eqs. (71) and(72) and the stars are the ANSYS simulation results. As expected, the voltage is an odd function of z and thecurrent an even function of z . The voltage developed forthe incident field in Figure 19 (Eq. 71) has the value ofthe normalized current developed for the incident field inFigure 15 (Eq. 68) and viceversa. -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z I m ag ( I ) z/ λ Fig. 21. Same as Figure 20, for frequency 60MHz or L/λ = 1 / . -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.2 -0.1 0 0.1 0.2 Z I m ag ( I ) z/ λ Fig. 22. Same as Figure 20, for frequency 120MHz or L/λ = 1 / . 2) Non matched transmission line: We compare here sev-eral unmatched cases for the − b x polarised plane wave from θ = π/ , and ϕ = π/ , shown in Figure 15 for frequency60 MHz. The matched solutions are given in Eqs. (67) and(68), and for this frequency ( kL = π/ ) are: V ( z ) = 1 − e − jπ/ cos( kz ) (73) Z I ( z ) = je − jπ/ sin( 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 ) = Γ L e − jkz + Γ R e jkz − j Γ L Γ R cos( kz ) √ R Γ L ) (75) Z ∆ I ( z ) = Γ L e − jkz − Γ R e jkz − L Γ R sin( kz ) √ R Γ L ) (76)Figures 23-25 show the voltage and normalized current forthe cases: Z L = Z / and Z R = 2 Z , Z L = Z R = Z / and Z L = Z R = 2 Z , respectively. B. Compatibility with previous works We show in this subsection that our analytic results arecompatible 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 ∆ Vdz + jk ( Z ∆ I ) = 0 ; Z d ∆ Idz + jk ∆ V = 0 , (77)which is clear, because the induced sources depend only onthe incident field and not on the loads.Therefore, it is enough to show that Eqs. (4) and (5) aresatisfied 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 ≃ , y = 0 and − L ≤ z ≤ L , which comesout e jkz cos θ , (78) -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z I m ag ( I ) z/ λ Fig. 23. Real and imaginary parts of the voltage V ( z ) and normalized current Z I ( z ) for the plane wave incidence shown in Figure 15, at frequency 60MHz(or L/λ = 1 / ), but for loads Z L = Z / and Z R = 2 Z , or Γ L = − / and Γ R = 1 / . The continuous line is the analytic solution, i.e. Eq. (73) pluscorrection (75) and Eq. (74) plus correction (76) and the stars are the ANSYSsimulation results. -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z I m ag ( I ) z/ λ Fig. 24. Same as Figure 23 for Z L = Z R = Z / or Γ L = Γ R = − / . -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 R ea l ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 I m ag ( V ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z R ea l ( I ) z/ λ -1.5-1-0.500.511.5 -0.1 -0.05 0 0.05 0.1 Z I m ag ( I ) z/ λ Fig. 25. Same as Figure 23 for Z L = Z R = 2 Z or Γ L = Γ R = 1 / . so the incident E x and H y fields along the TL are E x ( z ) = E x e jkz cos θ ; H y ( z ) = H y e jkz cos θ , (79)where the values E x and H y are understood as the values atthe origin. Using the unit vectors identity b x = b r sin θ cos ϕ + b θ cos θ cos ϕ − b ϕ sin ϕ , given the incident E field has only θ and ϕ components, the x component of the electric field (atorigin) is E x = E θ cos θ cos ϕ − E ϕ sin ϕ = E (cos α cos θ cos ϕ − sin α sin ϕ ) (80)where for the second expression we used Eq. (3). UsingEqs. (11), (79) and (80), Z I s in Eq. (7) comes out Z I s = jkV + (sin α sin ϕ − cos α cos θ cos ϕ ) e jkz cos θ , (81)To derive the H y component of the incident field we note thatthe H components of the incident plane wave are: H θ = E ϕ η ; H ϕ = − E θ η (82)Using the unit vectors identity b y = b r sin θ sin ϕ + b θ cos θ sin ϕ + b ϕ cos ϕ , given the incident H field has only θ and ϕ components, the y component of H (at origin) is H y = H θ cos θ sin ϕ + H ϕ cos ϕ =1 η [ E ϕ cos θ sin ϕ − E θ cos ϕ ] , (83)where in the second expression we used Eq. (82). UsingEqs. (3) and (11), this can be rewritten as η H y = V + [sin α cos θ sin ϕ − cos α cos ϕ ] , (84)Using Eqs. (79), V s in Eq. (6) comes out V s = jkV + [sin α cos θ sin ϕ − cos α cos ϕ ] e jkz cos θ , (85)Here it is left to show that Eqs. (81) and (85) equal to theLHS of Eqs. (5) and (4) respectively, applied on the matchedsolution V ( z ) , I ( z ) given in Eqs. (37), (53). We start with twointermediate results for the functions f ( z ) and f ( z ) definedin Eq. (22): df /dz + jkf = k cos ( θ/ e jkz cos θ df /dz − jkf = − k sin ( θ/ e jkz cos θ , (86)and obtain dVdz + jk ( Z I ) = − jkV + [cos ( θ/ 2) cos( ϕ + α )+sin ( θ/ 2) cos( ϕ − α )] e jkz cos θ (87) Z dIdz + jkV = jkV + [sin ( θ/ 2) cos( ϕ − α ) − cos ( θ/ 2) cos( ϕ + α )] e jkz cos θ (88)Using simple trigonometric identities one finds that results (87)and (88) are identical to Eqs. (85) and (81), respectively. V. T HE POWER TRANSFERRED TO THE LOADS We calculate here the power transferred to the loads for thegeneral case of loads Z L and Z R as shown in Figures 1 and 2.We need first the voltage and current for the matched at bothteminations case at the terminations ( z = ± L ). From Eqs. (37)and (53) we obtain V ( − L ) = jV + e − jkL sin[2 kL sin ( θ/ ϕ − α ) (89) V ( L ) = − jV + e − jkL sin[2 kL cos ( θ/ ϕ + α ) (90)and Z I ( − L ) = − V ( − L ) ; Z I ( L ) = V ( L ) (91)satisfy the boundary conditions. From Eqs. (89)-(91), thepowers on the “left” and “right” loads for the matched at bothteminations case, are P ( − L ) ≡ − V ( − L ) I ∗ ( − L ) = | V + | Z sin [2 kL sin ( θ/ ( ϕ − α ) (92) P ( L ) ≡ V ( L ) I ∗ ( L ) = | V + | Z sin [2 kL cos ( θ/ ( ϕ + α ) (93)We now generalise the powers for a TL with any loads.We set z = ± L in Eqs. (49), (54), or simpler, use directlyEqs. (47), (48), (C.28) and (C.29), obtaining V NM ( − L ) = (1 + Γ L ) Γ R e − j kL V ( L ) + V ( − L )1 − Γ L Γ R e − j kL (94) V NM ( L ) = (1 + Γ R ) Γ L e − j kL V ( − L ) + V ( L )1 − Γ L Γ R e − j kL (95)and Z L I NM ( − L ) = − V NM ( − L ) ; Z R I NM ( L ) = V NM ( L ) (96)satisfy the boundary conditions.Now we express the power on the “left” load Z L by P NM ( − L ) = Re {− V NM ( − L ) I ∗ NM ( − L ) } and on the “right”load Z R by P NM ( L ) = Re { V NM ( L ) I ∗ NM ( L ) } . Using Eqs. (94)-(96) one obtains P NM ( − L ) = 1 − | Γ L | | − Γ L Γ R e − j kL | [ P ( − L ) + | Γ R | P ( L ) − P mix Re { Γ R e − j kL } ] (97) P NM ( L ) = 1 − | Γ R | | − Γ L Γ R e − j kL | [ P ( L ) + | Γ L | P ( − L ) − P mix Re { Γ L e − j kL } ] . (98)The generalised results for the powers are expressed interms of the matched-TL powers P ( − L ) and P ( L ) (givenin Eqs. (92)-(93)) plus an additional “mixed” term P mix givenby P mix ≡ V ( L ) I ∗ ( − L ) ≡ − V ( − L ) I ∗ ( L ) = | V + | Z sin[2 kL sin ( θ/ kL cos ( θ/ 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 proportionalto sin [2 kL sin ( θ/ , according to the radiation pattern ofa TL carrying a forward wave only and maximised for thepolarisation radiated in this case: b p + ( θ, ϕ ) = b θ cos ϕ + b ϕ sin ϕ ⇒ α = ϕ, (100)see Section 2.1 in [1]. Also, we see P ( − L ) = 0 for apolarisation orthogonal to b p + . This suggests that P ( − L ) , i.e.the power into the left load for the matched on both sides TLis closely related to transmitting properties of a source at theleft termination, issuing a forward wave.Similarly, P ( L ) is proportional to sin [2 kL cos ( θ/ , ac-cording to the radiation pattern of a TL carrying a backwardwave only and maximised for the polarisation radiated in thiscase: b p − ( θ, ϕ ) = b θ cos ϕ − b ϕ sin ϕ ⇒ α = − ϕ, (101)see Section 2.2 in [1], and P ( L ) = 0 for a polarisationorthogonal to b p − . This suggests that P ( L ) , i.e. the powerinto the right load for the matched on both sides TL is closelyrelated to transmitting properties of a source at the righttermination, issuing a backward wave.The general non matched case given in Eqs. (97)-(98)is affected by both radiation patterns and polarisations, andtheir combination found in the mixed term in Eq. (99). Weremark that for a polarisation orthogonal to either b p + or b p − , P mix = 0 and either P ( − L ) or P ( L ) are 0, in which case thereceived powers are either according to sin [2 kL cos ( θ/ or sin [2 kL sin ( θ/ .The formal connection between the transmitting and receiv-ing properties of TL is analysed the next section.VI. C ONSISTENCY BETWEEN TRANSMITTING - RECEIVINGRADIATION PROPERTIES To analyse the connection between the results of receivingelectromagnetic radiation shown in this work and the resultsof transmitting (radiating) electromagnetic radiation in [1], wehave to summarise below some results from [1] and rewritethem in a convenient form.The far electric field radiated by a TL carrying a forwardwave only as in the upper panel of Figure 26, is given in Fig. 26. Upper panel: radiating TL fed by a current source I at z = − L ,matched at z = L . Lower panel: the same TL in receive mode, with avoltmeter at z = − L measuring the open circuit voltage V oc . Section 2.1 of [1] in terms of the value of I + in the middle of the TL at z = 0 . We rewrite it here in terms of the currentsource I in the standard form for a radiating antenna fed bya current E + = jkη G ( r ) I l + eff , (102)where l + eff is the effective antenna length for the radiation of aTL carrying a forward wave only, is given by l + eff = 2 jde − jkL sin[2 kL sin ( θ/ b p + (103)where the polarisation vector b p + is defined in Eq. (100).The radiation resistance for this case (given in Eq.(36) of[1]) is r rad = η π ( kd ) [1 − sinc(4 kL )] , (104)which does not need a “+” superscript, being valid also for abackward wave only. The radiation pattern (directivity) for aforward wave only is D + ( θ ) = η k πr rad | l + eff | = 2 sin [2 kL sin ( θ/ − sinc(4 kL )] , (105)as given in Section 2.1, Eq. (13) in [1]. When dealing withradiation from TL, which is usually small relative to the powercarried by the TL, we may use the term “radiation losses” forthe power lost to radiation. So in the upper panel of Figure 26,the radiated power is | I | r rad , and the power transmitted bythe TL is | I | Z , and clearly r rad ≪ Z . But in the contextof the radiating properties we have to consider the radiationefficiency, defined as the radiated power, divided by the totalpower into the antenna. In our case this is e r = r rad r rad + Z ≃ r rad Z . (106)which is small, meaning that the (matched) TL is not anefficient antenna. Eq. (106) is valid for both forward only orbackward only wave, hence does not need a superscript. Theantenna gain is the directivity multiplied by the efficiency: G + ( θ ) = D + ( θ ) e r = ( kd ) η sin [2 kL sin ( θ/ πZ , (107)Now, in receive mode, calculating the scalar product be-tween the incident plane wave (Eq. (2)) and the effective lengthin Eq. (103) yields the open circuit voltage measured in thelower panel of Figure 26 V oc = E · l + eff = 2 jE de − jkL sin[2 kL sin ( θ/ ϕ − α ) , (108)which is compared with the open end voltage obtained fromthe formalism developed in this work. So using Eq. (94) with Γ R = 0 and Γ L = 1 (open) yield exactly the result (108)where we identify V + = E d according to Eq. (11). Also, theabsolute value of V oc is 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 Z L (left). Eq. (97), reduces for Γ R = 0 to P NM ( − L ) = (1 − | Γ L | ) P ( − L ) , so maximum power isobtained for Γ L = 0 ( Z L = Z ), yielding Eq. (92). Using thematched polarisation ( α = ϕ ) in (92) results in the maximumreceived power: P rec = | V + | Z sin [2 kL sin ( θ/ . (109) Dividing this by the Poynting vector S = E /η , results in theeffective receiving cross section area A + (for receiving intothe left termination - the equivalent of transmitting a forwardwave) A + = P rec S = d η Z sin [2 kL sin ( θ/ . (110)which equals exactly to λ π G + .A similar analysis for the backward wave, using the currentsource at the z = L termination (pointing upward) and thematched load at z = − L in Figure 26 yields the field radiatedby a backward wave current: E − = jkη G ( r ) I l − eff , (111)where l − eff is the effective antenna length for the radiation of aTL carrying a backward wave only, is given by l − eff = − jde − jkL sin[2 kL cos ( θ/ b p − , (112)and the polarisation vector b p − is defined in Eq. (101). For thiscase, the gain is the directivity function D − given in Eq. (17)of [1] multiplied by the efficiency in Eq. (106), yields G − ( θ ) = ( kd ) η sin [2 kL cos ( θ/ πZ , (113)The effective receiving cross section area A − comes out A − = d η Z cos [2 kL cos ( θ/ , (114)satisfying the relation A − = λ π G − .VII. C ONCLUSIONS We derived in this work the voltage and differential currentdeveloped on an ideal two-conductors TEM transmission line(TL) of any small electric cross section, connected to passiveloads and hit by a monochromatic plane wave, as shown inFigure 1.For this derivation we used our knowledge on the radiationproperties of TL [1] to build S matrices which describe theradiating system and used the reciprocity to derive the currentand voltage induced on the TL. This methodology allowedus to derive first the voltage and current on a matched atboth teminations TL, yielding the relatively simple expressionsgiven in Eqs. (37) and (53). The generalisation to any loadsis then obtained, using the S matrix.We validated our analytic results in Section IV-A for boththe matched on two sides TL case and the non matchedcase for different plane wave incidences. We also showed inSection IV-B that our analytic results are compatible with themethodology used in previous works [2]–[6]. The simplicityof this proof, requiring the application of Eqs. (4) and (5) only on 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 andshowed the connection between those powers and the radiatingproperties of the TL. For the matched at both teminations TL,the power on the “left” load has the same spatial depencenceon the incident plane wave direction as the radiation pattern of a forward wave, and is maximised for the polarisation of aradiating forward wave b p + (see Eq. (100)). Similarly, for thematched at both teminations TL, the power on the “right” loadhas the same spatial depencence on the incident plane wavedirection as the radiation pattern of a backward wave, and ismaximised for the polarisation b p − (see Eq. (101)). The generalnon matched case is affected by both radiation patterns andpolarisations, and their combination found in a mixed term.In Section VI we showed that the formal relations betweenreceiving cross section area and antenna gain are satisfied forthe TL. A PPENDIX AF AR RADIATED E FIELD AND ITS NORMALIZATION To derive the S matrices we need the contribution of currentsshown in Figure A.1 to the far field. Based on [1], a currentelement Id in the x direction (where I is the current and d the separation distance in the twin lead representation), at z = z w (w=wire) contributes the following E field in θ and ϕ polarisations (see left panel of Figure A.1). Fig. A.1. Currents contributing to radiation. The left panel shows a x directedcurrent I at location z w on the TL (which may represent a parallel port ora 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 h carrying a forward and/orbackward current wave, I ± evaluated at the middle location z m , contributingto the far E field according to Eq. (A.3). E θ ( x ) = − jkG ( r ) η dIe jkz w cos θ cos θ cos ϕ (A.1) E ϕ ( x ) = jkG ( r ) η dIe jkz w cos θ sin ϕ, (A.2)where the subscript ( x ) shows that this is a x directed currentcontribution.A TL section of lenght h around z = z m (m=middle)carrying a forward and/or backward current contributes the fol-lowing far θ polarised E fields (see right panel of Figure A.1). E θ ( z ) = − kG ( r ) η dI ± ( z m ) e jkz m cos θ (1 ± cos θ ) cos ϕ sin[ kh (1 ∓ cos θ ) / (A.3)where the upper and lower signs are for forward and backwardwaves respectively, and the values used for the forward orbackward currents are in the middle of the line at z m , bothcurrents defined in the + z direction. The subscript ( z ) showsthat this is a z directed current contribution.In the final result, E θ = E θ ( x ) + E θ ( z ) and E ϕ = E ϕ ( x ) .To work out the S matrices (Section II and Appendix C), oneneeds to scale the radiated E θ and E ϕ fields to outgoing voltages V − θ and V − ϕ at the ports θ and ϕ (Section II),or to outgoing currents I − θ and I − ϕ (Appendix C). Thoseports have been defined for the impedance Z , which is thecharateristic impedance of the analysed TL. To make thoseports matched, we use the antenna model shown in Figure A.2.The model assumes a real antenna impedance Z A matched Fig. A.2. The antenna model assumes a real antenna impedance Z A matchedby an ideal transformer to the θ or ϕ port of impedance Z . via an ideal transformer to the θ or ϕ port of impedance Z .Using the effective length of the antenna l eff (for the givenincidence direction), one expresses the open circuit voltage onthe antenna [9] V oc = E l eff , (A.4)where E is E θ or E ϕ . The θ or ϕ port being matched, theantenna “sees” a matched load, so that the actual voltage onthe antenna terminals is V oc / , and the outgoing voltage at theport is V − = 12 E l eff p Z /Z A . (A.5)Now to activate the antenna in transmit mode, we feed the θ or ϕ port by V + , obtaining the current feeding the antenna I A = V A /Z A = V + / p Z A Z (A.6)Using the well known formula for the far E field radiated bya dipole [9] E = jkη G ( r ) I A l eff (A.7)where l eff is the effective dipole length discussed before, intothe same direction [9]. The field radiated from the antenna(s)has been named E to distinguish it from the previouslydiscussed field travelling toward the antenna(s). Expressingit as function of V + , we obtain E = jkη G ( r ) l eff V + / p Z A Z (A.8)This field E radiated from the far antenna(s), is the fieldincident on the TL (or one of its components), mentioned inFigure 3. We want to normalise it so that E d = V + (seeEq. (11)), considering the field incident on the TL as a planewave. Imposing condition (11) on Eq. (A.8), results in l eff = √ Z A Z jkη G ( r ) d (A.9)Given the far antennas are only tools to build the S matricesand from them infer the voltage and current along the TL, the“actual” value of l eff is of no interest, so to express the fields radiated by the TL E θ and E ϕ as voltages, we set l eff from(A.9) into Eq. (A.5), obtaining V − θ or ϕ = Z E θ or ϕ jkη G ( r ) d (A.10)For the case we need to express the fields radiated by theTL E θ and E ϕ as outgoing currents (Section C), we use I − = − V − /Z , obtaining: I − θ or ϕ = − E θ or ϕ jkη G ( r ) d (A.11)A PPENDIX BG ENERALISED SCATTERING MATRIX In this appendix we define the generalized scattering ma-trix and explain its reciprocity properties. The generalizedscattering matrix [7] for an arbitrary network (as shown inFigure B.1) is defined by the following matrix equation Fig. B.1. An N ports arbitrary network interfaced to transmission lines ofcharacteristic impedances Z , Z , ... Z N . V − = SV + , (B.1)where V ± are column vectors for the incoming and outgoingvoltage waves at the ports of the network, where each porthas 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)where Z is a diagonal matrix of the characteristic impedancesin Figure B.1: Z ij = Z i δ ij , (B.3)so one can easily apply functions on them, as follows.Normalising the voltages and currents at each port accordingto a ± ≡ √ Z − V ± ; b ± ≡ √ ZI ± , (B.4)and setting into Eq. (B.1), result in a − = √ Z − S √ Za + , (B.5) which define the (ordinary) scattering matrix, for which weuse a lower case “s” s ≡ √ Z − S √ Z . (B.6)It is known [7]–[12] that the reciprocity condition for theordinary scattering matrix is s ij = s ji . (B.7)Using this in Eq. (B.6) results in the reciprocity condition forthe generalized scattering matrix S ij Z j = S ji Z i (B.8)Scattering matrices are usually defined for voltage waves, butin this work we need to define a generalized scattering matrixfor current waves as follows I − = S I I + , (B.9)and we name it S I to distinguish it from the regular scatteringmatrix in (B.1). Using Eq. (B.2), one obtains V − = − ZS I Z − V + . (B.10)Comparing it with (B.1) we get the relation S = − ZS I Z − . (B.11)Setting it in (B.6) results in s = −√ ZS √ Z − , (B.12)and using (B.7) we obtain S I ij Z i = S I ji Z j (B.13)A PPENDIX CD ERIVATION OF THE CURRENT ALONG THE TLWe define a system of M + 2 ports, the first M ports areserial ports on the TL (shown in Figure 6) and 2 additionalports are far antennas at polarisations b θ and b ϕ .Our purpose is to determine the TL current for an incidenceof a plane wave, we therefore calculate the matrix S definedby I − = SI + , (C.1)although this is not the usual definition of the scattering matrix(see Appendix B). For simplicity, we still call it S , but take intoconsideration that any reflection coefficient has an oppositesign.Feeding port < n < M , located at z n with a forward(entering) current I + n and terminating the other TL ports ≤ i ≤ M by the impedances defined for those ports (i.e. Z for ports 1 and M , and Z T for the middle ports) results in aforward wave from z = z n to z = L and a backward wavefrom z = z n to z = − L , because the waves encounter at theintermediate ports a very small serial impedance Z T → .The reflection coefficient at port n is S n,n = Z T − Z Z T +2 Z ≃− Z T Z , so the port current I n = I + n (1 + S nn ) ≃ I + n Z T Z .The current waves at ports i = n give rise to outgoing currentsat the ports given by: I − i = n = I n e − jk ∆ z | i − n | = I + n Z T Z e − jk ∆ z | i − n | . (C.2) and for i = n I − n = I + n S n,n = I + n (cid:18) − Z T Z (cid:19) (C.3)This results in the following (partial) n column of the Smatrix S ≤ i ≤ M , 2) + f sin ( θ/ ϕ (C.8)Now we add the x and z directed currents contributions to I − θ = I − θ ( x ) + I − θ ( z ) , using − cos( θ ) = 2 sin ( θ/ and θ ) = 2 cos ( θ/ , after some algebra we obtain: I − θ = − jI + n ( Z T /Z )[ f + f ] cos ϕ (C.9) I − ϕ has only the contribution of the x directed currents, givenin Eq (C.7), so that I − ϕ = I − ϕ ( x ) (C.10)Results (C.9) and (C.10) define the S θ,n and S ϕ,n matrixelements respectively, for the columns < n < M : S θ, Microwave Engineering , Wiley India Pvt., 2009[8] Collin, Robert E. Antennas and radiowave propagation , McGraw-Hill,1985.[9] Orfanidis S.J., Electromagnetic Waves and Antennas , ISBN: 0130938556,(Rutgers University, 2002)[10] S. Ramo, J. R. Whinnery and T. Van Duzer, Fields and Waves inCommunication Electronics , 3rd edition, Wiley 1994[11] E. C. Jordan and K. G. Balmain, Electromagnetic Waves and RadiatingSystems , 2nd edition, Prentice Hall 1968[12] C. A. Balanis,