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4From Eqs.(7c) and (7d) it follows that U 2 a /U 1 b = I 2 a /I 1band Eqs.(5) and (7) yieldAAabI=Ia2b1I=Ia1b2( Z( ZThe attenuation will be independent of the choice ofsource port and load port if A a /A b = 1. In all other casesA a /A b ≠ 1 and relative impedance values will determinethe choice of the source and load port of the filter [19].The condition A a /A b = U a 2 /U b 1 = I a 2 /I b 1 = 1 is met ifI b 2 =I a 1 and/or Z g = Z L . It is well known that the conditionI b 2 = I a 1 , that simultaneously makes I a 2 = I b 1 , is met in thecase of a purely symmetrical filter; the reciprocitytheorem was not needed to demonstrate this. However,the condition Z g = Z L resulting in A a /A b = 1 is sometimesoverlooked as noted in Section 6.3. Moreover, it is thiscondition that explains why the engineer will measureno difference in the attenuation if source and load portare interchanged when using a 50Ω measuring system.Of course, another path could also have beenfollowed to arrive at the two conditions representingA a = A b . The Kirchhoff network can be characterized bya T-network and U a b2 /U 1 can be expressed in thenetwork impedances Z 1 , Z 2 and Z 3 (see Fig.1). Afterstraight forward calculations it then follows thatUUa2b1Z2Z=Z Z1LL(8)(9)where Z 2 = Z 1 Z 2 +Z 2 Z 3 +Z 3 Z 1 +Z 3 Z g +Z g Z L +Z 3 Z L . The conditionZ 1 = Z 2 then represents the purely symmetricalfilter and, again, the condition Z g = Z L represents the50Ω measuring system.3.3. DM/CM conversionThe application of a reciprocity theorem may change anunsuccessful measurement into a successful one. Anexample is the measurement of the conversion of adifferential-mode (DM) voltage into a common-mode(CM) current to characterize the emission properties ofa telephone subscriber line or a power line to which adigital-signal is applied. The DM voltage U DM issupplied by the digital signal, and the resulting CMcurrent I CM is a direct measure of the radiated emissioncapability of such a line. This measure can be expressedin a conversion admittance Y CM = I CM /U DM .To measure this conversion over a certain frequencyrange, e.g. 0.1 MHz − 30 MHz, it seems rather obviousto apply a DM voltage to the line and to measure theresulting CM current. However, the line is also arelatively efficient receiving antenna and ambient fieldswill induce CM currents that may be of the same orderof magnitude as the CM currents being measured, orpossibly even larger. Another problem might be thecorrect measurement of a DM voltage at the higher endof the frequency range. Both problems can becircumvented by applying a CM voltage U CM to the lineand measuring the resulting DM current I DM , i.e. bydetermining the conversion admittance Y DM = I DM /U CM .The reciprocity theorem can now be used to find thecondition leading to Y CM = Y DM .gg+ Z Z+ Z12− Z− ZZggLL+ Z+ Z) −U) −U22ggZ gU gCM-portU CMDM-portI DMa-stateb-stateFig.5 The two states when discussing the DM/CMconversionThe line being characterized can be considered as a twoportnetwork where port 1 is the CM port and port 2 theDM port, see Fig.5. The following choice ofterminations in the a- and b-state is now possible. Instate a, representing the determination of Y DM , port 1 isterminated by the voltage generator and a highimpedance FET probe measures U a 1 = U CM . Port 2 isshort-circuited (U a 2 = U DM = 0) and a current probearound the short circuit measures I a 2 = I DM . In state b,representing the determination of Y CM , port 2 isterminated by the generator and port 1 by the shortcircuit. In this case Eq.(5) reduces to U a 1 I b 1 = U b 2 I a 2 , sothatICMIDMY(10)CM= = = YDMU UDM UCM= 0CM-portI CMCM U DM = 0DM-portU DMA practical way of measuring Y DM is to use Macfarlane’sprobe [20]. As described in [21], the CMvoltage is applied to the CM input of the probe. Thevoltage is measured via a high impedance FET probe atthis input and the current is measured in the short circuitbetween the probe’s two DM terminals. Reference [21]also gives measured Y DM data.3.4. Site attenuationAlthough only voltages and currents have beenconsidered above, it does not mean that the reciprocitytheorem does not apply if electric and magnetic fieldsplay a role in the signal transfer. This should already beclear from the given examples, as field couplings playan important role in each of these examples.Another example is the determination of the siteattenuation in which the signal transfer between twoantennas above a reflecting plane is considered.Fortunately, the signal transfer can be modelled into apassive two-port network [22] containing linearimpedances as already described by Brown and King in1934 [23]. Using a 50Ω measuring system, we canconclude from the discussion in Section 3.2 that it is notimportant which port is connected to the generator andwhich one to the receiver as long as the antennas havefixed positions. However, as noted in Section 6.2, thisconclusion might not always be correct if the so-callednormalized site attenuation is considered4. Electromagnetic fieldsThis section considers the Lorentz reciprocity theoreminterrelating the electromagnetic fields in two states thatcan occur in one and the same domain in space. UsingCarson’s formulation [24] and today’s notation thereciprocity theorem for fields reads:Z gU g
5“If E a , H a are the field vectors due to a periodicdisturbance from a source A 1 located at O 1 and E b , H bare the corresponding field vectors due to adisturbance originating in A 2 from a source located atO 2 , then∫1+2( Ea× Hb) ⋅ds =1+2( E× H) ⋅ds(11)the surface integrals as indicated by the subscripts 1+2being taken over closed surfaces 1 and 2 surroundingthe sources A 1 and A 2 respectively.”Not directly given in [24] is the additional equality alsofollowing from the reciprocity considerations∫D( Ea⋅ Jb(12)where D is the volume containing the sources A 1 and A 2represented by the source vectors J a and J b , respectively.In Eq.(11) the surface of D is indicated by thesubscripts 1+2. Equation (12) will be used in thederivation of the hybrid reciprocity theorem in Section5. As will be noted below, Eqs.(11) and (12) areconnected to a special case of the general reciprocityrelation Eq.(19).So the theorem combines two field states (a and b),comparable to the discussed reciprocity theorem for anN-port Kirchhoff network. The derivation of Eqs. (11)and (12) starts from sets of two Maxwell’s equationsexpressed in the frequency domain for the states a andb:aacurlE= − jω µ Haa a(13a,b)curlH= + jωεE + JandbbcurlE= − jω µ Hbb b (14a,b)curlH= + jωεE + JNow the trick is to remember the existence of the vectorrelation div(X×Y)=Y⋅curlX−X⋅curlY and to writediv(div(EEab× H× Hba) = H) = H(15)(16)An expression for H b ⋅curlE a in Eq.(15) can be found bymultiplying the left and right hand members of Eq.(13a)by H b (be careful with the order of the variables)H(17)In a similar way, expressions can be found for the otherterms on the right hand side of Eqs.(15) and (16) bymultiplying Eqs.(13b), (14a) and (14b) by the propervectors. After substitution of all resulting expressions inEqs.(15) and (16) and subtracting the two finalequations, we find thatdiv(18)This equation, in which the ω-terms no longer appear, issometimes called the local form of the reciprocitytheorem. To find the general form, Eq.(18) has to be∫∫b a)dv = ( E ⋅ J ) dvbaD⋅curlE⋅curlEba⋅ curlE= − jω µabHb− E− Ebab⋅ Ha⋅curlH⋅curlHa b b a b a a b( E × H − E × H ) = E ⋅ J − E ⋅ Jabaintegrated over the volume D containing all sourcesrepresented by J a and J b , yieldinga b b a∫ div( E × H − E × H ) ⋅Da b b a∫ ( E × H − E × H ) ⋅dsSb a a b∫ ( E ⋅ J − E ⋅ J ) ⋅dvDd v =(19)We had to expect an integration as the Maxwell equationsconsider field derivatives whereas an expressionfor the fields is needed. The most left hand member ofEq.(19) has been converted from an integral over avolume D into an integral over the surface S of D byapplying Gauss’s theorem. This action also ‘removes’the div operation. Equation (19) is the general form ofthe Lorentz reciprocity theorem, a form that can even beextended [14], although this extension is not needed inthe context of this paper.A special case is the situation in which the relationbetween the E and the H vector of the field is fixed,such as in the far-field of an antenna. Then the outcomeof the integrals in Eq.(19) is equal to zero, a value givenby Lorentz as he considered the propagation propertiesof light. In the far-field, the field propagates as a planeor quasi-plane wave so that the E and the H vector ofthe field are perpendicular to each other and have aconstant ratio, η=√(µ/ε), the wave impedance (377 Ω inair). In vector notation the latter means that ν×E= ηH,where ν is the unit vector perpendicular to the planeformed by E and H. After application of this relation toE a ×H b and to E b ×H a and after application of the vectorrelation X×(Y×Z)= Y⋅(X⋅Z)−Z(X⋅Y) we find thata b b aE × H = E × H = η(E(20)a b b aso that E × H − E × H = 0 and, consequently,the integrals in Eq.(19) are equal to zero and Eqs.(11)and (12) given at the start of this section automaticallyfollow.5. The hybrid reciprocity theoremThe experimentalist can only humbly lift his hat at themathematical fireworks presented in Section 4 and thenpass to the order of the day. An application is what isneeded to catch his or her interest. As mentioned in theintroduction, a very useful application was alreadygiven in 1929 by Ballantine [10], referring to the workof Wilmotte [11].This application, resulting in thehybrid reciprocity theorem, can also be found in Section4.5 of a more recent textbook [16]. The theorem givesan expression for the voltage U i induced by an incidentfield E i in an antenna or structure acting as an antenna,e.g. an EUT (Equipment Under Test) and its connectedcables. Since the transmission and reception ofelectromagnetic waves is of interest in EMC fieldmeasurements, the two states a and b will now bedenoted by t (of transmission) and r (of reception).In the following, two wire antennas A 1 and A 2 areconsidered (see Fig.6). Each wire antenna consists ofb⋅ E=a)
Page 2 and 3: 2you bear in mind that in their day
Page 6 and 7: 6I 0U iFig.6 The states (a) transmi
Page 8 and 9: 8space antenna factor of one and th
Page 10 and 11: 10Eq.(30), although the magnetic di
Page 12: 12Spherical Shells”, Trans. IEEE

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