Physical meaning of the dipole radiation resistance in lossless and lossy media
11 Physical meaning of the dipole radiation resistancein lossless and lossy media
M. S. Mirmoosa,
Member, IEEE , S. Nordebo, and S. A. Tretyakov,
Fellow, IEEE
Abstract —In this tutorial, we discuss the radiation from aHertzian dipole into uniform isotropic lossy media of infiniteextent. If the medium is lossless, the radiated power propagates toinfinity, and the apparent dissipation is measured by the radiationresistance of the dipole. If the medium is lossy, the power expo-nentially decays, and the physical meaning of radiation resistanceneeds clarification. Here, we present explicit calculations of thepower absorbed in the infinite lossy host space and discuss thelimit of zero losses. We show that the input impedance of dipoleantennas contains a radiation-resistance contribution which doesnot depend on the imaginary part of the refractive index. Thisfact means that the power delivered by dipole antennas tosurrounding space always contains a contribution from far fieldsunless the real part of the refractive index is zero. Based on thisunderstanding, we discuss the fundamental limitations of powercoupling between two antennas and possibilities of removing thelimit imposed by radiation damping.
Index Terms —Absorbed power, Hertzian dipole, lossy media,radiation resistance, radiated power.
I. I
NTRODUCTION
The radiation resistance of a dipole antenna is a simpleclassical concept which is used to model the power radiatedinto surrounding infinite space, as explained in many antennatextbooks, e.g. [1]. However, its physical meaning is not al-ways easy to grasp. Indeed, if the surrounding space is lossless,the energy cannot be dissipated. Yet, if the space around theantenna is infinite, the radiation resistance is nonzero, which istantamount to absorption of power. One can perhaps say thatthe radiated energy is transported all the way to infinity. On theother hand, if the surrounding medium is lossy, the radiatedpower is exponentially decaying, and it is sometime assumedthat the usual definition of the radiation resistance does notapply, because at the infinite distance from the antenna theradiated fields are zero. One can perhaps say that all the poweris dissipated in the antenna vicinity. Within this interpretation,we have a confusing “discontinuity”: If the medium is lossless, all the radiated power is transported to infinity, but if theloss factor is nonzero (even arbitrarily small), no power istransported to infinity.Although there is extensive classical literature on antennasin absorbing media (see the monograph [2] and e.g. Refs. [3]–[7]), we have found only limited and sometimes even con- This work has been partly supported by the Swedish Foundation forStrategic Research (SSF) under the programme Applied Mathematics and theproject Complex Analysis and Convex Optimization for EM Design.M. S. Mirmoosa and S. A. Tretyakov are with the Department of Electronicsand Nanoengineering, Aalto University, P.O. Box 15500, FI-00076 Aalto, Fin-land (email: mohammad.mirmoosa@aalto.fi and sergei.tretyakov@aalto.fi).S. Nordebo is with the Department of Physics and Electrical Engineering,Linnæus University, 351 95 V¨axj¨o, Sweden (email: [email protected]). tradictory discussions on the definition and physical meaningof key parameters of antennas in absorbing media, such asthe input impedance and radiation resistance [4], [8]–[12].In this tutorial paper we carefully examine the notion ofradiation resistance of a Hertzian dipole in infinite isotropichomogeneous media and discuss the physical meaning of thismodel in both lossless and lossy cases. We explicitly calculatethe absorbed power in the infinite space and show that theradiation resistance as a model of power transported to infinityis also nonzero for dipoles in lossy media. Subsequently, westudy the limit of zero loss. Finally, we discuss implications ofthis theory for understanding and engineering power transferbetween antennas in media.From the applications point of view, antennas embeddedin lossy media have received attention in geophysics, marinetechnology [3], [13], [14], medical engineering [15]–[17], etc.Hence, it is worthy to develop a thorough understandingof radiation from the Hertzian dipole which is placed in alossy medium. These results will also help to understandsub-wavelength emitters or nanoantennas immersed in dis-sipative media, since we can make an analogy between theHertzian dipole and the sub-wavelength emitter/nanoantennathat operates in the optical range. The investigation of lightinteraction with plasmonic or all-dielectric nanoantennas ismainly conducted in the assumption that the host medium isnot dissipative [18]–[20]. However, from the practical pointof view, in most applications nanoantennas (nanoparticles) areplaced in absorptive environments (e.g. see Refs. [21], [22]).This review is also relevant to the studies of absorption andscattering by small particles in lossy background [23]–[30].If a point source dipole is embedded in a lossy host, thereis also a theoretical problem of singularity of power absorbedin the medium, which is due to singularity of the dipole fieldsat the source point. This issue has been discussed e.g. in [7].The problem of source singularity should be addressed takinginto account the final size and shape of the antenna. Here, wefocus our discussion on radiation phenomena, studying powerabsorbed outside of a small sphere centred at the source point.II. R
ADIATION FROM H ERTZIAN DIPOLES IN ISOTROPICHOMOGENEOUS LOSSY MEDIA
Let us consider a Hertzian dipole antenna located in aninfinite homogeneous space, as illustrated in Fig. 1. Thecurrent amplitude in the dipole is fixed and denoted by I (we consider the time-harmonic regime, assuming e jωt timedependence). The dipole length is l . For simplicity, we assumethat the medium is non-magnetic ( µ = µ ), which does not a r X i v : . [ phy s i c s . c l a ss - ph ] D ec 𝑅 Fig. 1. The Hertzian dipole is radiating in a lossy medium. compromise the generality of our discussion. We characterizethe background medium by its complex relative permittivity (cid:15) = (cid:15) (cid:48) − j(cid:15) (cid:48)(cid:48) = (cid:15) (cid:48) − jσ/ ( ω(cid:15) ) , where σ is the effectiveconductivity. We will also use the complex refractive index n = √ (cid:15) = n (cid:48) − jn (cid:48)(cid:48) (the square root branch is defined alongthe positive real axis so that n (cid:48)(cid:48) ≥ ).The power delivered from the ideal current source I isusually written in terms of the equivalent resistance R in , as P rad = 12 R in | I | . (1)In fact, the resistance R in is the real part of the ratio ofthe voltage to electric current at the input terminals of thedipole [1]. Hence, R in has the meaning of the real part of thedipole input impedance. We remind that the Hertzian dipolemodel assumes that the antenna current is fixed, that is, it is anideal current source without any dissipation inside the antennastructure. If the infinite medium surrounding the dipole islossless, such as free space, the input resistance is identical tothe radiation resistance. In this case, the usual calculation ofthe Poynting vector flux through a fictitious spherical surface(which is easy to evaluate in the far zone, see, e.g. [1]) resultsin R in (lossless) = R rad = η ( k l ) π n (cid:48) , (2)in which R rad represents the radiation resistance. Also, η and k are the free-space intrinsic impedance and wave number,respectively. Note that the refraction index n = n (cid:48) is a real,non-negative number . However, if the medium is dissipative,the situation may remarkably change. The input resistance R in may differ extremely from the radiation resistance since it mustalso model the dissipation in the near zone. In the following,this difference will be discussed in detail. In isotropic non-magnetic media characterized by relative permittivity (cid:15) ,the choice of the square root branch for n = √ (cid:15) according to the passivitycondition n (cid:48)(cid:48) ≥ ensures that n (cid:48) ≥ . Thus, the resistance (2) is alwaysnon-negative. The real part of the refraction index n (cid:48) is negative in double-negative media, where the real parts of both permittivity and permeability arenegative. However, also in that case the resistance value in (2) is non-negative,because for lossless media we get R in = η k l ) π µn , where both µ and n are negative. A. Time-averaged power and energy conservation
If the host medium is lossy and absorbs power, the calcu-lated “radiated” power depends on the radius of the fictitiousspherical surface, and careful considerations are necessary.Since the fields of the point source are singular at the sourcelocation, we consider the outward power flux through thesurface of a small sphere of the radius R , with the dipoleat its center, as shown in Fig. 1 using yellow color. The lawof energy conservation tells that all the outward flowing powermust be absorbed by the exterior infinite volume ( r > R inFig. 1). In other words, P rad = P abs , (3)where P rad = 12 (cid:73) S Re (cid:2) E × H ∗ (cid:3) · d S (4)and P abs = 12 (cid:90) V σ | E | d V. (5)Here, S denotes the surface of radius R and V representsthe volume of the spherical shell extending from r = R to r → ∞ . Furthermore, σ is the finite conductivity of the lossymedium. Since Eq. (3) must always hold for a lossy medium,it should be also true in the limiting case when the dipole islocated in a lossless medium. It may be difficult to perceivebecause the conductivity would be zero ( σ = 0 ) in this caseand the right-hand side of Eq. (3) seems to vanish ( P abs = 0 ).However, the left-hand side is apparently not zero ( P rad (cid:54) = 0 ).Thus, care should be taken in considering the limit of zerolosses. Let us first evaluate the power radiated from the sphereof radius R , given by Eq. (4) for the general case of lossymedia when σ (cid:54) = 0 , and then study the case when σ tends tozero. This surface integral was calculated in Ref. [7]. If wesubstitute the known expressions of the electric and magneticfields [1] E r = − j Il πω(cid:15) (cid:15) cos θ (cid:18) r + j kr (cid:19) exp( − jkr ) ,E θ = − j Il πω(cid:15) (cid:15) sin θ (cid:18) r + j kr − k r (cid:19) exp( − jkr ) ,H φ = Il π sin θ (cid:18) r + j kr (cid:19) exp( − jkr ) , (6)generated by the Hertzian dipole into Eq. (4) considering that k = k n , and calculate the integral, the result reads P rad = η ( k l ) | I | π n (cid:48) (cid:20) n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) k R + 4 n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) k R + 2 n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) k R + 1 (cid:21) e − k n (cid:48)(cid:48) R . (7)Derivation of the above equation is straightforward (see Sup-plementary Information). However, calculating P abs does notappear to be that simple. Paper [7] contains a statement thatequality (3) holds, but calculation of the volume integral (5)is not given, and the limit of σ → is not discussed. To calculate the integral (5), we firstly write the square ofthe absolute value of the electric field components, which gives | E r | = | I | l cos θ π ω (cid:15) ( n (cid:48) + n (cid:48)(cid:48) ) · (cid:18) r + 2 k n (cid:48)(cid:48) r + k ( n (cid:48) + n (cid:48)(cid:48) ) r (cid:19) e − k n (cid:48)(cid:48) r , (8)and | E θ | = | I | l sin θ π ω (cid:15) ( n (cid:48) + n (cid:48)(cid:48) ) · (cid:18) r + 2 k n (cid:48)(cid:48) r + k (3 n (cid:48)(cid:48) − n (cid:48) ) r + 2 k n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) r + k ( n (cid:48) + n (cid:48)(cid:48) ) r (cid:19) e − k n (cid:48)(cid:48) r . (9)Using the above expressions, we find the square of the absolutevalue of the electric field, and next we can evaluate the volumeintegral in Eq. (5) to calculate the power P abs absorbed inthe exterior infinite spherical shell. Upon integration over theangles, we find that (see Supplementary Information) P abs = | I | l πω(cid:15) n (cid:48) n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) (cid:90) ∞ R f ( r ) e − k n (cid:48)(cid:48) r d r, (10)where f ( r ) = f ( r ) + f ( r ) + f ( r ) , with f ( r ) = 1 r + 2 k n (cid:48)(cid:48) r + k ( n (cid:48) + 5 n (cid:48)(cid:48) )3 r ,f ( r ) = 2 k n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) )3 r ,f ( r ) = k ( n (cid:48) + n (cid:48)(cid:48) ) . (11)Here, we have used the relation between the imaginary partof the permittivity and the corresponding conductivity (cid:15) (cid:48)(cid:48) = 2 n (cid:48) n (cid:48)(cid:48) = σω(cid:15) , (12)which gives σ = 2 ω(cid:15) n (cid:48) n (cid:48)(cid:48) . (13)Note that the term f , which does not depend on the distance,comes from the /r terms in the expression for the electricfield. To simplify Eq. (10), we use partial integration, andderive the following identity: (cid:90) ∞ R r m e − κr = m − (cid:88) i =1 ( − i − κ i − e − κR P ( m − , i ) R m − i + ( − m − κ m − ( m − (cid:90) ∞ R e − κr r d r, (14)where P ( x, y ) refers to the permutation formula. The function f ( r ) consists of three terms and for each of them we can usethe expression given by Eq. (14). Hence, (cid:90) ∞ R e − κr r dr = (cid:20) R − k n (cid:48)(cid:48) R + 2 k n (cid:48)(cid:48) R (cid:21) e − κR − k n (cid:48)(cid:48) (cid:90) ∞ R e − κr r d r, (15) k n (cid:48)(cid:48) (cid:90) ∞ R e − κr r d r = (cid:20) k n (cid:48)(cid:48) R − k n (cid:48)(cid:48) R (cid:21) e − κR + 4 k n (cid:48)(cid:48) (cid:90) ∞ R e − κr r d r, (16)and k ( n (cid:48) + 5 n (cid:48)(cid:48) )3 (cid:90) ∞ R e − κr r dr = k ( n (cid:48) + 5 n (cid:48)(cid:48) )3 R e − κR − k n (cid:48)(cid:48) ( n (cid:48) + 5 n (cid:48)(cid:48) )3 (cid:90) ∞ R e − κr r d r, (17)where we have introduced the notation κ = 2 k n (cid:48)(cid:48) . We cancombine the above results (15–17) and write (cid:90) ∞ R f ( r ) e − k n (cid:48)(cid:48) r d r = (cid:20) R + 2 k n (cid:48)(cid:48) R + k ( n (cid:48) + n (cid:48)(cid:48) )3 R (cid:21) e − κR − k n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) )3 (cid:90) ∞ R e − κr r d r. (18)Comparing Eq. (18) with Eq. (11) we find that the last termin Eq. (18) (the coefficient before the integral) is the same asfunction f ( r ) but with the opposite sign. Therefore, only thefirst three terms inside the square brackets in Eq. (18) remainand, interestingly, the last term vanishes. Thus, (cid:90) ∞ R (cid:2) f ( r ) + f ( r ) (cid:3) e − κr d r = (cid:20) R + 2 k n (cid:48)(cid:48) R + k ( n (cid:48) + n (cid:48)(cid:48) )3 R (cid:21) e − κR . (19)The last step in this long derivation of P abs in Eq. (10) is theevaluation of the integral for function f ( r ) , which gives (cid:90) ∞ R f ( r ) e − κr d r = k ( n (cid:48) + n (cid:48)(cid:48) ) n (cid:48)(cid:48) e − κR . (20)Using Eqs. (19) and (20) and the identity η /k = 1 / ( ω(cid:15) ) , wefinally arrive to the following formula for the power absorbedin the exterior environment (in the spherical shell extendingfrom R to infinity): P abs = η ( k l ) | I | π (cid:20) n (cid:48) n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) ( k R ) + 4 n (cid:48) n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) ( k R ) + 2 n (cid:48) n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) )( k R ) + n (cid:48) (cid:21) e − κR . (21)Now, if we compare Eq. (21) with Eq. (7), we see that they areidentical, meaning that Eq. (3) holds for any non-zero valueof the conductivity (or the imaginary part of the permittivity),including the limiting case as σ tends to zero.Considering the expression (21) for the absorbed power, wenote that there are several terms which are proportional to n (cid:48)(cid:48) and singular at R → . They can be interpreted as the powerabsorbed in the near vicinity of the dipole. The singularity isdue to the fact that the fields of a point dipole are singular atthe dipole position. Thus, if the medium is lossy ( n (cid:48)(cid:48) (cid:54) = 0 ), theabsorbed power diverges. Naturally, these terms tend to zeroif n (cid:48)(cid:48) → . B. Radiation resistance
The last term in Eq. (21), η k l ) | I | π n (cid:48) e − k n (cid:48)(cid:48) R , dependson n (cid:48)(cid:48) and R only in the exponential factor, in contrast tothe singular near-field terms. The exponential factor tends tounity when either R or n (cid:48)(cid:48) tends to zero. Because this lastterm is not singular, we can let R → and interpret this termas the dipole radiation resistance, which measures the “powerdelivered to infinity”. Therefore, R rad = 2 lim R → (cid:104) η ( k l ) | I | π n (cid:48) e − k n (cid:48)(cid:48) R (cid:105) | I | = η ( k l ) π n (cid:48) . (22)Importantly, this term results from integration of /r (“wave”)terms in the expression for the dipole fields, and it is exactlythe same as the power radiated into infinite lossless media(see Eq. (2)). Thus, the radiation resistance of electric dipolesin lossy media does not depend on the imaginary part of therefractive index, and it is expressed by the same formula as forlossless background. This formula for the radiation resistancewas given in paper [9] (the first term in Eq. (10)), derivedon the basis of re-normalizing the power flow density, whichis not offering clear physical insight. It also agrees with theresult presented in paper [10], Eq. (2) , which is basicallyequivalent to substituting complex-valued material parametersinto the formula derived for the free-space background. Onthe other hand, contradictory expressions can be also found inthe literature, for example, in Refs. [11], [12].III. D ISCUSSION
As we see from Eq. (21), the absorbed power calculatedas the volume integral (5) is not zero even in the limit ofzero conductivity (lossless background). This result seeminglycontradicts to the fact that for σ = 0 the integrand of (5) isidentically zero. However, we cannot conclude that in thiscase the integral is zero, because for lossless backgroundthe integral (cid:82) V | E | dV diverges. Care should be taken inconsidering the limit of vanishing absorption if the absorptionvolume is infinite.Let us discuss this limit. The expression for the power ab-sorbed in the medium has a form of a product of conductivity,which tends to zero in the lossless limit, and the integral of theenergy density, which diverges when we extend the integrationvolume to infinity. Thus, we should write the power balancerelation considering absorption in a sphere of a finite radius R and then properly calculate the limit. For this consideration,the energy conservation relation (3) is written as P rad = 12 (cid:90) V R \ V R σ | E | d V + 12 (cid:73) S R Re { E × H ∗ } · d S , (23)where σ = ω(cid:15) (cid:15) (cid:48)(cid:48) = 2 ω(cid:15) n (cid:48) n (cid:48)(cid:48) , and V R and V R denotespheres of radii R and R , respectively. The first term onthe right-hand side of (23) is the power dissipated in the Note the misprint in Eq. (2) of [10]: the right-hand side should read Z ( nω, (cid:15) ) /n . surrounding medium between spherical radii R and R , andthe second term is the power radiated away beyond radius R .Now we can accurately calculate the limit for R → ∞ andvanishing σ (and n (cid:48)(cid:48) ). Calculating the integral (10) with a finiteupper limit R and considering vanishing n (cid:48)(cid:48) and infinitelygrowing R , the power balance relation (23) takes the form P rad = η ( k l ) | I | π n (cid:48) (cid:18) − lim n (cid:48)(cid:48) → , R →∞ e − k n (cid:48)(cid:48) R (cid:19) + η ( k l ) | I | π n (cid:48) (cid:18) lim n (cid:48)(cid:48) → , R →∞ e − k n (cid:48)(cid:48) R (cid:19) . (24)Here, the first line gives the power absorbed in the infinitespace, and the second line is the power crossing the sphericalsurface of the infinite radius. Clearly, the value of the double-limit depends on the order in which the two limits are taken.Thus, there are two possible interpretations of the absorptionin the lossless infinite space.1. Taking first the limit R → ∞ and then the limit ofvanishing loss factor n (cid:48)(cid:48) → , in the spirit of the principle ofvanishing absorption [31], we can say that in this interpretationall the radiated power is dissipated in the vacuum and there isno radiation to infinity (because the expression in the secondline tends to zero).2. Taking first the limit n (cid:48)(cid:48) → and then extending thevolume integration over the whole space letting R → ∞ , theexpression in the first line (absorption in the medium) tendsto zero. This is a very common interpretation of the losslessinfinite space. There are no losses in the “vacuum”, and all theradiated power is radiated away throughout the infinite space,beyond any finite radius R (in this sense “transported all theway to infinity”).However, we stress that in the expression for the total“absorbed power” (24) the two double-limit expressions cancelout, meaning that whatever is the interpretation, the radiationresistance defined as P rad / | I | is a well-behaving, continuousfunction of σ or n (cid:48)(cid:48) , and based on the first interpretation abovewe can use the expression (21) for both lossy and losslessbackgrounds .Another, perhaps more important, implication of this consid-eration is that the input resistance given by Eq. (2) for lossy background contains exactly the same term as the radiationresistance of Hertzian dipole antenna in lossless media or vac-uum. Thus, it is misleading to assume that if the backgroundmedium is lossy, there is no radiation resistance as such, asthe fields exponentially decay. We see that the total inputresistance is the sum of the radiation resistance, proportionalto n (cid:48) , and the near- and intermediate-zone loss resistance,proportional to n (cid:48)(cid:48) . Importantly, as already discussed, the termproportional to n (cid:48) does not vanish if n (cid:48)(cid:48) becomes zero, and ithas exactly the same form in both lossy and lossless cases.The only scenario where there is no radiation resistance atall is when the real part of the refraction index n (cid:48) vanishes. Actually, there can be infinitely many intermediate “intepretations” sincethis double-limit expression can take any value from zero to unity dependingon the way of taking the limit.
Perhaps counter-intuitively, this case corresponds to lossless background media. Indeed, calculating (cid:15) = n = n (cid:48) − n (cid:48)(cid:48) − jn (cid:48) n (cid:48)(cid:48) , (25)we see that if n (cid:48) = 0 , then (cid:15) = − n (cid:48)(cid:48) is purely real, meaningthat the medium is lossless. We also note that the permittivityis negative that is the case when wave propagation is notpossible.Formula (21) can be written in terms of the contributionto the real part of the input impedance of the antenna dueto dissipation in the surrounding medium (upon dividing by | I | / ): R in = η ( k l ) π n (cid:48) + η ( k l ) π n (cid:48) n (cid:48)(cid:48) (cid:20) n (cid:48) + n (cid:48)(cid:48) ) ( k R ) +4 n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) ( k R ) + 2( n (cid:48) + n (cid:48)(cid:48) )( k R ) (cid:21) . (26)Here, we have assumed that k R is very small so that e − κR ≈ . As mentioned before, the first term is the sameas the radiation resistance in lossless media, and can beinterpreted as the radiation resistance in lossy media. Thesecond term, singular at R → , is due to dissipation inthe near and intermediate zones. This second term naturallyvanishes if n (cid:48)(cid:48) = 0 . Thus, R in = R rad + R nrad , (27)where the non-radiative resistance is given by R nrad = η ( k l ) π n (cid:48) n (cid:48)(cid:48) (cid:20) n (cid:48) + n (cid:48)(cid:48) ) ( k R ) +4 n (cid:48)(cid:48) ( n (cid:48) + n (cid:48)(cid:48) ) ( k R ) + 2( n (cid:48) + n (cid:48)(cid:48) )( k R ) (cid:21) . (28)This consideration clarifies the physical meaning of radiationresistance in the general case of isotropic background media:It vanishes only if propagation is not possible . In other words,the radiation resistance is not zero even in lossy media,because the propagation constant is not equal to zero and theoutward power flux is not zero.In terms of applications, this conclusion is important forunderstanding of ultimate limits for power transported fromone antenna to another. Let us position a receiving Hertziandipole antenna of length l in the field created by our radiatingdipole antenna (in an infinite isotropic medium characterizedby the refractive index n = n (cid:48) − jn (cid:48)(cid:48) ). The current induced inthe receiving antenna I rec is proportional to the electric field E created by the transmitting antenna at the receiver positionand inversely proportional to the impedance: I rec = ElR load + jX + R in . (29)Let us assume that the receiving antenna is loaded by a resistor R load at its center. The power delivered to the load reads P = 12 R load | I rec | . (30)Obviously, to maximize the delivered power we should bringthe antenna to resonance, making the total reactance X = 0 , which is always possible. The ideal scenario where the de-livered power can be arbitrarily high corresponds to R in = 0 (then we have P = | E | l /R load which diverges for R load → ). Now we should look at the expression for the effectiveinput resistance and find under what conditions this expressiongives the smallest value. Apparently, conventional losslessbackground ( n (cid:48)(cid:48) = 0 ) is better than the lossy one, but even inthat case the received power is limited, because the radiationresistance (2) is not zero. This consideration brings us to thewell-known limit of the effective absorption cross section forany dipole antenna/scatterer in lossless background [32]. Butnotice an important special case of n (cid:48) = 0 . In this case theinput resistance is zero, and the delivered power has no upperbound.Accordingly, we reach an enlightening conclusion. If twoantennas are in an environment which does not allow wavepropagation, the power delivered from one antenna to the othercan be arbitrarily high. In contrast, if wave propagation isallowed, the delivered power is fundamentally limited by theultimate absorption cross section of a dipole antenna, which, inturn, is determined by its non-zero radiation/input resistance.Another case when the power delivered to a load from anideal current source is limited only by the parasitic resistanceof conductors, is the case of zero frequency (DC) circuits.Interestingly, the reason why there is no fundamental limit isthe same: There is no radiation loss at DC.The scenario of such unlimited-capacity power deliverychannel corresponds, for instance, to the case of two dipoleantennas inside a lossless-wall waveguide below cut-off. Thetwo antennas are coupled only by reactive fields, and at theresonance of the antenna pair inside the waveguide, all theeffective reactances are compensated. At this frequency, thereceiving antenna is effectively connected to the ideal currentsource feeding the transmitting antenna by a zero-impedancelink. Here, the power delivered from one antenna to the otheris limited only by the power available from the source, andby parasitic losses in the waveguide walls and in the antennawires. Another example is coupling between two antennas inlossless plasma. IV. C ONCLUSIONS
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