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The friis transmission formula extends the

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The Friis transmission formula extends the reciprocity theorem to the case of potentially lossy and improperly matched antennas: P rx = ( 1 − | ρ r | 2 ) λ 2 4 π G r ( θ r r ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright A eff r ( 1 − | ρ t | 2 ) P t 4 πr 2 G t ( θ t t ) where the r and t subscripts refer to the receiver and transmitter, respectively. Note that this formula is still incomplete and does not account for polarization mismatches. In chapter 9, we will see that Faraday rotation in anisotropic media can destroy reciprocity. 2.3 Practical wire antennas The elemental dipole antenna is an idealization, but practical wire antennas can be viewed as assemblies of such elements. This provides an approach for analyzing wire antennas that are physically long compared to the wavelength. In principle, one must know the current distribution on the wires to solve for the radiation fields. In practice, it is often acceptable to guess the distribution and then calculate the fields using superposition. Keeping track of the phases of the contributions from the radiating elements is the main difficulty. Rigorous computational solutions agree reasonably well with the solutions presented below and with measurements. We consider the far zone fields only in what follows. We will analyze the long wire (length L ) dipole antenna in Figure 2.5 using the results from the elemental dipole analysis. We expect only E θ and H φ field components in the far field. It is sufficient to solve for the former, since the latter is related by the impedance since E θ /H φ = Z . From the elemental dipole I ( z ) dz , we expect the differential contribution dE θ = jZ k 4 π I ( z ) dz sin θ ( z ) e jkR ( z ) R ( z ) Integrating the contributions from all the elemental dipoles that constitute the long wire antenna will then yield the total observed electric field. Note that the harmonic time dependence exp ( jωt ) is implicit. We next make a number of assumptions to simplify the line integral: 1. Take r λ , L . The first assumption restates the far field limit. The second is something new and will be examined closely later. 2. We can consequently replace θ ( z ) with θ . 35
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z=0 z=L/2 z=-L/2 r R(z) I(z)dz E θ ,H φ (r, θ,φ ) θ θ( z 29 Figure 2.5: Long wire antenna geometry. The observation point is ( r,θ,φ ). 3. In the denominator, we can let R ( z ) r . 4. More care must be taken with the phase term, since even small variations in R (comparable to a wavelength) with z can produce significant changes. Regarding all the rays from the antenna to the observation point as parallel permits us to replace R ( z ) r z cos θ in the phase term. The differential contribution from each elemental dipole to the observed electric field then becomes: dE θ = jZ k 4 πr I ( z ) dz sin θe jkr e jkz cos θ What is needed next is a specification of the current distribution on the wire antenna, I ( z ) . Calculating this self consistently is very demanding, even computationally. Intuition can provide a reliable guide, however. We expect the current to nearly vanish at the ends of the wire in view of the demands of current continuity. Elsewhere on the wire,
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