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Unformatted text preview: Note that the harmonic time dependence exp ( jÏ‰t ) is implicit. We next make a number of assumptions to simplify the line integral: 1. Take r â‰« Î» , 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 Î¸ ( z ) with Î¸ . 35 z=0 z=L/2 z=L/2 r R(z) I(z)dz E Î¸ ,H Ï† (r, Î¸,Ï† ) Î¸ Î¸( z 29 Figure 2.5: Long wire antenna geometry. The observation point is ( r,Î¸,Ï† ). 3. In the denominator, we can let R ( z ) â†’ 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 ) â†’ r âˆ’ z cos Î¸ in the phase term. The differential contribution from each elemental dipole to the observed electric field then becomes: dE Î¸ = jZ â—¦ k 4 Ï€r I ( z ) dz sin Î¸e âˆ’ jkr e jkz cos Î¸ 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, we may expect a standing wave pattern. A distribution that satisfies these conditions is: I ( z ) = I â—¦ sin( k [ L/ 2 âˆ’ z  ]) This turns out to be a reasonably good approximation for the actual current distribution on a long wire antenna, except that the actual distribution never exactly vanishes at the feedpoint, as this one can. Substituting I ( z ) and integrating yields the following expression for the observed electric field: E Î¸ = jkZ â—¦ sin Î¸I â—¦ 4 Ï€r e âˆ’ jkr bracketleftBigg integraldisplay âˆ’ L/ 2 sin k ( L/ 2 + z ) e jkz cos Î¸ dz + integraldisplay L/ 2 sin k ( L/ 2 âˆ’ z ) e jkz cos Î¸ dz bracketrightBigg = jkZ â—¦ sin Î¸I â—¦ 2 Ï€r e âˆ’ jkr bracketleftBigg cos( k L 2 cos Î¸ ) âˆ’ cos( k L 2 ) k sin 2 Î¸ bracketrightBigg = jZ â—¦ I â—¦ 2 Ï€r e âˆ’ jkr bracketleftBigg cos( k L 2 cos Î¸ ) âˆ’ cos( k L 2 ) sin Î¸ bracketrightBigg = Z â—¦ H Ï† 36 The Poynting flux can then be calculated. P r = 1 2 â„œ ( E Ã— H âˆ— ) Â· Ë† r = 1 2 â„œ ( E Î¸ H âˆ— Ï† ) = Z â—¦ 2  I â—¦  2 (2 Ï€r ) 2 bracketleftBigg cos( k L 2 cos Î¸ ) âˆ’ cos( k L 2 ) sin Î¸ bracketrightBigg 2 Like the elemental dipole, this antenna does not radiate power in the direction of the wire (despite the sin 2 Î¸ term in the denominator) and has a maximum in the equatorial plane. 2.3.1 Halfwave dipole An illustrative and practical case is that of a halfwave dipole antenna with L = Î»/ 2 . Now, kL/ 2 = Ï€/ 2 and the second cosine term in the numerator vanishes. This leaves: P r =  I â—¦  2 Z â—¦ 8 Ï€ 2 r 2 bracketleftbigg cos( Ï€ 2 cos Î¸ ) sin Î¸ bracketrightbigg 2 We will see that trigonometric functions of trigonometric functions are rather common expressions in antenna theory....
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 Spring '13
 HYSELL
 The Land, power density, Solid angle

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