# Artículo de Friss/Índice/Área efectiva de un dipolo

Ir a la navegación Ir a la búsqueda
Español Inglés

# Área efectiva del dipolo:

$A_{dip}$ =${\frac {P_{r}}{P_{0}}}={\frac {3\lambda ^{2}}{8\pi }}=0.1193\lambda ^{2}$ [0.3cm] $P_{o}=E^{2}/120\pi$ $Pr={\frac {E^{2}a^{2}}{4(80\pi ^{2}a^{2}/\lambda ^{2})}}$ = ${\frac {\lambda ^{2}E^{2}a^{2}}{320\pi a^{2}}}$ = ${\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}$ $A_{dip}={\frac {\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}{\frac {E^{2}}{120\pi }}}={\frac {120\pi \lambda ^{2}E^{2}}{320\pi ^{2}E^{2}}}={\frac {3\lambda ^{2}}{8\pi }}$ # Small Dipole with No Heat Loss

For a small uniform current element the available output power is equal to the induced voltage squared divided by four times the radiation resistance. Thus

$P_{r}={\frac {E^{2}a^{2}}{4R_{rad}}}$ where
E=effective value of the electric field of the wave.
a=length of the current element.
$R_{rad}=80\pi ^{2}a^{2}/\lambda ^{2}$ Since the power flow per unit area is equal to the electric field squared divided by the impedance of free space,
i.e., $P_{o}=E^{2}/120\pi$ , we have

$A_{dip}={\frac {P_{r}}{P_{0}}}={\frac {3\lambda ^{2}}{8\pi }}=0.1193\lambda ^{2}$ The effective area of a half-wavelength dipole with no heat loss is only 9.4 per cent, 0.39 decibels,2 larger than the effective area of the small dipole. Therefore

$A_{0.5\lambda }=0.1305\lambda ^{2}$ The area of a rectangle with one-half wavelength and one-quarter wavelength sides is $0.125\lambda ^{2}$ and it is, therefore, a good approximation for the effective areas of small dipoles and half-wavelength dipoles.


$A_{0.5}=0.1305\lambda ^{2}$ $0.1305\lambda ^{2}=\pi r^{2}$ ${\frac {0.1305}{\pi }}\lambda ^{2}=r^{2}$ $0.04154\lambda ^{2}=r^{2}$ ${\sqrt {0.04152\lambda ^{2}}}={\sqrt {r^{2}}}$ $r=0.2038\lambda$ 