Artículo de Friis/Índice/Área efectiva de un dipolo

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] $P_{o}=E^{2}/120\pi$ $Pr={\frac {E^{2}a^{2}}{4(80\pi ^{2}a^{2}/\lambda ^{2})}}$ = ${\frac {\lambda ^{2}E^{2}a^{2}}{320\pi a^{2}}}$= ${\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}$ $A_{dip}={\frac {\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}{\frac {E^{2}}{120\pi }}}={\frac {120\pi \lambda ^{2}E^{2}}{320\pi ^{2}E^{2}}}={\frac {3\lambda ^{2}}{8\pi }}$ Área efectiva de una circunferencia

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