Artículo de Friis/Índice/Área efectiva de un dipolo

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