Modelo Dos Rayos/Índice/Ecuaciones 3.45 y 3.48

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Ecuaciones 3.45-3.48

Equations 3.45-3.48 If the received E-field is evaluated at some time, say at $t=d{″}^{}/c$, equation (3.39) can be expressed as $E_{TOT}(d,t=\frac{d{″}^{}}{c})=\frac{E_0{d}_0}{d{\prime }^{}}cos(w_c(\frac{d{″}^{}-d{\prime }^{}}{c}))-\frac{E_0{d}_0}{d{″}^{}}cos{0}^{\circ }$ $=\frac{E_0{d}_0}{d{\prime }^{}}cos{\theta }_{\mathrm{\Delta }}-\frac{E_0{d}_0}{d{″}^{}}$ $=\frac{E_0{d}_0}{d}[cos{\theta }_{\mathrm{\Delta }}-1]$ where d is the distance over a flat earth between the bases of the transmitter and receiver antennas. Referring to the phasor diagram of Figure 3.9 which shows how the direct and ground reflected rays combine, the electric field (at the receiver) at a distance d from the transmitter can be written as $|E_{TOT}(d)|=\sqrt{(\frac{E_0{d}_0}{d}{)}^2(cos{\theta }_{\mathrm{\Delta }}-1{)}^2+(\frac{E_0{d}_0}{d}{)}^{2}si{n}^2{\theta }_{\mathrm{\Delta }}}$ or $|E_{TOT}(d)|=\frac{E_0{d}_0}{d}\sqrt{2-2cos{\theta }_{\mathrm{\Delta }}}$ Using trigonometric identities, equation (3.47) can be expressed as $|E_{TOT}(d)|=2\frac{E_0{d}_0}{d}sin(\frac{{\theta }_{\mathrm{\Delta }}}{2})/3.8)$ Phasor diagram (Figure 3.9