Question

A 50 W transmitter at 900 Mhz is radiating in to free space using a lineally polarized 12 dB omni directional antenna. Then calculate the power density (S) and electric field intensity (E) at a distance of 10 km from the antenna along the direction of the main beam?

• A. $S={10}^{-6}W/m^{2}E=2V/m$
• B. $S=4.5\times {10}^{-5}W/m^{2}E=30\times {10}^{-3}V/m$
• C. $S=6.3\times {10}^{-7}W/m^{2}E=2.18\times {10}^{-2}V/m$
• D. $S=5.2\times {10}^{-6}W/m^{2}E=2.5V/m$

Answer 1

The correct option is C $S=6.3\times {10}^{-7}W/m^{2}E=2.18\times {10}^{-2}V/m$

Expressing the gain of the transmit antenna as a ratio

$G=10\frac{G_t.dB}{10}={10}^{12}=15.85$

The power density at a distance 'R' from an antenna having a gain of $G_t$ and transmitting a power of $P_t$ is

$S=\frac{P_t{G}_t}{4\pi R^2}$

$\therefore P_t=50W,G_t=15.85,\&R=10km$

Substituting these values we get

$S=\frac{50\times 15.85}{4\pi \times (10\times {10}^3{)}^2}=6.3065\times {10}^{-7}W/m^2$

$\therefore$ The antenna is linearly polarized, with out loss of generality we can

$chooseE=E_{\theta }{\hat{a}}_{\theta }\&henceH=\frac{E_{\theta }}{\eta }{\hat{a}}_t‘(weknowthat\frac{E_{\theta }}{H_{\theta }}=\eta )$

$\therefore$ The average power density is

$S=\frac{1}{2}Re\{\overline{E}\times \overline{H}\}=\frac{1}{2}Re\left\{E_{\theta }\frac{E_{\theta }}{\eta }\right\}{\hat{a}}_{\theta }$

$=\frac{1}{2}\frac{|E_{\theta }{|}^2}{\eta }{\hat{a}}_x=S{\hat{a}}_f$

$\therefore |E_{\theta }|=\sqrt{2S\eta }$

$=\sqrt{2\times 6.3068\times {10}^{-7}\times 120\pi }$

$=2.1798\times {10}^{-2}V/m$