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(C) For satisfactory communication in Line-of-Sight $(LOS)$ mode, the maximum distance $d_{	ext{max}}$ between a transmitting antenna of height $h_1$

Vedclass · Accepted Answer

$12.8$

Vedclass · Answer

(C) For satisfactory communication in Line-of-Sight $(LOS)$ mode, the maximum distance $d_{	ext{max}}$ between a transmitting antenna of height $h_1$ and a receiving antenna of height $h_2$ is given by:$d_{	ext{max}} = \sqrt{2Rh_1} + \sqrt{2Rh_2}$Given that $h_1 = h_2 = d 	ext{ meters}$ and $d_{	ext{max}} = 2d 	ext{ kilometers} = 2d 	imes 1000 	ext{ meters}$.Substituting these values into the formula:$2d 	imes 1000 = \sqrt{2Rd} + \sqrt{2Rd}$$2000d = 2\sqrt{2Rd}$$1000d = \sqrt{2Rd}$Squaring both sides:$(1000d)^2 = 2Rd$$1,000,000 d^2 = 2 	imes 6400 	imes 1000 	imes d$Dividing both sides by $d$ (assuming $d 
eq 0$):$1,000,000 d = 2 	imes 6400 	imes 1000$$1000 d = 2 	imes 6400$$d = \frac{12800}{1000} = 12.8 	ext{ m}$

$A$ transmitting and receiving antenna have a height of $d$ meters each. The maximum distance between them for satisfactory communication in Line-of-Sight $(LOS)$ mode is $2d$ kilometers. If the radius of the Earth is $6400 \text{ km}$, then the value of $d$ is: (in $\text{ m}$)

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