If a transmitting antenna at the top of a tower has a height of $$49$$ m and the height of receiving tower is $$100$$ m . Calculate the maximum distance between them for satisfactory communication in LOS (Line$$ Of Sight $$) mode $$($$ $$R=6400km$$ $$)$$ is z×$$103m$$ . Then z $$is(nearly)$$

Question

If a transmitting antenna at the top of a tower has a height of $$49$$ m and the height of receiving tower is $$100$$ m . Calculate the maximum distance between them for satisfactory communication in LOS (Line$$ Of Sight $$) mode $$($$ $$R=6400km$$ $$)$$ is z×$$103m$$ . Then z $$is(nearly)$$

Infinity Learn · Accepted Answer

Given  :  $$hT=49m$$ $$hR=100m$$ $$R=6400$$×$$103m$$  Using, $$d=dT+dR$$ ⇒$$d=2RhT+2RhR$$         $$=2$$×$$6400$$×$$1000$$×$$49+2$$×$$6400$$×$$1000$$×$$100$$ After simplification                                  $$d=60$$×$$103m$$

Propagation of electromagnetic waves

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If a transmitting antenna at the top of a tower has a height of 49 m and the height of receiving tower is 100 m . Calculate the maximum distance between them for satisfactory communication in LOS (Line Of Sight ) mode ( $R = 6400 k m$ ) is $z \times 10^{3}$ m . Then z is(nearly)

$G i v e n : h_{T} = 49 m h_{R} = 100 m R = 6400 \times 10^{3} m U \sin g, d = d_{T} + d_{R} \Rightarrow d = \sqrt{2 R h_{T}} + \sqrt{2 R h_{R}} = \sqrt{2 \times 6400 \times 1000 \times 49} + \sqrt{2 \times 6400 \times 1000 \times 100} A f t e r s i m p l i f i c a t i o n d = 60 \times 10^{3} m$

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