Question

In a LOS communication, consider d = 40km, the requirement is to make two antennas (transmitter and receiver) such that the height of one antenna should be three times of the other. Considering this, find the appropriate heights of these two antennas.

Answer 1

Answer:

Height of two antennas are

$h_1 = 16.7 m$

$h_2 = 50.2 m$

Explanation:

As we know that the distance between transmitter and receiver is given as

$d = \sqrt{2h_1R} + \sqrt{2h_2R}$

here we know that

$h_2 = 3h_1$

now we have

$d = \sqrt{2h_1R} + \sqrt{6h_1R}$

$40 km = \sqrt{h_1}(\sqrt{2R} + \sqrt{6R})$

R = 6400 km

$40 = \sqrt{h_1}(\sqrt{12800} + \sqrt{38400})$

$h_1 = 16.75 m$

$h_2 = 50.2 m$

#Learn

Topic : Height of antenna

https://brainly.in/question/2894373

Answer 2

The heights of these two antennas are 16.74 m and 50.22 m.

Explanation:

Given that,

Distance d= 40 km

Height of first antenna h₁= 3 times of h₂

We need to calculate the height of second antenna

Using formula of distance

$d=\sqrt{2h_{1}R}+\sqrt{2h_{2}R}$

Put the value of height into the formula

$d=\sqrt{2\times3h_{2}R}+\sqrt{2h_{2}R}$

$d=\sqrt{2\times h_{2}R}(1+\sqrt{3})$

Squaring both sides

$d^2=2h_{2}R(1+\sqrt{3})^2$

$h_{2}=\dfrac{d^2}{2\times R\times(1+\sqrt{3})^2}$

Put the value into the formula

$h_{2}=\dfrac{(40\times10^{3})^2}{2\times6400\times10^3\times(1+\sqrt{3})^2}$

$h_{2}=16.74\ m$

We need to calculate the height of first antenna

Using given statement

$h_{1}=3h_{2}$

Put the value into the formula

$h_{1}=3\times16.74$

$h_{1}=50.22\ m$

Hence, The heights of these two antennas are 16.74 m and 50.22 m.

Learn more :

Topic : height of antenna

https://brainly.in/question/11899658