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  • A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver.

A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be? 

 

Answers (1)

The values given in the question are,

Velocity of waves = 3 \times 10^{8} m/s

Time to reach a receiver = 4.04 \times 10^{-3} s

Height of satellite, h = 600\: km

Radius of earth = 6400\: km

Size of transmitting antenna= hT

By using the formula = distance travelled by wave/time = velocity

The final velocity of waves = 606\: km

After using the Pythagoras theorem,

d = 85.06\: km

Thus, the distance between source and receiver= 2d = 170\: km

So, the maximum distance from the transmitter to the ground would by the EM waves would be

= h_t = 565\: m    

Posted by

infoexpert21

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