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Assuming that signals are conveyed at 3.06108 m s^–1 regardless of any variations in the atmosphere with increasing height, estimate the maximum propagation delay that would be incurred when a HAP operating at a height of 20 km above the Earth’s surface retransmits data that is directed to it from a ground transmitter directly below, to receiving antennas
spread across a circular area directly below and which is 60 km in diameter. Indicate your
reasoning and any assumptions that you make in your answer. Compare this with typical delays when using a geo-stationary satellite and comment upon the result.

2007-10-19 16:11:16 · 2 answers · asked by ABD A 1 in Science & Mathematics Engineering

2 answers

I will assume you meant the signals are propagated at 3.0 * 10^8 m s^-1.

The time for the signal to get from the ground to the HAP is (20 * 10^3 m) / (3.0 * 10^8 m s^-1) = 66.667 *10^-6 s

The minimum time it would take for the signal to reach back to the ground would be in a straight line back to the transmitter: another 66.67 microseconds for a total of 133.3 microseconds.

The maximum time the signal would take, would be to reach the edge of the large circle the HAP 'sees' on the earth. As a first-order approximation (an assumption) you could say the earth is flat in that 60 km diameter circle, so a right-triangle could be made with:
- one side as the 20 km distance from the earth to the HAP;
- one side as the 30 km radius if the circle, and
- the 3rd side as the hypoteneuse of the triangle.

Using Pythagorus' Theorem, the hypoteneuse would be 36.0 km.

The amount of time it takes for the signal to propagate is:
(36 * 10^3 m) / (3.0 * 10^8 m s^-1) = 118.8 microseconds
The total maximum time would be 185.46 microseconds.

In reality there would be some delay time through the HAP (we assumed zero delay). Also at the edge of the 60 km circle the earth is curved *away* from the HAP, so it would take a little longer (I assumed a flat earth for this problem).

As for regular TV communication satellites, they orbit at about 35900 km which is much further that the HAP so, the delays will be much larger. At this distance, the propagation delay "up" will be essentially the same as going "down". I think by now you can figure out the number for the delay time.

.

2007-10-23 05:54:11 · answer #1 · answered by tlbs101 7 · 0 0

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2017-01-03 22:45:04 · answer #2 · answered by Anonymous · 0 0

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