Problem 1: Suppose there is a 10Mbps microwave link between a geostationary sate
ID: 3606927 • Letter: P
Question
Problem 1: Suppose there is a 10Mbps microwave link between a geostationary satellite and its base station on Earth. Every minute the satellite takes a digital picture (of the earth) and sends it to the base station. Assume that propagation speed is 2.99x 108m/s. a) What is the propagation delay? b) What is the bandwidth-delay product, R * dpro c) Let X be a the size of the photo. What is the minimum value for x that keeps the link d) If the pictures are half of such size, what is the maximum interval of time between pictures e) What is the average queueing time if I take a burst of 100 pictures of size S? What is the f) How wide is a bit over the channel? continuously transmitting? that keeps the link busy? maximum S such that I finish transmitting them before the next scheduled picture is taken g) What is the size of a picture such that it starts to be received before the transmitter stops sending it? And what is the maximum amount of bits that can be on the channel at the same time?Explanation / Answer
Solution:
a, b, c, and d are solved, please repost others.
Propagation Delay Tprop = D/ S
Let, distancebetween satellite and base station be D = 36000km
a)
PropagationSpeed, S = 2.99 * 10^8 meters/second
Therefore, propagation delay = D / S = (3.6 *10^7 / 2.99) *10^(-8)seconds= 0.12040133779 seconds
(b)
Bandwidth delay product = R *Tprop
= 10 *10^6 * 0.12040133779
= 1204014 bits
(c)
Size of the file be X
transmission rate = 10^7bits/second
Time taken = 1 minute or 60seconds
X= 10^7 * 60 or 6 * 10 ^8
= 600 Mb
d)
If the size of the picture is half i.e. 300 Mb
Maximum interval of time= 300/10= 30 seconds
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