10. Calculate the checksum for blocks of data with the following byte sum. (a) 1220 (b) 950
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Answer:
t= 8.7*10⁻⁴ sec.
Explanation:
If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.
As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.
Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:
If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:
⇒ t = 8.7*10⁻⁴ sec.