Propagation Delay
With all forms of transmission medium (electrical or optical) there is a short time delay for the signal to travel (propagate) through the medium. For examples :
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In free space speed approximates to 3*108 m/s
In cables speed approximates to 2*108 m/s
Transmission Delay
Transmission delay is the time taken to transmit a block or frame at the specified bit rate.
For examples :
-
1000 bit frame at a transmission rate of 10kb per second.
Tx = 1000/(10 * 103) secs = 0.1 secs
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Or for Ethernet -
Max packet size of 1500 bytes
R = 10Mbps (approx)
Tx = (1500 * 8)/(10 * 10-6) = 1200 * 10-6 = 1.2 * 10-3 = 1ms (approx)
Satellite Problem
A satellite lies in a geostationary orbit 35800km above the Earths surface. A data link makes use of the satellite to transmit information at a data rate of 50Mbps.
- Estimate the time taken for the signal to propagate from one station on Earth to another via the satellite.
v = d/t
3 * 108 = 35800000/t
t = 358/3*104 = 0.117
The signal will take 0.234 seconds to travel from station A to the satellite and back to station B. - Estimate the quantity of data (in transit) on the link at any instance of time. (Assume propagation velocity is 3*108 ms-1.)
R = N/Tx
N = R * Tx = 50 * 106 * 0.23
= 11.7Mbit
Link Efficiency (Utilization)
Idle RQ:
If seach frame was simply transmitted immediately after the previous one then utilization would be 100%. However, Idle RQ also uses various control signals such as ACKnowledgements.
- Assuming no errors
The time before a new frame is transmitted is the time taken to transmit the data plus the propagation delay, then the time to receive the acknowledgement and its propogation delay.
i.e. TIx + TP + TackX + TP
The time taken to acknowledge the data is insignificant compared to the time taken to transmit it, so we can disregard it in our calculations. (We also ignore the time taken to check each frame as its received, perhaps for CRC errors.)
means the propagation delay is dominant.
means the transmission delay is dominant. - With errors
Where Nr is the number of attempts to transmit each frame. The value of Nr can be obtained from the bit error rate P of the link. If the link operates with frames containing Ni bits, then for values of NiP < 1 the probability that frame contains an error is :
NiP
Hence the probability a frame has no error is :
1 - NiP
Hence the number of transmission attempts required to transmit a good frame is on average :
For example, if error ratio P is 1 in 10000 then for 1000 bit frames, the probability of an error in frame is 0.1.
Hence 1 frame in 10 is, on average, corrupted. Average number of transmission attempts required to achieve successful transfer is : 1/(1-0.1) = 1.1
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