Re: [Linuxptp-users] Doubts about PTP standard
PTP IEEE 1588 stack for Linux
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From: Namit A. <nam...@gm...> - 2016-01-26 10:33:53
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Namit Agrawal <namitagrawal88@...> writes: But the widely varying residence time can be taken care of in a simple (t2 - t1) and (t4 - t3). Because no matter what, the difference will always tell you the time it took to propagate including the residence time. I think the point that nobody explains to user is that when you are calculating the delay and offset by just dividing them by 2, you are assuming that propagation on both sides is the same which is not. So you cannot take average of something which is not equal on both sides. That's why you have to take into consideration the residence time. But this leads me to another question which is regarding the calculation done in the appendices of the IEEE Std. 1588 paper from page 201 to 219. Q. How and why is assumed mean path delay equal to calculated mean path delay?? Q. Why do we have to calculate separately mean path delay when a simple assumed mean path delay is giving us the answer?? |