Karl Reichert wrote: > I'm sorry but I still don't get it. And sorry for the -20/-30 mistake. > >>>>> Hello, >>>>> >>>>> I read the paper "RTnet -- A Flexible Hard Real-Time Networking >>>> Framework" at >> http://www.rts.uni-hannover.de/rtnet/download/RTnet-ETFA05.pdf and I >>>> have a question concerning chapter 3.2, formula 3. >>>>> T_slot = T_sched + t_slot - t_offset >>>>> >>>>> As far as I understand this formula, the slaves calculate their own >>>> starting point T_slot from T_sched (the planned transmission time of >> master), >>>> t_slot (slot duration?) and t_offset (offset between master and slave >>>> clock). To my mind, this doesn't makes any sence. Maybe I misunderstand >> one of >>>> the variables, but let me give a small example: >>>>> T_sched = 15 (absolute time point) >>>>> t_slot = 10 (duration) >>>>> t_offset = 20 >>>>> >>>>> Then, slave's starting point would be 15+10-20=5 >>>>> But how is it possible, if the master is the first one sending? I >> don't >>>> understand this formula. >>>> >>>> Oops, you stumbled over an impreciseness of the variables in that >> paper. >>>> They must read T_sched,master and T_slot,slave, i.e. the first one is >>>> related to the master view of time while the second one was transformed >>>> to the slave time base. >>>> >>>> Jan >>>> >>> But still, this doesn't makes sense to me. As I understand the paper, >> formula 2 and 3 should fix potential scheduling jitters (the time difference >> between scheduled and "real" sending time on the master). But how can this >> be fixed, if the formula deals with the T_master_sched and not the >> T_master_real (=T_master_xmit)? >>> >>> Let me give an example: >>> at the same point of time (the moment the master sends sth), master >> clock=10, slave clock=30 >>> the master jitter may be 2, which means that T_master_sched=10 and >> T_master_xmit=12 >>> t_travel may be 3 >>> t_slot (Slot-Length) may be 5 >>> >>> t_offset=12+3-35=-30 >> Hmm, my calculator prints "-20"... ;) >> > Okay, I agree, sorry for the mistake. > >>> T_slave_slot=10+5+30=45 >> And that would give 35 then, which is precisely the expected value (5 us >> >from the scheduled cycle start 30 us in slave time). 8) >> > But if slave starts sending at 35 (which means 15 from master's point of > view) then it would send at the same time the master does. The slot is 5, so > if master would start exactly at 10, this would work. But because of the > jitter, it starts sending at 12, so if it uses his whole 5 ticks long slot, > it would end at 17 and this would clash with slave, which starts sending > already at 15.
I think I see the misunderstanding: t_slot is not the slot size of the master slot here. It is the slot _offset_ (relative to the cycle start) of some arbitrary slave (or the master - then t_offset would just be 0). That offset must account for a) the time the preceding slot requires at least and b) the potential jitter of the preceding slot user. Thus, if 5 us is already the required size for slot n-1, then slot n would rely on jitter-free sending of frames in slot n-1 - no good idea in practice. Hope that helped to get the idea. Jan
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