On Fri, 2005-06-17 at 09:34 +0200, Marcus Metzler wrote: > Andreas> (orbital_position << 22) | (polarization << 20) | > Andreas> frequency_mhz > > I looked for some examples in my current dvbrc which is for 5 orbital > positions. With 207 transponders I found 32 cases of the same TSID > involving 117 of the transponders, i.e. between 3 aand 4 transponders > per TSID. If you exclude the cases where you have different orbital > positions to distinguish between the transponders you still have 28 > cases with 108 transponders. If you use the polarisation to > distinguish the transponders you get 29 cases (15 for V with 49 > transponders and 14 for H with 41 transponders) > and still 100 transponders. > If you take the first 4 digits of the frequency, all these remaining > cases can be resolved. > So a combination of transport ID, orbital position and frequency > should be enough. Polarisation does not really help much. So how do you want to distinguish the following two transponders without including the polarization? How does the bogus transport stream ID help? Can you give an example where my proposal creates the same ID for two different multiplexes? 1.0W 11.654 GHz H, TSID 0x0001 ONID 0x0001 1.0W 11.654 GHz V, TSID 0x0001 ONID 0x0001 Regards, Andreas
