Ben Greear <greearb@xxxxxxxxxxxxxxx> writes:At least some out-of-tree drivers seem to be able to do this. For instance, these guys:
I think if I could support these scenarios below, I would have
everything I need:
* Configure T1 as unchannelized bitstream, bridge entire thing to
second T1.
I think it should be easy with such card, though I think the drivers
can't currently do that.
My assumption for bridging a bitstream is that timeslot sync is not absolutely critical. However, IF* Configure channels 1-5 as a bitstream and bridge that to channels
1-5
of a second T1. (random proprietary bit-streaming protocol,
I think the hardware would permit that. Probably needs additional
driver support and I'm not sure about timeslot synchronization (for
HDLC, sync doesn't matter).
The key for me is that if you ever miss a slot in bit-stream mode, you can never make it up becausewould probably bridge HDLC just fine, but handling
HDLC as frames would be more efficient I think.)
Not sure, maybe yes (less data to bridge due to bit stuffing, flags etc.),
maybe not (variable length of HDLC frame).
I'd be happy with a software approach. In fact, if I could get a framed packet (ie, I know thatchannels 6-10 configured as an HDLC interface, bridged as HDLC
frames to channels 6-10 of a second T1. (PPP & other protocols over
HDLC)
channels 10-24 each configured as a separate bit-stream, bridged to
channels 10-24 on the second T1. (Voice)
I think the above could be a problem - I think common T1/E1 cards
can do only one stream at once.
I wonder if it can be done in software - the hardware framer would
have to pass all data transparently, and it would be demultiplexed,
processed and then multiplexed by the driver. Quite complicated,
but I think at 2 Mbps it wouldn't be a CPU performance problem.
Excellent. I actually want to write the bridge logic myself in user-space..I just need the driver* Configure entire T1 as HDLC transport, bridge HDLC frames from one
T1 to the other.
Easy even with existing drivers I think (no bridge support but it's
trivial).