Re: Full Duplex and Hub


Robert Redelmeier wrote:
q_q_anonymous@xxxxxxxxxxx <q_q_anonymous@xxxxxxxxxxx> wrote in part:
I guess that if a comp tries to send while it's receiving
(i..e. acts full duplex) , then it'll prob cause a collision
in the hub, propagated to all comps, but undetected by comps
'cos they're running full duplex.

Possibly, but I think that would depend on the the ethernet
chip and drivers software. Since the JAM signal that indicates a
detected collision is not supposed to exist on full-duplex networks
(filtered at switch), the card/driver might [silently?] fall-back
to half duplex.

After all, almost all ethercards are capable of full-duplex,
but not all networks are.

-- Robert

Getting philosophical, but not so much about cards - but network
interfaces. THerefore not so much distinction between ethercards or
interfaces, and the network being FDX or HDX. Since Hubs, switches,
any network device, has network interfaces, whether portions of network
is FDX or HDX is totally a function of that.

Regarding the Jam signal
my understanding and memory of what seifert has written previously on
usenet, is that the jam signal is a continuation of the frame, it is to
lengthen the frame, nothing unique that is detected. And also, it's not
even necessary nowadays.
Since a jam is part of the frame, it's not "filtered by a switch".

Maybe it was necessary in the days of really long coax. To do with
ensuring that the frame stretched from one end of the coax to the
other, - I guess it should anyway - but the jam signal (extending the
frame) is for some reason only necessary in the context of a
collision.


Switches can operate in FDX of HDX. If you attach a hub to a switch
port, it has to operate HDX. (at least at that port) (I guess that
either - Other ports will be FDX and the switch will buffer accordingly
so the HDX port won't receive(from the switch) whilst that HDX port
sends outside the switch. OR - any port communicating with a HDX port
will act HDX - a form of auto-negotiation will take place, though that
seems like it might be inefficient ).

.

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