Date: Fri, 1 Nov 1996 14:04:40 +0100 (MET) From: Christoph Kukulies <kuku@gilberto.physik.rwth-aachen.de> To: grog@lemis.de (Greg Lehey) Cc: questions@FreeBSD.ORG Subject: Re: Any ISDN-BRI cards work under FreeBSD? Message-ID: <199611011304.OAA09714@gilberto.physik.rwth-aachen.de> In-Reply-To: <199611010818.JAA19813@freebie.lemis.de> from Greg Lehey at "Nov 1, 96 09:18:21 am"
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> Terry Lambert writes: > >>>> done it yet. They currently only run single channel connections (you > >>>> can have two different connections to different destinations). The > >>>> theoretical maximum throughput is 8 kB/s (64,000 bps), which is > >>>> somewhat less than the theoretical maximum of the 115.2 kbps lines > >>>> (11.52 kB/s). > >> > >> Why do you divide by 8 in the one case and by 10 in the other? > > > > My guesses: > > > > 8) 8*8k = 64k; conversion is for sync framing > > > > 10) 1 start + 8bits + 1 stop = 10bits; conversion is for > > async framing > > I'm surprised you need to guess. Yes, that's correct. People seem to > have problems understanding this one, so I'll go into a little more > detail: > > Synchronous transmission is block oriented. Various techniques are > used to recognize the beginning and end of the block, and all the data > in a block are sent without any delay between the bits. When > transmitting 8 bit bytes (octets), there's a ratio of 8 bits per > octet, so 64 kbps becomes 8 kB/s. But in HDLC you have bit stuffing (8th bit or nineth?). Doesn't this have to be taken into account? > > Asynchronous transmission is character oriented. Each character > starts with a start bit, then come the data bits, then one or more > stop bits. Nowadays there is only one stop bit, but that expands each > byte to 10 bits. The advantage is that you don't need any specific > timing between characters (thus the term asynchronous), so it's quite > well suited to things like keyboard input. In sync transmission, each > input character would have to be made into a block. The disadvantage > is the significantly lower data rate. > > > A more interesting question might be 64k + 64k = 128k. 128k != 115.2k. > > Yes, this is where we came in. Currently, the driver can't do that. > Somewhere in the back of my head I have a recollection that TCP can do > it, though: you just set up two routes, and it should be able to pass > packets down the route with the shorter output queue. Can anybody > expand on this? > > Of course, 128k != 128k as well if one is sync and the other is > async. 128k sync is 160k async. > > Greg > --Chris Christoph P. U. Kukulies kuku@gil.physik.rwth-aachen.de
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