Date: Mon, 5 Oct 92 16:17:30 CDT
From: rfranken@cs.umr.edu
Newsgroups: comp.dcom.telecom
Subject: Re: Confused About T1 Bandwidths
Organization: TELECOM Digest
X-Submissions-To: telecom@eecs.nwu.edu
X-Administrivia-To: telecom-request@eecs.nwu.edu
X-Telecom-Digest: Volume 12, Issue 763, Message 3 of 6

> Data Comm Folks,

> I am confused about the T1 usable rates.  My understanding is that the
> T1 lines can come in two speeds:

> 1. 1,344 Kbps i.e. 24 channels @ 56 Kbps each;  OR
> 2. 1,536 Kbps i.e. 24 channels @ 64 Kbps each.
 
> Now, what do I need to do in order to get the higher 1,536 speed: Is
> this the function of the terminating CSUs, the phone company circuit
> or both? The difference is 192 Kbps which is quite a lot and we would
> like to take advantage of this extra bandwith if possible.
 
> Do I need to tell the phone company when I order the circuit that I
> want the 1,536 Kbps line?
 
> My understanding also is that if the two CSUs are connected back to
> back via an in-house circuit less then 6,000 long they can indeed get
> the 1,544 Kbps usable bandwith rates since no framing bits are needed.

A T1 line represents 1's by alternating between high and low voltage
(called Alternate Mark Inversion) -- that is, 0's are represented by 0
volts, and 1's are represented by +voltage (if the previous 1 was
represented by a negative voltage) or by -voltage (if the previous 1
was represented by positive voltage).  This allows repeaters, which
must get their timing from the line, to remain in sync over long
streams of 1 bits (since there will be a voltage change between two
consecutive 1's).  However, any more than 14 straight zeros, and the
repeater may lose sync (since there is no voltage change between
consecutive zeros).  In a T1 carrying voice, 1's can be inserted where
needed to maintain 1's density (since you won't likely notice the
difference), but that is obviously not acceptable in a digital
high-speed data circuit. There are two ways around this:
 
 1) Use only 56KBps of each 64KBps DS0 channel.  This means seven out
of every eight bits are used.  The eighth bit is set to 1, ensuring
1's density.
 
 2) Use B8ZS coding.  What this does is replace any 00000000 byte with
10011010 (I don't think thats the correct bit pattern, but it will
suffice for this explanation).  Since no 00000000 streams occur,
sufficient 1's density will be maintained.  But, when the receiver
received 10011010, how does it know if the original byte really was
10011010 or 00000000? When 10011010 is sent in place of 00000000 the
sender deliberately causes a bi-polar violation on the fifth bit.
Normally, 10011010 should be sent (voltage wise) as +00-+0-0 or
-00+-0+0.  If a substitution for 00000000 is occurring, then +00--0+0
or -00++0-0 is sent.  The receiver then detects this bipolar violation
and substitutes 00000000.

 (A third method is to guarantee that your source will never send two
many zeros consecutively.  This is generally not possible with
computer data, which could be anything.)
 
So, you need to get a circuit from the telephone company that is set
up for B8ZS.  (If the circuit is a direct line connected by repeaters,
then you can probably use B8ZS, as most repeaters don't care about
anything but a voltage transition at least once every 15 bits.  But,
this is rarely the case.  Usually the telco will have some other
equipment in the line that will show an error on the line (becasue of
all the bipolar violations) if it is not configured for B8ZS.
 
For a fast serial line, framing bits are never needed (the computers
take care of figuring out where a byte starts).  If nothing inbetween
the sender and receiver cares about framing, and you had a CSU that
would support it, then you could use the full 1.544 (most repeaters
don't care about framing, but, for example, a DACS probably would).


Brett    (rfranken@cs.umr.edu)