[36059] in North American Network Operators' Group
Re: Clear Channel on a T1
daemon@ATHENA.MIT.EDU (Ron Buchalski)
Fri Mar 23 12:57:09 2001
From: "Ron Buchalski" <rbuchals@hotmail.com>
To: vandry@TZoNE.ORG
Cc: nanog@merit.edu
Date: Fri, 23 Mar 2001 17:51:42
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Good point, Phil. I believe that the modern T1 spec states that a maximum
or 45 zeros can be sent before the CSU will inject a '1'. I don't have a
spec handy to verify it.
Of course, with B8ZS, you'll never approach this long string of zeros on the
line, so it's a moot point. B8ZS replaces eight zeros with 4/4 ones/zeros,
in a known bipolar violation pattern. This will insure that ones' density
is met on the line, and the circuit endpoints can replace the injected ones
with eight zeros and restore the bit patterns of the modified DS0s.
It's also the source of grief when a telco provisions one segment of a T1
circuit without B8ZS. When this non-B8ZS segment sees too many zeros, ones
will be injected in the stream, corrupting the data.
Of course, this NEVER happens anymore...right? ;-)
-rb
>From: Phillip Vandry <vandry@TZoNE.ORG>
>To: "Ron Buchalski" <rbuchals@hotmail.com>
>CC: nanog@merit.edu
>Subject: Re: Clear Channel on a T1
>Date: Thu, 22 Mar 2001 18:04:46 -0500 (EST)
>
>
>Please correct me if I'm wrong (because I'm not _that_ certain of this) but
>I think you still need the 1's density to maintain timing even in the
>absence of repeaters. If you have a whole bunch of zero bits in a row the
>voltage on the line will remain 0 for the length of the series. You get no
>indication of when one bit ends and another starts. If the string of zeroes
>is very long, the receiver might miscount the number of 0 bits that the
>sequence represents (more than intended if its clock is running faster than
>the sender's, less than intended otherwise). You don't have this problem
>with a long string of 1 bits because the voltage transitioning from + to -
>or back at every bit boundary.
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