Im gonna have to disagree with you Priscilla, although you are "technically"
correct. <g> Let me explain.

T1 rules state that there must be a 15% ones (marks) density, or no more
than 15 consecutive zeros (spaces). Although AMI/B8ZS is only a line coding
protocol (how marks and spaces are represented), In my experience telcos
will force the least significant bit (LSB) in each channel to be a 1, when
the coding is AMI, except RBS frames obviously.

Data is notorious for having long strings of zeros, and apparently is not an
issue for voice. I dont know what the bits are doing when there is not an
active voice conversation on a channel. So if 2 consecutive DS0's have the
robbed bit a 0, I don't know if this breaks the T1 rules. Bunches of zeros
aren't a problem with B8ZS, but does cause timing issues in AMI. Hence the
telcos forcing the LSB to be a 1. The other thing that I have not verified,
is what the actual throughput is on a AMI DS0, so I am bot 100% confident in
my theory. :)

Scott Meyer
CCNA, CCDA, MCSE, etc
[EMAIL PROTECTED]

-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of
Priscilla Oppenheimer
Sent: Tuesday, July 25, 2000 3:59 PM
To: [EMAIL PROTECTED]
Subject: Re: Clear channel question?



At 02:12 PM 7/25/00, Lonnie Paschall wrote:
>Clear channel is a t1 with B8ZS (Binary Eight Zero Substitution) line
>coding. AMI line coding steals 1 bit from every 8 for timing. Which results
>in 56Kbps instead of 64Kbps. B8ZS coding allows you to use the full DS0
rate
>of 64Kbps by using the BPV's in the signal for timing. I may not be 100%
>technically right but it gives you a good picture of whats going on.

It's true, you aren't quite technically right, I don't think. ;-)

Whether you use B8ZS or AMI bit encoding won't affect whether you get 56
Kbps or 64 Kbps. It's the signalling method that robs bits and causes a
single DS0 to be only 56 Kbps instead of 64 Kbps. If you're not using
robbed-bit signalling, then you have a "clear channel."

Alternate Mark Inversion (AMI) simply means that each logical 1 bit is
transmitted as a positive or a negative pulse, after which the line voltage
always returns to zero. A logical 0 bit is transmitted as a zero voltage.
This format is called AMI because each pulse, or mark, is of opposite
polarity from the previous pulse.

Binary Eight Zero Substitution (B8ZS) is an enhancement that deals with the
problem with AMI whereby a long sequence of zeros provides no activity on
the line and is indistinguishable from a loss of signal. With B8ZS, any
sequence of eight consecutive zeros is replaced on the line by: 4 zeros, a
bipolar violation, a valid pulse, a zero, another violation, and a valid
pulse.

(A bipolar violation is not a mental disease. It's an intentional breaking
of the rule that says if the last pulse was negative then the next pulse
must be positive. ;-)

Robbed bits are used to transmit voice signalling information, such as
on-hook, off-hook, etc. Voice signaling is placed in the least significant
bit position of every DS0 in the 6th and 12 frame of every superframe, and
the 18th and 24th frame of an extended superframe (ESF).

If you use all four bits, then they are called the A,B,C, and D bits, and
they provide 16 possible value for voice signalling. If these bits are
being robbed, then a T1 or fractional T1 cannot carry transparent channels
of 64 Kbps each. In the U.S., carriers offer some number of 56-Kbps
channels, using only seven bits in every octet in order to avoid the data
corruption that would result from the insertion of signalling bits. If you
don't want this "feature," you could ask the carrier to provide you a set
of transparent 64 Kbps channels. This is sometimes called "clear channel."

I must admit, after getting to this point, though, that the term "clear
channel" could have other meanings as well. We would have to know the
context to give you more help with your question

Priscilla





>Lonnie
>
>""Gert Jan"" <[EMAIL PROTECTED]> wrote in message
>8lkd3a$j8q$[EMAIL PROTECTED]">news:8lkd3a$j8q$[EMAIL PROTECTED]...
> > On the Foundation Routing and Switching there was a question What is a
>clear
> > channel?
> > Anyone met this qeustion and can provide me some explanation?
> >
> > Thanks,
> > Gert jan
> >


________________________

Priscilla Oppenheimer
http://www.priscilla.com

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