Priscilla,
As you mentioned that the robbed bits in the super
frames are only for voice signals, does this mean, if
I am only transfering data, I should tell me carrier
not use the robbed bits? Still a bit confused about
the clear channels, does this mean, there is not bits
robbed in this channel? are they still using ESF? if
it is esf, by default every 6, 12, 18 and 24 bit are
robbed, if these bit are part of 64k and they are
robbed, how can we still have 64k?
Thanks
Kent
--- Priscilla Oppenheimer <[EMAIL PROTECTED]> wrote:
>
> 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
>
>
>
>
>
>
>
> >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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