Adam,
If the 6th, 12th, 18th and 24th bit of the
extended superframe are robbed, does this mean the
data part of a T1 will be 20x64k+4x56k=1504K(1.504M)
in stead of 1.536M?
Hope I am not missing something here.
Thanks
Kent
--- Subramanian Nallasivam <[EMAIL PROTECTED]> wrote:
> Hi Adam,
>
> Thanks for your clarification . I am having one
> more doubt. Where comes the
> concept of red and yellow alaram.
>
> Thanks,
> -Subbi.
>
> Adam Quiggle wrote:
>
> > Hey Ole,
> >
> > It really is 8000 samples. The number 8000 comes
> from Nyquist's
> > theorem which in a nutshell states that in order
> to properly represent
> > an analog signal in digital form we must sample
> twice the frequency
> > range. Since an analog signal is represented from
> 0-4Khz we must sample
> > an analog signal (2*4000) times a second in order
> to transport it with
> > minimal loss. Thus we get 8000 samples a second.
> Each sample is encoded
> > with 8 bits to represent an analog signal at
> anyone time, thus we get:
> >
> > 8 bits/samples * 8000 samples/sec = 64,000
> bits/sec (look familiar?)
> >
> > Now, if we look at a single frame of a channelized
> T1 we can see that
> > we have twenty-four 8-bit timeslots plus a single
> bit for framing or:
> >
> > Data Channels (23B+1D)
> > 24 channels * 8 bits/sample * 8000 samples/second
> = 1,536,000bps
> >
> > Framing Bit
> > 1 bit/sample * 8000 samples/second = 8000 bps
> >
> > So, we can see that for data we have 1,536,000 bps
> and 8,000 bps for framing,
> > which equals 1,544,000 bps. This is why people
> think of a T1 as 1.544Mb
> > but it's true throughput is only 1.536Mb.
> >
> > So, what is the framing bit used for? Framing
> allows the CSU/DSU to find
> > the timeslots in the channelized T1. Remember,
> that 8000 T1 frames are
> > flying out of the circuit every second, and the
> CSU/DSU must know how
> > to find the T1 frames in this circuit (don't ask
> me why, but I can't stop
> > thinking of Lucille Ball in the bon/bon episode
> :-)
> >
> > In order to maintain sync, the CSU/DSU must grab
> every 193rd bit off of
> > the line, string them together and see if it forms
> the "pattern" it is looking
> > for. The type of framing used describes the
> pattern to look for.
> >
> > When using Superframing twelve of these bits are
> strung together and must
> > form the following pattern of "100011011100". If
> it doesn't form this
> > pattern, the CSU/DSU is said to slip and it must
> then resync its
> > timing to the signal.
> >
> > Extended Superframing is an extension of
> Superframing and instead of
> > trying to collect twelve of these framing bits, it
> collects 24 of these
> > bits for syncing to the signal.
> >
> > So, what about robbed bit signaling (RBS)? RBS is
> not a function of
> > what type of framing you are using. RBS also
> known as Channel Associated
> > Signaling (CAS) is a technique in which
> supervisory and signaling information
> > is sent by removing the 8th bit of the 6th and
> 12th Superframe or the
> > 8th bit of the 6th, 12th, 18th and 24th bit of the
> extended superframe.
> >
> > The overall affect for those using Channel
> Associated Signaling is that you
> > don't need a D channel, however all your B
> channels use 56kbps.
> >
> > You can find more information on the following
> topics in McGraw Hill BCRAN:
> >
> http://www.bookpool.com/.x/p9zwbh4yt6/sm/0072124806
> >
> > Channelized T1/E1 frame format: 208-213
> > Framing and linecoding: 226-228
> > Robbed bit signaling/Channel Associated Signaling:
> 230-231
> >
> > The purpose of this book was to go beyond the
> Ciscopress material, which
> > focuses on the BCRAN exam material only and add
> information that seems
> > to have been ignored by Ciscopress's BCRAN. You
> will note that in this
> > book, when
> > a topic is not covered on the BCRAN exam, there is
> a note indicating such,
> > so that you don't review material not required for
> the exam....ok shameless
> > plug over... :-)
> >
> > Adam Quiggle
> >
> > At 05:31 PM 7/11/00, Ole Drews Jensen wrote:
> > >Hi Subbi,
> > >
> > >You are dragging me out in orbit now, but I'll
> try to hang on.
> > >
> > >To get back to the ISDN PRI...
> > >
> > >The 8000 samples, are you sure that it's not 8
> k-samples (8*1024) = 8192
> > >samples???
> > >
> > >That would make sense since the 192 bits (24*8)
> multiplied with 8192 equals
> > >1572864 which again equals 1536 kbps
> (1572864/1024). That brings us back to
> > >24 channels where the last one is only used for
> signaling - not data.
> > >
> > >Ole
> > >
> > >~~~~~~~~~~~~~~~~~~~~~~~~~~
> > > Ole Drews Jensen
> > > Systems Network Manager
> > > CCNA, MCSE, MCP+I
> > > RWR Enterprises, Inc.
> > > [EMAIL PROTECTED]
> > >~~~~~~~~~~~~~~~~~~~~~~~~~~
> > >
> > >-----Original Message-----
> > >From: Subramanian Nallasivam
> [mailto:[EMAIL PROTECTED]]
> > >Sent: Tuesday, July 11, 2000 4:14 PM
> > >To: Ole Drews Jensen
> > >Cc: [EMAIL PROTECTED]
> > >Subject: Re: Some ISDN PRI questions
> > >
> > >
> > >Hi Ole,
> > >
> > > Let me take the case of T1. In T1 all
> the 24 channel are used for
> > >sending data. There will be one framing bit which
> constitiute to 193 bits.
> > >So
> > >193 * 8000 = 1544 kbps. And one bit from every
> sixth frame in case of
> > >superframe(which consists of 12 T1/DS1 frames)
> will be robbed for
> > >signalling.
> > >The robbed bit will be the least significant bit
> and this concept is known
> > >as
> > >robbed-bit signalling. If I am wrong then please
> correct me.
> > >
> > >Thanks,
> > >-Subbi.
> > >
> > >
> > >Ole Drews Jensen wrote:
> > >
> > > > Hi Subbi,
> > > >
> > > > That would still only leave 184 * 8000 =
> 1472000 bps or (1472000/1024)
> > > > 1437.5 kbps for data.
> > > >
> > > > Ole
> > > >
> > > > ~~~~~~~~~~~~~~~~~~~~~~~~~~
> > > > Ole Drews Jensen
> > > > Systems Network Manager
> > > > CCNA, MCSE, MCP+I
> > > > RWR Enterprises, Inc.
> > > > [EMAIL PROTECTED]
> > > > ~~~~~~~~~~~~~~~~~~~~~~~~~~
> > > >
> > > > -----Original Message-----
>
=== message truncated ===
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