stuck with this ....any one have a clue?

Posted on 2007-09-29
Last Modified: 2013-11-13
A character-interleaved time division multiplexer is used to combine the data streams
of a number of 110 bps asynchronous terminals for data transmission over a 2400-bps
digital line. Each terminal sends asynchronous characters consisting of 7 data bits, 1
parity bit, 1 start bit, and 2 stop bits. Assume that one synchronization character is
sent every 19 characters and in addition, at least 3% of the line capacity is reserved
for pulse stuffing to accommodate speed variations from the various terminals.
(a) Determine the number of bits per character.
(b) Determine the number of terminal that can be accommodated by the multiplexer.
(c) Sketch a possible framing pattern for the multiplexer.
Question by:ultra_mick
    LVL 28

    Expert Comment

    by:Bill Bach
    Sounds like a homework problem from an old Data Communications textbook...

    Author Comment

    yes its homework...i am currently persuing a Bach. Deg and i am stuck with this tutorial question....tried to research with the WWW but cant find my answer.....the problem is i dont understand the question. The reason i post this is not just to get a free answer....i hope experts in here can teach me / enlighten me how to derive with an answer. I AM not looking for someone to do my tutorial....but some one to Teach me how. :))
    LVL 28

    Accepted Solution

    So which part do you have a question about?

    Part A should be easy enough -- 7 + 1 + 1 + 2 = 11 bits per character.

    Part B & C:
    The line bandwidth is 2400 bits per second, and the data is coming from the terminals at 110 bits per second.  Simple math would indicate 2400/110, or 21.8 terminals can be multiplexed on the line.  However, we certainly can't use a fraction of a terminal, so we'd assume 21, but this still doesn't take into account the overhead required.

    We must first save off 3% of the bandwidth for pulse stuffing.  So, we should really start at 2400 * (1 - .03) = 2328, which would allow for 21.16 terminals.  However, for every 19 bytes sent, we need a SYN byte.  My question is "is it every 19th character, or do we send a SYN after 19 characters have been sent?"  The wording makes it hard to tell, but I'll assume that a SYN is sent AFTER each 19th data byte. So, there is 1 byte of overhead for each 19 data bytes, or 1/20th of all traffic is overhead for synchronization, so we have to get rid of another 5%.  Luckily, the way the math works out, we can simply say that we'll get 20 terminals on the line with the extra overhead.

    What would this look like?  In the first second, it would look like this:
       SYN - Slot1 - Slot2 - Slot3 - .... - Slot20 - SYN
    In the next seconds, it would look like:
       Slot1 - Slot2 - Slot3 - .... - Slot20 - SYN - Slot1
       Slot2 - Slot3 - .... - Slot20 - SYN - Slot1 - Slot2
       Slot3 - .... - Slot20 - SYN - Slot1 - Slot2 - Slot3

    The only thing I'm guessing on is the pulse stuffing, and on the wording of "every 19 characters", but you can make the necessary translations if needed.

    If you need a more detailed explaination, try this link:

    Author Comment

    thank you so have pointed me towards the light......... with what you have explained i can carry on with the rest of the tutorial on my own........thousands of thank you !

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