A data source produces 7-bit IRA characters. Derive an expression of
the max effective data rate ove a B-bps line for the following:
  e.  Asynchronous transmission, with a 1.5-unit stop element and a parity bit.
  f.  Synchronous transmission, with a frame consisting 48 control
bits and 128 information bits.  The information field contains 8-bit
(parity included) characters.
  g.  Same as b), except that the information fiels is 1024 bits.

For each case, compute the fraction g of transmitted bits that are
data bits. Then the maximum effective data rate R is: R=gB, where B is
the data rate on the line.

Hi cool2!!

Thank you for asking to Google Answers.

This is the solution that I found:

First of all remember the following:

·In asynchronous transmission, 1 start bit (0) was sent at the
beginning and 1 or more stop bits (1s) was sent at the end of each
byte. There may be a gap between each byte.

·In synchronous transmission, bits was sent one after another without
start/stop bits or gaps. It is the responsibility of the receiver to
group the bits.

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· Asynchronous transmission, with a 1.5-unit stop element and a parity bit.

There are 7 data bits, 1 start bit, 1.5 stop bits, and 1 parity bit, then:

g = 7/(1 + 7 + 1 + 1.5) = 
  = 7/10.5 = 
  = 0.666 

R = 0.666B

---------------------------------------------------------

· Synchronous transmission, with a frame consisting 48 control bits
and 128 information bits.  The information field contains 8-bit
(parity included) characters.


Each frame contains 48 control bits + 128 information bits, so each
frame will have a total of:

48 + 128 = 176 bits


The number of characters is 128/8 = 16, then the number of data bits will be:

16 * 7 = 112

then:

g = 112/176 =
  = 0.636

then:

R = gB = 0.636B

-----------------------------------------------------------

· Synchronous transmission, with a frame consisting of 48 control bits
and 1024 information bits. The information field contains 8-bit
(parity included) IRA characters.


Total bits = 48 + 1024 = 1072 bits

Number of characters = 1024/8 = 128

Number of data bits = 128 * 7 = 896

g = 896/1072 =
  = 0.836

R = gB = 0.836B

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For additional reference see the following document (look for
"Physical Interface" section at page 6):
http://www.cs.ust.hk/faculty/hamdi/Class/Training-M-Sol.ps


To see this document you need a PostScript viewer:
GSview 4.6:
Read carefully the instructions, note that GSview requires
Ghostscript. You must download Ghostscript separately, there is a link
in the page.
http://www.cs.wisc.edu/~ghost/gsview/get46.htm


If you don't want to install this software, just go after the
following link and look for the highlighted text " maximum effective
data rate":
http://66.102.9.104/search?q=cache:WbIGtxqNadQJ:www.cs.ust.hk/faculty/hamdi/Class/Training-M-Sol.ps+%22maximum+effective+data+rate+%22+b-bps&hl=es

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I hope that this helps you. Please request for any clarification
needed before rate this answer, I will gladly respond your requests.


Best Regards.
livioflores-ga