Computer networks :Nyquist's theorem

Discussion in 'Homework Help' started by tavati, Jan 26, 2009.

  1. tavati

    Thread Starter New Member

    Jan 26, 2009
    i am reading a book by Tanenbaum on computer networks. i have some basic doubt about the max. data rate(MDR).

    the eq'n is : MDR = 2H * log 2 (V) samples per second

    if V uses m bits then MDR = 2H * log 2 (2^m) bits per second
    = 2H * m

    However, a line in this book on page 127 says, " if the symbol consists of 0 volts for a logical 0 and 1 volt for logical 1, the bit rate 2400bps. ( at a baud of 2400 bauds). if, however, if the voltages 0,1,2,3 are used, every symbol consists of 2 bits, ...... data rate of 4800 bps.

    this last example of four voltages does not gel with the concept. it has to be other way i.e. if we choose 2 bits to form a symbol, then there will be four symbols and hence four levels .

    rest all fits in but why Tanenbaum wrote like this ?
  2. mik3

    Senior Member

    Feb 4, 2008
    Maybe the book means represent 0, 1, 2 and 3 in binary form.
  3. tavati

    Thread Starter New Member

    Jan 26, 2009
    dear sir,

    i have one more question.

    when we talk about sending data in bits and talk about symbols- i want to just know the big picture here.

    suppose i have to transmit letter 'A' and suppose the binary is 11000110. so every bit is transmitted as a symbol ( i.e. a combination of bits which we decide as above ) or is a 1 transmitted as 1 only followed by another 1 followed by three zeroes etc.

    to say it other way is there a mapping of 0 and 1 with symbols as designed by engineers at the physical layer or have i mixed it up.
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    Oh dear you are getting mixed up.

    The equation you refer to is about binary sequences. Binary means there are only two levels or states available.

    Conventionally we designate these 1 and 0.

    So every time we send a piece of information (bit) it can represent either one of two pieces of information 1 or 0.

    No more no less.

    Your book then goes on to say that if we now have a system where we have four characters, say A, B, C, D or 0, 1, 2, 3 then every time send a single character it can represent one of four pieces of information.

    Four is obviously twice two as 4800 is twice 2400.

    In answer to the question in your other thread

    Yes there are standards. The most important is probably ASCII.

    Normally we don't write out the whole binary but consider the 1s and 0s in groups of 8 (octets)
  5. Skeebopstop

    Active Member

    Jan 9, 2009
    Baud rates can be defined over voltage thresholds. So lets map it out:

    0 volts = 00 binary (0 decimal)
    1 volts = 01 binary (1 decimal)
    2 volts = 10 binary (2 decimal)
    3 volts = 11 binary (3 decimal)

    so as you can see, we can send any 2 bit combination in a single line transmission, hence 2400*2 = 4800. This could be used to send the character 'A' at 4800 bps.

    01 00 00 01 = 1 volt 0 volt 0 volt 1 volt

    With 4 signals I transmitted 8 bits, 8/4 = 2.
  6. Skeebopstop

    Active Member

    Jan 9, 2009
    Textbooks confuse things often times. They must attach a formula to everything which can be daunting. Hell it even took me a minute of staring at your equation to know what Tanenbaum was after.