Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shannon Limit on 64QAM

  1. Nov 11, 2011 #1
    I have a simple qusetion regarding how to interpret Shannon's limit. I am looking at this for fun and am not a system/theoretical engineer so please excuse my errors as I'm sure there a lot. I believe Shannon's limit can be written as:

    Capacity =log2(1+Br/Wn * Eb/No)

    where Br is the bit rate and Wn is the noise bandwidth.A 64QAM system has a theoretical capacity of 6bit/hz. Assuming the noise bandwidth is the same as the channel bandwidth, then Br/Wn=6 (6bits per symbol). The required Eb/No is therefore about 10. I believe this is the accepted answer. My question is if I look at the equation, it seems to me that one can increae the bit rate Br by over sampling and increase Br/Wn, thereby reducing the required Eb/No. This makes sense as over sampling and averaging is a common technique to to increase S/N. If this is true, then the equation says that if one over sample fast enough (make Br really large), one can bring Eb/No down to 1. I know this is wrong. Real 64QAM systems need a lot of Eb/No. So what did I miss in my interpertation of the equation.

    Thanks for the help
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?