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From: hramcha on 19 Mar 2010 08:00
>On Mon, 26 May 2008 09:16:45 -0500, "zoncolan" <imarote(a)hotmail.com>
>wrote:
>
>>Hi,
>>
>>I'm working on Turbo Codes about DVB-SH standard. This employ QPSK and
>>16QAM. I've done several simulations and now I have a graphic BER vs
Eb/No
>>for code rates 1/5, 2/9, ... , 1/2 and 2/3.
>>
>>I know that for BPSK one of the expressions for the Shannon limit is
>>
>> Eb/No >= 1/2r * [2^(2r)-1],
>>
>>and if we have r=1/2, Eb/No>=1 or Eb/No>=0db.
>>
>>My question is, does anyone know a similar expression for QPSK and 16QAM
>>(with the term code rate in it)?
>>
>>Thanks,
>>
>>Iván.
>
>I'm not certain that I completely understand the question, but there's
>a chance that what you're looking for is pretty simple.
>
>The usual expression for Capacity solves for C, in bps/Hz with SNR as
>an input variable. With a known modulation (e.g., 16-QAM) and code
>rate, the modulation density or bps/Hz is already known. So
>rearranging the equation to solve for SNR provides the "capacity" in
>terms of SNR for that modulation and code rate.
>
>Converting from SNR to Eb/No or Es/No is then straightforward.
>
>Hope that helps a bit, if not, feel free to clarify.
>
>
>Eric Jacobsen
>Minister of Algorithms
>Abineau Communications
>http://www.ericjacobsen.org
>
>Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
>

I have a similar question for calculating capacity for 16 QAM.

Is the following math correct when deriving equivalent SNR for capacity
value.

r = log2(16)*code_rate

SNR_lin = (2^r-1)/r


Based on this for code rate 2/3 and 16 QAM, SNR_lin = 2 or 3 dB,

IS this correct ?









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