Es/No and Eb/No in OFDM system

hi all,

I"m working on a simulation of comparing OFDM with different modulation
schemes (QPSK,QAM) etc'...
My question is about the conversion from Es/No to Eb/No
I'm not sure if I need to take the cyclic prefix under consideration
for instance:

Es/No=Eb/No+10log10(log2M * Nc) + 10log10(CodeRate) + ...
.. 10log10(Tinf/(Tinf+Tcp))

is the last part correct?


where
Tinf - is the information period
Tcp - is the added Cyclic prefix period
Ttot = Tcp+Tinf


- thanks

On May 11, 4:27 pm, "partypooper" <sshango...@hotmail.com> wrote:
> hi all,
>
> I"m working on a simulation of comparing OFDM with different modulation
> schemes (QPSK,QAM) etc'...
> My question is about the conversion from Es/No to Eb/No
> I'm not sure if I need to take the cyclic prefix under consideration
> for instance:
>
> Es/No=Eb/No+10log10(log2M * Nc) + 10log10(CodeRate) + ...
> .. 10log10(Tinf/(Tinf+Tcp))
>
> is the last part correct?
>
> where
> Tinf - is the information period
> Tcp - is the added Cyclic prefix period
> Ttot = Tcp+Tinf
>
> - thanks

I didn't check your equation, but I think it is appropriate to include
the cyclic prefix as overhead in your Eb/No calculation. When I see Eb/
No, I think "energy per information bit over noise power spectral
density". This would suggest that the energy spent on the cyclic
prefix doesn't count, much how you must account for the code rate in
your calculation.

Jason