# Shannon Limit on 64QAM

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

anorlunda
Staff Emeritus
sophiecentaur
Gold Member
ping @sophiecentaur , another communications question Sophie.
You really can make my brain hurt, at times!!
Shannon doesn't actually commit to the amount of processing needed to reach his limit; it's very much a theoretical limit. This could be the problem. I am very rusty in that direction, sorry.

anorlunda
Staff Emeritus
You really can make my brain hurt, at times!!
Well, that's good isn't it? :-) Thanks for your help.

I hope you don't mind that I send these your way. You are my communications go to.

• sophiecentaur
marcusl
This is not a correct statement of Shannon's capacity formula. It is missing the leading factor of bandwidth$$C=W\log_2(1+SNR)$$in bits/s, where W is bandwidth in Hz. The SNR may be expressed as $$SNR=\frac{P_s}{N_0W}=\frac{B_rE_b}{N_0W}$$ with P_s the signal power and N_0 the noise power spectral density, so this part is ok. However, $\frac{B_r}{W}$ is not bits/symbol--there is no reference to symbols in this equation.