Shannon Limit on 64 QAM 1

[zmoose]

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 Eb/No. This makes sense as over sampling and averaging is a common technique to reduce noise. It seems to me that 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 can't be done. Real 64QAM systems need a lot of Eb/No. So what did I miss in my interpertation of the equation.

Thanks for the help

 

[2dye4]

I may not fully understand but here a try.

Oversampling doesn't get you anywhere unless there is more bandwidth to be sampled.

The bandwidth is already limited to to the razors edge by the modulation method to reduce the noise and oversampling cannot get any additional information.

Remember the Shannon theorem states that sampling at the required rate is sufficient to perfectly capture the entire waveform.
Oversampling is performed in real life to compensate for imperfect hardware and is a practical not theoretical issue.

 

[IRstuff]

Sampling at the limit mearely guarantees that there's sufficient sampling to receover the signal. Phasing is anathema to this problem. Oversampling ensures that phasing is not an issue.

TTFN

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