Shannon Limit on 64QAM

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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
 

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

To get to the question, the OP proposes to keep the bandwidth constant and increase SNR by oversampling and averaging. This belief is a commonly held and unfortunate mistake. Shannon's sampling theorem states that a continuous waveform may be exactly reconstructed by sampling at a rate = twice the bandwidth. Sampling at a higher rate produces data points, containing both signal noise, that are correlated with each other and contain no independent information. (If they added information, then the reconstruction would improve, in contradiction to the sampling theorem.) Since the noise samples are correlated, averaging them cannot reduce the noise power--hence SNR remains the same. This is the problem with the OP's logic and the answer to it.
 
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