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LDPC and Shannon limit

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Bratija

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shannon limit

Hi!
I know similar questions have been around, I searched the forum but I couldn't find the answer. I'm running some LDPC on Matlab, AWGN channel and BPSK modulation. If I have code rate and SNR-BER graph (data), how can I calculate and plot Shannon limit? Thanks in advance.
 

shannon limit qpsk

Shannon limit for gaussian channel is C = ΔF log (1+S/N)
 

ldpc ber awgn

ring0 said:
Shannon limit for gaussian channel is C = ΔF log (1+S/N)
Click to expand...

For what does ΔF stand for? If it's bandwidth, than I'm at the beginning because I don't know it (Matlab simulation, BPSK, AWGN)
 
shannons limit

For BPSK, you may substitute ΔF with the symbol rate.
 

ldpc shannon

Can you please be more specific? For example, I have code from matrix 600x4000, from that code rate is 0.85 and message length is 3400 bits. I modulate coded message (which length is 4000 bits) with BPSK and send it through AWGN channel for different SNR and calculate BER. Then I plot it and get this.
2ebwln7.jpg


And my question is how can I plot here Shannon limit? Or have I mixed up something?
 

shannon limit graph

^^
I knew that it doesn't depend on BER, but I thought it depends on code rate. I have found -1.6 dB before, but I thought there was more to it. Thank you!
 

shannon limit matlab

that's what i needed! thank you very much!

Added after 2 hours 14 minutes:

only thing, i must use continuous output channel becaus bpsk decoder gives out LLR (log like-hood ratio)... last check, for this system (LDPC encoder -> BPSK modulator -> AWGN channel -> BPSK demodulator -> LDPC decoder) Eb/N0 and SNR which i use in AWGN channel are same, right?
 

dear bratija i am currently working on ldpc. i am facing the problem in ber curve. can u send me the codes which u have written.i think its very help ful to me.
 
Re: shannon ldpc

Here, R is the coding rate defining R=k/n, with bits/bits. What's the difference between this coding rate R and Shannon Capacity C ? Why can "Substitue C by R" ?
Another question, what's the difference between this coding rate R and the "R" in R≤C=Blog2(1+R*Eb/N)? While, the "R" in R≤C=Blog2(1+R*Eb/N) is with "transmitting R bits/sec" .

Thanks,

mathuranathan said:
Hi,
Yes.. For each code rate there is a shannon limit. For each code rate, there is a minimum Eb/N0 above which a low error transmission is guarenteed.

You can find this limit for each rate by following equation...

For binary input, binary outpur AWGN channel the Shannon Capacity equation is,
C = 1 - H(P) -------> (1)

where H(P) is the entropy of the system given by,
H(P) = -Plog2(P) -(1-P)log2(1-P) ------->(2)
where P = 0.5 erfc(sqrt(R*Eb/N0)) ; R => code rate

Substitue C by R (code rate) and Eb/No by (Eb/No)min,

so the equation becomes,

R = 1+Plog2(P)+(1-P)log2(1-P); with P = 0.5 erfc(sqrt(R*(Eb/N0)min))

For each value of code rate R (0.85 in your case), find the minimum Eb/No that satisfies this equation.
This becomes your shannon limit

Regards,
Mathuranathan

Added after 5 minutes:

attached the pdf which details this procedure...
Click to expand...
 

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