How to explain this BER vs SNR behaviour?

Question

I wrote a script in MATLAB to simulate an LDPC created using an algorithm of a colleague of mine. The parity check-matrix is regular and has dimensions $1012 \times 1518$ and the code rate is $R=1/3$.

The channel I'm using is a BPSK one with added Gaussian white noise. The mapping of the bits I used were the values $-1$ and $1$. The decoding method is sum-product (or belief propagation).

To perform the simulation, I used as a guide the Example 2.6 in the 36th page of this paper. I'm getting this curve:

This is really different from what I would have expected. The fact that the function is no monotonically decreasing was very shocking. Also, the fact that there is a BER of approximately $1$ for a positive SNR doesn't make any sense to me.

I'm thinking that, maybe, what is wrong is how I defined the a priori LLRs or the SNR. Given that I used $E_b = 1$ to map the bits, then I used some formulas from the section 3.2 of this paper:

$$R\cdot E_b = E_s$$ $$N_0 = 2\sigma^2$$

So, using the formula mentioned in the Example 2.6 of the first paper, we get that

$$r_i = 4y_iR\frac{E_b}{N_0}$$

I believe that this weird behaviour is coming from that formula, maybe a difference of definitions between the two papers that I didn't take into account. I don't really know, what do you think is the reason of the function having that shape?

Extra info:

• The SNR axis in the plot corresponds to $R\frac{E_b}{N_0}.$
• I've read here a question that could solve my problem, but I'm not sure if it applies directly. Here, I don't have any information about the data bit rate that is mentioned in that question. I don't really know what that quantity they call $R_b$ stands for, either.

Answer 1

In a BPSK system, the SNR at the output of the receiver's matched filter is $$\text{SNR}=\frac{E_b}{\sigma^2_n}.$$ In this formula, $E_b$ is the energy used to transmit one bit, and $\sigma_n^2$ is the power of the noise at the filter output.

Assuming an AWGN channel, the noise process at the filter input has constant PSD; this value is usually denoted $N_0/2$. The Wiener-Khinchine theorem relates the PSD and $\sigma_n^2$: $$\sigma_n^2=\frac{N_0}2,$$ so the SNR may be written as $$\text{SNR}=\frac{2E_b}{N_0}.$$

Note that the convention is to plot the bit error rate vs $E_b/N_0$ (note the missing $2$ in the numerator). The reason is that for quadrature systems, the total noise power in the receiver is $N_0$.

In a coded system, each information bit is transmitted using several coded bits. For example, a Hamming (7,4) code uses 7 coded bits to transmit 4 information bits; the code rate is $R=4/7$. In gneral, $R=k/n$, for a code that transmits $k$ information bits using $n$ coded bits.

To make a fair comparison between different codes, we usually keep $E_b$ constant. This means that the energy of each code bit depends on the code rate: $$E_c=RE_b,$$ or $E_b=E_c/R$.

In simulation, it is often convenient to set $E_c=1$. Then, we have $$\frac{E_b}{N_0}=\frac{E_b}{2\sigma_n^2}=\frac{E_c}{2R\sigma_n^2}=\frac{1} {2R\sigma_n^2}.$$