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pretty much new here.

This question comes from an Online course quiz which i have already completed but cant seem to get a good sleep over, just because i cant figure it out.

Below is the question in the Image. 32-PAM assignmnet

Given the range of the sample distribution and the error probability, from what I know,

$P_{err} = erfc(G/\sigma)$

IMO, $\sigma$ is assumed to be the error energy, which we were told is given by

$\sigma = \Delta^2 / 12$

where $\Delta = (B - A) / 2^R$

$B - A = (100 - (-100)) = 200$

and $R$ is the Range of the various intervals which I at one point chose to be 32 and at another point chose to be 5.

From some programmed online calculator, I got the inverse error function of $P_{err}$ given by $erfc^{-1}(0.01)$ (which corresponds to $(G/\sigma)$) to be 1.821, but this is where it all goes bad, as I keep getting wrong values for $G$ which I presume is caused by the wrong results from the computation of $\sigma$.

I know i might be doing it all wrong, and that's why am here.

Thanks in advance.

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  • $\begingroup$ One would hope that the question would ask for the minimum spacing that guarantees an error probability of at most $10^{-2}$ instead of at least $10^{-2}$ !! $\endgroup$ – Dilip Sarwate May 7 at 2:37
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I am not sure if you can use $P_{err}=\text{erfc}(G/\sigma)$ because noise is not gaussian distributed. Here is my take on it.

Assuming uniform probability of transmission for all 32 symbols $x_i$,the received signal $y=x_i+n$ so given that $x_i$ was transmitted $y$ is also uniformly distributed in the interval $[-100+x_i,100+x_i]$. Suppose say the transmitted symbol was $3G$. The range of $y$ is $[-100+3G,100+3G]$. If $G \ge 100$, there would be no issue even if noise occurs. You would always detect the correct symbol if you use appropriate boundaries ($|y-x_i| \le 100$). Suppose say $G \lt 100$ so these boundaries overlap. If $G=75$, what would happen if we receive value $y=150$? The transmitted symbol could have been either $G$ or $3G$. So we can choose either $G$ or $3G$ with probability of $0.5$. Similarly, on the other side if $y \gt 275$, you can choose $5G$ as transmitted symbol with probability $0.5$ So the correct decision will be taken when $175 \le y \le 275$, so $P_{err,x_i=3G}=0.5$.

So if you want $P_e=0.01$, for the symbols having 2 neighbors, you can distribute the error probability evenly on both sides ($0.01$ with each of probability 0.5). If youur transmit symbol was $G$, the overlap of regions will be at $G+100-0.005=99.995$ which will be your $2G$. So $G \ge 49.9975$.

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  • $\begingroup$ Oh good point Jithin! (that it's not a Gausisan tail probability)- I see now in the fine print of the question that the distribution and B and A are all specified, I missed that...deleting my incorrect answer. $\endgroup$ – Dan Boschen May 5 at 18:09
  • $\begingroup$ Wow, stuffs like this were never mentioned in the lecture. I will sit with this tomorrow and absorb it all and then proceed to check if this is right. Thanks a lot for the help guys. $\endgroup$ – Dhavids May 6 at 19:14
  • $\begingroup$ @Dhavids I am curious to know which online course is this. Because I have hardly come across pure digital communication courses online with quiz and exams. $\endgroup$ – jithin May 8 at 17:27
  • $\begingroup$ @jithin This is a Cousera 8 weeks dsp course, you can access it here. It covers mostly the basics and there are weekly quizzes as well as jupyter notebook assignments. It was a fun ride for someone like me who just wanted a taste of what DSP is all about. $\endgroup$ – Dhavids May 10 at 21:14
  • $\begingroup$ @jithin I Just plugged in both answers (49.99 and 50) and both were deemed incorrect. Although i can get a good sleep over it now, i still want to understand how is done. I will probably mail one of the instructors. Thanks a lot. $\endgroup$ – Dhavids May 10 at 21:30
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You need to distinguish between:

  1. The error rate for the inner symbols - the error is half of the overlapped segments between the symbol to its closest neighbors (by symmetry we consider one and multiply by 2). $$P_{e1} = Pr(|n|>100-G) = 2*Pr(n>100-G) = 2 \frac{100 - G}{200}$$
  2. The error rate for the points 31G and -31G - same as above bu only for one segment: $$P_{e2} = Pr(n>100-G) = \frac{100 -G}{200}$$

It remains to solve $$\frac{30}{32} P_{e1} +\frac{2}{32} P_{e2} = 0.01 $$

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