Minimum value of G (Amplitude) that guarantees an error probability of at least $10^{-2}$ in a 32-PAM transmission system

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.

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.

• 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 $10^{-2}$ !! – Dilip Sarwate May 7 '20 at 2:37

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 $$G \ge 49.9975$$.

• 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. – Dan Boschen May 5 '20 at 18:09
• 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. – Dhavids May 6 '20 at 19:14
• @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. – jithin May 8 '20 at 17:27
• @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. – Dhavids May 10 '20 at 21:14
• @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. – Dhavids May 10 '20 at 21:30

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 $$\frac{30}{32} P_{e1} +\frac{2}{32} P_{e2} = 0.01$$