Discussion For Homework5

Question: In problem 4 you ask us to do problem 4.11 from Proakis 4th Edition. But isn't that what we have done in previous homework?

Answer[Krishna]: What is the MAP decision on what you think I meant assuming that could have been some errors in typing?

Notice the difference between an MAP and ML decision for this. If you use an ML detector, you would not factor the prior probability and think I really meant 4.11. Given that 4.11 was already on a homework, the prior probability of this being on the next homework is zero and hence the MAP detector would not pick this problem. It would think it is more likely that 5.11 was what was meant and it turned in to an error while typing. As you can see I am in a light hearted mood!

Question: In prob 2, it is written that: P(e/Sn) = P(e/Sn,x)P(x<x0) + P(e/Sn,x)P(x>x0). If we substitute in this equation with the previously concluded results, the P(e/Sn) will be function of x which is a variable How could it be?

Answer[Krishna]: You are correct. I am sorry I did not write this carefully. What I meant was

(1)
\begin{align} P(e|s_n) = \int P(e,x|s_n) dx = \int_{x < x_0} P(e,x|s_n) dx + \int_{x \geq x_0} P(e,x|s_n) dx \end{align}

which can be written as

(2)
\begin{align} P(e|s_n) = \int_{x < x_0} P(e|x,s_n) P(x|s_n) dx + \int_{x \geq x_0} P(e|x,s_n) P(x|s_n) dx \end{align}

Now use the two different values of $P(e|x,s_n)$ in the two different integrals. Since $x$ gets
integrated out, the result will only be a function of $x_0$ and hence you can optimize over $x_0$.

Question: In problem 1, you want us to calculate the probability of error for an M-ary FSK scheme. Should we assume it is orthogonal?

Answer[Krishna]: Yes, you can assume they are orthogonal.

Qestion: In Problem 6: Isn't there any dimensions or spacing between the different points of the given 8 AMPM constellations?

Answer[Krishna]: Assume that the nearest neighbors in each dimension are separated by $2A$.

Question: In problem 2: What does it mean: showthat M ary signalling is optimal for low spectral efficiency regime.

Answer[Krishna]: It means show that $\lim M \rightarrow \infty$ for any $\frac{E_b}{N_0}$ >-1.59 dB, the probability of error $\rightarrow 0$. This is what is predicted by information theory as the best possible $\frac{E_b}{N_0}$ in the low spectral efficiency regime and, hence, this is optimal.

Question: In Problem 5: Does each user transmit 4 possible wave forms $\{ s_1(t), -s_1(t), s_2(t), -s_2(t) \}$ so the signal constellation has 4 points? Is it even necessary to find a basis and create the signal constellation to solve the problem? Are $a_1, a_2$ random or deterministic?

Answer[Krishna:] Well, user one transmits either $s_1(t)$ or $-s_1(t)$ and user 2 transmits either $s_2(t)$ or $-s_2(t)$. That is, $a_1,a_2$ are random and equally likely to be $\pm 1$

question: In problem 1: It is required to calculate the prob of error of the M ary QAM signals. Is that for the square QAM signals? i.e. the 4 QAM, 16 QAM, 64 QAM only … or this is general for all the possible rectangular and square QAM constellations

Question: In problem 5: Is it required to find a closed form for the exact prob of error, or just to write the integration form?

Answer[Krishna:] Closed form for M-PAM, M-QAM. Leave it as a expression for M-PSK

Question: In the above question, I meant problem 5 of the multi usser communications

Answer[Krishna:] For the exact probability of error, leave it as an integral.

Question: In prob 7 in which we need to find a constellation for which the neighborhood is neither upper bound nor a lower bound. I am confused on how to show that. Should I find a constellation for which the neighborhood bound is an upper bound or some SNR and is lower bound for some others? Is that possible?

Answer[Krishna:] This means find a constellation for which when you apply the bound, it is not obvious whether it is an upper bound or lower bound. For example, it could turn out when you take the sum of a few pairwise error probabilities, there is overlap in some decision regions and some regions are not covered by the terms considered. Now, you cant be sure whether this is an upper bound or lower bound. This is what I meant.