Comments[grader]: Thank Mr. Alaparthi very much for his effort in uploading the scanned pages for problems 4.5 and 4.10 from Wozencraft and Jacobs. The typed version is also available now.

Question: In problem 1 of Homework 4, it is required to prove that the correlation in baseband is equivalent to the correlation in the passband. How could this be true if the basis functions in baseband are totally different from those of the passband. They can alo be different in their number, so we cannot achieve the same number of vectors.

Comment[student]: The problem 1 does not ask us to prove that correlation in both the cases is the same. Instead it says that decision reached by the demodulator after comparing the correlations is the same, regardless of the processing being done in baseband or passband.

Comment[Krishna:] The correlation is also the same up to a scale factor of 1/2.

Answer[Krishna:] Even though the basis functions are different for the passband signals and the baseband signals, nevertheless there is a valid vector representation for the passband signal and equivalent baseband signal. Hence, there is a valid correlation operation or inner product operation. Similarly, there is an energy for the passband and baseband signal as well. hence we can compute $Re<y,s_m> - \frac{1}{2} E_m$ both in passband and baseband.

I dont quite understand what you mean by they can also be different in their number, so we cannot achieve the same number of vectors. As I understand what you have written, it is not correct. The number of vectors will be the same whether you think of passband signals or baseband signals and will be exactly M (the total number of signals). The dimensionality of the space can however be different and complex instead of real, that is all.

question: It was mentioned that the translation and rotation properties don't affect the probavility of error. Is this valid only for AWGN noise or it is valid for any additive distributions?

Answer[Krishna]: Do you remember the argument about why translation and rotation do not affect the probability of error? The best thing to do is to look at the two codeword error probability where the two codewords are not along the X-axis. In this case, we used a rotation matrix to rotate the axis to make the signals lie along the x-axis. The rotation matrix also multiplied the noise vector and then we argued that if $n$ is a noise vector, then $T \ n$ also has the same distribution as $n$, where $T$ is a rotation matrix. Think about whether this will be true for an arbitrary distribution. The best way to think about this is to see if you can produce a counter example to this. This exercise is usually very helpful in figuring the crux of the matter. Also remember when uncorrelatedness implies independence.

Do the same exercise with translation and see if you can produce a counter example for any additive channel, where the noise is independent of the transmitted signal. How about if the noise is dependent on the transmitted signal?

Question: In prob 2 of the homework, which is 4.5 from Wozencraft's: It is ststed: The channel may be discarded without affecting P(e)min if P(s1)>= q. Evaluate q.

What is meant by: the channel is discarded?

Also, hat is meant by P(e)min?

Comments[student]: As I understand, P(e)min is the probability of error obtained by using the optimal decision regions. The condition for "the channel is discarded" should be "P(e)min >= 1-q", where the optimal decision is simply $s_1$.

Answer[krishna]: Correct. Channel is discarded means the output from the channel are not taken in to account in forming the decision. If one were to ignore the output of the channel, then the optimal decision is to choose the signal with the larger a priori probability of being sent and the probability of error in this case will be $1-q$.

Question: In prob 4 of the HW, which is 4.10 from wozencraft's. In the last requirement, it is required to find the minimum attainable error for specific cases with a 2 point constellations.

what is meant by the minimum attainable error. I guess it is not a lower bound, because it is a 2 point constellation in a 2-D basis, so, we can easily find the exact probability of error for AWGN channels

Answer[Krishna:] It is not the lower bound, rather the exact probability of error. They call this the "minimum prob. of error" because this is the minimum over all possible receiver structures and decision rules, i.e., this is the probability of error for an optimum receiver.

Question: In problem 3 do we have to prove that maximal ratio combining provides an optimal receiver, or can we assume that this is true and just calculate the probability of error for this scheme?

Answer[Krishna:] You cannot assume that maximal ratio combining is optimal.

Question: In problem 1, It is required to sow that 2 correlation demodulators are equivalent, one in the passband and the other in the baseband. These correlators will be based on correlating the received signal with the main signalls S1(t), S2(t) … in the pass band and SL1(t), SL2(t) … in the baseband? Or should we make this correlation with the basis functions?

Answer[Krishna]: You can take them to be the signals. But if you think a bit, it doesnt really matter since basis functions are also signals.

Question: In problem 1, Is the channel AWGN?

Comment[student]: Do we need this information that whether the channel is AWGN or not. What am I missing if dont assume it to be AWGN, please explain.

Answer[Krishna]: No, you do not need the channel to be AWGN. Let me ask you why do you think the channel has to be AWGN? Where do you think the Gaussianity or White property of the noise will come in to play?

However, you need to assume that the transmitted signals are all strictly bandlimited to a small bandwidth around the carrier frequency, which is something we have always assumed. You can also assume that the received signal is first sent through a bandpass filter which retains only the frequency components of the received signal in the signal bandwidth.

Question: In problem 3, are we sending pulses of energy Es in each of the L channels or we are sending pulse of energy Es/sqrt(L) in each of the channels?

Answer[Krishna]: We are sending pulses of energy $E_s$ in each of the L channels. An example of this is in wireless communications when you have one transmit antenna and several receive antennas. So there is actually only one input to the channel but many outputs.

Question: Inproblem 4, part c, it is required to find a certain fillter and its esponse to p(t). Does that mean that we need to make the full convolution between the filter and p(t) for all t?

Answer[Krishna]: Yes, you have to perform the full convolution

Question: Back to one of the above questions. In Problem 1,should I assume AWGN? I am asking this, because te correlators receivers were proved to be optimal for AWGN.

Answer[Krishna]: Okay, now I realize your concern. You can assume that the noise is AWGN since we have explicitly talked about a correlation receiver.

Question: In problem 3, it is required to design the optimal detector. Does the detector required should come after the demodulator that was shown in the figure, or should be a part of it?

Answer[Krishna]: After the demodulator.