MICRO-461: Exercise 2 Problem 3

Dear Prof. Enz, dear TAs,

May I ask a few questions?

I have some about Exercise 2 Problem 3, equations 1, 2, 3, 4 and 8.

1) I don't understand how can I calculate the expression of Em for equations 1 & 2?

2) I don't understand how to get the expressions of equations 3 & 4. More specifically how to obtain the expression under the square root in Q? Where does the factor 2 for Q come from in eq. 4?

Thank you,

Alexandre

Re: Exercise 2 Problem 3

by Hung-Chi Han - Thursday, 23 June 2022, 07:55

Hi Alexandre,

1) (1) is just a average energy for 4-ASK modulation. However, there is a typo in (2), E_s should be replaced by T_s. To calculate the average energy, you just need to give the corresponding amplitude to the equation given by problem 3. For instance, amplitude is A/2 for inner points and is 3A/2 for the outer points.

2) I think there is a mistake in the derivation. Using the P_e = Q(\sqrt(2E/N_0)), It should be:

P_e,in (probability of error per inner constellation point) = Q(\sqrt(A^2 T_s/ (4 N_0))), since E_b = A^2 T_s / 8
P_e,out (probability of error per outer constellation point) = Q(\sqrt(9A^2 T_s/ (4 N_0))), since E_b = 9A^2 T_s / 8

I hope it answers your question.

Best,
Hung-Chi

Re: Exercise 2 Problem 3

by Joël Lingg - Thursday, 23 June 2022, 10:35

Hi Hung-Chi,

Thanks for your answer.

A short follow-up: Shouldn't we consider the differences in energy between the signal points for 2)?

$Q(\sqrt(9A^2 T_s/ (4 N_0)))$ would also denote the error probability of a BPSK with amplitude

$3A/2$ but wouldn't we already do errors for much lower noise?

As an example: For the left-most signal with energy

$-3A/2$ we are making an error if the noise changes the signal above the blue line. This line is at a distance of

$A/2$ from the left-most signal. In the case of BPSK, we are looking at the case when the noise changes the signal to be on the other side of 0, meaning it has an amplitude larger than A.

Hence we should be able to use the original formula of BPSK

$P_e=Q(\sqrt{A^2 T_s/ (N_0)})$ and change

$A\rightarrow A/2$ which gives us the formula as in the solution. For the points with amplitude

$A/2$ , we have 2 of those points (e.g. -A and 0); hence, the error probability is doubled.

Am I missing something?

Best,

Joël

Re: Exercise 2 Problem 3

by Alexandre Nicolas Lechartier - Saturday, 25 June 2022, 14:08

Dear Dr. Hung-Chi,

Thank you for all your answers!

2) Should there be a factor 2 in front of Q for P_e,out as in the correction? If yes, may you explain why?

Best,
Alexandre