Discussion For Homework2

Question: In the first question in the assignment, shouldn't it be

(1)
\begin{align} N(t) = X(t) \cos(2\pi f_c t) - Y(t) \sin(2 \pi f_c t) \end{align}

instead of what's given as both X(t) and Y(t) are the I and Q components??

Comment[Student]: Yes, I believe this is a typo. Problem one of Homework 2 is just problem 4.3 from the 4th ed. of the text. In problem 4.3 it referes to eqn. 4.1-38 which is what you have posted above. However this statement is rather transparent in the working out of the problem as you never have to directly deal with N(t). Just solve the problem on the basis that N(t) is a zero-mean, stationary, narrowband process with autocorrelation of the equivalent low-pass process, Z(t), as defined in the problem.

Question: In problem 4, what does "down convert" mean exactly?

Comment[Student] : In Question 4, 'down convert' refers to the passband-to-baseband conversion of the Signal Waveform.

Answer[Krishna]: That is correct. It is called down convert because it brings the frequency "down" from passband to baseband

Question: The down conversion from the passband to the baseband … does it mean to just multiply the passband by exp(jwt) or take the real part of the previous product?

Answer[Krishna]: Down conversion should mean finding the equivalent baseband waveform.

It is probably best to think in terms of how the book does this in Chapter 2, i.e. think of the analytic signal (one with only positive frequencies) and then multiply by $e^{j \omega t}$ to bring it to baseband. Notice that this is not just multiplying by $e^{j \omega t}$ and taking the real part. Remember, the baseband signal is complex in general so taking just the real part would obviously not work always.

You can also think of this as multiplying the passband signal by $e^{j \omega t}$ and then low pass filtering the result. This low pass filtering is important.

Question: How to calculate the energy of a piece of complex signal $Z(t)$? Is it

(2)
\begin{align} \int Z^\ast(t)Z(t){\rm d}t, \end{align}

or

(3)
\begin{align} \frac{1}{2}\int Z^\ast(t)Z(t){\rm d}t? \end{align}

Thanks.

Answer[Krishna]: Lets define it as without the 1/2.

Question: If anybody wants to share the gain they are getting over 16-QAM for the last problem (just the gain without any hints about what to do), we can have a running counter of the best possible gain.

Comment[student]: I have a gain of about 0.58 dB over 16-QAM.

Comment[grader]: To the student who get 0.58db gain: could you please send me your matlab file for this problem? My email is ude.umat.ece|ehgniqud#ude.umat.ece|ehgniqud. Thanks.

Comment[grader]: I found there are some caculation errors in the solutions for the "hexagon design" in last problem (the descriptoin is fine). The solutions have been corrected and re-uploaded.

page revision: 23, last edited: 16 Feb 2009 01:40