Homework 6

Homework 6#

Chapter 6 (6.7-6.9)

Due NLT 4 Nov (Lesson 31) by 2359 on Gradescope#

Directions:#

  • Collaboration is authorized as noted in the syllabus Collaboration policy.

  • All homework must be complete, error-free, and neatly organized. Points may be deducted for sloppy and illegible work.

  • Answers should be clearly indicated by a box.

  • Use engineering notation with proper units.

  • Submit work to Gradescope with pages assigned to each problem. Scans/uploads must be legible and neat without excessive margins.

Problems:#

  • 6.7-2

  • 6.7-4 –> You can figure out Part (a). In Part (b), you want to compare the transmitted power for the binary case to that of the 16-ary case. Transmitted power is equal to the average energy per bit multiplied by the pulse rate. For the binary case, let the average bit energy be \(E_b\) and the pulse rate be \(R_b\) - since in the binary case the bit rate equals the pulse rate. Then \(P_{binary}=E_bR_b\).

    It can be shown (in Problem 6.7-3 if you are interested!) that the average energy per bit for \(M\)-ary PAM is given by

    \[E_{av}=\frac{M^2-1}{3}E_b,\]

    where we assume all \(M\) levels are equally likely. Then the transmitted power is \(E_{av}\) multiplied by the pulse rate.

  • 6.7-5 –> Don’t confuse the number of bits per sample with the \(\log_2(M)\) number of bits per pulse! Steps to follow:

    1. Find the number of bits/sample and bit rate (bits/sec), \(R_b\).

    2. Use the available bandwidth and roll-off factor to find the pulse rate (pulses/sec), \(R_s\). This can also be called the symbol rate - hence the subscript \(s\).

    3. Calculate the number of bits per pulse.

  • 6.8-1 –> Part (a): remember binary PSK is equivalent to polar signaling. Part (b): use Carson’s Rule, where \(\Delta f\) is the peak frequency deviation from the center point between the two frequencies and \(B\) is the bandwidth of the baseband signal.

  • 6.8-1 Add-on Part (c) –> If the modulator generates a 16-PSK signal, what is the bandwidth of the modulated output? What about for a 16-QAM signal?

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