ECE 445S Real-Time Digital Signal Processing Lab - Homework 6 Hints

• Please fully justify your answers by including your Matlab code, Matlab plots, and mathematical analysis, as well as your interpretation of the code, plots and analysis.

• Please keep all of the work for each problem together. This will make it easier for the grader to grade them and hence assign appropriate partial or full credit for problems.

• Problem 6.1. In a receiver, the carrier phase can vary with time due to many factors, such as (1) time-varying channel response, and (2) mismatch in the carrier frequency used in downconversion to the carrier frequency used in the transmitter for upconversion. For the latter, consider a transmitter carrier frequency of fc and a receiver carrier frequency of fc + df:

  cos(2 pi (fc + df) t ) = cos(2 pi fc t + 2 pi df t)
  

  The time-varying phase theta(t) is 2 pi df t, where df is the difference between the receiver and transmitter carrier frequencies. The phase 2 pi df t is slowly varying with time because df is a much smaller frequency compared to fc.

  Carrier phase recovery notes and marker board notes for the steepest descent version of the basic Costas loop.

  For the Costas Loop, you're asked to show that the update equation for theta[k] in Fig. 10.9 on page 208 is the same as that for Fig. 10.12 on page 211.

  In solving this problem, you'll need to use trigonometric identities and an assumption that the phase theta[k] is slowly varying with time. The latter assumption implies that cos(theta[k]) is approximately constant, which allows the switching of the order of modulation and lowpass filtering in each branch.

  It is a common assumption in phase-locked loops that the phase being tracked is varying slowly with time.

• Problem 6.2. For a simpler problem involving developing a steepest ascent algorithm for symbol timing recovery, please see midterm #2 spring 2021 problem 2.1(d)

  Homework problem 6.2 is related to the block diagram in Fig. 12.1 on page 252.

  The symbol index is k and the symbol time is T. The sampling time T_s is equal to the symbol time T divided by the number of samples per symbol M.

  The symbol timing offset, tau, is the offset to the actual start of the current symbol. An offset of tau - T and tau + T would be at the start of the previous and next symbol, respectively. It is okay if a symbol timing recovery algorithm converges to a timing offset of tau, tau - T or tau + T. For the latter two cases, we could adjust the timing offset by adding or subtracting T, respectively.

  During the kth symbol, we can estimate the symbol timing offset, and use it to delay (if positive) or advance (if negative) the time that the (k+1)st symbol is sampled.

  This specific problem uses a different objective function J_FP(tau) which is a version of the fourth-order objective function that is plotted in Fig. 12.11 on page 264. The concavity of the fourth-order objective function depends on the value of the rolloff parameter, beta. For beta values of 0, 0.1, 0.2 and 0.3, the concavity matches the concavity in Fig. 12.11, whereas for beta values of 0.4, 0.5 and 1.0, the concavity is the opposite of that in Fig. 12.11. I have adapted a Matlab script from the Johnson, Sethares and Klein book to plot the concavity: myclockreccost.m. In this problem, please use beta = 0.2, and proceed with minimizing the objective function.

  To set up the adaptive update equation for the symbol timing offer tau, we'll need to take the first derivative of J_FP(tau) with respect to tau. The problem asks to use the approximation of the first derivative of x(tau) with respect to tau given in JSK problem 12.16(b).

• Problem 6.3. QAM is the sum of two upconverted PAM signals, with each upconverted PAM signal carrying a message. We had simulated Analog QAM on homework problem 3.2, and Analog QAM is also mentioned in lecture slides 1-15 and 15-6. Lecture 15 covers QAM transmitters and lecture 16 covers QAM receivers. The Digital QAM transmitter and receiver are shown on QAM Receiver lecture slide 16-3.

  There are three parts to this question.

  Please see Fig. 16.12 on page 381 and read the accompanying text. Plotting symbol error rate vs. signal-to-noise ratio is a very common first step in analyzing communication system performance. This curve plots the lower bound from a formula. Another way to use this graphical representation is to simulate a communication system for different SNR settings and scatter plot the results. This could allow comparison of two equalization methods, two timing recovery methods, etc. Superimposing the lower bound from a formula on the plot shows how close (or far away) the methods are from the ideal answer.

  Marker Board Notes.

  Part (a). In digital QAM, the baseband signal is bandpass; the bandpass band is close to zero frequency; and the DC component is zero. The baseband QAM carrier frequency is small in value but greater than the lowpass pulse shape's bandwidth to create a bandpass spectrum. For a raised cosine pulse shape, the bandwidth is pulse shape is (1 + alpha) W where W is half of the symbol rate, as mentioned on Pulse Shaping & Interpolation lecture slide 7-13.

  Part (b). Average and peak transmit power for 4-QAM in lecture is available on QAM Transmitter slide 15-18. For this part, please provide calculations for average and peak transmit power for 4-QAM and 16-QAM, and comment on the differences. Please compute the peak-to-average-power ratio for 4-QAM and 16-QAM.

  As an example, here's power analysis for an 8-QAM constellation. With QAM symbol amplitudes of the form i[n] + j q[n], 8-QAM amplitudes could be -3d + j d, -3d - j d, -d + j d, -d - j d, d + j d, d - j d, 3d + j d, 3d - jd. The power over one symbol period is proportional to the square of the absolute value of the symbol amplitude: 10 d², 10 d², 2 d², 2 d², 2 d², 2 d², 10 d², 10 d². Taking the absolute value is needed because each symbol amplitude is complex-valued.

  1. 8-QAM: peak power is proportional to 10 d².
  2. 8-QAM: average power is proportional to 6 d².
  3. 8-QAM: peak-to-average-power ratio is 1.67.

  Part (c). What other tradeoffs are made when moving from a 4-QAM to a 16-QAM modulation scheme? Please compare the probability of symbol error vs. signal-to-noise ratio for 4-QAM and 16-QAM. Where is the signal-to-noise ratio measured in the receiver? Please see the paragraph on the homework assignment.

bevans@ece.utexas.edu