# EE445S Real-Time Digital Signal Processing Lab - Homework 6 Hints

Homework #6: Assignment in Word and PDF formats.

• Please keep all of the work for each problem together. Please don't put the writeup in one place and plots in another. That makes it much more difficult for the grader to grade them.

• Problem 6.1. This phase locked loop uses the squaring block as employed for pre-emphasis in problem 5.3.

Many phase locked loop algorithms can make a 90-degree mistake. In part (a), the correct phase offset is -0.8 rad. Adding pi/2, or 1.57 rad, and one gets 0.77 rad, which is the value being tracked in part (a).

For part (a), you might try an initial guess of -0.2 rad to see what happens.

In practice, a receiver can in fact detect when its phase locked loop is 90 degrees out of phase-- bit errors increase like crazy as flagged by error detection and correction and other checks such as checksum in the medium access control layer.

In part (g), the error surface is "basically a plot of the objective as a function of the variable that is being optimized" (page 118). Examples of error surfaces are in Fig. 6.17 on page 119. In this problem, the objective function is given by J(theta).

Please note that in the problem statement, the equation

u(k Ts) = rp(k Ts) cos(4 pi f0 k Ts + theta)

should have been

u(k Ts) = rp(k Ts) cos(4 pi f0 k Ts + 2 theta)

That is, there should have been a factor of 2 in front of theta. The factor of "2 theta" is used in the derivation of the update equation for theta[k+1]. That is, the update equation is correct.

The mistake in the equation for u(k Ts) does not affect parts (a)-(f) because the update equation is correct. However, the mistake in the equation for u(k Ts) may affect your solution in part (g).

• Problem 6.3. 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.

Part (a). In digital QAM, the baseband signal has zero content at zero frequency, and is centered at a low carrier frequency. The transmission bandwidth depends on the low carrier frequency and bandwidth of the pulse shape used in the transmitter. The bandwidth of the pulse shape is half of the symbol rate times any excess bandwidth. The bandwidth for a raised cosine is (1 + alpha) W where W is half of the symbol rate. Please see lecture slide 16-4.

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

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. The peak power is proportional to 10 d2 and the average power is proportional to 6 d2.

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.

Last updated 04/10/14. Send comments to bevans@ece.utexas.edu