EE445S Real-Time Digital Signal Processing Lab - Homework 3 Hints

Homework #3: 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 3.1: A bandpass filter is commonly used in a communications receiver to reduce noise and interference outside of the transmission band. The transmission band in this problem is from 3000 Hz to 4000 Hz.
For the simulation of the narrowband interferer, you could use either a sinusoid or bandpass noise.
As an example of simulating a sinusoidal interferer outside of the transmission band, one can replace the line
```
n=0.25*randn(1, time/Ts );
```
in improvesnr.m with
```
t = Ts : Ts : time;
f0 = 1000;
n = cos(2*pi*f0*t);
```
By starting at t=Ts and ending at t=time, we ensure that signal n would have the same number of samples as signal x.
Bandpass noise can be generated by passing spectrally flat noise through a bandpass filter. The bandpass filter would have a passband width that is much smaller than the transmission bandwidth of 1000 Hz, e.g. 100 Hz or less. The bandpass filter may either be a finite impulse response filter or an infinite impulse response filter.
Problem 3.2: An analog QAM receiver must perfectly track the carrier frequency and carrier phase to recover the transmitted messages perfectly (same goes for a digital QAM receiver) assuming an ideal channel. This problem asks you to explore the degradation in the received messages if the carrier frequency has been perfectly tracked in the receiver but the carrier phase tracking has a slight error in it.
In this problem, there are two baseband message signals m₁(t) and m₂(t). Please use different baseband message signals. It simplifies the calculations if you use the same linear phase lowpass filter for the two filters in the receiver.
With the phase error, energy in transmitted message signal #1 will leak into received message signal #2, and vice-versa, which is known as cross-interference. Let's do the math. Assume that there is a phase error of theta at the receiver.
```
s₁(t) = LPF{ m₁(t) cos(2 pi f_c) cos(2 pi f_c + theta) - m₂(t) cos(2 pi f_c) sin(2 pi f_c + theta) }

s₁(t) = LPF{ 1/2 m₁(t) ( cos(theta) + cos(4 pi f_c + theta ) ) - 1/2 m₂(t) ( sin(-theta) + sin(4 pi f_c + theta ) ) }

s₁(t) = 1/2 cos(theta) m₁(t) + 1/2 sin(theta) m₂(t)
```
When theta = 0, i.e. when there is no phase error,
```
s₁(t) = 1/2 m₁(t)
```
which recovers the in-phase component up to a scalar gain.
The lowpass filter in the equations above was assumed to be ideal with no delay. In practice, the input signal experiences delay through the filter by the time it reaches the output.
The delay through any LTI filter is its group delay (see slide 5-11). The group delay is equal to the negative of the slope of the phase response For a linear phase FIR filter, the group delay n₀ is a constant equal to (N-1)/2 samples, where N is the number of filter coefficients. Please use an odd-length FIR filter so that the group delay through the FIR filter is an integer. The code in AM.m uses an FIR filter of order 100, which has 101 coefficients.
To compute the cross-interference in the upper channel, you'll need to align transmitted message #1 and received message #1 in time. Here are two ways:
- Delay the transmitted signal by n₀
  -OR-
- Remove the first n₀ samples and the last n₀ samples of the LPF output
Then, sum the squares of the samples in the difference signal to compute the amount of energy leakage. Finally, divide the energy leakage by the energy in transmitted message signal #1 to find out the fraction of energy leakage to get
```
                                                  2 
                          Sum ( m [n-n ] - s [n] )  
                           n     1    0     1
Fraction Energy Leakage = -------------------------
                                     2               
                                Sum m [n-n ]           
                                 n   1    0         
```
For the time alignment discussed above, you'll need to change the Matlab code for AM.m by replacing
```
m = 2 * filter(b, 1, x);
```
with
```
m = 2 * conv(b, x);
```
to obtain the output samples of the LPF for all input samples.
Please note the reciprocal of the fraction of energy leakage is a measure of signal-to-noise ratio at the LPF output:
```
      Signal Power       Average Power of Message #1 at LPF Output
SNR = ------------ = ------------------------------------------------
      Noise Power    Average Power of Cross-Interference at LPF Output
```
You could try different phase offset values to find the phase offset that gives a different signal that has 1% of the energy of the in-phase channel when there isn't any phase offset.
Note: In the receiver in Figure 5.10, the term sin(2 pi fc t) should be -sin(2 pi fc t) to match the negation of the lower path in the transmitter.
Problem 3.3: In the pull-down menu in fdatool for choosing a filter structure, SOS means second-order sections, i.e. a cascade of biquads.
Once you've designed your filter using the Filter Design and Analysis (FDA) tool in Matlab, you can export the filter coefficients. In the Design Filter View of the FDA tool, the upper left window has the text "Current Filter Information". Below this text, you can right click on the text "Structure" and pick either cascade of biquads or single section.
Cascade of biquads. If you leave the filter as a cascade of biquads, then for each biquad, expand the factored form of the transfer function and report the feedforward coefficients (b0, b1 and b2) and the feedback coefficients (a1 and a2). Here's how to connect the poles, zeros and gain with the coefficients using the transfer function:
```
                  -1          -1                     -1          -2
         (1 - z0 z  )(1 - z1 z  )     1 - (z0 + z1) z   + z0 z1 z  
H(z) = C ------------------------ = C ------------------------------
                  -1          -1                     -1          -2
         (1 - p0 z  )(1 - p1 z  )     1 - (p0 + p1) z   + p0 p1 z  
```
So,
```
b0 = C
b1 = -C (z0 + z1)
b2 = C z0 z1
a1 = p0 + p1
a2 = - p0 p1
```
Single section. Then, under the File menu, you can click "Export". The "Export" command will by default export the filter coefficients to the Matlab workspace. The numerator coefficients will be in the Num vector, and the denominator coefficients will be in the Den vector. Then, in Matlab, you can type Num to see the numerator coefficients, and type Den to see the denominator coefficients. Moreover, you can apply the filter to a signal in Matlab using
```
outputSignal = filter(Num, Den, inputSignal);
```

Last updated 02/22/14. Send comments to (Mailbox)

bevans@ece.utexas.edu