EE313 Linear Systems & Signals - Mini-Project #2 Hints

Report Guidelines. Please see the report guidelines on the homework page including the organization.
Radar System. Please see the handsketch of the radar system we are simulating.
utplotspec Matlab function The original version of utplotspec used the obsolete phase function to compute the phase response. In the updated version, phase has been replaced with a way to compute unwrapped phase using the Matlab commands unwrap and angle.
General. Use the parameters in Table 10.1 on page 320 throughout mini-project #2 except in Exercise 1.1(d).
Exercise 1.1(a). Complex-Valued Chirp Signal. We first analyze the continuous-time complex chirp
s(t) = exp(j pi W t² / T) for -T/2 ≤ t ≤ T/2.
in MATLAB using the parameters in Table 10.1 on page 320.
```
T = 25E-6;    %% pulse length 25us
W = 2E6;      %% swept bandwidth 2MHz
fs = 20E6;    %% sampling rae 20 MHz
Ts = 1/fs;    
t = (-T/2) : Ts : (T/2);
s = exp(j*pi*W*(t.^2)/T);
```
which gives a time-bandwidth product TW = 50.
Since s(t) is complex-valued, we could plot the real part:
```
figure;
plot(t, real(s));
xlabel( 't' );
```
As continuous time increases from -T/2 to 0, the chirp decreases in frequency from -W/2 to 0, and then increases in frequency from 0 to W/2 from time 0 to T/2.
The next step is to connect s(t) for -T/2 ≤ t ≤ T/2 and s[n] for 0 ≤ n ≤ N-1 where
s[n] = exp(j 2 pi alpa (n - N/2)²)
With a sampling time of T_s, the N samples in discrete time span T seconds of continuous time, i.e. T = N T_s.
The problem asks us to find correct formulas for alpha and N in terms of p and T W. We know the following:
f_s = p W
T_s = 1 / f_s = 1 / (p W)
T = N T_s = N / (p W)
From the last equation, we can get the relationship for N in terms of p and T W:
T W = N / p
N = p T W
Here, N has to be an integer.
The text has the following note in part (a):
(Note: It may not be possible to make the discrete-time chirp symmetric, depending on how the sampling times are defined. Starting at t_n = -T/2 may not be the best strategy if t = 0 is not included in the sampling gird.)
When sampling s(t), the first sample is taken at -T/2 and the subsequent N-1 samples are taken at increments of Ts. The origin is included. However, the most positive sample time is at -T/2 + (N-1) Ts, which does not include the sample at time T/2 because that would correspond to s[N]. The value of s[0] = s(-T/2) = exp(pi T W / 4) is not zero, which means that s[n] is not conjugate symmetric about its midpoint.
To find the value for alpha in terms of p and T W, there are a couple of different ways:
- Compare s[n] and s(n Ts), equate the n² terms in the complex sinusoids, and express the answer for alpha in terms of p and T W
- Compare s[0] and s(-T/2), equate the n² terms in the complex sinusoids, and and express the answer for alpha in terms of p and T W
Please show your work in your report. For reference, my intermediate answer is
alpha = T W / (2 N²)
and from there, you can substitute information from above to express alpha in terms of p and T W.

Exercise 1.1(b). Code to synthesize a discrete-time chirp Here's the MATLAB function to synthesize a discrete-time chirp signal. Please place the following function in a file called dchirp.m.

function s = dchirp( TW, p )
% DCHIRP   generate a sampled chirp signal
%   usage  s = dchirp( TW, p )
%          s : samples of a discrete-time "chirp" signal
%              exp(j pi (W/T) t^2 )   for -T/2 <= t <= T/2
%          TW : time-bandwidth product
%          p : sample at p times the Nyquist rate (W)
N = p*TW;
alpha = TW / (2*N^2);
n = 0 : N-1;
s = exp(j*2*pi*alpha*(n - N/2).^2);

Exercise 1.1(d). Low oversampling factor. The problem says to use a sampling rate fs = 1.2 W. This means the oversampling factor p = fs / W = 1.2.
Computing and Plotting the Fourier transform. Please use the code utplotspec.m which implements the command
```
utplotspec(x, Ts);
```
where x is the discrete-time signal and Ts is the sampling time. Please use the value for Ts for the chirp signal being analyzed.
The utplotspec function will plot a magnitude and phase of the signal's Fourier transform vs. frequency in Hz and return the frequency-domain samples computed by the FFT algorithm and their corresponding frequencies in Hz. Please see Appendix A in the Mini-Project #2 assignment for more information on the connection between the discrete-time Fourier transform and the FFT.
Exercise 1.1(e). Skip this exercise.
Exercise 1.1(f). Skip this exercise.
Exercise 1.2(a). Skip this exercise.
Exercise 1.2(b). You can use the program utplotspec.m described above to plot the magnitude spectrum. From the magnitude spectrum,
- Verify that the chirp pulse sweeps frequencies from -W/2 to W/2, which would be from -1 MHz to 1 MHz according to the value for W in Table 10.1.
- Compare the plot against the plot in Fig. 10.1 on page 321 in shape. The plot is Fig. 10.1 uses W = 12 MHz.
The last sentence, which says "Scale the spectrum so that its magnitude is correct at DC", appears to relate to part (a), so you can ignore the sentence.
Exercise 1.2(c). Skip this exercise.
Exercise 1.2(d). Skip this exercise.
Exercise 2.1(a). Skip this exercise.
Exercise 2.1(b) Hint #1 Matched Filtering. Our goal is to find the distance to a metallic object in an enviroment as part of locating the object.
To find the distance, a radar system will transmit a discrete-time complex chirp pulse signal s[n] as a continuous-time signal through its antenna. After that, we process the received signal to try to detect the chirp after it bounces off a metallic object in the environment and returns to the radar. The processing applies the matched filter to the received signal and searches for a peak in the absolute value of the output signal of the matched filter. Based on the time at which the peak in the response occurs, we can estimate the round-trip delay from the radar transmission to the radar reception, and use the round-trip delay to find the distance of the object.
The matched filter is a linear time-invariant (LTI) finite impulse response (FIR) filter whose impulse response is directly related to the pulse shape to be detected. In discrete time, for a discrete-time complex-valued pulse shape s[n], the matched filter would have an impulse response
h[n] = s^*[-n]
where ^* indicates complex conjugation and the argument -n means to flip the signal in time. The relevant Matlab commands are conj to conjugate the elements of the vector of signal values and fliplr to flip the vector of signal values. You could conjugate the vector of signal values and then flip the resulting vector, or vice-versa. The answer will be the same either way.
By using h[n] = s^*[-n], the matched filter will correlate the received signal against the pulse shape s[n]. When the correlation is high, i.e. the output of the matched filter is high enough in absolute value, the pulse has been detected. See the comments below on Exercise 2.1(c) concerning correlation.
This problem introduces a constant G without explaining what it represents. It could represent the attentuation experienced by the propagating pulse. Please set it to one.
The phrase "mainlobe" refers to the shape of the real part of the chirp pulse in the time domain around its middle values, i.e. at t = 0 for the continuous-time chirp and n = (N-1)/2 for the discrete-time chirp.
Please see Exercise 2.1(b) Hint #2 next.
Exercise 2.1(b) Hint #2 Delaying the Chirp by a Continuous-Time Delay. We can realize a continuous-time delay by T_d seconds in the transmitted chirp signal by using the continuous-time chirp signal s(t) defined in equation (1-1):
s(t) = exp(j pi W t² / T) for -T/2 ≤ t ≤ T/2.
In MATLAB, using the parameters in Table 10.1 on page 320, a vector of samples for s(t) can be defined as
```
T = 25E-6;    %% pulse length 25us
W = 2E6;      %% swept bandwidth 2MHz
fs = 20E6;    %% sampling rae 20 MHz
Ts = 1/fs;    
t = (-T/2) : Ts : (T/2);
s = exp(j*pi*W*(t.^2)/T);
```
To delay a signal by T_d seconds, we would write mathematically
s(t - T_d)
For the continuous-time chirp signal, the range of continuous-time values would be
-T/2 ≤ t - T_d ≤ T/2
which is equivalent to
(-T/2) + T_d ≤ t ≤ (T/2) + T_d
In the MATLAB code, we could change the range of t being evaluated in the second-to-last line of the code to be
```
t = (-T/2) + Td : Ts : (T/2) + Td;
```
In addition, please see Exercise 2.1(b) Hint #1 above.
Exercise 2.1(c). Connection between Correlation and Convolution. Correlation measures the similarity between two signals, with a higher absolute value meaning a higher similarity. Here's the definition of correlation R_xy[k] between discrete-time signals x[n] and y[n]:
```
              oo         *
    R  [k] =  Sum  x[n] y [n+k] 
     xy      n=-oo
```
The outer discrete-time variable k indicates the shift in time between the two discrete-time signals. That is, correlation involves sliding y^*[n] across x[n] and summing the result for each value of the shift, k.
Correlation is very similar to convolution:
```
              oo         *          oo         *
    R  [k] =  Sum  x[n] y [n+k] =   Sum  x[n] y [k+n]
     xy      n=-oo                 n=-oo
```
In convolution, the y^*[k+n] term would be y^*[k-n]. Note that the roles of k and n are reversed in correlation. We can write correlation as
```
                     *
    R  [k] = x[n] * y [-n]
     xy
```
That is, prior to convolution, we'll flip y^*[n] and convolution will flip it back. That way, we'll "fool" convolution to perform sliding y^*[n] across x[n].
Autocorrelation measures how similar a signal x[n] is vs. shifts in time of itself:
```
               oo        *     
    R  [k] =  Sum  x[n] x [k+n]
     xx      n=-oo             
```
The maximum value for R_xx[k] occurs when k = 0, i.e. when x[n] is aligned with x^*[n]. For a complex-valued signal x[n], the product x[n] x^*[n] is the magnitude squared of x[n] because the phases subtract out.
The built-in Matlab command for correlation is xcorr. Using xcorr with one argument will compute the autocorrelation.
Exercise 2.1(d). Skip this exercise.
Exercise 2.3(a). Please see Exercise 2.1(b) Hint #2 Delaying the Chirp by a Continuous-Time Delay.
Exercise 2.3(b). Skip this exercise.
Exercise 2.3(c). Skip this exercise.
Exercise 2.3(d). Skip this exercise.
Exercise 2.3(e). Skip this exercise.

Last updated 11/01/23. Send comments to (Mailbox)

bevans@ece.utexas.edu