EE313 Linear Systems and Signals - Final Exam

The final will be an open book, open notes, open laptop, comprehensive exam that is scheduled to last the entire final exam period. The laptop must have all external networking connections disabled. Because the final exam is an open book, notes and laptop exam, you'll likely need to move quickly through the exam but also need to think deeply about possible ways to solve the problems. To this end, having a week of regular sleep, eating, and exercise will be very helpful.

The final exam will be on Wednesday, December 19th, 2:00-5:00pm, in the following rooms:

Current enrollment is 72 students. In both rooms, there should be an aisle or three empty seats on either side of you.

Office Hours and Review Session

Previous Exams

Here are several example final exams: Also, please visit the Web pages for the following old exams:

Coverage

You will be responsible for the following sections of the Signal Processing First textbook by McClellan, Schafer and Yoder:

You will also be responsible for the material covered on all homework assignments, hints and solutions, as well as in all handouts and lectures:

There will be 8-10 questions on the final exam. There won't be any questions about Matlab.

Other Questions

Fall 1999 Final Exam

Problem 2 on Fall 1999 Midterm #2

This problem gives the step response of an LTI system, which we'll call ystep(t). It then asks you to find the response to a new signal x(t). Here are two different approaches for solving this problem:

Fall 2003 Final Exam

Problem 6 on Fall 2003 final exam. On problem 6(b) of the Fall 2003 final exam, the solution set (which I didn't write) is making the improper substitution
f(w) d(w) = f(0) d(w)
where d(w) is the Dirac delta functional. That is,
10 - j 2 w           10
---------- pi d(w) = -- pi d(w)
 4 + j w              4
which again is not valid.

Now, later in the problem, an inverse Fourier transform will be taken, and at that time, one can apply the property

       oo
 1    /            j w t        1
---   | f(w) d(w) e      dw = ---- f(0)
2 pi  /                       2 pi
     -oo
Under integration, we can simplify expressions involving the Dirac delta, and one should obtain the same answer that is given in the solution set.

Spring 2009 Final Exam

Problem 1 on Spring 2009 final exam.

Differential Equation Rhythm. The initial conditions are not zero, so the system is not LTI. See Section 16-11.1 on page 50 of the supplemental Chapter 16 on The Laplace Transform in the Signal Processing First book for a similar example.

Problem 2 on Spring 2009 final exam.

The system is LTI, which means that all initial conditions must be zero. The initial conditions are y(0-) and y'(0-).

(a) Take the Laplace transform of both sides of the differential equation:

s^2 Y(s) + 7 s Y(s) + 10 Y(s) = X(s)
(s^2 + 7s + 10) Y(s) = X(s)
Y(s)         1
---- =  ------------- = H(s)
X(s)    s^2 + 7s + 10
(b) Poles are the roots of the denominator.
s^2 + 7s + 10 = (s + 2)(s + 5)
Poles are at s = -2 and s = -5. Zeros are the roots of the numerator. No zeros.

(c) Region of convergence for H(s) is the intersection of the region of convergence for the pole at s = -2 which is Re(s) > -2 and the region of convergence for the pole at s = -5 which is Re(s) > -5. The intersection gives Re(s) > -2.

(d) Step response means the system response to a unit step u(t) input. The output is y(t) = h(t) * u(t) or equivalently Y(s) = H(s) X(s). When x(t) = u(t), X(s) = 1/s. So, Y(s) = 1 / ( s (s+2) (s+5) ). Region of convergence is Re(s) > 0. We can apply partial fractions decomposition on Y(s):

              1        1/10   1/6   1/15
Y(s) = ------------- = ---- - --- + ----
       s (s+2) (s+5)    s     s+2   s+5
Then take the inverse Laplace transform:
y(t) = (1/10) u(t) - (1/6) exp(-2 t) u(t) + (1/15) exp(- 5 t) u(t)

Problem 7 on Spring 2009 final exam.

Discrete-Time Filter Design. A notch filter is needed to remove the interfering frequency at 2 pi f0 / fs = pi / 2 and of course -pi /2. Pole radii of 0.9 would work.

Fall 2010 Final Exam

Problem 3 on Fall 2010 final exam.

(a) The transfer function for the LTI system is H(s) = s^2 / ((s+1)(s+2)).

(b) There are two zeros at s = 0. The poles are s = -1 and s = -2.

(c) ROC: Re{s} > 1. The ROC is the intersection of Re{s} > -2 and Re{s} > -1 due to the poles.

(d) Because the ROC includes the imaginary axis, we can substitute s = j w into the transfer function H(s) to obtain the frequency response:

Hfreq(w) = H(j w) = (j w)^2 / ((j w + 1)(j w + 2))

(e) There many possible approaches.

Approach #1: Compute magnitude responses for a few frequency values. w = 0 rad/s is a good place to start. Based on the denominator, w = 1 rad/s and w = 2 rad/s look interesting, too.

| Hfreq(0) | = 0

| Hfreq(1) | = | -1 / ((1 + j)(2 + j)) | = 1 / ( |1 + j| |2 + j| ) = 1 / ( sqrt(2) sqrt(5) ) = 1 / sqrt(10) = 0.3162

| Hfreq(2) | = | -4 / ((1 + 2j)(2 + 2j)) | = 4 / sqrt(40) = 0.635

Looks highpass.

Approach #2: Start evaluating the magnitude response by evaluating different values of w:

Hfreq(w) = |j w - 0|^2 / ( |j w - (-1)| |j w - (-2) | )

The term |j w - s0| is the Euclidean distance between the point j w and the point s0 in the complex s plane. That is, it's the length of the vector whose head is j w and whose tail is s0. Highpass filter.

Approach #3: Map the zeros and poles from the s plane to the z-plane using z = exp(s T) where T is the sampling time and then evaluate the magnitude response. In the z-domain, the zeros are at z = 1. The pole locations go to the positive real line in the z-domain at exp(-T) and exp(-2 T). Highpass filter.

Problem 4 on Fall 2010 final exam.

Discrete-Time Stability. Involves the determination of the range of a free parameter K to keep the discrete-time system bounded-input bounded-output (BIBO) stable. Different values of K that give BIBO stability will also give different frequency responses for the system. Analyze the transfer function to determine what values of K keep both poles inside the unit circle. The answer is 0.6 < K < 1. Values of K from 0.6 to 0.7 or so will yield a lowpass system. As K increases, the frequency response becomes bandpass.

Problem 5 on Fall 2010 final exam.

(a) When the input is x(t) = cos(2 π t), the output is y(t) = 1⁄2 + 1⁄2 cos(4 π t). The input has only one frequency at 1 Hz, but the output has frequencies at 0 Hz and 2 Hz present. A linear time-invariant system cannot create new frequencies. The only frequencies on the output must be present on the input.The system cannot be linear and time-invariant.

(b) For an input of cos(w0 t), the output is 1/2 + 1/2 cos(2 w0 t). In the Fourier domain, cos(w0 t) is a pair of Dirac deltas at frequencies -w0 and w0, each with an area of pi. Also, 1/2 + 1/2 cos(2 w0 t) has Dirac deltas at frequencies -2 w0, 0, and 2 w0, with areas of pi/2, pi and pi/2, respectively. The input signal cos(w0 t) has been modulated by cos(w0 t). However, the system can't be multiplication by cos(w0 t) due the step response of the system, which is a step function. The output of the system is the square of its input.

Is the region of convergence in the Laplace transform affected when there is a time shift?

The region of convergence is not affected by a time shift. A time shift simply causes the Laplace transform to have a multiplication term in the form on exp(-s t0) where t0 is the time shift.

Also, if it is a unilateral Laplace transform, do we need to worry about how to find the region of convergence (in lecture slide 11-10 it states that there is no need to specify a region of convergence)?

The region of convergence is not necessary if one is computing the inverse Laplace transform and the time-domain signal that will result is causal. However, for other reasons, such as BIBO stability checking for transfer functions and converting transfer functions to frequency responses, knowing the region of convergence is critical.


Last updated 12/22/18. Send comments to bevans@ece.utexas.edu.