The final exam will be on Monday, December 18th, 9:00am-12:00pm, in EERC 1.516 and EERC 1.518. Each room has an identical layout of tables with 122 seats. Each seat has a power outlet.

- EERC 1.518 if your last name starts with A-L
- EERC 1.516 if your last name starts with M-Z

- Fall 2017 without solutions and with solutions
- Summer 2016 without solutions and with solutions
- Fall 2010 without solutions and with solutions
- Spring 2009 without solutions and with solutions
- Fall 2005 without solutions and with solutions
- Fall 2003 without solutions and with solutions
- Fall 1999 without solutions and with solutions

- Thursday, December 14th, 12:00-2:00 pm, EERC 6.882, by Prof. Evans
- Saturday, December 16th, 12:00-2:00 pm, EERC 0.814E, by Ms. Ghosh

- Chapter 1: Introduction
- Chapter 2: Sinusoids
- Chapter 3: Spectrum Representation
- Chapter 4: Sampling and Aliasing
- Chapter 5: FIR Filters
- Chapter 6: Frequency Response of FIR Filters
- Chapter 7: z-Transforms
- Chapter 8: IIR Filters
- Chapter 9: Continuous-Time Signals and Systems
- Chapter 10: Frequency Response
- Chapter 11: Continuous-Time Fourier Transform
- Section 12-3 on Sampling and Reconstruction
- Chapter 16: Laplace transform
- Appendix A: Complex Numbers

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.

**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(where d(w) d(w) = f(0) d(w)

10 - j 2 w 10 ---------- pi d(w) = -- pi d(w) 4 + j w 4which 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 -ooUnder integration, we can simplify expressions involving the Dirac delta, and one should obtain the same answer that is given in the solution set.

**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 7 on Fall 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.

**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:

*H*_{freq}(*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.

| *H*_{freq}(0) | = 0

| *H*_{freq}(1) | = | -1 / ((1 + j)(2 + j)) | = 1 / ( |1 + j| |2 + j| ) = 1 / ( sqrt(2) sqrt(5) ) = 1 / sqrt(10) = 0.3162

| *H*_{freq}(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*:

*H*_{freq}(*w*) = |j *w* - 0|^2 / ( |j *w* - (-1)| |j *w* - (-2) | )

The term |j *w* - *s*_{0}| is the Euclidean distance between the point j *w* and the point *s*_{0} in the complex s plane. That is, it's the length of the vector whose head is j w and whose tail is *s*_{0}. 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(w_{0} t), the output is 1/2 + 1/2 cos(2 w_{0} t). In the Fourier domain, cos(w_{0} t) is a pair of Dirac deltas at frequencies -w_{0} and w_{0}, each with an area of pi. Also, 1/2 + 1/2 cos(2 w_{0} t) has Dirac deltas at frequencies -2 w_{0}, 0, and 2 w_{0}, with areas of pi/2, pi and pi/2, respectively. The input signal cos(w_{0} t) has been modulated by cos(w_{0} t). However, the system can't be multiplication by cos(w_{0} t) due the step response of the system, which is a step function. The output of the system is the square of its input.

Last updated 01/03/18. Send comments to bevans@ece.utexas.edu.