EE313 Linear Systems and Signals - Final

Coverage

The final exam in EE 313 is comprehensive.

The final exam will be on Friday, December 10th, 2:00-5:00 pm, in ACA 1.104, which seats 121. There are 36 students enrolled.

Prof. Evans will be holding office hours for final exam preparation as follows:

• Wednesday, December 8th, 12:00-2:00 pm
• Thursday, December 9th, 12:00-2:00 pm
• Friday, December 10th, 12:00-1:00 pm

You will be responsible for the following sections of Robert's Signals and Systems book:

• Chapter 1: all sections except 1.4
• Chapter 2: all sections except 2.14
• Chapter 3: all sections
• Chapter 4: sections 4.1, 4.2 and 4.3
• Chapter 5: sections 5.1, 5.2, 5.3, 5.4 and 5.5
• Chapter 6: sections 6.1, 6.2, 6.3 and 6.9
• Chapter 7: section 7.1, 7.2 and 7.3
• Chapter 9: all sections
• Chapter 10: sections 10.1, 10.2, 10.3 and 10.4
• Chapter 11: sections 11.1, 11.2, 11.3, 11.4 and 11.5
• Chapter 12: sections 12.1, 12.2, 12.3, 12.4, 12.5, 12.6 and 12.7
• Appendix A, D, E, F, G, H and I

You will also be responsible for the material covered on all homework assignments, as well as the posted hints and solution sets for those homework assignments.

You will also be responsible for all of the material in the reader.

You are responsible for what I said during lecture and what is on the course Blackboard Web site (e.g. homework hints).

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

If you have decided to buy the optional text DSP First, then the first eight chapters would be relevant to what we will have studied in EE 313.

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.

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 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.

Last updated 08/14/16. Send comments to bevans@ece.utexas.edu