Mock up Final Exam - Rhea

Sample Final Exam

ECE 438

Fall 2016

Instructor: Prof. Mireille Boutin

1. Compute and Sketch the graph of the DTFT of $x_d[n] = cos(2 \pi 250 \frac{1}{400} )$

2. If x(t) is $\frac{1}{8} sinc(\frac{t-15}{7})$ find $\chi(f)$ and plot $|\chi(f)|$

3. A continuous time signal is such that $\chi(f)$ is 0 when |f| > 3 KHz. You would like a cutoff at 2KHz and a gain of 5.

a. A sample exists with a sampling rate of 9000 samples/sec. Can you process this signal to make a band limited interpolation? If no, explain why. If yes, explain how.

b. If this sample is downsampled by a factor of 3, can you still make the reconstruction?

4. Find the DFT of $x[n] = e^{j \frac{\pi}{2} n} * cos( \frac{\pi}{5} n)$

5. Find the N-Point inverse DFT x[n] of $e^{j \frac{\pi}{3} k}$

6. Compare the total number of operations for a N= 64 point DFT via summation formula, decimation by two, and radix two DFT. Which makes more sense for N=64. How about for N = 65?

7. Find the Z transform of $x[n] = 2^{n} * u[n-10]$

8. Find the frequency response in two different ways of $$ y[n] + 2y[n-1] + 7y[n-3] = x[n] + 3x[n-2] $$ Is this an implicit or an explicit system? What does this tell us about the plot on the Z plane?

9. The sampling rate of a vocal recording is 10KHz. There exists a large formant at 2KHz, medium sized formants at 3 KHz and 1.5 KHz, and a small formant at 4KHz. Below, plot the poles of the frequency response of this vocal recording.

10. Consider a discrete space system defined by $$ g[m,n] = h[m,n] ** f[m,n] $$ with $h[m] = \frac{1}{25} * \begin{bmatrix} 1 & 3 & 1 \\ 3 & 9 & 3\\ 1 & 3 & 1 \end{bmatrix}$ What is the response to the following input (assume 0-boundry condition):

$\begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$

SOLUTION

1. $\chi_d(w) = 400 rep_2\pi (\frac{\delta(\frac{400}{2\pi} -250) + \delta(\frac{400}{2\pi} +250)}{2})$ However, aliasing takes place and the deltas experience a shift to + and - $3\pi/4$ as shown:

2. By the known relationship between the rect and sinc signals, the $\chi(f) = \frac {7} {8} rect(7f) e^{-j 17 \pi f}$ The graph is shown below:

3. a. Yes! This can be accomplished by passing the signal through a low-pass filter with a gain of 5 and a cutoff at $\frac {2 \pi 2000} {9000}$ to obtain the desired output.

b. No! The sampling rate is now 3KHz, which is below the Nyquist sampling rate of the highest frequency that may be present (i.e. 3 KHz which would have a Nyquist requirement of a 6 KHz sampling rate at least).

4. $x[n] = e^{j \frac{\pi}{2} n} * cos( \frac{\pi}{5} n) = e^{j \frac{\pi}{2} n} * \frac {1}{2} (e^{j \frac{2 \pi}{10} } + e^{-j \frac{2 \pi}{10} }) = \frac {1}{2} * e^{j \frac{2 \pi}{4} n} * (e^{j \frac{2 \pi}{10} } + e^{-j \frac{2 \pi}{10} }) = \frac {1}{2} * e^{j \frac{10 \pi}{20} n} * (e^{j \frac{4 \pi}{20} } + e^{-j \frac{4 \pi}{20} }) = \frac {1}{2}(e^{j \frac{14 \pi}{20} } + e^{-j \frac{6 \pi}{20} })$

Compare to the IDFT formula $x[n] = \sum_{k=0}^{19} (\frac{X_20[k]}{20} e^{j \frac{2 \pi}{20} kn}$

Thus $$ X_20[k] $$ is 10 when k = 6 and 14 and 0 when k = 0,1,2,3,4,5,7,8,9,10,11,12,13,15,16,17,18, and 19 It is periodic with period 20.

5. $e^{j \frac{\pi}{3} k} = e^{j \frac{2\pi}{6} k} = e^{- j \frac{2\pi}{6} 5k}$ Compare to DFT formula $X_6[k] = \sum_{k=0}^{5} (x[n] e^{-j \frac{2 \pi}{6} kn}$ Thus x[n] is 1 when n = 5 and 0 when n = 0,1,2,3,and 4. It is periodic with period 6.

6. For N = 64: Summation formula: Complex Ops = $2N^{2} - N$ = 8128 Decimation by two: Complex Ops = $N^{2} + N$ = 4160 Radix two DFT: Complex Ops = $$ N log_2(N) $$ = 384

For N = 65: Summation formula: Complex Ops = $2N^{2} - N$ = 8385 Decimation by two: Complex Ops = $N^{2} + N$ = 4290 For radix 2, assume N = 128 and fill indices 64-125 with zeros. Radix two DFT: Complex Ops = $$ N log_2(N) $$ = 896 This is notable because increasing N by one significantly increases the Radix two value but it is still far fewer complex operations than the other two.

7.