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1. Compute and Sketch the graph of the DTFT of | 1. Compute and Sketch the graph of the DTFT of | ||
| − | <math> x_d[n] = cos(2 \pi 250 | + | <math> x_d[n] = cos(2 \pi 250 \frac{1}{400} )</math> |
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| − | 6. Compare the total number of operations for a N= 64 point DFT via summation formula, decimation by | + | 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? |
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| − | 1. <math> \chi_d(w) = 400 rep_2\pi (\frac{\delta(\frac{400}{2\pi} -250) + \delta(\frac{400}{2\pi} +250}{2}) </math> | + | 1. <math> \chi_d(w) = 400 rep_2\pi (\frac{\delta(\frac{400}{2\pi} -250) + \delta(\frac{400}{2\pi} +250)}{2}) </math> |
| − | However, aliasing takes place and the deltas experience a shift to + and - 3\pi/4 as shown: | + | However, aliasing takes place and the deltas experience a shift to + and - <math> 3\pi/4 </math> as shown: |
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| − | 2. By known | + | 2. By the known relationship between the rect and sinc signals, the <math> \chi(f) = \frac {7} {8} rect(7f) e^{-j 17 \pi f} </math> |
| + | 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 <math> \frac {2 \pi 2000} {9000} </math> | ||
| + | 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. <math> 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} }) </math> | ||
| + | |||
| + | Compare to the IDFT formula <math> x[n] = \sum_{k=0}^{19} (\frac{X_20[k]}{20} e^{j \frac{2 \pi}{20} kn} </math> | ||
| + | |||
| + | Thus <math> X_20[k] </math> 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. <math> e^{j \frac{\pi}{3} k} | ||
| + | = e^{j \frac{2\pi}{6} k} | ||
| + | = e^{- j \frac{2\pi}{6} 5k} </math> | ||
| + | Compare to DFT formula <math> X_6[k] = \sum_{k=0}^{5} (x[n] e^{-j \frac{2 \pi}{6} kn} </math> | ||
| + | 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 = <math> 2N^{2} - N </math> = 8128 | ||
| + | Decimation by two: Complex Ops = <math> N^{2} + N </math> = 4160 | ||
| + | Radix two DFT: Complex Ops = <math> N log_2(N) </math> = 384 | ||
| + | |||
| + | For N = 65: | ||
| + | Summation formula: Complex Ops = <math> 2N^{2} - N </math> = 8385 | ||
| + | Decimation by two: Complex Ops = <math> N^{2} + N </math> = 4290 | ||
| + | For radix 2, assume N = 128 and fill indices 64-125 with zeros. | ||
| + | Radix two DFT: Complex Ops = <math> N log_2(N) </math> = 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. | ||
Revision as of 22:02, 23 April 2017
Sample Final Exam
ECE 438
Fall 2016
Instructor: Prof. Mireille Boutin
All questions below are derived from the homework of Prof. Mireille Boutin, all rights reserved by 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.
