Homework5_sol.pdf - EE311 Discrete Systems Homework 5 Solution Read the end of chapter 17 solved problems class examples and lecture notes 1 17 15(a

Homework5_sol.pdf - EE311 Discrete Systems Homework 5...

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EE311: Discrete Systems 1 Homework 5 Solution Read the end of chapter 17 “solved problems , class examples, and lecture notes. 1. 17. 15 (a) and (d), and plot |X(k)|. By using the DFT formula, we could obtain X(k)’s for k=0,1,2,3, 4 as follows: Plots of |X(k)| are skipped. 2. 17.19 (a) and (b). b) Given that: and the IDFT formula: We have: 3. 17.23. You may use fft() in matlab to obtain the DFTs and plot their magnitudes. (hint: example Matlab code in the lecture notes.)
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EE311: Discrete Systems 2 N=64 is skipped. Clearly, as N increases, the frequency domain resolution of |X(k)| is finer.
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EE311: Discrete Systems 3 4. 17.36 (b): (1) Calculate the N-point DFTs; (2) Let N=8, plot |X(k)| vs k. (3)How many peaks do you observe from the plot in (2)? (4)Is the spectrum |X(k)| consistent with the time-domain signal?
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Unformatted text preview: Solution: For b) 1) 2) For N=8, X(k)’s are: X(k) =8/(2j) [d(k-1)-d(k-7)] = [0, 4/j, 0,0,0,0,0,-4/j] |X(k)|=[0,4,0,0,0,0,0,4] 3) There are 2 peaks on the magnitude plot. However, the 2 nd peak at k=7 is redundant to the 1 st peak at k=1 given that the signal is a real-valued signal and its DFT satisfies the conjugate symmetric property. This is a one-sided spectrum. 4) Yes, the spectrum is consistent with the time-domain signal. The signal is a single tone sinusoidal and we observe a single component from the plot in (2). Note that the spectrum plot in (2) is a one-sided plot so there are 2 peaks (the 2 nd is redundant to the 1 st one). The spectrum actually corresponds to one single peak at k=1, i.e., one frequency component....
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