Designing Information Devices and Systems II Fall 2018 Elad Alon and Miki Lustig Homework 12

Transcription

1 EECS 1B Designing Information Devices and Systems II Fall 01 Elad Alon and Miki Lustig Homework 1 This homework is due on Friday, December 7 at 11:59 pm. Self grades are due on Monday, December 10 at 11:59 pm. 1. DFT (a) Compute the DFT coefficients of x 1 [n] = cos( π n) where n {0,1,...,5}. ( ) π cos n = 1 e j π n + 1 e j π n (1) u k [n] = 1 e j π kn () x 1 = ( u 1 + u 5 ) (3) X 1 [k] = k = 1,5. 0 k 1,5. (4) (b) Plot the time domain representation of x 1. Plot the magnitude, X[n], and plot the phase, X[n], for its DFT-basis representation. EECS 1B, Fall 01, Homework 1 1

2 EECS 1B, Fall 01, Homework 1

3 (c) Compute the DFT coefficients of x [n] = cos( 4π n) where n {0,1,...,5}. x = ( u + u 4 ) (5) X [k] = k =,4. 0 k,4. () (d) Plot the time-domain representation and the magnitude and phase for the DFT-basis representation of x. EECS 1B, Fall 01, Homework 1 3

4 EECS 1B, Fall 01, Homework 1 4

5 (e) How about the general case, x p [n] = cos( π pn), where n {0,1,...,5}? x p = ( u p + u p ) (7) X p [k] = k = p, p. 0 k p, p. () [ T (f) Compute the DFT coefficients of s = ]. s = ( u 0 + u 3 ). (9) S[k] = k = 0,3. 0 k 0,3. (10) (g) Compute the DFT coefficients of y 1 [n] = cos( π n π) where n {0,1,...,5}. EECS 1B, Fall 01, Homework 1 5

6 y 1 = ( u 1 u 5 ) (11) Y 1 [k] = k = 1,5. 0 k 1,5. (1) EECS 1B, Fall 01, Homework 1

7 . DFT Sampling Matching Select the correct answer from the multiple choice options provided and give some justification. (a) A sampled time domain signal and its DFT coefficients are given below: Now given the following time domain signal, which of the options below shows the correct DFT coefficient magnitudes? The correct DFT coefficients are shown in B. The new signal completes 3 full cycles during the discrete sequence. Its amplitude is the same as the first sequence. EECS 1B, Fall 01, Homework 1 7

8 (b) Given the time domain signal below, which of the options below shows the correct DFT coefficient magnitudes? The correct DFT coefficients are shown in A. The DFT coefficients of a unit impulse are 1 all N. This impulse has been shifted, so the DFT coefficients have varying phase and are not purely real; however, their magnitudes are still uniformly 1 N. EECS 1B, Fall 01, Homework 1

9 (c) Given the magnitude of the time domain signal below, which of the options below shows the correct DFT coefficient magnitudes? Hint: this is a trick quesiton! Since we are only given the magnitude of x[n], any four of the responses can be the correct DFT coefficients. That is because if we modulate x[n], the DFT coefficients will shift without changing the magnitude of x[n]. EECS 1B, Fall 01, Homework 1 9

10 3. Denoising signals using the DFT Professor Alon is sad. He just managed to create a beautiful audio clip consisting of a couple pure tones with beats and he wants Professor Lustig to listen to it. He calls Professor Lustig on a noisy phone and plays the message through the phone. Professor Lustig then tells him that the audio is very noisy and that he is unable to truly appreciate the music. Unfortunately, Professor Alon has no other means of letting Professor Lustig listen to the message. Luckily, they have you! You propose to implement a denoiser at Professor Lustig s end. (a) In the IPython notebook, listen to the noisy message. Plot the time signal and comment on visible structure, if any. The plot is in the notebook. There is no visible structure. It looks too noisy. (b) Take the DFT of the signal and plot the magnitude. In a few sentences, describe what the spikes you see in the spectrum are. Hint: take a look at the documentation for numpy.fft.fft. Note the norm="ortho" option. We know the Professor Alon generated pure tones. In the DFT basis, these pure tones are represented by coefficients each. Note the conjugate symmetry. (c) There is a simple method to denoise this signal: Simply threshold in the DFT domain! Threshold the DFT spectrum by keeping the coefficients whose absolute values lie above a certain value. Then take the inverse DFT and listen to the audio. You will be given a range of possible values to test. Write the threshold value you think works best. Any threshold from 0 onwards works well. Yay, Professor Alon is no longer sad! 4. DFT Exam Question This question comes from Fall 017 s final exam: (a) Let h[n], n = 0,,7 be an N = length sequence. Let H[k], k = 0,,7 be the DFT of h[n]. H[k] = [1,0, 1,0,1,0, 1,0] Use the properties of DFT and compute X[k], which is the DFT of x[n], where: x[n] = h[n] + e ( j πn + j π ) h[n] Since the DFT transformation is LTI, if we can find the DFT coefficients of each term of x[n], we can just add them at the end. The first term of x[n] is h[n], which we were given the DFT coeffieicents of already. Looking at the nd term of x[n]: e j π e j πn h[n] = jw n h[n] Since h[n] is being scaled by a constant j, H[k] is also scaled by j. h[n] is also being multiplied by W n, which modulates the signal. This causes the DFT coefficients to circularly shift by 1 to the right, which means the DFT coefficients of the nd term is [0, j,0, j,0, j,0, j]. Adding the two term s DFT coefficients together, we get: X[k] = [1, j, 1, j,1, j, 1, j] EECS 1B, Fall 01, Homework 1 10

11 (b) Consider a real valued sequence x[n], n = 0,1,,3. The norm of this signal: x =. Let X[k], k = 0,1,,3 be the DFT 4 of x[n]. We know that X[0] = 0, X[3] = 1 + j. Provide one possible set of values for the missing entries X[1], X[] that satisfy the above constraints. Is the solution unique? Since x[n] is purely real, the DFT coefficients must have conjugate symmetry, which means X[1] = X[3] = 1 j. By Parseval s: This means X[] = 0, giving us: X[k] = x[n] = x = 4 X[k] = [0,1 j,0,1 + j] This solution is unique. 5. Properties of DFT An N-sample, possibly complex signal x[n] is bounded, so that x[n] < 1 for all n. (a) What is the largest value possible for X[k], the magnitude of the DFT of x[n]? Consider the case where x[n] = 1. In that case, X[k] = Nδ[k]. According to Parseval, N 1 x[n] = N 1 X[k] = N, so the largest possible is N. n=0 k=0 Since x[n] is strictly less than 1, max{x[k]}< N (b) Find an expression for all the x[n] sequences which achieve this maximum. For an integer d, x[n] = e jπnd/n results in X[k] = Nδ[k d] which achieves the maximum.. Inverse DFT We have a signal x[n], n = 0,,7. Let X[k],k = 0,,7 be the DFT of x[n]. We know the first five DFT coefficients are 1, + j, j, and 0. In other words: If x[n] is purely real, find x[n] X[k] = [1, + j, j,,0,?,?,?] Since x[n] is purely real, the DFT coefficients must have conjugate symmetry, which means: X[k] = [1, + j, j,,0,, j, j] EECS 1B, Fall 01, Homework 1 11

12 Now that we have all the DFT coefficients, we can find x[n] by saying: x = F X e j π e j 4π e j π e j π e j 10π e j 1π e j 14π 1 + j 1 e j 4π e j π e j 1π e j 1π e j 0π e j 4π e j π j x = 1 1 e j π e j 1π e j 1π e j 4π e j 30π e j 3π e j 4π 1 e j π e j 1π e j 4π e j 3π e j 40π e j 4π e j 5π 0 1 e j 10π e j 0π e j 30π e j 40π e j 50π e j 0π e j 70π 1 e j 1π e j 4π e j 3π e j 4π e j 0π e j 7π e j 4π j 1 e j 14π e j π e j 4π e j 5π e j 70π e j 4π e j 9π j (1 + j) j ( 1 + j) 1 ( 1 j) j 1 (1 j) + j 1 j 1 j 1 j 1 j x = 1 1 ( 1 + j) j (1 + j) 1 (1 j) j j ( 1 j) ( 1 j) j (1 j) 1 (1 + j) j ( 1 + j) 1 j 1 j 1 j 1 j j 1 (1 j) j ( 1 j) 1 ( 1 + j) j (1 + j) j j + j + + j + j 1 + (1 + j)(1 + j) ( 1 + j) + 3 ( 1 j) 1 + (1 j)(1 j) 1 + ( j ) j j + j + j + ( j ) x = ( 1 + j)(1 + j) (1 + j) + 3 (1 j) ( 1 j)(1 j) 1 ( + j) + j j ( j) 1 + ( 1 j)(1 + j) (1 j) + 3 (1 + j) 1 + ( 1 + j)(1 j) 1 j + j + j j + j + j (1 j)(1 + j) ( 1 j) + 3 ( 1 + j) (1 + j)(1 j) x = () Write Your Own Question And Provide a Thorough Solution. Writing your own problems is a very important way to really learn material. The famous Bloom s Taxonomy that lists the levels of learning is: Remember, Understand, Apply, Analyze, Evaluate, and Create. Using what you know to create is the top level. We rarely ask you any homework questions about the lowest level of straight-up remembering, expecting you to be able to do that yourself (e.g. making flashcards). But we don t want the same to be true about the highest level. As a practical matter, having some practice at trying to create problems helps you study for exams much better than simply counting on solving existing EECS 1B, Fall 01, Homework 1 1

13 practice problems. This is because thinking about how to create an interesting problem forces you to really look at the material from the perspective of those who are going to create the exams. Besides, this is fun. If you want to make a boring problem, go ahead. That is your prerogative. But it is more fun to really engage with the material, discover something interesting, and then come up with a problem that walks others down a journey that lets them share your discovery. You don t have to achieve this every week. But unless you try every week, it probably won t ever happen. EECS 1B, Fall 01, Homework 1 13